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Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 4 Another Peek Beyond the Point MCQ improves accuracy in objective exams.

MCQ on Another Peek Beyond the Point Class 7

Another Peek Beyond the Point MCQ Class 7

Class 7 Maths Another Peek Beyond the Point MCQ

Question 1.
Which digit is in the thousandths place in the number 48.50736?
(a) 5
(b) 0
(c) 7
(d) 3
Solution:
(c) 7
In 48.50736, on the right side of the decimal point, the place values of digits are tenths (5), hundredths (0), thousandths (7) and so on.

Question 2.
Which of the following is the correct expansion of 32.406?
(a) 30 + 2 + 0.4 + 0.006
(b) 30 + 2 + 0.04 + 0.006
(c) 30 + 2 + 0.4 + 0.06
(d) 30 + 0.4 + 0.006
Solution:
(a) 30 + 2 + 0.4 + 0.006
The expanded form of 32.406 is given as
32.406 = 3 × 10 + 2 × 1 + 4 × \(\frac{1}{10}\) + 0 × \(\frac{1}{100}\) + 6 × \(\frac{1}{1000}\)
= 30 + 2 + 4 × 0.1 + 0 × 0.01 + 6 × 0.001
= 30 4- 2 + 0.4 + 0.006

Question 3.
The value of 0.385 + 4.07 + 12 is:
(a) 16.455
(b) 16.450
(c) 16.40
(d) 15.455
Solution:
(a) 16.455
Adding after aligning the decimal points in given numbers:

Question 4.
Which of the following is equal to 4.8 × 2.5?
(a) 10.2
(b) 12.0
(c) 11.75
(d) 9.5
Solution:
(b) 12.0
Multiplying the decimal numbers after converting them into fractions:
48 × 25 = \(\frac{48}{10} \times \frac{25}{10}=\frac{48 \times 25}{10 \times 10}=\frac{1200}{100}\) = 12

Question 5.
Cost of a pencil is ₹7.25. What is the cost of 12 pencils?
(a) ₹87
(b) ₹88.50
(c) ₹90
(d) ₹72.50
Solution:
(a) ₹87
Cost of one pencil = ₹7.25
∴ Cost of 12 pencils = 12 × 7.25
= (12 × 7) + (12 × 0.25)
= 84 + 12 × \(\frac{1}{4}\) = ₹87

Question 6.
7 + \(\frac{5}{10}+\frac{3}{100}+\frac{1}{1000}\) =
(a) 0.7531
(b) 75.31
(c) 7.531
(d) 0.7531
Solution:
(c) 7.531
7 + \(\frac{5}{10}+\frac{3}{100}+\frac{1}{1000}\)
= 7 + 0.5 + 0.03 + 0.001 = 7.531

Question 7.
A car covers a distance of 14.75 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
(a) 147.5 km
(b) 1475 km
(c) 1.475 km
(d) 14.075 km
Solution:
(a) 147.5 km
The distance covered by car in one litre of petrol = 14.75 km
The distance covered by car in 10 litres of petrol = 14.75 × 10 = 147.5 km

Question 8.
A shop packs sugar in 0.375 kg packets. If one full box contains 24 such packets, then the total weight of one full box is:
(a) 9.0 kg
(b) 8.75 kg
(c) 9.100 kg
(d) 8.25 kg
Solution:
(a) 9.0 kg
Weight of 1 packet of sugar = 0.375 kg
Since a box contains 24 packets of sugar,
Weight of one box = 0.375 kg × 24
= \(\frac{375}{1000} \times 24=\frac{9000}{1000}\)
= 9.0 kg

Question 9.
48.6 ÷ 10 =
(a) 4.86
(b) 0.486
(c) 48.06
(d) 486
Solution:
(a) 4.86
48.6 ÷ 10 = \(\frac{486}{10} \div 10=\frac{486}{10} \times \frac{1}{10}=\frac{486}{100}\) = 4.86

Question 10.
Which of the following division gives a quotient less than the dividend?
(a) 12.5 ÷ 0.5
(b) 7.29 ÷ 9
(c) 4.8 ÷ 0.08
(d) 18.6 ÷ 0.3
Solution:
(b) 7.29 ÷ 9
We know that in decimal division, if divisor is greater than 1, the quotient is less than the dividend.
Now, 9 > 1, so on dividing 7.29 by 9 we will get a quotient (0.81) less than 7.29.

Question 11.
Which of the following statements are correct with respect to the multiplication of the decimal number 8.879 by 89?
(i) 8.879 × 89 = 8.87 × 89 + 0.009 × 89
(ii) 8.879 × 89 = 8 × 89 + 0.8 × 89 + 0.79 × 89
(iii) 8.879 × 89 = 8 × 89 + 0.8 × 89 + 0.07 × 89 + 0.009 × 89
(iv) 8.879 × 89 = 8 × 89 + 0.879 × 89
Choose the correct option from the following:
(a) Only (i), (ii) and (iii)
(b) Only (iii)
(c) Only (i), (iii) and (iv)
(d) All of the above
Solution:
(c) Only (i), (iii) and (iv)
The expanded form of 8.879 is:
8.879 = 8 × 1 + 8 × \(\frac{1}{10}\) + 7 × \(\frac{1}{100}\) + 9 × \(\frac{1}{1000}\)
= 8 + 0.8 + 0.07 + 0.009
Now, 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = 8 × 89 + 0.8 × 89 + 0.07 × 89 + 0.009 × 89
or 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = (8.87 + 0.009) × 89 = 8.87 × 89 + 0.009 × 89
or 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = (8 + 0.8 + 0.079) × 89 = 8 × 89 + 0.8 × 89 + 0.079 × 89
or 8.879 × 89 = (8 + 0.8 + 0.07 + 0.009) × 89 = (8 + 0.879) × 89 = 8 × 89 + 0.879 × 89
∴ The statements (i), (iii) and (iv) are correct.

Question 12.
Which of the following multiplications of decimal numbers are correct?
(i) 0.5 × 0.05 = 0.025
(ii) 1.2 × 100 = 1200
(iii) 2.5 × 4 = 10
(iv) 0.1 × 0.1 = 0.1
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iii)
(d) (ii), (iii) and(iv)
Solution:
(a) (i) and (iii) only
Statements (i) and (iii) are correct as 25
0.5 × 0.05 = \(\frac{5}{10} \times \frac{5}{100}=\frac{25}{1000}\) = 0.025 and
2.5 × 4 = 2 × 4 + 0.5 × 4 = 8 + 2 = 10
But statements (ii) and (iv) are incorrect as 1.2 ×x 100 = 120 and 0.1 × 0.1 = 0.01

Question 13.
Which of the following years are not leap years?
(i) 2024
(ii) 2100
(iii) 2000
(iv) 1800
Choose the correct option from the following:
(a) (ii) and (iv)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (i) and (iii)
Solution:
(a) (ii) and (iv)
A year divisible by 4 is a leap year; a year divisible by 100 is not a leap year; however, a year divisible by 400 is a leap year.
The year 2024 is divisible by 4 and not divisible by 100, so it is a leap year.
The year 2100 is divisible by 100 but not divisible by 400, so it is not a leap year.
The year 2000 is divisible by 400, so it is a leap year.
The year 1800 is divisible by 100 but not divisible by 400, so it is not a leap year. Therefore, the years that are not leap years are 2100 and 1800.

Another Peek Beyond the Point Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): 32 × 3 tenths = 32.3
(R): Moving the decimal point to the right makes the number bigger.
Solution:
(d) A is false but R is true.
32 × 3 tenths = 32 × \(\frac{3}{10}=\frac{32 \times 3}{10}\)
= \(\frac{96}{10}\) = 9.6
So, Assertion (A) is false.
We know that multiplying a number by 10 shifts the decimal point one place to the right, making the number bigger.
For example, 1.2 × 10 = 12
So, Reason (R) is true.

Question 2.
(A): When 56.3 is multiplied by 1000, the result is 56300.
(R): When multiplying a decimal number by 10, 100, or 1000, the decimal point shifts to the left by as many places as there are zeroes in the multiplier.
Solution:
(c) A is true but R is false.
We know that, when multiplying a decimal number by 10, 100, or 1000, the decimal point shifts to the right (not left) by as many places as there are zeroes with one.

By using the above logic, when 56.3 is multiplied by 1000, the result is 563,00.
∴ Assertion (A) is true, but Reason (R) is false.

Question 3.
(A): The area of a square with a side length of 0.5 cm is 2.5 sQuestion cm.
(R): The area of a square of side a is calculated using the formula a × a.
Solution:
(d) A is false but R is true.
We know that the area of a square of side a is calculated using the formula a × a.
So, if the side length, a = 0.5 cm, then
area = a × a = 0.5 cm × 0.5 cm = 0.25 sq. cm
∴ Assertion (A) is false, but Reason (R) is true.

Question 4.
(A): The year 3100 is a leap year.
(R): A century year that is not divisible by 400 is not a leap year.
Solution:
(d) A is false but R is true.
Since 3100 is a century year, it must be divisible by 400 to be a leap year.
But 3100 is not divisible by 400, so it is not a leap year
Therefore, Assertion (A) is false, but Reason (R) is true.

Another Peek Beyond the Point Class 7 Fill in the Blanks

Question 1.
In the blanks below, fill ‘<’, ‘>’ or ‘ = ’ after analysing the expressions on the LHS and RHS.
(i) 17.89 × 5.9 ______ 185.2 × 0.57
(ii) 5.6 × 12.72 _____ 89.04 × 0.8
Solution: <, =
(i) LHS = 17.89 × 5.9
= \(\frac{1789}{100} \times \frac{59}{10}=\frac{105551}{1000}\) = 105.551
RHS = 185.2 × 0.57 = \(\frac{1852}{10} \times \frac{57}{100}=\frac{105564}{1000}\)
= 105.564
The whole number parts of both are equal i.e. 105.
Also, the digit in the tenths place is the same for both i.e. 5.
However, the digit in the hundredths place of the LHS i.e. 5 is less than that of the RHS i.e. 6.
Since 5 < 6, we conclude:
105.551 < 105.564
So, 17.89 × 5.9 < 185.2 × 0.57.

(ii) LHS = 56 × 12 72 = \(\frac{56}{10} \times \frac{1272}{100}=\frac{71232}{1000}\)
= 71.232
RHS = 8904 × 08 = \(\frac{8904}{100} \times \frac{8}{10}=\frac{71232}{1000}\)
= 71.232
Since all digits match, we conclude:
5.6 × 12.72 = 89.04 × 0.8

Question 2.
When a decimal number is multiplied by 10000, the decimal point shifts to the right by ______ places.
Solution: four
We know that the number of zeroes in the multiplier (10, 100, 1000, …) determines how many places the decimal point shifts to the right.
Since the multiplier (10000) hag 4 zeroes, the decimal point shifts four places to the right.

Question 3.
6.7 × 10 + 6.7 × 100 + 0.67 × 1000 = _______
Solution: 1407
6.7 × 10 + 6.7 × 100 + 0.67 × 1000
= 67 + 670 + 670 = 1407

Question 4.
Evaluate the following using the information 256 × 12 = 3072.
(i) 25.6 × 1.2 = _______
(ii) 307.2 ÷ 1.2 = _______
(iii) 30.72 ÷ 25.6 = ______
(iv) 0.256 × 0.12 = _____
Solution: 30.72, 256, 1.2, 0.03072
(i) 25.6 × 1.2 = \(\frac{256}{10} \times \frac{12}{10}\)
= \(\frac{256 \times 12}{100}=\frac{3072}{100}\) = 30.72

(ii) 307.2 ÷ 1.2 = \(\frac{3072}{10} \div \frac{12}{10}\)
= \(\frac{3072}{10} \times \frac{10}{12}=\frac{256 \times 12}{12}\) = 256

(iii) 30.72 ÷ 25.6 = \(\frac{3072}{100} \div \frac{256}{10}\)
= \(\frac{256 \times 12}{100} \times \frac{10}{256}=\frac{12}{10}\) = 1.2

(iv) 0.256 × 0.12 = \(\frac{256}{1000} \times \frac{12}{100}\)
= \(\frac{256 \times 12}{100000}=\frac{3072}{100000}\) = 0.03072

Question 5.
Fill in the blanks:
(i) 5.5 km = _____ m
(ii) 125.5 mL = _________ L
(iii) 14.5 cm = ______ mm
(iv) 68 g = _______ kg
Solution: 5500, 0.1255, 145, 0.068
(i) As 1 km = 1000 m
∴ 5.5 km = 5.5 × 1000 = 5500 m

(ii) As 1 mL = \(\frac{1}{1000}\) L
∴ 125.5 mL = \(\frac{125.5}{1000}\) = 0.1255 LHS

(iii) As 1 cm = 10 mm
∴ 14.5 cm = 14.5 × 10 = 145 mm

(iv) As 1 g = \(\frac{1}{1000}\) kg
∴ 68 g = \(\frac{68}{1000}\) = 0.068 kg

Go through BSE Odisha Class 8 Science Solutions Chapter 10 Light: Mirrors and Lenses Question Answer to understand textbook questions more clearly.

Class 8 Science Curiosity Chapter 10 Question Answer

Class 8 Science Ch 10 Light: Mirrors and Lenses Question Answer

Class 8 Science Chapter 10 Light: Mirrors and Lenses Question Answer

Probe and Ponder Questions

Question 1.
Can we make mirrors which can give enlarged or diminished images?
Answer:
Yes, we can make mirrors that produce ehnlarged or diminished images. Specially, concave mirrors can form both enlarged and diminished images depending on the object’s distance from the mirror, while convex mirrors always produce diminished images. Concave mirrors give magnified images when the object is close (between the pole and the focal point) and diminished images when the object is far (beyond the centre of curvature). Convex mirrors always form virtual, erect and similar images.

Question 2.
On side-view mirrors of vehicles, there is a warning that says “Objects in mirror are closer than they appear”. Why is this warning written there?
Answer:
The warning “Objects in mirror are closer than they appear” is written on the side-view mirrors of vehicles because these mirrors are typically convex. Convex mirrors make objects appear smaller and therefore seem farther away than they really are. The warning reminds drivers to allow extra distance when judging how near other vehicles or objects are.

Question 3.
Why is there a curved line on some reading glasses?
Answer:
The curved line on reading glasses is due to the convex shape of the lens. These lenses are thicker in the middle and thinner at the edges, causing light rays to converge (bend inward) and improve near vision for individuals with presbyopia. This curvature is essential for correcting the vision by focusing light properly on the retina.

Question 4.
Share your questions ………….
Answer:
Here are some fun questions you might think of from the chapter:

• How does a magnifying glass make things look bigger?
• Why do spoons act like funny mirrors?
• Can lenses in our eyes change shape?
• What happens if you use a concave mirror to focus sunlight?

InText Questions

Question 1.
How can we distinguish between concave and convex mirrors? (Page 155)
Answer:
To distinguish between concave and convex mirrors, observe how they reflect light and the images they form. Concave mirrors curve inward, causing parallel light rays to converge at a focal point and can form both real (inverted) and virtual (erect) images depending on object distance.

Convex mirror curve outward, causing parallel light rays to diverge and always form virtual, erect and diminished images. A simple test is to hold the mirror near an object and move it; a concave mirror can magnify objects when close while a convex mirror always makes them appear smaller.

Question 2.
Do you remember learning about the use of telescope in the chapter ‘Beyond Earth’ in Curiosity, Grade 6? (Page 156)
Answer:
Yes, some laws govern image formation by mirrors, including plane, concave and convex mirrors. These laws are based on the laws of reflection, which state that the angle of incidence (the angle at which light hits the mirror) is equal to the angle of reflection (the angle at which light bounces off the mirror). Additionally, the incident ray, the reflected ray, and the normal (an imaginary line perpendicular to the mirror’s surface at the point of incidence) all lie in the same plane.

Question 3.
We have observed images formed by three types of mirrors-plane, concave, and convex. But are there any laws which govern the image formation? (Page 157)
Answer:
Yes, the two laws of reflections which govern the image formation.

Question 4.
Are laws of reflection applicable to spherical mirrors also? (Page 160)
Answer:
Yes, the laws of reflection apply to spherical mirrors. The same rules-angle of incidence equals angle of reflection and the incident ray, reflected ray and normal lie in the same plane – hold at every point on a spherical mirror’s surface. Using these laws and simple geometry we can predict where an image will form for concave and convex spherical mirrors.

Question 5.
We explored the images of an object formed by curved mirrors. But how do objects look when viewed through transparent materials with curved surfaces? (Page 162)
Answer:
When we view the printed letters by a magnifying glass which is a convex lens, it appear bigger in size. In case of concave lens the image is smaller than object.

Question 6.
What changes can be seen in the objects when viewed through lenses? (Page 163)
Answer:
Image formed by a convex lens can be enlarged, diminished or of the same size as the object and it may be erect or inverted, depending upon the distance of the object from the lens. But image formed by a concave lens is always erect and diminished in size.

Question 7.
Do lenses also converge or diverge the light beam? Can it also burn a paper ? (Page 163)
Answer:
Yes, lenses can both converge and diverge light beams. Convex (convering) lenses bring parallel rays together at a focus, while concave (diverging) lenses cause parallel rays to spread out as if they came from a virtual focus.

Question 8.
Since a convex lens converges a light beam, can it also burn a paper? (Page 164)
Answer:
Yes. A convex lens can concentrate sunlight to a small spot and raise the temperature there enough to scorch or even ignite paper. This is the same principle used in a magnifying glass to focus sunlight.

Question 9.
Where all are the lenses used ? (Page 165)
Answer:
Lenses are used in many devices such as eyeglasses, cameras, microscopes, telescopes, projectors, magnifying glasses and medical instruments. Different lens types (convex or concave) are chosen depending on whether the light needs to be converged or diverged.

Light: Mirrors and Lenses Class 8 Questions and Answers

Keep the Curiosity Alive (Pages 166-169)

Question 1.
What is the angle made by the reflected ray with the mirror?

(i) 40°
(ii) 50°
(iii) 45°
(iv) 60°
Answer:
(ii) If the incident ray makes a 40° angle with the normal, then :
Angle of incidence (i) =40°
According to the law of reflection,
Angle of reflection (r) = Angle of incidence
(i) = 40°
Now, Angle between reflected ray and mirror = 90° angle of reflection
= 90°-40°=50°

Question 2.
The figure shows three different situations where a light ray falls on a mirror:

(i) The light ray falls along the normal.
(ii) The mirror is tilted, but the light ray still falls along the normal to the tilted surface.
(iii) The mirror is tilted, and the light ray falls at an angle of 20° from the normal.
Draw the reflected ray in each case (Use a ruler and protractor for accurate drawing). What is the angle of reflection in each case?
Answer:
(i) Light ray falls along the normal Angle of incidence = 0°
Angle of incidence =0°
Angle of reflection =0°
The ray retraces its path. It reflects straight back along the same line.

(ii) The mirror is tilted, but the light ray still falls along the normal to the tilted surface.
Even though the mirror is tilted, the ray is still perpendicular to the surface.
Angle of incidence = 0°
Angle of reflection = 0°
The ray again retraces its path – the tilt doesn’t affect the reflection if the ray is normal to the surface.

(iii) The mirror is tilted, and the light ray falls at 20° from the normal.
Angle of incidence =20°
By the law of reflection:
Angle of reflection =20°
The reflected ray will make a 20° angle with the normal, on the opposite side.

Question 3.
In the figure, the cap of a sketch pen is placed in front of three types of mirrors.

Match each image with the correct mirror.

Answer:

Image	Mirror
(i) (ii) (iii)	Convex mirror Concave mirror Plane mirror

Question 4.
In the Figure, the cap of a sketch pen is placed behind a convex lens, a concave lens, and a flat transparent glass piece- all at the same distance.

Match each image with the correct type of lens or glass.

Answer:

Image	Lens/glass type
(i) (ii) (iii)	Convex lens Concave lens Flat transparent glass piece

Question 5.
When the light is incident along the normal on the mirror, which of the following statements is true:
(i) Angle of incidence is 90°
(ii) Angle of incidence is 0°
(iii) Angle of reflection is 90°
(iv) No reflection of light takes place in this case
Answer:
(ii) When light hits a surface at a 90° angle (perpendicular), it is considered to be incident normally.
Angle of incidence = Angle of reflection: In reflection, the angle of the reflected light is always equal to the angle of the incident light. Since the light is hitting the mirror normally (at a 0° angle of incidence), the reflected light will also be at a 0° angle of reflection.

Question 6.
Three mirrors-plane, concave, and convex, are placed in the figure. Based on the images of the graph sheet formed in the mirrors, identify the mirrors and write their names above the mirrors.

Answer:

Question 7.
In a museum, a woman walks towards a large convex mirror (Figure). She will see that :
(i) her erect image keeps decreasing in size.
(ii) her inverted image keeps decreasing in size.
(iii) her inverted image keeps increasing in size, and eventually it becomes erect and magnified.
(iv) her erect image keeps increasing in size.

Answer:
The correct answer is (i) her erect image keeps decreasing in size. Because a convex mirror always forms a virtual, erect, and diminished image.

Question 8.
Hold a magnifying glass over the text and identify the distance at which you can see the text bigger than they are written. Now move it away from the text. What do you notice? Which type of lens is a magnifying glass?
Answer:
A magnifying glass uses a convex lens to make text appear larger. When we hold the magnifying glass close to the text and then slowly move it away, the image of the text will first appear larger and sharper, then progressively get larger and more blurred, and finally invert and appear smaller again.

Question 9.
Match the entries in Column I with those in Column II.

Answer:

Column I	Column II
(i) Concave mirror	(a) Spherical mirror with a reflecting surface that curves inwards.
(ii) Convex mirror	(b) It forms an image which is always erect and diminished in size.
(iii) Convex lens	(c) An object placed behind it may appear inverted at some distance.
(iv) Concave lens	(d) The object placed behind it always appears diminished in size.

Question 10.
The following question is based on Assertion/Reason.
Assertion: Convex mirrors are preferred for observing the traffic behind us.
Reason: Convex mirrors provide a significantly larger view area than plane mirrors.
Choose the correct option:
(i) Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion.
(ii) Both Assertion and Reason are correct, but Reason is not the correct explanation for Assertion.
(iii) Assertion is correct, but Reason is incorrect.
(iv) Both Assertion and Reason are incorrect.
Answer:
(i) Both Assertion and Reason are correct, and Reason is the correct explanation for Assertion.

Question 11.
In the Figure, note that O stands for object, M for mirror, and I for image.

Which of the following statements is true?
(i) Figure (a) indicates a plane mirror, and Figure (b) indicates a concave mirror.
(ii) Figure (a) indicates a convex mirror and Figure (b) indicates a concave mirror.
(iii) Figure (a) indicates a concave mirror and Figure (b) indicates a convex mirror.
(iv) Figure (a) indicates a plane mirror, and Figure (b) indicates a convex mirror.
Answer:
(ii) Figure (a) indicates a convex mirror and Figure (b) indicates a concave mirror.

Question 12.
Place a pencil behind a transparent glass tumbler (Figure a). Now fill the tumbler halfway with water (Figure b). How does the pencil appear when viewed through the water? Explain why its shape appears changed.

Answer:
When we place a pencil behind a transparent glass tumbler and fill the tumbler halfway with water, the pencil will appear to be bent or broken at the point where it enters the water. The submerged portion of the pencil might also appear slightly thicker or shifted from its actual position. This phenomenon is called refraction of light.

Why it Happens
Change in medium: Light travels at different speeds in different materials (media). Air is an optically rarer medium, meaning light travels faster in it, compared to water, which is an optically denser medium where light travels more slowly.

Bending of Light (Refraction): When light rays from the pencil travel from the water (denser medium) into the air (rarer medium) and then into our eyes, their speed changes. This change in speed causes the light rays to bend or deviate from their original path. Specifically, when light goes from a denser medium to a rarer medium, it bends away from the normal (an imaginary line perpendicular to the surface at the point where light enters the new medium).

Apparent Position: Because our brains interpret light rays as traveling in straight lines, the bending of light at the water-air interface makes the submerged part of the pencil appear to be at a different location than its actual position, creating the illusion of a bent or broken pencil.

The extent of this bending depends on the angle at which we view the pencil and the difference in optical density between the two media (water and air). If we view the pencil straight from above, we won’t observe the bending as the light rays would be traveling perpendicular to the surface, and refraction would be minimal.

Class 8 Science Chapter 10 Question Answer

Activity 1.

Let us explore
Aim: To study the image of an object formed by a curved reflecting surface.
Materials Required: A highly polished large stainless steel spoon.

Procedure:

• Take a highly polished large stainless steel spoon and hold its curved surface close to your face. Can you see your image in it?
• You will notice that the image of your face is different from the image you see in a plane mirror.
• Now, observe the image, by moving spoon slowly away from your face. Do you observe any change in the image?
• Repeat the same steps by flipping the spoon.

Observations:

• When we looked at the inner side of spoon which is curved inwards, we observed that the image was inverted.
• When we looked at the outer side of the spoon which is bulges outwards, the image of our face was erect but smaller in size.

Inferences:

• Curved mirrors are a part of hollow glass sphere.
• The reflecting surface of the spherical mirror may be curved inwards or outwards.

Activity 2.

Let us distinguish

Aim: To distinguish between concave and convex mirror.
Materials Required: A concave mirror, a convex mirror.

Procedure:

• Keep a concave and a convex mirrors on a table whose reflecting surfaces facing upwards.
• Now look at them from the side, keeping your eye at their level, so that you can identify whether the reflecting surface is curved inwards or outwards.

Observations:

• The Reflecting surface of one mirror is curved inwards.
• The Reflecting surface of other mirror is curved outwards.

Inference:

Concave mirror	Convex mirror
1. Reflecting surface is curved inwards.	1. Reflecting surface is curved outwards.
2. A thin layer of aluminium is coated on outer curved surface.	2. A thin layer of aluminium is coated on inner curved surface.
3. It forms both type of images: erect and inverted.	3. It forms only erect images at all distances.

Activity 3.

Let us explore
Aim: To study the characteristics of images formed by spherical mirrors.
Materials Required: A concave mirror, a convex mirror, two small wooden blocks, a small toy or some other object.

Procedure:

• Take two mirrors and hold them side by side in an upright position on a table.
• Keep the object in front of them at a small distance (3-4 cm away) as shown in Figure (a).
• What kind of images do you see in each mirror? Are the images of the same size as the object? Are they erect? Do you see lateral inversion in the images?
• Note down your observation.
• Now, move the object gradually from the mirror.
• What changes do you see in the images in both the mirrors? Do the images become smaller or larger? Do they continue to be erect? Again, note down your observations.
• Repeat the above steps with each mirror separately.

Observations:

1. When object is close to the mirror.

Concave	Convex
(i) Enlarged size (Larger)	(i) Diminished size (Smaller)
(ii) Erect	(ii) Erect
(iii) Lateral inversion is seen.	(iii) Lateral inversion is seen.

2. When object is moved away:

Concave	Convex
(i) The size decreases as the distance increases.	(i) Smaller at all distance.
(ii) Inverted	(ii) Remains erect
(iii) Lateral inversion is seen.	(iii) Lateral inversion is seen.

Inference :
For a Convex Reflecting Surface:

• An erect and smaller image is formed.
• The image remains erect and smaller at all distances.

For a Concave Reflecting Surface:

• At smaller distance, the image is erect and enlarged.
• At larger distances, the image formed is inverted. The size of the image decreases as the distance from the spoon increases.

Conclusion: Common in Both: Lateral inversion (left-right reversal) is observed in the images.

Activity 4.

Let us experiment
Aim: To verify the first laws of reflection.
Materials Required: A drawing board, a white sheet of paper, a comb, a torch, a strip of plane mirror.

Procedure:

• Fix a white sheet of paper on a drawing board.
• Take a comb and close all its openings except one in the middle.
• Hold the comb perpendicular to the sheet of paper.
• Throw light from a torch through the opening of the comb from one side. With slight adjustment of the torch and the comb you will see a ray of light along the paper on the other side of the comb.
• Keep the comb and the torch steady.
• Place a strip of plane mirror in the path of the light ray.
• What do you observe ?
  After striking the mirror, the ray of light is reflected in another direction.
• After striking the mirror, the ray of light is reflected in another direction. The light ray, which strikes any surface, is called the incident ray. The ray that comes back from the surface after reflection is known as the reflected ray.
• Daw lines showing the position of the plane mirror, the incident ray and the reflected ray on the paper. Remove the mirror and the comb. Draw a line making an angle of 90° to the line representing the mirror at the point where the incident ray strikes the mirror. This line is known as the normal to the reflecting surface at that point.
• The angle between the normal and incident ray is called the angle of incidence (∠i). The angle between the normal and the reflected ray is known as the angle of reflection (∠r).
• Measure the angle of incidence and the angle of reflection.
• Repeat the activity several times by changing the angle of incidence.

Observations:

S.No	Angle of incidence (∠i)	Angle of Reflection (∠r)
1.	30°	30°
2.	35°	35°
3.	45°	45°
4.	50°	50°
5.	60°	60°

Inference: Angle of incidence is always equal to the angle of reflection. This is known as the law of reflection.
Incident ray of light: Ray of light which falls on a polished surface.
Reflected ray of light: Ray of light which gets reflected from a polished surface.
Normal is a line at right angle to the reflecting surface at the point of incidence.
Angle of incidence: Angle made by incident ray with the normal.
Angle of reflection: Angle made by reflected ray with the normal.

Laws of Reflection: There are two laws of reflection :
Law 1: The angle of reflection (∠r) is equal to the angle of incidence (∠i)
Law 2: The incident ray, the reflected ray and the normal at the point of incidence all lie in the same plane.

Activity 5.

Let us experiment

Aim: To verify the second law of reflection, i.e. the Incident ray, Reflected ray and the normal all lie in the same plane.
Materials Required: A mirror strip, holder, ray box.

Procedure:

• Fix a white sheet of paper on a drawing board in such a way that a small portion of it project a little beyond the edge of the drawing board.
• Place the plane mirror strip on the sheet of paper and hold it vertically with a mirror stand.
• Throw a ray of light on the mirror using a ray box.
• Look at the reflected ray. The reflected ray extends the projected portion of the paper.
• Make a cut in the middle of the projected portion of the sheet.
• Bend that portion of the projected sheet on which the reflected light falls.
• The reflected ray of light is not seen on the bent portion of the sheet.

Inference: The entire sheet fixed on to the drawing board represents a plane. The incident, reflected ray and the normal lie in the plane of paper.

Activity 6.

Let us explore
Aim: To observe the reflection of multiple parallel rays fall on the plane mirror and spherical mirrors.
Materials Required: A plane mirror, a concave mirror, a convex mirror, stands for mirrors, a torch, a comb, a paper clip.

Procedure:

• Arrange the setup as mentioned in Activity 4 again.
• Instead of a single slit, you have to leave many opening of the comb uncovered so that you may obtain multiple parallel beams of light.
• Now, we will fall these multiple parallel beams of light upon the plane mirror, concave mirror, and convex mirror, respectively. Observe the reflected beams. Is your observation same as what is shown in figure (b), (c) and (d)?

Observations:

Case I: Multiple reflected beams are also parallel [Figure (b)].
Case II: When multiple beams of light fall upon a concave mirror, the multiple refiected beams get closer, that is, they converge [Figure (c)].
Case III: In the case of a convex mirror, the multiple reflected beams spread, that is, they diverge [Figure (d)].

Inference:

• Spherical mirror also follows the laws of reflection.
• A concave mirror converge (get closer) the multiple beam of light after reflection.
• A convex mirror diverge (spread) the multiple beam of light after reflection.
• But in a plane mirror multiple reflected beam of light remains parallel.

Activity 7.

Let us explore

Aim: To show that concave mirror is a converging mirror.
Materials Required: A concave mirror, a sheet of thin paper or newspaper.

Procedure:
1. Take a concave mirror and move its reflecting surface towards the Sun. Let the light of the Sun be reflected by the mirror on the sheet of paper.

2. We should adjust the distance of the paper so that you can get a sharp bright spot on it see in figure.
3. You must hold the mirror and the sheet of paper contineously for a few minutes. Does the paper start to burn producing smoke?
Observation: The bright spot is formed on the paper and paper get ignited after few minutes.

Inference:

• Bright spot is due to concentration of reflected light from the sun. i.e., reflected light converged at a point after reflection.
• This produce sufficient heat at this point which can ignite the paper.
• Bright spot is actually the image of the sun.

Precaution:

• You should always perform this activity under the guidance of a teacher or an adult.
• We should not look towards the Sun or into the mirror reflecting the Sun.
• We should focus the reflected light only on a piece of paper, not towards anyone’s face or eyes.

Activity 8.

Let us explore
Aim: To read the printed letters through glass/plastic strip.
Materials Required: A flat strip of glass or clear plastic, such as a flat scale, few drops of oil, dropper, water and a paper or book.

Procedure:

• Take a glass or plastic strip.
• Spread few drops of oil on it and rub it to leave a very thin coating. We can also use wax instead of oil.
• Now with the help of a dropper or our finger, put a few drop of water on the oiled/waxed spot. (The oil/wax will help the water form a nice round drop.)
• Observe the water drop. What is the shape of its surface? Is it flat or curved inward or curved outward?
• Now, observe the paper underneath the glass/plastic strip. The paper or book should be directly under the water drop (see figure).
• Now, look down through the water drop at the text below. Do you find some change in the size of the letters just below the water drop? Do they look enlarged or smaller?

Observation:

• The surface of the water drop is curved outside.
• The letters under the water drop appear than the letters nearby.

Inference:

• The curved drop of water acts like a simple lens.
• This is similar to magnifying glass which is also a lens that helps in reading small print making the letters appear larger.

Activity 9.

Let us experiment
Aim: To show the formation of images by a lens when the objects are placed at different distances.
Materials Required: A convex lens, a concave lens, a lens holder or stand and a smaller object.

Procedure:
(A) For a convex lens

• Take a convex lens and mount it on a lens stand. Place it on a table.
• Place a lighted candle at a known distance (say 50 cm) from the lens.
• Place a white paper screen on the other side of the lens.
• Move the screen towards or away from the lens to get a sharp image of the candle flame.
• Since the image is obtained on the screen, therefore the image formed is real. Note down the positions of the lens, the candle, and that of the screen.
• Now move the candle by 10 cm towards the lens and try to obtain its image on the screen by moving the screen. Note down the positions of the lens, the candle and the screen.
• Repeat the experiment by changing the position of the candle and that of the screen.

Observation:

Inference:

• Convex lens forms a real and inverted image of the object when placed away from the lens.
• Convex lens forms virtual and erect image when the object is placed very near to the lens.

Procedure:

• Mount a concave lens on a lens stand and keep it on a table.
• Place a lighted candle at a known distance (say 50 cm) from the lens.
• Place white paper sheet on the other side of the lens.
• Move the screen towards or away from the lens and look for the image of the candle flame. Image of the candle flame is not seen on the screen.
• Now fix the position of the screen and move the candle towards the lens. Image of the candle flame is not seen on the screen.
• Remove the screen and look into the concave lens from that side. An erect virtual and smaller image is seen through the concave lens.
• Look into the concave lens for different distances from the lens and record your observations.
  (i) When the object is far off from the lens.
  (ii) When the object is nearer to the lens.

Observations and Inference:

• The image formed by a concave lens cannot be obtained on a screen. Therefore, the images formed by a concave lens are virtual.
• The image formed by a concave lens for all distances from the lens is erect and smaller in size than the object.

Activity 10.

Let us investigate
Aim: To show that convex lens is a congerging lens and concave lens is a diverging lens.
Materials Required: A thin transparent glass plale, a convex lens, a concave lens, a torch and a comb, a paper clip, two books of same size, white sheet of paper.

Procedure:

• Hold a comb upright with a paper clip to obtain multiple parallel beams of light.
• With the help of two books placed adjacent to each other, fix the glass plate or lens upright in between them see in figure and spread paper sheets on both books.
• Now pass the multiple parallel beams of light through the thin glass plate, convex lens, and concave lens one by one as shown in figure. Does the parallel beam of light pass through without any change in its direction in all three cases?
• Note down and analyse your observations.

Observation:

• The beam of light passes through the thin glass place without any deviation (i.e., no change in its path).
• In the convex lens, beam of light converges after passing through it (i.e., come closer).
• In concave lens, beam of light diverges after passing through it (i.e., beam spread).

Inference: A convex lens is called a converging lens and a concave lens is called a diverging lens. We draw the following figures for above Activity-10.

Activity 11.

Let us investigate
Aim: To show that a convex lens converges sunlight at a point and can burn a piece of paper.
Materials Required: A convex lens, a sheet of thin paper or newspaper.

Procedure:

• Hold a comb upright with a paper clip to obtain multiple parallel beams of light.
• With the help of two books placed adjacent to each other, fix the glass plate or lens upright in between them see in figure and spread paper sheets on both books.
• Now pass the multiple parallel beams of light through the thin glass plate, convex lens, and concave lens one by one as shown in figure. Does the parallel beam of light pass through without any change in its direction in all three cases?
• Note down and analyse your observations.

Observation:

• The beam of light passes through the thin glass place without any deviation (i.e., no change in its path).
• In the convex lens, beam of light converges after passing through it (i.e., come closer).
• In concave lens, beam of light diverges after passing through it (i.e., beam spread).

Inference:

• Bright spot is due to concentration of reflected light from the sun. i.e., reflected light converged at a point after reflection.
• This produce sufficient heat at this point which can ignite the paper.
• Bright spot is actually the image of the sun.

Precautions: Do not look at the Sun directly or through the lens as it may damage your eyes.

Light: Mirrors and Lenses Class 8 Extra Questions and Answers

Short Answer Type Questions

Question 1.
State the characteristics of the image formed by a plane mirror.
Answer:

• Plane mirror forms an erect image.
• It forms a virtual image.
• Size of the image is same as that of the object.
• Image gets formed at the same distance behind the mirror as the object stands in front of it.
• Image formed is a laterally inverted image. i.e. right hand side of the object seems to be the left hand side and vice versa.

Question 2.
What is a virtual image ? Give one situation where a virtual image is formed.
Answer:
The image which cannot be taken on a screen is called a virtual image. When some object is placed very close to the concave mirror we do not get any image of that object on the white screen placed in front of the mirror. Such image is called a virtual image.

Question 3.
What is the difference between virtual images produced by concave, plane and convex mirrors ?
Answer:
Virtual image produced by concave mirror is magnified, that produced by plane mirror is of the same size and the virtual image produced by convex mirror is diminished.

Question 4.
State one way in which the image formed in a convex mirror is similar to that in plane mirror and one way in which it is different.
Answer:
Similarity: Both convex mirror and plane mirror always form a virtual and erect image of an object.
Difference: Convex mirror always forms an image which is smaller than the object but a plane mirror always forms an image which is of the same size as the object.

Question 5.
Which type of mirror is used as a shaving mirror ? Support your answer with reason.
Answer:
Concave mirrors are used as shaving mirrors to see a large image of the face. This is because when the face is held within the focus of a concave mirror, then an enlarged image of the face is seen in the concave mirror. This helps in making a smooth shave.

Long Answer Type Questions

Question 1.
How can you distinguish between plane mirror, convex mirror and concave mirror by merely looking at the image formed in each case ?
Answer:
To distinguish between a plane mirror, a convex mirror and a concave mirror, the given mirror is held near the face and the image is seen.

• If the image is upright and same size as the object and, so it does not change in size when the mirror is moved, then the mirror is a plane mirror.
• If the image is upright magnified and it becomes inverted when the mirror is moved away from the face, then the mirror is a concave mirror.
• If the image is upright and diminished and it remains upright when the mirror is moved away from the face, then the mirror is a convex mirror.

Question 2.
Write the uses of solar concentrator.
Answer:

• Solar concentrators are devices which can concentrate sunlight to a small area with the help of mirrors and lenses.
• The concentrated sunlight is utilised to heat a liquid to produce steam which can be used to generate electricity or for providing heat for various purposes, such as large scale cooking or for solar furnaces.
• Solar furnaces are even used for melting steel! Do you remember learning in an earlier chapter, about electric furnaces for melting steel?

Case-Study Based Questions

1. Read the following passage carefully and answer the questions that follow :

If the reflecting surface of a spherical mirror is concave, it is called a concave mirror. If the reflecting surface is convex, then it is a convex mirror. The inner surface of a spoon acts like a concave mirror, while its outer surface acts like a convex mirror. The image of an object formed by a plane mirror cannot be obtained on a screen. Let us investigate if it is also true for the image formed by a concave mirror. The image formed by a plane mirror could not be obtained on a screen. Such an image is called a virtual image.

(i) Which of the following can be used to form a real image ?
(a) Concave mirror only
(b) Plane mirror only
(c) Convex mirror only
(d) Both concave and convex mirrors
Answer:
(a) Concave mirror only

(ii) Which of the statement is ture :
(a) If reflecting surface is convex then this a covex mirror
(b) Inner surface of spoon acts like concave
(c) Outer surface is spoon acts like convex mirror
(d) All of these
Answer:
(d) All of these

(iii) Which of the statement is false :
(a) The image formed by plane mirror can be obtained on screen
(b) Plane mirror image can not be obtained on screen
(c) An inverted image can be obtained by plan mirror
(d) None of these
Answer:
(b) Plane mirror image can not be obtained on screen

(iv) A spherical mirror having reflecting surface curved outward is a :
(a) plane mirror
(b) concave mirror
(c) convex mirror
(d) either concave or convex
Answer:
(c) convex mirror

Picture Based Questions

I. Observe the following picture and answer the questions.

(i) Angle of incidence is equal to angle of reflection. This is one of the :
(a) law of refraction
(b) normal law
(c) both (a) and (b)
(d) angle of reflection
Answer:
(d) angle of reflection

(ii) The angle between the normal and the reflected ray is called :
(a) angle of refraction
(b) angle
(c) angle of incidence
(d) angle of reflection
Answer:
(d) angle of reflection

(iii) Define angle of incidence?
Answer:
The angle between the normal and the incident ray is called angle of incidence.

Light: Mirrors and Lenses Class 8 MCQ

Multiple Choice Questions (Mcqs)

Question 1.
A virtual image: ………..
(a) can be formed on the screen
(b) cannot be formed on the screen
(c) is formed only by the plane mirror
(d) is formed only by the convex mirror
Answer:
(b) cannot be formed on the screen

Question 2.
Which of the following would you prefer to use while reading small letters found in a dictionary ?
(a) A concave mirror
(b) A concave lens
(c) A convex mirror
(d) A convex lens
Answer:
(d) A convex lens

Question 3.
The image of an object formed by a plane mirror is:
(a) virtual
(b) real
(c) diminished
(d) upside down
Answer:
(a) virtual

Question 4.
A diverging mirror is:
(a) a plane mirror
(b) a convex mirror
(c) a concave mirror
(d) none of the above
Answer:
(b) a convex mirror

Question 5.
The image formed by spherical mirror is virtual. The mirror will be:
(a) concave
(b) convex
(c) either concave or convex
(d) none of the above
Answer:
(c) either concave or convex

Assertion and Reasoning

These questions consist of two statements, each printed as Assertion (A) and Reason (R). While answering these questions, you are required to choose any one of the following four responses.

(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.
(c) Assertion (A) is correct but the Reason (R) is wrong.
(d) Assertion (A) is wrong but the Reason (R) is correct.

1. Assertion (A): When the object is placed very close to the lens, the image formed is virtual, erect and magnified.
Reason (R): This happens because the convex lens can form real and inverted image when the object place very close.
Answer:
(a) Assertion (A) and Reason (R) both are correct and reason is correct explanation for assertion.

2. Assertion (A): The light ray, which strikes any surface, is called the incident ray.
Reason (R): The ray that comes back from the surface after reflection is known as the reflected ray.
Answer:
(b) Assertion (A) and Reason (R) both are correct and reason is not correct explanation for assertion.

Fill in the blanks

1. An image that cannot be obtained on a screen is called ………….
Answer:
virtual image

2. Light travels in ………… lines.
Answer:
straight

3. Convex mirrors are ………… in the middle than at the edges whereas concave lenses are ………… in the middle than at the edges.
Answer:
thicker, thinner

4. The rear view mirror/side mirror in automobiles is a ………… mirror.
Answer:
convex

5. The inner surface of a steel spoon acts as a ………… mirror.
Answer:
concave

True or False

1. A concave lens can be used to produce an enlarged and erect image.
Answer:
False

2. A convex lens always produces a real image.
Answer:
False

3. The sides of an object and its image formed by a concave mirror are always interchanged.
Answer:
True

4. An object can be seen only if it emits light.
Answer:
False

5. A concave mirror can not be used as a magnifying mirror.
Answer:
False

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 3 Finding Common Ground MCQ improves accuracy in objective exams.

MCQ on Finding Common Ground Class 7

Finding Common Ground MCQ Class 7

Class 7 Maths Finding Common Ground MCQ

Question 1.
The correct prime factorisation of 168 is:
(a) 2 × 2 × 2 × 21
(b) 2 × 4 × 3 × 7
(c) 2 × 2 × 6 × 7
(d) 2 × 2 × 2 × 3 × 7
Solution:
(c) 2 × 2 × 6 × 7
Prime factorisation of number 168 using division method:

Thus, 168 = 2 × 2 × 2 × 3 × 7
Hence, the prime factorisation of 168 is 2 × 2 × 2 × 3 × 7

Question 2.
If the number 198 can be written as product of prime numbers as 198 = 2 × 3 × 11 × ______. The
missing prime factor is:
(a) 1
(b) 2
(c) 3
(d) 11
Solution:
(c) 3

Prime factorisation of number 198 using division method:
Thus, 198 = 2 × 3 × 3 × 11
Hence, the missing prime factor is 3.

Question 3.
The multiple of 5 × 3 × 7 is:
(a) 41
(b) 420
(c) (5 × 3 × 7) × 2
(d) Both (b) and (c)
Solution:
(d) Both (b) and (c)
The multiple of 5 × 3 × 7 is a number that contains all its factors.
Here, (5 × 3 × 7) × 2 and (5 × 3 × 7) × 4 i.e. 420, are multiples of 5 × 3 × 7.

Question 4.
21 mango trees, 42 apple trees and 56 orange trees have to be planted in rows such that each row contains the same number of trees of one variety only. Minimum number of rows in which the trees may be planted is:
(a) 3
(b) 15
(c) 17
(d) 20
Solution:
(c) 17
Maximum number of plants in each row
= HCF of 21, 42 and 56 = 7

Number of rows of mango trees = 21 ÷ 7 = 3
Number of rows of apple trees = 42 ÷ 7 = 6
Number of rows of orange trees = 56 ÷ 7 = 8
∴ Required number of rows = 3 + 6 + 8
= 17

Question 5.
All the factors of the number 96 are:
(a) 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48
(b) 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
(c) 1, 2, 4, 8, 16, 32, 64
(d) 1, 2, 3, 6, 8, 12, 24, 48, 96
Solution:
(b) 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96
Prime factorisation of 96 is given as
96 = 2 × 2 × 2 × 2 × 2 × 3
Taking combinations of prime factors, all factors of 96 are:
1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96

Question 6.
The HCF of numbers 144, 216 and 360 is:
(a) 72
(b) 24
(c) 36
(d) 12
Solution:
(a) 72
Prime factorisations of 144, 216 and 360 are:
144 = 2 × 2 × 2 × 2 × 3 × 3
216 = 2 × 2 × 2 × 3 × 3 × 3
360 = 2 × 2 × 2 × 3 × 3 × 5
∴HCF (144, 216, 360) = 2 × 2 × 2 × 3 × 3 = 72

Question 7.
The LCM of 57 and 84 is:
(a) 1584
(b) 1596
(c) 4688
(d) 4788
Solution:
(b) 1596
Using prime factorisation method, we have
57 = 3 × 19 and
84 = 2 × 2 × 3 × 7
∴ LCM = 2 × 2 × 3 × 7 × 19 = 1596

Question 8.
The LCM and HCF of 117, 693 and 819 are:
(a) LCM = 9007, HCF = 9
(b) LCM = 9009, HCF = 7
(c) LCM = 9009, HCF = 9
(d) LCM = 9007, HCF = 7
Solution:
(c) LCM = 9009, HCF = 9
Prime factorisations of 117, 693 and 819 are:
117 = 3 × 3 × 13
693 = 3 × 3 × 7 × 11
819 = 3 × 3 × 7 × 13
Thus, HCF = 3 × 3 = 9
The LCM is the product of the highest occurrence of each prime factor, i.e.
LCM = 3 × 3 × 7 × 11 × 13 = 9009

Question 9.
The HCF of two numbers is 40, and their LCM is 8400. If one of the numbers is 240, what is the other number?
(a) 1200
(b) 1400
(c) 1600
(d) 1800
Solution:
(b) 1400
Let the unknown number be x.
We know,
Product of two numbers = HCF × LCM
⇒ 240 × x = 40 × 8400
⇒ x = \(\frac{40 \times 8400}{240}=\frac{8400}{6}\) = 1400

Question 10.
The HCF of two numbers is 8. Which one of the following can never be their LCM?
(a) 24
(b) 48
(c) 56
(d) 60
Solution:
(d) 60
We know, LCM of numbers is a multiple of their HCF.
Since 60 is not a multiple of 8, it can never be the LCM.

Question 11.
There are some sweets in a jar. The sweets can be put into boxes of 3 or 7 with no sweets left over. When the sweets are put into boxes of 20, there are 5 sweets left over. What is the smallest possible number of sweets in the jar?
(a) 48
(b) 105
(c) 125
(d) 165
Solution:
(b) 105
If the sweets can be packed in boxes of 3 or 7 with no remainder, then the number of sweets in the jar is a multiple of LCM of 3 and 7, i.e. 21.
Clearly, the required number is multiple of 21 which leaves a remainder of 5 when divided by
20, i.e. 105.

Question 12.
Which of the following statements are correct?
(i) Prime factorisation of 120 is 2 × 2 × 2 × 3 × 5.
(ii) 18 is a factor of 120.
(iii) Prime factorisation of 126 is 2 × 3 × 3 × 11.
(iv) 42 is a factor of 126.
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Solution:
(a) (i) and (iv)
(i) Prime factorisation of 120 is given as:
120 = 2 × 2 × 2 × 3 × 5
∴ (i) is correct.

(ii) Prime factorisation of 18 is 2 × 3 × 3
For 18 to be a factor of 120, the product 2 × 3 × 3 must be the part of the prime factorisation of 120.
Since 3 × 3 does not appear in the prime factorisation of 120, 18 is not a factor of 120.
∴ (ii) is not correct.

(iii) Prime factorisation of 126 is given as:

Thus, 126 = 2 × 3 × 3 × 7
∴ (iii) is not correct.

(iv) All factors of 126 are 1,2, 3,6, 7,9, 14, 18, 21, 42, 63 and 126.
∴ (iv) is correct.

Question 13.
A rectangular plot measures 168 metres by 196 metres. Find which of the following correctly represent the side length (in metres) of possible square tile that can be used to pave the entire plot without cutting any tiles.
(i) 3
(ii) 14
(iii) 28
(iv) 6
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Solution:
(d) (ii) and (iii)
The possible square tiles that can be used to pave the entire plot without cutting any tiles includes HCF of the dimensions of the rectangular plot and its all factors.
HCF of 168 and 196 using common factor method:

∴ HCF (168, 196) = 2 × 2 × 7
All the factors of 28 are 1, 2, 4, 7, 14 and 28.
Thus, 14 and 28 are the possible tile sizes to cover the floor completely.

Question 14.
Which of the following statements are correct?
(i) 14 and 20 have exactly three common factors.
(ii) The LCM of 14 and 20 is 140.
(iii) 4 × 5 × 7 is a multiple of both 14 and 20.
(iv) 40 is a factor of LCM (4, 5, 10).
Choose the correct option from the following:
(a) (i) and (iv) only
(b) (i), (iii) and (iv)
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Solution:
(c) (ii) and (iii) only
(i) As 14 = 2 × 7; 20 = 2 × 2 × 5
Common factors of 14 and 20 are 1 and 2 only.

(ii) We have, 14 = 2 × 7; 20 = 2 × 2 × 5
LCM of 14 and 20 = 2 × 2 × 5 × 7 = 140

(iii) As, 14 = 2 × 7 and 20 = 2 × 2 × 5
4 × 5 × 7 is the LCM of 14 and 20.
Thus, it is a multiple of both 14 and 20.

(iv) Prime factorisations of 4, 5 and.10 are:
4 = 2 × 2; 5 = 5; 10 = 2 × 5
The LCM is the product of the highest occurrence of each prime factor, i.e.
LCM (4, 5, 10) = 2 × 2 × 5 = 20
So, 40 is a multiple of 20 (not a factor).
Hence, statements (ii) and (iii) are correct.

Finding Common Ground Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The HCF of numbers 31 and 43 is 1.
(R): HCF of two prime numbers is always 1.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that a prime number has two factors only i.e. 1 and the number itself.
∴ HCF of two prime numbers is always 1.
Here, 31 and 43 are prime numbers.
∴ HCF(31, 43) = 1
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The HCF of numbers 313 and 314 is 1.
(R): The HCF of any two consecutive numbers is always 1.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know, the HCF of two consecutive natural numbers is always 1 because they have no common factor other than 1.
∴ HCF (313, 314) = 1
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): Two traffic lights change simultaneously every 60 seconds and 90 seconds. They will both change together again after 6 minutes.
(R): The time after which repeating events coincide is a multiple of the LCM.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Prime factorisations of 60 and 90 are:
60 = 2 × 2 × 3 × 5; 90 = 2 × 3 × 3 × 5
LCM is the product of the highest occurrence of each prime factor, i.e.
LCM (60, 90) = 2 × 2 × 3 × 3 × 5 = 180
So, both traffic lights will change together again after 180 seconds, i.e. 3 minute.
Thus, both lights will change together again after 3 minutes, 6 minutes, 9 minutes… (Multiples of 3).
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): If two numbers are co-prime, their HCF is 1 and their LCM is equal to their product.
(R): Co-prime numbers have no common factors except 1.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
When two numbers are co-prime, they do not their highest common factor (HCF) is 1.

Since there are no common factors to be shared by two numbers, all the prime factors of both numbers, taken to their highest occurrence, are included while finding the LCM. As a result, the LCM becomes equal to the product of the two numbers.

Hence, both Assertion (A) and Reason (R) are true, and Reason (R)

Finding Common Ground Class 7 Fill in the Blanks

Question 1.
The HCF (336, 378) has the prime factorisation: 2 × 3 × ________ .
Solution: 7
Prime factorisations of 336 and 378 using division method:
336 = 2 × 2 × 2 × 2 × 3 × 7
378 = 2 × 3 × 3 × 3 × 7

∴ HCF (336, 378) = 2 × 3 × 7

Question 2.
Three mugs contain 145 mL, 235 mL and 495 mL of water respectively. The maximum capacity of a mug that can exactly measure the water of three mugs is _______ mL.
Solution: 5
We have, 145 = 5 × 29 ; 235 = 5 × 47 ;
495 = 5 × 9 × 11
Required capacity = HCF of 145, 235 and 495 = 5
Hence, the maximum capacity of a mug that can exactly measure the water of three mugs is 5 mL.

Question 3.
546 = 2 × 3 × ________ × 13.
Solution: 7
Prime factorisation of 546 is given as

Thus, 546 = 2 × 3 × 7 × 13

Question 4.
If the LCM of x and y is 720, their HCF is 12, and x is 144, then y is _____ .
Solution: 60
We know, product of two numbers = HCF × LCM
⇒ x × y = HCF × LCM
⇒ 144 × y = 12 × 720
⇒ y = \(\frac{12 \times 720}{144}\) ⇒ y = \(\frac{720}{12}\) = 60

Question 5.
_______ is the smallest number which, when divided by 12 and 18, leaves a remainder of 5 in each case.
Solution: 41
Required smallest number = LCM(12, 18) + 5
The LCM of 12 and 18 using prime factorisation method:
12 = 2 × 2 × 3
18 = 2 × 3 × 3
The LCM is the product of the highest occurrence of each prime factor.
Hence, LCM (12, 18) = 2 × 2 × 3 × 3 = 36
Thus, the smallest number = 36 + 5 = 41

Question 6.
If the product of two numbers is 600 and their HCF is 10, then their LCM is ______.
Solution: 60
Let two numbers be A and B, then according to question A × B = 600 and HCF =10.
Using the property: A × B = HCF × LCM
⇒ 600 = 10 × LCM
⇒ LCM = \(\frac{600}{10}\) = 60

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 5 Prime Time Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 5 Prime Time Solutions

Ganita Prakash Class 6 Chapter 5 Solutions

Class 6 Maths Ganita Prakash Chapter 5 Solutions Prime Time

Question 1.
Who am I?
(a) I am a number less than 40. One of my factors is 7. The sum of my digits is 8.
(b) I am a number less than 100. Two of my factors are 3 and 5. One of my digits is 1 more than the other.
Solution:
(a) The numbers less than 40 whose one of the factor is 7 are 7, 14, 21, 28, 35. Out of these numbers the sum of digits of 35 is 8.
Hence, the number less than 40 whose one of the factors is 7 and sum of digits equals to 8 is 35.

(b) The number less than 100 whose two factors 3 and 5 are 15, 30, 45, 60, 75, 90. Out of these numbers, the number 45 has one of its digits 1 more than the other.
Hence, the number less than 100 whose two factors are 3 and 5 and one of its digits is 1 more than the other is 45.

Question 2.
In the diagram below, Guna has erased all the numbers except the common multiples. Find out what those numbers could be and fill in the missing numbers in the empty regions.

Solution:

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48.
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36 and 72.
Common factors of 24, 48 and 72 are 1, 2, 3, 4, 6, 8, 12, 24.
So, in multiples of 6 and 8, common multiples are 24, 48 and 72.
Other possibilities of the numbers are 3 and 8, 3 and 24, 4 and 6, 4 and 24, 6 and 24, etc.

Question 3.
Find the smallest number that is a multiple of all the numbers from 1 to 10.
Solution:
To find the smallest number that is a multiple of all numbers from 1 to 10, we need to determine least common multiple (LCM) of the numbers from 1 to 10.
Prime factorisation of numbers from 1 to 10.
1 = 1; 2 = 2; 3 = 3; 4 = 2 × 2; 5 = 5; 6 = 2 × 3; 7 = 7; 8 = 2 × 2 × 2; 9 = 3 × 3; 10 = 2 × 5
LCM (1 to 10) = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2520 Thus, the required smallest number is 2520.

Question 4.
Write three pairs of prime numbers less than 20 whose sum is a multiple of 5.
Solution:
Prime numbers less than 20 are 2, 3, 5, 7, 11, 13, 17 and 19.
Now, 2 + 3 = 5 (multiple of 5)
2 + 13 = 15 (multiple of 5)
7 + 13 = 20 (multiple of 5)
Thus, three pairs are (2, 3), (2, 13) and (7, 13).

Question 5.
The numbers 13 and 31 are prime numbers. Both these numbers have same digits 1 and 3 . Find such pairs of prime numbers up to 100.
Solution:
Pairs of prime numbers having same digits upto 100 are:
17 and 71 (both have digits 1 and 7);
37 and 73 (both have digits 3 and 7)
79 and 97 (both have digits 7 and 9)

Question 6.
Twin primes are pairs of primes having a difference of 2. For example, 3 and 5 are twin primes. So are 17 and 19. Find the other twin primes between 1 and 100.
Solution:
Twin primes between 1 and 100 are
(3, 5) → 5 – 3 = 2;
(5, 7) → 7 – 5 = 2;
(17, 19) → 19- 17 = 2;
(29, 31) → 31 – 29 = 2;
(59, 61) → 61 – 59 = 2;
(71, 73) → 73 – 71 = 2

Question 7.
Observe that 3 is a prime number and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 give another prime? Find atleast five such examples.
Solution:
Five such examples are:
2 × 2+ 1 = 5 (prime number)
2 × 5 + 1 = 11 (prime number)
2 × 11 + 1 = 23 (prime number)
2 × 23 + 1 = 47 (prime number)
and 2 × 29 + 1 = 59 (prime number)

Question 8.
What is the smallest number whose prime factorisation has
(a) three different prime numbers?
(b) four different prime numbers?
Solution:
(a) The smallest three prime numbers are 2, 3 and 5.
Thus, the smallest number with exactly three different prime factors = 2 × 3 × 5 = 30

(b) The smallest four prime numbers are 2, 3, 5 and 7.
Thus, the smallest number with exactly four different prime factors = 2 × 3 × 5 × 7 = 210

Question 9.
The first number has prime factorisation 2 × 3 × 7 and the second number has prime factorisation 3 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
Prime factorisation of two numbers are 2 × 3 × 7 and 3 × 7 × 11.
Both the numbers have common factors 3 and 7. Therefore, they are not co-prime.
Since the prime factors of one number are not the prime factors of another number, therefore one of them does not divide the other.

Question 10.
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
Solution:
The smallest 4-digit palindrome is 1001 .
Checking divisibility by 4:
1001 (not divisible by 4);
1111 (not divisible by 4);
1221 (not divisible by 4)
1331 (not divisible by 4);
1441 (not divisible by 4);
1551 (not divisible by 4);
1661 (not divisible by 4);
1771 (not divisible by 4);
1881 (not divisible by 4);
1991 (not divisible by 4);
2002 (not divisible by 4);
2112 (divisible by 4)
Therefore, die smallest 4-digit palindromic number divisible by 4 is 2112 .
The largest 4-digit palindrome is 9999.
Checking divisibility by 4:
9999 (not divisible by 4);
9669 (not divisible by 4);
9339 (not divisible by 4);
9009 (not divisible by 4):
9889 (not divisible by 4);
9559 (not divisible by 4);
9229 (not divisible by 4);
8998 (not divisible by 4);
9779 (not divisible by 4);
9449 (not divisible by 4);
9119 (not divisible by 4);
8888 (divisible by 4)
Therefore, the largest 4 -digit palindromic number divisible by 4 is 8888.

Question 11.
The teacher asked if 14560 is divisible by all of 2, 4, 5, 8 and 10. Guna checked for divisibility of 14560 by only two of these numbers and then declared that it was also divisible by all of them. What could those two numbers be?
Solution:
If a number is divisible by 8 and 5, then it will also be divisible by 2, 4 and 10.
Divisibility by 8 ensures it is always divisible by 2 and 4.
Divisibility by 5 ensures it ends with 0 or 5. But since the number also needs to be divisible by 8 then it must end with 0.
A number ending with 0 is also divisible by 10.
Therefore, 8 and 5 are the required numbers.

InText Questions

Question 1.
Look at the table below. What do you notice?

(i) Is there anything common among the shaded numbers?
(ii) Is there anything common among the circled numbers?
(iii) Which numbers are both shaded and circled? What are these numbers called?
Solution:
(i) All shaded numbers are multiples of 3.
(ii) All circled numbers are multiples of 4.
(iii) Numbers (both shaded and circled) are 36, 48 and 60.
They are called ‘common multiples of 3 and 4’.

Question 2.
Grumpy and Jumpy are playing treasure finding game. Treasures are kept on two numbers. Jumpy gets the treasures only if he is able to reach both the numbers with the same jump size. Also, a jump size of 1 is not allowed.
Where should Grumpy place the treasures so that Jumpy cannot reach both the treasures? Check if these pairs are safe:
(a) 15 and 39
(b) 4 and 15
(c) 18 and 29
(d) 20 and 55
Solution:
Grumpy should place the treasures at co-prime numbers.
(a) Pair (15 and 39) is not safe because 15 and 39 are not co-prime as 3 is their common factor.
(b) Pair (4 and 15) is safe because 4 and 15 are co-prime.
(c) Pair (18 and 29) is safe because 18 and 29 are co-prime.
(d) Pair (20 and 55) is not safe because 20 and 55 are not co-prime as 5 is their common factor.

Question 3.
Observe the following thread art. The first diagram has 12 pegs and the thread is tied to every fourth peg (we say that the thread-gap is 4). The second diagram has 13 pegs and the thread- gap is 3. What about the other diagrams? Observe these pictures, share and discuss your findings in class. In some diagrams, the thread is tied to every peg. In some, it is not. Is it related to the two numbers (the number of pegs and the thread- gap) being co-prime?

Make such pictures for the following:
(a) 15 pegs, thread-gap of 10
(b) 10 pegs. thread-gap of’ 7
(c) 14 pegs, thread-gap of 6
(d) 8 pegs, thread -gap of 3
Solution:
Yes, when two numbers are co-prime, the thread is tied to every peg.

Question 4.
Find numbers between 330 and 340 that are divisible by 4. Also, find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4. What do you observe?
Solution:
Numbers between 330 and 340 that are divisible by 4 are 332 and 336
Numbers between 1730 and 1740 that are divisible by 4 are 1732 and 1736
Numbers between 2030 and 2040 that are divisible by 4 are 2032 and 2036
We observe that if number formed by last two digits of a number is divisible by 4, then the number is divisible by 4.

Question 5.
Find numbers between 120 and 140 that are divisible by 8. Also find numbers between 1120 and 1140, and 3120 and 3140, that are divisible by 8. What do you observe?
Solution:
Numbers between 120 and 140 that are divisible by 8 are 128 and 136.
Numbers between 1120 and 1140 that are divisible by 8 are 1128 and 1136.
Numbers between 3120 and 3140 that are divisible by 8 are 3128 and 3136.
We observe that if last three digits of a number is divisible by 8, then the number is divisible by 8.

Question 6.
Fill the grid with prime numbers only so that the product of each row is the number to the right of the row and the product of each column is the number below the column.

Solution:
(i)

7	5	3	105
2	5	2	20
2	5	3	30
28	125	18

(ii)

2	2	2	8
3	5	7	105
5	7	2	70
30	70	28

Prime Time Class 6 Extra Questions

Prime Time Class 6 Very Short Question Answer

Question 1.
Write all factors of each of the following numbers:
(i) 24
(ii) 32
Solution:
(i) We can write 24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
∴ 1,2,3, 4, 6, 8, 12 and 24 are the factors of 24.

(ii) We can write 32 = 1 × 32 = 2 × 16 = 4×8
1, 2, 4, 8, 16 and 32 are the factors of 32.

Question 2.
Write first five multiples of each of the following numbers.
(i) 9
(ii) 13
Solution:
(i) To obtain first five multiples of 9, we multiply it by 1,2, 3, 4 and 5 respectively.
9 × 1 = 9;
9 × 2 = 18;
9 × 3 = 27;
9 × 4 = 36;
9 × 5= 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45.

(ii) To obtain first five multiples of 13, we multiply it by 1, 2, 3, 4 and 5 respectively.
13 × 1 = 13;
13 × 2 = 26;
13 × 3 = 39;
13 × 4 = 52;
13 × 5 = 65
Hence, the first five multiples of 13 are 13, 26, 39, 52 and 65.

Question 3.
Show that 17 is a factor of 170017 without actual division.
Solution:
We can write
170017 = 170000 + 17 = 17 × 10000 + 17 × 1 = 17 × (10000 + 1) = 17 × 10001
Hence, 17 is a factor of 170017.

Question 4.
Write any three numbers that are multiples of 15 but not of 30.
Solution:
Multiples of 15 are 15, 30, 45, 60, 75, 90, … Multiples of 30 are 30, 60, 90, …
Thus, three numbers that are multiples of 15 but not of 30 are 15, 45 and 75.

Question 5.
Find the common factors of 16 and 28.
Solution:
We can write 16 = 1 × 16 = 2 × 8 = 4 × 4
∴ The factors of 16 are 1, 2, 4, 8 and 16.
And, 28 = 1 × 28 = 2 × 14 = 4 × 7
∴ The factors of 28 are 1, 2, 4, 7, 14 and 28.
Hence, the common factors of 16 and 28 are 1, 2 and 4.

Question 6.
Write five prime triplets such that the difference between the greatest and the smallest prime number is 6.
Solution:
{7, 11, 13}, (11, 13, 17}, {13, 17, 19}, {17, 19, 23} and {37, 41, 43}.

Question 7.
Write all prime numbers less than 100.
Solution:
The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97.

Question 8.
Write five pairs of co-prime numbers.
Solution:
Two numbers are said to be co-prime if they do not have a common factor other than 1. Thus, five pairs of co-primes are (2, 3), (4, 5), (5, 6), (9, 10), (10, 21).

Question 9.
Write two numbers whose product is 10000. The two numbers should not have 0 as the units digit.
Solution:
We can write 10000 = 100 × 100
= (2 × 2 × 5 × 5) × (2 × 2 × 5 × 5)
= (2 × 2 × 2 × 2) × (5 × 5 × 5 × 5)
= 16 × 625

Question 10.
Check the divisibility 100100 by 10:
Solution:
Since 100100 ends with the digit 0, it is divisible by 10.

Question 11.
Check the divisibility 412030 by 10:
Solution:
Since 412030 ends with the digit 0, it is divisible by 10.

Question 12.
Check the divisibility 718005 by 10:
Solution:
Since 718005 does not end with the digit 0, it is not divisible by 10.

Question 13.
Check the divisibility 123456 by 4:
Solution:
The number formed by the last two digits of 123456 is 56.
Since 56 = 4 × 14, it is divisible by 4.
∴ 123456 is divisible by 4.

Question 14.
Check the divisibility 725976 by 4:
Solution:
The number formed by the last two digits of 725976 is 76.
Since 76 = 4 × 19, it is divisible by 4.
∴ 725976 is divisible by 4.

Question 15.
Check the divisibility 985442 by 4:
Solution:
The number formed by the last two digits of 985442 is 42.
Clearly, 42 is not divisible by 4.
∴ 985442 is not divisible by 4.

Question 16.
Give an example of a number which is divisible by 2 but not by 4
Solution:
6

Question 17.
Give an example of a number which is divisible by 4 but not by 8.
Solution:
20

Question 18.
Give an example of a number which is divisible by both 4 and 8 but not by 32.
Solution:
48

Question 19.
Express each of the following as a sum of three unique prime numbers.
(i) 10
(ii) 25
(iii) 35
(iv) 49
Solution:
(i) 10 = 2 + 3 + 5
(ii) 25 = 5 + 7 + 13
(iii) 35 = 5 + 7 + 23
(iv) 49 = 13 + 17 + 19

Question 20.
Express each of the following as a sum of two unique prime numbers:
(i) 15
(ii) 30
(iii) 54
(iv) 80
Solution:
(i) 15 = 2 + 13
(ii) 30 = 7 + 23
(iii) 54 = 13 + 41
(iv) 80 = 19 + 61

Question 21.
Express each of the following as a sum of twin primes:
(i) 12
(ii) 36
(iii) 60
(iv) 84
Solution:
(i) 12 = 5 + 7
(ii) 36 = 17 + 18
(iii) 60 = 29 + 31
(iv) 84 = 41 + 43

Question 22.
Check the divisibility of the following numbers by 5:
(i) 235965
(ii) 783120
(iii) 415812
Solution:
(i) Since 235965 ends with the digit 5, it is divisible by 5.
(ii) Since 783120 ends with the digit 0, it is divisible by 5.
(iii) Since 415812 does not end with the digits 0 or 5, it is not divisible by 5.

Prime Time Class 6 Short Question Answer

Question 1.
Write all factors of each of the following numbers:
(i) 60
(ii) 420
Solution:
(i) We can write
60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15
= 5 × 12 = 6 × 10
∴ 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60 are the factors of 60.

(ii) We can write
420 = 1 × 420 = 2 × 210 = 3 × 140 = 4 × 105 = 5 × 84 = 6 × 70 = 7 × 60 = 10 × 42
= 12 × 35 = 14 × 30 = 15 × 28 = 20 × 21
∴ 1, 2, 3, 4, 5, 6, 7, 10, 12, 14, 15, 20, 21, 28, 30, 35, 42, 60, 70, 84, 105, 140, 210 and 420 are the factors of 420.

Question 2.
The product of two numbers is 42. Their sum is 17. What are the numbers?
Solution:
We can write 42 = 1 × 42 = 2 × 21 = 3 × 14 = 6 × 7.
∴ 1, 2, 3, 6, 7, 14, 21 and 42 are the factors of 42.
Out of all the above factors, only 3 and 14 add up to a total of 17.
Hence, the required numbers are 3 and 14.

Question 3.
Find the smallest number that is a multiple of all the numbers from 1 to 10 except 7.
Solution:
We know, 8 is a multiple of 1, 2, 4 and 8.
9 is a multiple of 1, 3 and 9.
8 × 9 is a multiple of 6.
5 × 8 × 9 is a multiple of 5 and 10.
[As 5 × 8 = 40 is a multiple of 5 and 10.]
∴ Required smallest number is 5 × 8 × 9 = 360.

Question 4.
Find the common factors of 12, 18 and 24.
Solution:
We can write 12 = 1 × 12 = 2 × 6 = 3 × 4
∴ The factors of 12 are 1, 2, 3, 4, 6 and 12.
18 = 1 × 18 = 2 × 9 = 3 × 6
∴ The factors of 18 are 1, 2, 3, 6, 9 and 18.
24 = 1 × 24 = 2 × 12 = 3 × 8 = 4 × 6
∴ The factors of 24 are 1,2, 3, 4, 6, 8, 12 and 24.
Hence, the common factors of 12, 18 and 24 are 1, 2, 3 and 6.

Question 5.
Find the common multiples of 6, 9 and 12.
Solution:
The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, …
The multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, ……
The multiples of 12 are 12, 24, 36, 48, 60, 72, 84, ………
Hence, the common multiples of 6, 9 and 12 are 36, 72, 108, … .

Question 6.
A number is divisible by both 6 and 14. By which other number will that number be always divisible?
Solution:
Since the number is divisible by 6 and 14, the number is a common multiple of 6 and 14. So, the number will always be divisible by the smallest common multiple of 6 and 14.
Now, the multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, … .
And, the multiples of 14 are 14, 28. 42, 56, 70, 84,
So, the common multiples of 6 and 14 are 42, 84, 126,….
∴ The smallest common multiple of 6 and 14 is 42.
Hence, 42 is the required number.

Question 7.
How many numbers between 1 and 100 have exactly three factors?
Solution:
As we know, prime numbers have exactly two factors, 1 and itself. So, squares of prime numbers will have exactly three factors.
2² = 4 has factors 1, 2 and 4.
3² = 9 has factors 1,3 and 9. .
5² = 25 has factors 1,5 and 25.
7² = 49 has factors 1, 7 and 49.
Thus, four numbers i.e. 4, 9, 25 and 49 between 1 and 100 have exactly three factors.

Question 8.
The product of two numbers is 36. Their difference is 9. What are the numbers?
Solution:
We can write 36 = 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6.
∴ 1, 2, 3, 4, 6, 9, 12, 18 and 36 are the factors of 36.
The difference between factors 12 and 3 is 9.
Hence, the required numbers are 3 and 12.
Note: The difference between factors 18 and 9 is also 9, but their product is not equal to 36.

Question 9.
Without actual division, show that 13 is a factor of each of the following numbers:
(i) 1313
(ii) 13013
(iii) 131313
Solution:
(i) We can write
1313 = 1300 + 13 = 13 × 100 + 13× 1 = 13(100 + 1) = 13 × 101
Hence, 13 is a factor of 1313.

(ii) We can write
13013 = 13000 + 13 = 13 × 1000 + 13 × 1 = 13(1000 + 1) = 13 × 1001
Hence, 13 is a factor of 13013.

(iii) We can write
131313 = 130000 + 1300 + 13 = 13 × 10000 + 13 × 100 + 13 × 1
= 13(10000 + 100 + 1) = 13 × 10101
Hence, 13 is a factor of’ 131313.

Question 10.
Find the common factors of:
(i) 12 and 18
(ii) 35 and 63
(iii) 60 and 210
Solution:
(i) We can write 12 = 1 × 12 = 2 × 6 = 3 × 4
∴ The factors of 12 are 1, 2, 3, 4, 6 and 12.
And, 18 = 1 × 18 = 2 × 9 = 3 × 6
∴ The factors of 18 are 1,2, 3, 6, 9 and 18.
Hence, the common factors of 12 and 18 are 1, 2, 3 and 6.

(ii) We can write 35 = 1 × 35 = 5 × 7
∴ The factors of 35 are 1, 5, 7 and 35.
And, 63 = 1 × 63 = 3 × 21 = 7 × 9
∴ The factors of 63 are 1. 3, 7, 9, 21 and 63.
Hence, the common factors of 35 and 63 are 1 and 7.

(iii) We can write 60 = 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10
∴ The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
210 = 1 × 210 = 2 × 105 = 3 × 70 = 5 × 42 = 6 × 35 = 7 × 30 = 10 × 21 = 14 × 15
∴ The factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
Hence, the common factors of 60 and 210 are 1, 2, 3, 5, 6, 10, 15 and 30.

Question 11.
Find first three common multiples of:
(i) 10 and 20
(ii) 25 and 50
(iii) 40 and 60
Solution:
(i) The multiples of 10 are 10, 20, 30, 40, 50, 60,
The multiples of 20 are 20, 40, 60, 80, …
Hence, the first three common multiples of 10 and 20 are 20, 40 and 60.

(ii) The multiples of 25 are 25, 50, 75, 100, 125, 150, …
The multiples of 50 are 50, 100, 150, 200, …
Hence, the first three common multiples of 25 and 50 are 50, 100 and 150.

(iii) The multiples of 40 are 40; 80, 120, 160, 200, 240, 280, 320, 360 …
The multiples of 60 are 60, 120, 180, 240, 300, 360 …
Hence, the first three common multiples of 40 and 60 are 120, 240 and 360.

Question 12.
A number is divisible by both 8 and 12. By which other numbers will that number be always divisible?
Solution:
Since the number is divisible by 8 and 12, the number is a common multiple of 8 and 12. So, the number will always he divisible by the smallest common multiple of 8 and 12.
Now, the multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72 …
The multiples of 12 are 12, 24, 36, 48, 60, 72, …
So, the common multiples of 8 and 12 are 24, 48, 72, …
∴ The smallest common multiple of 8 and 12 is 24.
Hence, the number is divisible by 24 and hence by all factors of 24, i.e., 1, 2, 3, 4, 6, 8, 12 and 24.

Question 13.
Write all pairs of twin primes between 1 and 100.
Solution:
The prime numbers less than 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83,89 and 97.
The required pairs of twin primes are (3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61) and (71, 73).

Question 14.
Write all pairs of prime numbers less than 30 whose sum is a multiple of 5.
Solution:
The prime numbers less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Now, 2 + 3 = 5, 2 + 13 = 15, 2 + 23 = 25, 3 + 7=10,
3 + 17 = 20, 7 + 13 = 20, 7 + 23 = 30, 11 + 19 = 30,
11 + 29 = 40, 13 + 17 = 30, 17 + 23 = 40
∴ The required pairs are (2, 3), (2, 13), (2, 23), (3, 7), (3, 17), (7, 13), (7, 23), (1 1, 19), (11, 29), (13, 17) and (17, 23).

Question 15.
Determine the prime factorisation of the following numbers:
(i) 3094
(ii) 5082
(iii) 10010
Solution:
(i) The prime factorisation of 3094 is 2 × 7 × 13 × 17.
(ii) The prime factorisation of 5082 is 2 × 3 × 7 × 11 × 11.
(iii) The prime factorisation of 10010 is 2 × 5 × 7 × 11 × 13.

Question 16.
Find prime factorisation of the following numbers without multiplying first:
(i) 48 × 60
(ii) 56 × 75
(iii) 81 × 25
Solution:
(i) Prime factorisation of 48 = 2 × 2 × 2 × 2 × 3
Prime factorisation of 60 = 2 × 2 × 3 × 5
Thus, prime factorisation of 48 × 60 is 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5.

(ii) Prime factorisation of 56 = 2 × 2 × 2 × 7
Prime factorisation of 75 = 3 × 5 × 5
Thus, prime factorisation of 56 × 75 is 2 × 2 × 2 × 3 × 5 × 5 × 7.

(iii) Prime factorisation of 81 = 3 × 3 × 3 × 3
Prime factorisation of 25 = 5 × 5
Thus, prime factorisation of 81 × 25 is 3 × 3 × 3 × 3 × 5 × 5.

Question 17.
The first number has prime factorisation 2 × 5 × 7 and the second number has prime factorisation 5 × 7 × 11. Are they co-prime? Does one of them divide the other?
Solution:
Since 5 and 7 are the common prime factors of two numbers, they are not co-prime.
Clearly, 2 is a prime factor of first number but not a prime factor of second number.
Hence, second number is not divisible by first number.
And, 1 1 is a prime factor of second number but not a prime factor of first number.
Hence, first number is not divisible by second number. So, none of them divide the other number.

Question 18.
Check the divisibility of the following numbers by 8:
(i) 6245826
(ii) 2727272
(iii) 2491664
Solution:
(i) The number formed by the last three digits of 6245826 is 826.
Clearly, 826 is not divisible by 8.
∴ 6245826 is not divisible by 8.

(ii) The number formed by the last three digits of 2727272 is 272.
Since 272 = 8 x 34, it is divisible by 8.
∴ 2727272 is divisible by 8.

(iii) The number formed by the last three digits of 2491664 is 664.
Since 664 = 8 X 83, it is divisible by 8.
∴ 2491664 is divisible by 8.

Prime Time Class 6 Long Question Answer

Question 1.
Who am I?
(i) I am a number less than 50. One of my factors is 11. The sum of my digits is 8.
(ii) I am an even number less than 100. One of my factors is 7. One of my digits is 1 more than the other.
Solution:
(i) Numbers less than 50 and having 11 as a factor are
11 × 1 = 11, 11 × 2 = 22, 11 × 3 = 33, 11 × 4 = 44
Out of these numbers, only 44 is the number whose sum of digits is 8.
Hence, the required number is 44.

(ii) Even numbers less than 100 and having 7 as a factor are
7 × 2 = 14, 7 × 4 = 28, 7 × 6 = 42, 7 × 8 = 56, 7 × 10 = 70, 7 × 12 = 84, 7 × 14 = 98
Out of these numbers, only 56 and 98 are the numbers whose one digit is 1 more than the other.
Hence, the required number is either 56 or 98.

Question 2.
Using prime factorisation, check whether the following pairs of numbers are co-prime or not:
(i) 154 and 195
(ii) 132 and 225
(iii) 357 and 286
Solution:
(i) Prime factorisations of 154 and 195 are as follows:
154 = 2 × 7 × 11 and 195 = 3 × 5 × 13
Since there are no common prime factors, 154 and 195 are co-prime.

(ii) Prime factorisations of 132 and 225 are as follows:
132 = 2 × 2 × 3 × 11 and 225 = 3 × 3 × 5 × 5
Since 3 is a common prime factor, 132 and 225 are not co-prime.

(iii) Prime factorisations of 357 and 286 are as follows:
357 = 3 × 7 × 17 and 286 = 2 × 11 × 13
Since there are no common prime factors, 357 and 286 are co-prime.

Question 3.
Using prime factorisation method, determine whether the first number is divisible by the second number or not:
(i) 420 and 105
(ii) 693 and 78
(iii) 990 and 90
Solution:
(i) Prime factorisations of 420 and 105 are as follows:
420 = 2 × 2 × 3 × 5 × 7 and 105 = 3 × 5 × 7
Clearly, all prime factors of 105 are prime factors of 420 and prime factorisation of 105 is included in the prime factorisation of 420.
We can write, 420 = (3 × 5 × 7) × 2 × 2 = 105 × 4
Hence, 420 is divisible by 105.

(ii) Prime factorisations of 693 and 78 are as follows:
693 = 3 × 3 × 7 × 11 and 78 = 2 × 3 × 13 Clearly, 13 is a prime factor of 78 but not a prime factor of 693.
Hence, 693 is not divisible by 78.

(iii) Prime factorisations of 990 and 90 are as follows:
990 = 2 × 3 × 3 × 5 × 11 and 90 = 2 × 3 × 3 × 5
Clearly, all prime factors of 90 are prime factors of 990 and prime factorisation of 90 is included in the prime factorisation of 990.
We can write, 990 = (2 × 3 × 3 × 5) × 11 = 90 × 11
Hence, 990 is divisible by 90.

Question 4.
Using prime factorisation, check whether the following pairs of numbers are co-prime or not:
(i) 308 and 585
(ii) 396 and 450
(iii) 1071 and 1430
Solution:
(i) Prime factorisations of 308 and 585 are as follows:
308 = 2 × 2 × 7 × 11 and 585 = 3 × 3 × 5 × 13
Since there are no common factors in these prime factorisations, 308 and 585 are co-prime.

(ii) Prime factorisations of 396 and 450 are as follows:
396 = 2 × 2 × 3 × 3 × 11 and 450 = 2 × 3 × 3 × 5 × 5
Since 2 and 3 are common factors in these prime factorisations, 396 and 450 are not co-prime.

(iii) Prime factorisations of 1071 and 1430 are as follows:
1071 = 3 × 3 × 7 × 17 and 1430 = 2 × 5 × 11 × 13
Since there are no common factors in these prime factorisations, 1071 and 1430 are co-prime.

Question 5.
Using prime factorisation method, determine whether the first number is divisible by the second number or not:
(i) 1260 and 315
(ii) 7623 and 198
(iii) 3780 and 126
Solution:
(i) Prime factorisations of 1260 and 315 are as follows:
1260 = 2 × 2 × 3 × 3 × 5 × 7 and 315 = 3 × 3 × 5 × 7
Clearly, all prime factors of 315 are prime factors of 1260 and prime factorisation of 315 is included in the prime factorisation of 1260.
We can write, 1260 = (3 × 3 × 5 × 7) × 2 × 2 = 315 × 4
Hence, 1260 is divisible by 315.

(ii) Prime factorisations of 7623 and 198 are as follows:
7623 = 3 × 3 × 7 × 11 × 11 and 198 = 2 × 3 × 3 × 11
Clearly, 2 is a prime factor of 198 but not a prime factor of 7623.
Hence, 7623 is not divisible by 198.

(iii) Prime factorisations of 3780 and 126 are as follows:
3780 = 2 × 2 ×3 × 3 × 3 × 5 × 7 and 126 = 2 × 3 × 3 × 7
Clearly, all prime factors of 126 are prime factors of 3780 and prime factorisation of 126 is included in the prime factorisation of 3780.
We can write,
3780 = (2 × 3 × 3 × 7) × 2 × 3 × 5 = 126 × 30
Hence, 3780 is divisible by 126.

Question 6.
State whether the following statements are true or false. Give reasons.
(i) If a number is divisible by 4, then it must be divisible by 8.
(ii) If a number is divisible by 8, then it must be divisible by 4.
(iii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(iv) The sum of two consecutive odd numbers is always divisible by 4.
(v) Sum of two even numbers gives a multiple of 4.
(vi) Sum of two odd numbers gives a multiple of 4.
Solution:
(i) False.
For example, 28 is divisible by 4 but not by 8.

(ii) True.
If a number is divisible by 8, then it must be divisible by 4 because 4 is a factor of 8.

(iii) False.
For example, let us take 14 and 6. Their sum, 14 + 6 = 20, is divisible by 4. But neither 14 nor 6 are divisible by 4.

(iv) True.
Let 2n + 1 and 2n + 3 be two consecutive odd numbers, where n = 0, 1,2,3, … .
Then, their sum,
(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), is divisible by 4.
For example, 3 and 5 are two consecutive odd numbers, whose sum is 8 and 8 is divisible by 4.

(v) False.
For example, 12 and 6 are even numbers but their sum 18 is not a multiple of 4.

(vi) False.
For example, 13 and 5 are odd numbers but their sum 18 is not a multiple of 4.

Question 7.
Solve the prime puzzle given below:

Solution:
385 = 5 × 7 × 11;
190 = 2 × 5 × 19;
42 = 2 × 3 × 7;
285 = 3 × 5 × 19;
154 = 2 × 7 × 11;
70 = 2 × 5 × 7
∴ Solution of given prime puzzle is as follows:

Question 8.
Solve the prime puzzle given below:

Solution:
Prime factorisations of the given numbers are as follows:
231 = 3 × 7 × 11;
130 = 2 × 13 × 5;
170 = 2 × 5 × 17;
102 = 3 × 2 × 17;
182 = 2 ×7 × 13;
275 = 5 × 5 × 11
∴ Solution of given prime puzzle is as follows:

Question 9.
Solve the prime puzzle given below:

Solution:
Prime factorisations of the given numbers are as follows:
286 = 2 × 11 × 13;
30 = 2 × 3 × 5;
70 = 2 × 5 × 7;
42 = 2 × 3 × 7;
110 = 2 × 5 × 11;
130 = 2 × 5 × 13
∴ Solution of given prime puzzle is as follows:

Prime Time Class 6 Case Based Questions

Question 1.
A Greek mathematician Eratosthenes, who lived around 2200 years ago, gave a method to list the prime numbers. The method is called the Sieve of Eratosthenes. In the given figure, encircled numbers arc prime numbers.

Based on the above information, answer the following questions:
(i) How many prime numbers are there between 1 and 100?
(ii) Which digits can never appear in the units place of a prime number?
(iii) Which digit most frequently appears in the units place of prime numbers less than 100?
(iv) Write live consecutive composite numbers less than 100 so that there is no prime number between them.
Solution:
(i) There are 25 prime numbers between 1 and 100.
(ii) 0, 4, 6 and 8 never appear in the unit place of a prime number.
(iii) Digit 3 appears 7 times in the units place of prime numbers less than 100.
(iv) 24, 25, 26, 27, 28 or 32, 33, 34, 35, 36 or 48, 49, 50, 51, 52 or 54, 55,56,57, 58 or 62, 63, 64, 65, 66 or 74, 75, 76, 77, 78 or 84, 85, 86, 87, 88
or 90, 91,92, 93, 94 or 91,92, 93, 94, 95 or 92, 93, 94, 95,96

Question 2.
The school gardening club has 210 marigold plants and 150 sunflower plants. The club members want to create equal rows for each type of flower such that each row contains the same number of one kind of plant and no plants are left unused.
Based on the above information, answer the following questions:
(i) Can they arrange the marigold plants in rows of 2, 5 or 7?
(ii) What is the largest number of plants per row they can use for each type so that all plants are used and plants in each row are equal?
(iii) They decide to pack the flowers into minimum flower boxes, and each box must contain the same number of plants. Which number less than 10 can they choose so that both 210 and 150 divide evenly by it?
(iv) Perform prime factorisation of 150 and 210 and find the common prime factors.
Solution:
(i) It is given that the school gardening club has 210 marigold plants. For 210 marigold plants to be arranged in rows of 2, 5 or 7, 210 must be divisible by 2, 5 or 7.
As 210 ends with 0, it is divisible by 2 and 5 both.
Now, 210 ÷ 7 = 30. Thus, 210 is also divisible by 7.
Thus, the marigold plants can be arranged in rows of 2, 5 or 7.

(ii) We need the largest number of plants of the same type per row such that

• Each row has the same number of plants.
• There are no leftover plants.
• Thus, we are looking for the largest number that can:
• Evenly divide 210 (so marigold plants are used up completely)
• Evenly divide 150 (so sunflower plants are used up completely)

This number will be the highest common factor of 210 and 150.
Now, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
And, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150.
The highest common factor is 30. Thus, if we use 30 plants in each row we get 210 ÷ 30 = 7 rows of marigold plant and 150 ÷ 30 = 5 rows of sunflower plant and no plants are left.

(iii) It is given that each flower box must contain equal number of plants and the number must be less than 10. Thus, we need to find the common factors of 150 and 210 which are less than 10.
Now, the factors of 210 are 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.
And, the factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75 and 150.
Thus, the common factors less than 10 are 1,2, 3, 5 and 6. ‘
For minimum boxes, the club members must place maximum plants in each box.
Thus, club members can keep 6 plants in each box, so that the flower boxes are minimum and there are equal plants in each box.

(iv) Prime factorisation of 210 = 2 × 3 × 5 × 7.
Prime factorisation of 150 = 2 × 3 × 5 × 5.
Thus, the common prime factors of 210 and 150 are 2, 3 and 5.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 2 Operations with Integers MCQ improves accuracy in objective exams.

MCQ on Operations with Integers Class 7

Operations with Integers MCQ Class 7

Class 7 Maths Operations with Integers MCQ

Question 1.
If sum of two integers is – 5, and one of the integers is 2, then the other integer is:
(a) -2
(b) -3
(c) -7
(d) -8
Solution:
(c) -7
Given, sum of two integers is – 5 and one of the integers is 2.
Let the other integer be 6.
Then, 2 + 6 = – 5 ⇒ b = – 5 – 2 = – 7

Question 2.
A pair of integers that give a product of – 20 is:
(a) -4, -5,
(b) 4, -3
(c) -5, 4
(d) -2, -10
Solution:
(c) -5, 4
We know, ( + ) × (-) = (-) or (-) × (+) = (-)
Thus, one integer must be negative and other must be positive to get product as – 20.
Also, we know that 4 ⇒ 5 = 20.
Taking 5 as negative, we get 4 ⇒ (- 5) = – 20
Thus, the pair of integers is -5 and 4.

Question 3.
For a pair of numbers, if the sum is 10 and the difference is 4, then least number in the pair is:
(a) 5
(b) 6
(c) 7
(d) 3
Solution:
(d) 3
Let a and b be the two numbers. Then,
a + b = 10 …(i)
and a – b = 4 …(ii)
Now, (a + b) + (a – b) = 10 + 4
[On adding (i) and (ii) ]
⇒ 2 × a = 14 ⇒ a = \(\frac{14}{2}\) = 7
Substituting a = 7 in a + b = 10, we get
7 + b = 10 ⇒ b = 10 – 7 = 3
Hence, the numbers are 7 and 3, and the least number in the pair is 3.

Question 4.
The temperature at 12 noon at a certain place was 15°C. If it decreases at the rate of 3°C per hour at what time will it be – 12°C?
(a) 11 am
(b) 11 pm
(c) 10 am
(d) 9 pm
Solution:
(d) 9 pm
Given, temperature at 12 noon, i.e. 12 pm at certain place = 15°C
And, change in temperature in 1 hour = – 3°C
Suppose, – 12°C is the temperature x hours after 12 noon.
∴ x × (-3) = (- 12)- 15
⇒ x × (-3) = – 27 ⇒ x = 9
Thus, time after 9 hours
= 12 pm + 9 hours= 9 pm

Question 5.
What term is used to describe a pair of a positive and a negative token that cancel out each other in the token model?
(a) Additive inverse
(b) Opposite pair
(c) Zero pair
(d) Neutral set
Solution:
(c) Zero pair
A pair of a positive (green) and a negative (red) token is called zero pair.

Question 6.
(-3) × (-6) =
(a) 18
(b) -18
(c) 12
(d) -12
Solution:
(a) 18
We know (-) × (-) = ( + )
Thus, (- 3) × (- 6) = 3 × 6 = 18

Question 7.
If both integers are same, then sum = ____ × integer.
(a) 2
(b) 0
(c) 3
(d) 4
Solution:
(a) 2
Let both integers be a.
Then, sum = a + a = 2 × a
Hence, sum = 2 × integer

Question 8.
If -400 × 3 × 3 + 567 = – 3033, then 43 – 3600 + 567 = ?
(a) – 2990
(b) -2980
(c) -3000
(d) -2970
Solution:
(a) – 2990
Given, – 400 × 3 × 3 + 567 = – 3033
⇒ (- 3600) + 567 = – 3033
Adding 43 to both sides, we get
43 + (- 3600) + 567 = 43 + (- 3033)
⇒ 43 – 3600 + 567 = 43 – 3033 = – 2990

Question 9.
If the sum and difference for a pair of integers is the same, one of the integers, is:
(a) 1
(b) 2
(c) 3
(d) 0
Solution:
(d) 0
Let the first integer be a and the second integer be 6.
We are given that, sum = difference
∴ a + b = a – b
⇒ a + b – a + b = 0 ⇒ 2b = 0 ⇒ b = 0
Thus, one of the integers is 0.

Question 10.
What will be the sign of product if we multiply 37 negative integers and 63 positive integers?
(a) Sometimes positive
(b) Always negative
(c) Never positive
(d) Can’t be determined
Solution:
(b) Always negative
We know, product of odd number of negative integers is negative and product of any number of positive integers is positive.
∴ Product of 37 (odd) negative integers and 63 positive integers is always negative.

Question 11.
The value of (- 30) ÷ 10 is:
(a) 3
(b) -3
(c) 1
(d) 0
Solution:
(b) -3
(-30) ÷ 10 = -(30 + 10)
[∵ (-a) ÷ b = -(a ÷ b)]
= -(3) = -3

Question 12.
The value of 0 ÷ (- 12) is:
(a) 4
(b) -12
(c) 12
(d) 0
Solution:
(d) 0
0 ÷ (- 12) = -(0 ÷ 12)
[∵ a ÷ (-b) = -(a ÷ b)]
= -(0) = 0

Question 13.
An integer which when multiplied by – 9 gives 117, is:
(a) 13
(b) -1053
(c) -13
(d) 1053
Solution:
(c) -13
Let the required integer be x.
Then, x × (- 9) = 117
⇒ x = 117 ÷ (-9) = -(117 ÷ 9) = – 13
Thus, the required integer is – 13.

Question 14.
If 11 × (a + 4) = 11 × (- 3) + 11 × 4, then a is :
(a) -1
(b) -2
(c) – 3
(d) – 4
Solution:
(c) -3
11 × (a + 4) = 11 × a + 11 × 4
[Using distributive property of multiplication over addition]
Given, 11 × (a + 4) = 11 × (- 3) + 11 × 4
⇒ 11 × a + 11 × 4 = 11 × (-3) + 11 × 4
On comparing, we get a = (- 3).

Question 15.
The value of (-36) ÷ (-9) is:
(a) 4
(b) – 9
(c) 6
(d) -2
Solution:
(a) 4
(-36) ÷ (-9) = 36 ÷ 9[∵ (-a) ÷ (-b) = a ÷ b]
= 4

Question 16.
The value of [(- 7) + (- 5)] ÷ [(- 4) + 1] is:
(a) -7
(b) -5
(c) 7
(d) 4
Solution:
(d) 4
[(- 7) + (- 5)] ÷ [(- 4) + 1]
= (-7 – 5) ÷ (-4 + 1)
= (-12) ÷ (-3)
= 12 ÷ 3 [∵ (-a) ÷ (-b) = a ÷ b]
= 4

Question 17.
The value of [8 × (-5) + 15 × 2 + 2] ÷ [1 × (-4)] is:
(a) -15
(b) – 5
(c) 7
(d) 2
Solution:
(d) 2
[8 × (- 5) + 15 × 2 + 2] ÷ [1 × (- 4)]
= (-40 + 30 + 2) ÷ (- 4)
= (-40 + 32) ÷ (- 4)
= (- 8) ÷ (- 4)
= 8 ÷ 4 [∵ (-a) ÷ (-b) = a ÷ b]
= 2

Question 18.
Which of the following expressions are equal to -30?
(i) -20 – (-5 × 2)
(ii) (-6 × 10) + (6 × 5)
(iii) (-2 × 5) + (-4 × 5)
(iv) (-6) × 5
Choose the correct option from the following:
(a) (ii) and (iv) only
(b) (iii) and (iv) only
(c) (ii), (iii) and (iv)
(d) (i), (ii) and (iv)
Solution:
(c) (ii), (iii) and (iv)
(i) -20 – (-5 × 2) = -20 – (-10) = -20 + 10 = -10
(ii) (-6 × 10) + (6 × 5) = -60 + 30 = – 30
(iii) (-2 × 5) + (-4 × 5) = – 10 + (- 20)
= -10 – 20 = -30
(iv) (-6) × 5 = – 30
Thus, expressions (ii), (iii) and (iv) are equal to – 30.

Question 19.
Which of the following expressions result in a negative integer?
(i) (-2) × (-3)
(ii) 10 + 3 × (-7)
(iii) (-5) × 2 + 10
(iv) (-5) × 7
Choose the correct option from the following:
(a) (i) and (iv)
(b) (iii) and (iv)
(c) (i) and (ii)
(d) (ii) and (iv)
Solution:
(d) (ii) and (iv)
(i) (- 2) × (- 3) = 6, which is a positive integer.
(ii) 10 + 3 × (- 7) = 10 + (- 21) = 10 – 21
= -11, which is a negative integer.
(iii) (-5) × 2 + 10 = -10 + 10 = 0, which is neither a positive nor a negative integer.
(iv) (-5) × 7 = -35, which is a negative integer.
Thus, the expressions (ii) and (iv) result in a negative integer.

Question 20.
Which of the following expressions are equal to -25?
(i) – 20 – (5 × 7)
(ii) (-7 × 1) + (- 3 × 6)
(iii) (-5 × 10) + (5 × 5)
(iv) (-5) × 8
Choose the correct option from the following:
(a) (ii) and (iii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(a) (ii) and (iii)
(i) -20 – (5 × 7) = – 20 – 35 = – 55 + – 25
(ii) (-7 × 1) + (-3 × 6) = -7 + (-18) = -7 – 18 = -25
(iii) (-5 × 10) + (5 × 5) = -50 + 25 = – 25
(iv) (-5) × 8 = -(5 × 8) = -40 ≠ -25
Thus, expressions (ii) and (iii) are equal to – 25.

Question 21.
Which of the following expressions result in odd integer?
(i) (-7) × (-3)
(ii) 10 + 3 × (- 2)
(iii) (-4) × 2 + 8
(iv) (-3) × 9
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)
Solution:
(d) (i) and (iv)
(i) (-7) × (-3) = 21,
which is an odd integer.
(ii) 10 + 3 × (- 2) = 10 + (- 6) = 4, which is an even integer.
(iii) (-4) × 2 + 8 = -8 + 8 = 0, which is an even integer.
(iv) (- 3) × 9 = – 27, which is an odd integer.
Thus, the expressions (i) and (iv) result in odd integers.

Question 22.
Which of the following expressions are correct?
(i) 72 ÷ (- 8) = – 9
(ii) 36 ÷ (-12) = – 3
(iii) (-78) ÷ (-2) = 38
Choose the correct option from the following:
(a) (i), (ii) and (iii)
(b) (ii) and (iii) only
(c) (i) and (ii) only
(d) (i) only
Solution:
(c) (i) and (ii) only
(i) 72 ÷ (- 8) = – (72 ÷ 8) = -(9) = – 9
(Correct)
(ii) 36 ÷ (-12) = -(36 ÷ 12) = -3 (Correct)
(iii) (-78) ÷ (-2) = 78 ÷ 2 = 39 (Incorrect)

Operations with Integers Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The additive inverse of (-12) × (-3) + 4 is -40.
(R): The additive inverse of a is – a.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
(-12) × (- 3) + 4 = 36 + 4 = 40
We know, the additive inverse of a is – a.
∴ The additive inverse of 40 is – 40.
Hence, the additive inverse of (-12) × (-3) + 4 is – 40.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The value of (-3) × (-4) × 5 × 2 × (-1) is negative.
(R): The product of odd number of negative integers is positive.
Solution:
(c) A is true but R is false.
We know that the product of odd number of negative integers is always negative.
In product (-3) × (-4) × 5 × 2 × (-1), there are 3 (odd) negative integers.
So, the product is negative.
Thus, Assertion (A) is true, but Reason (R) is false

Operations with Integers Class 7 Fill in the Blanks

Question 1.
60 + 40 + ______ = 0
Solution: -100
Let the required integer be a.
Then, 60 + 40 + a = 0 ⇒ 100 + a = 0
We know that the sum of an integer and its additive inverse is 0.
∴ a = – 100
Thus, 60 + 40 + (-100) = 0.

Question 2.
(-1) × _____ = – 43
Solution: 43
We know, (-1) × a = -a, for all integers a.
Thus, (-1) × 43 = -43

Question 3.
(-9) × (-5) × 6 × (-3) = _______
Solution: -810
The product of odd number of negative integers is negative.
∴ (-9) × (-5) × 6 × (-3)
= -(9 × 5 × 6 × 3) = -810

Question 4.
If we multiply 17 positive integers and 92 negative integers, then the sign of the product is _________ .
Solution: positive
We know that the product of even number of negative integers is always positive.
Final product = (Product of 17 positive integers) × (Product of 92 negative integers)
= (Positive) × (Positive) = Positive
Thus, if we multiply 17 positive integers and 92 negative integers, then the sign of the product is
positive.

Question 5.
When two negative integers are added, we get a ______ integer.
Solution: negative
When two negative integers are added, we get a negative integer.

Question 6.
Find:
(i) -13 + 42 = _____
(ii) 47 + (-26) = ______
(iii) 91 – 19 = ______
(iv) -38 – (-29) = _____
Solution: 29, 21, 72, -9
(i) -13 + 42 = 29
(ii) 47 + (-26) = 21
(iii) 91 – 19 = 72
(iv) -38 – (-29) = – 9

Question 7.
(7 + 3) + (5 + 2) + ( _____ + 6)
= (3 + 6) + (5 + 7) + (2 + 9)
Solution: 9
RHS = (3 + 6) + (5 + 7) + (2 + 9)
= 3 + 6 + 5 + 7 + 2 + 9
= (7 + 3) + (5 + 2) + (6 + 9)
= (7 + 3) + (5 + 2) + (9 + 6) = LHS
∴ (7 + 3) + (5 + 2) + (9 + 6)
= (3 + 6) + (5 + 7) + (2 + 9)

Question 8.
-19 + 23 – 8 + _____ = 12
Solution: 16
Let the required number be x. Then
-19 + 23 – 8 + x = 12 ⇒ 23 – 27 + x = 12
⇒ -4 + x = 12 ⇒ x = 12 + 4= 16
∴ -19 + 23 – 8 + 16 = 12

Question 9.
(-8) × (-9) + (-3) × (-3) × (-3) = _________
Solution: 45, 9, -36
(-8) × (- 9) + (- 3) × (- 3) × (- 3)
= (8 × 9) – (3 × 3 × 3) = 72 – 27 = 45

Question 10.
(-4) × (10 – 5 – 4 + 8) = (-4) × ____ = ______
Solution: 9, -36
(-4) × (10 – 5 – 4 + 8) = (-4) × 9 = – 36

Question 11.
When one of the multiplier or the multiplicand is positive and the other is negative, their product is _______ .
Solution: negative
We know that, when one of the multiplier or the multiplicand is positive and the other is negative, their product is negative.

Question 12.
When both the multiplier and the multiplicand are negative, the product is __________ .
Solution: positive
When both the multiplier and the multiplicand are negative, the product is positive.

Question 13.
Fill in the blanks keeping in view the properties of multiplication and division of integers:
(i) (-2) × 5 = _____ × (-2)
(ii) (-4) × [(____) + (2)] = (-4) × (-5) + (______) × (2)
(iii) 100 × [(____) × (-45)] = [ ____ × (-4)] × (-45)
(iv) (-15) ÷ (_____) = (15) ÷ (-3)
Solution: 5, -5, -4, -4, 100, 3
(i) (-2) × 5 = 5 × (- 2)
[Using commutative law of multiplication]
(ii) (-4) × [(-5) + (2)] = (-4) × (- 5) + (-4) × (2)
[Using distributive law of multiplication]
(iii) 100 × [(-4) × (-45)] = [100 × (-4)] × (-45)
[Using associative law of multiplication]
(iv) (-15) ÷ (3) = (15) ÷ (-3)

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 4 Data Handling and Presentation Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 4 Data Handling and Presentation Solutions

Ganita Prakash Class 6 Chapter 4 Solutions

Class 6 Maths Ganita Prakash Chapter 4 Solutions Data Handling and Presentation

Question 1.
Shri Nilesh is a teacher. He decided to bring sweets to the class to celebrate the new year. The sweets shop nearby has jalebi, gulab jamun, gujiya, barfi, and rasgulla. He wanted to know the choices of the children. He wrote the names of the sweets on the board and asked each child to tell him their preference. He put a tally mark ‘ | ’ for each student and when the count reached 5, he put a line through the previous four and marked it as

Complete the table to help Shri Nilesh to purchase the correct numbers of sweets.
(a) How many students chose jalebi? _______
(b) Barfi was chosen by _______ students?
(c) How many students chose gujiya? _______
(d) Rasgulla was chosen by _______ students?
(e) How many students chose gulab jamun? _______
Solution:

(a) It is clear from the table, Jalebi was chosen by 6 students.
(b) It is clear from the table, Barfi was chosen by 3 students.
(c) Gujiya was chosen by 13 students.
(d) Rasgulla was chosen by 7 students.
(e) Gulab jamun was chosen by 9 students.

Question 2.
Magan Bhai sells kites at Jamnagar. Six shopkeepers from nearby villages come to purchase kites from him. The number of kites he sold to these six shopkeepers are given below.

Shopkeeper	Number of Kites sold
Chaman	250
Rani	300
Rukhsana	100
Jasmeet	450
Jetha Lai	250
Poonam Ben	700

Prepare a pictograph using the symbol to represent 100 kites.
Answer the following questions:
(a) How many symbols represent the kites that Rani purchased?
(b) Who purchased the maximum number of kites?
(c) Who purchased more kites, Jasmeet or Chaman?
(d) Rukhsana says Poonam Ben purchased more than double the number of kites that Rani purchased. Is she correct? Why?
Solution:
Required pictograph is given below.

(a) Since Rani purchased 300 kites, it is clear from the pictograph, 3 symbols represent the kites that Rani purchased.

(b) The shopkeeper who purchased the maximum number of kites is the one with the most symbols in pictograph. It is clear from the pictograph that Poonam Ben purchased the maximum number of kites (7 × 100 = 700 kites).

(c) Number of kites purchased by Jasmeet = 450 Number of kites purchased by Chaman = 250 Jasmeet purchased more kites.

(d) Number of kites purchased by Poonam Ben = 700 Number of kites purchased by Rani = 300
Hence, Rukhsana is correct as 700 kites is more than double of 300 kites (300 × 2 = 600).

Question 3.
Samantha visited a tea garden and collected data of the insects and critters she saw there. Here is the data she collected —

Help her prepare a bar graph representing this data.
Solution:
To prepare a bar graph representing the given data, follow the point given below.
(i) Draw a horizontal line labeled, ‘Insects and Critters’ with each type of insect or critter evenly spaced.

(ii) Draw a vertical line labeled ‘Number of insects and critters seen’ with numbers starting from 0 up to the maximum number seen (in this case, 10) evenly spaced.

(iii) For each insect or critter, draw a bar that rises to the number seen. For example, the bar of caterpillars should reach up to 10.
Required bar graph is given alongside.

Question 4.
Chinu listed the various means of transport that passed across the road in front of his house from 9 a.m. to 10 a.m.

(a) Prepare a frequency distribution table for the data.
(b) Which means of transport was used the most?
(c) If you were there to collect this data, how could you do it? Write the steps or process.
Solution:
(a) Frequency distribution table for the given data is given below:

Means of transport	Bike	Car	Bus	Auto rickshaw	Bicycle	Bullock Cart	Scooter
Frequency	13	6	4	8	8	2	9

(b) The bike has the highest frequency in the table, indicating it was the most common means of transport observed.

(c) To collect this data, you could follow the steps given below. ;
Observation Timeframe: I will choose a specific timeframe, such as 9 a.m. to 10 a.m., to observe the road traffic.
Recording Data: 1 will use a tally chart or counting app to record the type of transport passing by during that hour.
Categorisation: Then I will organise the data into categories (e.g., bike, car, scooter, bus, etc.).
Final Count: After the observation period, I will get the total the number of occurrences for each category.
Analysis: Finally, I prepare a frequency distribution table based on the recorded data for analysis.

Question 5.
The number of girl students in each class of a school is depicted by the pictograph.

Observe this pictograph and answer the following questions.
(a) Which class has the least number of girl students?
(b) What is the difference between the number of girls in Class 5 and 6?
(c) If two more girls were admitted in Class 2, how would the graph change?
(d) How many girls are there in Class 7?
Solution:
(a) The pictograph shows the least number of symbols for Class 8.
Hence, Class 8 has the least number of girl students.

(b) Girl students in Class 5 = 2.5 × 4 = 10 Girl students in Class 6 = 4 × 4 = 16
Hence, there are 6 more girls in Class 6 than in Class 5.

(c) Here, adding 2 more girls would increase the count by 2 and requires a half additional symbol.
The pictograph would show half additional symbol for class 2, assuming each symbol represent 4 girls.

(d) Girl students in Class 7 = 3 × 4 = 12.

Data Handling and Presentation Class 6 Extra Questions

Data Handling and Presentation Class 6 Very Short Question Answer

Question 1.
Define the following terms:
(i) Observations
(ii) Data
Solution:
(i) Observations are the individual pieces of information we collect.
For example, if we ask five friends their ages, each friend’s age is one observation.

(ii) Any collection of facts, numbers, measures, observations or other descriptions of things that convey information about those things is called data.

Question 2.
Define the following terms:
(i) Tabulation of data
(ii) Tally marks
Solution:
(i) Tabulation of data is the process of systematically arranging data into rows and columns within a table to make it easier to understand, compare and analyse.

(ii) Tally marks are short vertical lines used to count items. Each time we count one item, we draw one line (|). After four lines (| | | |), we draw a oblique line (\) across them to show five.

Question 3.
Define the following terms:
(i) Raw data
(ii) Array
Solution:
(i) Raw data is the information as we first collect it (original form), before we sort or arrange it.
For example, writing ages on a sheet of paper as friends tell you, without putting them in any order.

(ii) Organising the data by arranging it in ascending or descending order is called an array.
For example, Meenal arranged the shoe sizes of the students in ascending order as follows:
3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7

Question 4.
Define the following terms:
(i) Frequency of observation
(ii) Frequency distribution
Solution:
(i) Frequency of observation tells us how many times a particular value appears in the data. For example, if the age ‘12’ appears three times in the list, then the frequency of age ‘12’ is 3.
(ii) Frequency distribution is a table that shows each value (or group of values) and how often it appears (frequency). For example:

Age	Frequency
10	2
11	1
12	3

Question 5.
A die was thrown 20 times and the following outcomes were noted:
1, 2, 5, 1, 6, 2, 3, 4, 2, 4, 2, 5, 6, 6, 6, 2, 3, 1, 1, 5
Represent the above data in the form of frequency distribution.
Solution:

Outcome	Frequency
1	4
2	5
3	2
4	2
5	3
6	4

Question 6.
Enlist 1 advantage and 1 disadvantage of presentation of data through pictograph.
Answer:
1 advantage and 1 disadvantage of pictograph are as follows:

Advantage	Disadvantage
1. Pictorial representation makes it easier to understand the data.	1. Drawing a pictograph is time consuming.

Data Handling and Presentation Class 6 Short Question Answer

Question 1.
Given below is the data showing the number of children in 15 families of a colony.
2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
Arrange the above data in ascending order and then form a tally table.
Solution:
Given data (number of children in 15 families) is as follows: 2, 3, 1, 1, 2, 3, 2, 3, 3, 4, 1, 2, 2, 1, 3
(i) Data in ascending order: 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4.
(ii) Given data in tabular form is shown below:

Question 2.
The number of cricket bat and ball pairs, sold by a shopkeeper during a week are given below:

Day	Mon.	Tue.	Wed.	Thu.	Fri.	Sat.
Number of bat and ball pair sold	12	9	15	15	3	27

Decide the scaling and draw a pictograph to represent the given data.
Solution:

Question 3.
Total number of cows in five villages are as follows:

Village	A	B	C	D	E
Number of cows	40	60	20	100	120

Decide the scale and draw a pictograph to represent the given data.
Solution:

Question 4.
Following pictograph shows the data of the number of students who like a particular sport.

Answer the following questions:
(i) How many students like cricket?
(ii) How many more students like basketball than volleyball?
Solution:
(i) Number of symbols for cricket = 4
Students who like cricket = 4 × 3= 12

(ii) Number of symbols for basketball = 3
∴ Students who like basketball = 3 × 3 = 9
Number of symbols for volleyball = 2
∴ Students who like volleyball = 2 × 3 = 6
Difference = 9 – 6 = 3
Hence, 3 more students like basketball than volleyball.

Question 5.
The bar graph below shows the production of Kharif and Rabi crops (in tons) in 5 different states in India. Draw a table that represents the data in the bar graph.

Solution:
Table for the given bar graph can be created as follows:

States	Rabi crops (in tons)	Kharif crops (in tons)
West Bengal	30	60
Karnataka	45	40
Madhya Pradesh	20	54
Tamil Nadu	36	22
Uttar Pradesh	50	42

Data Handling and Presentation Class 6 Long Question Answer

Question 1.
The weights of newborn babies (in Kg) in a hospital on a particular day are as follows:
2.1, 2.5, 3.5, 2.7, 2.9, 3.1, 2.6, 2.5, 2.8, 2.3, 2.9, 3.3, 3.4, 2.7, 2.8
Answer the following questions:
(i) How many babies are born on that day?
(ii) Arrange the above data in descending order.
(iii) Determine the highest weight.
(iv) Determine the lowest weight.
(v) Determine the range of weight.
(vi) How many babies weigh below 2.5 kg?
Solution:
(i) There are 15 babies born on that day.

(ii) Writing the weights from highest to lowest:
3.5, 3.4, 3.3, 3.1, 2.9, 2.9, 2.8, 2.8, 2.7, 2.7, 2.6, 2.5, 2.5, 2.3, 2.1

(iii) The highest weight is the first weight in the descending order of weights.
Highest weight = 3.5 kg

(iv) The lowest weight is the last weight in the descending order of weights.
Lowest weight = 2.1 kg

(v) We know, range = highest value – lowest value
∴ Range of weight = highest weight – lowest weight
= 3.5 kg – 2.1 kg = 1.4 kg

(vi) The weights below 2.5 kg are 2.1 kg and 2.3 kg. So, the required number of babies is 2.

Question 2.
Study the following pictograph:

Answer the following questions:
(i) In which city was the rainfall maximum and how much?
(ii) In which city was the rainfall minimum and how much?
(iii) Which cities had rainfall of more than 50 cm?
Solution:
(i) Chennai has three full icons, which is maximum.
∴ Rainfall in Chennai = 3 × 50 cm = 150 cm

(ii) Delhi has one full icon, which is minimum.
∴ Rainfall in Delhi = 1 × 50 cm = 50 cm.

(iii) Mumbai has 2 full icons.
∴ Rainfall in Mumbai = 2 × 50 cm = 100 cm (which is greater than 50 cm)
Hyderabad has 1 full and 1 half icons.
∴ Rainfall in Hyderabad = 50 cm + 25 cm = 75 cm (which is greater than 50 cm) Chennai has 3 full icons.
∴ Rainfall in Chennai = 50 × 3 = 150 cm (which is greater than 50 cm)
Kolkata has 2 full and 1 half icons.
∴ Rainfall in Kolkata =100 cm + 25 cm =125 cm (which is greater than 50 cm)
Hence, Mumbai, Hyderabad, Chennai, and Kolkata had rainfall of more than 50 cm.

Question 3.
In a shop, there are 5 different types of toys. The number of each toy is given in the table below. Construct a bar graph representing this data. Use appropriate scaling.

Toys Name	Blocks	Cars	Dolls	Balls	Bikes
Number of Toys	40	48	60	35	30

Solution:
The required bar graph can be easily created using the following steps:
Step 1: On a graph sheet draw two mutually perpendicular lines, a horizontal and a vertical line.

Step 2: Label horizontal line as ‘Toys name’ & write names of toys from the table with equal gaps between them: Blocks, Cars, Dolls, Balls, Bikes. Label vertical line as ‘Number of Toys’.

Step 3: Scaling: Choose a suitable scale to show the number of toys.
Here, we take: 1 unit (1 big division) = 10 toys. Each big division is further divided into 10 sub-divisions, representing 1 toy.

Step 4: Calculate the height of each bar:
10 toys = 1 large divisions
∴ 1 toy = \(\frac{1}{10}\) large divisions
= 1 sub-division
Now, the height of the bar for Blocks
= 40 × (\(\frac{1}{10}\)) = 4 large divisions
The height of the bar for Cars
48 × (\(\frac{1}{10}\)) = 4 large divisions and 8 sub-divisions
The height of the bar for Dolls = 60 × (\(\frac{1}{10}\)) = 6 large divisions
The height of the bar for Balls = 35 × (\(\frac{1}{10}\)) = 3 large divisions and 5 sub-divisions
The height of the bar for Bikes = 30 × (\(\frac{1}{10}\))

Step 5: Now draw vertical bars for each toy using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 4.
The zookeeper in Delhi is preparing a presentation for higher officers on the number of animals in the zoo. He intends to create a bar graph for better visualisations. The number of each animal is given below in the table. Construct a bar graph for the table.

Animals	Number of Animals
Lion	8
Elephant	12
Gorilla	9
Zebra	15
Giraffe	7
Cheetah	6

Solution:
Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as ‘Animals’ and write the names of the animals from the table with equal gaps between them: Lion, Elephant, Gorilla, Zebra, Giraffe, Cheetah Label the vertical lines ‘ Number of Animals’.

Step 3 (Scaling) : Choose a suitable scale to show the number of animals.
Here, we take: 1 unit (1 large division)
= 2 animals
∴ 1 animal = \(\frac{1}{2}\) unit = 1 sub-division
This scale helps us draw the graph clearly.

Step 4: Calculate the height of each bar using the scale:
Height of the bar for the Lion = 8 ÷ 2 = 4 large divisions
Height of the bar for the Elephant = 12 ÷ 2 = 6 large divisions
Height of the bar for the Gorilla = 9 ÷ 2 = 4 large divisions and 1 subdivision
Height of the bar for the Zebra = 15 ÷ 2 = 7 large divisions and 1 sub-division
Height of the bar for the Giraffe = 7 + 2 = 3 large divisions and 1 subdivision
Height of the bar for the Cheetah = 6 + 2 = 3 large divisions

Step 5: Now draw vertical bars for each animal using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 5.
The following bar graph shows the revenue of your country from exports of various items in 1 year. (1 unit = 10 crore rupees)

(a) Write the information given by the bar graph in a table.
(b) What is the difference between the maximum revenue and the minimum revenue?
(c) What is the total revenue from exports?
Solution:
(a) The required table is shown below:

Items	Revenue (Rupees in crores)
Electronics	30
Foodgrains	70
Livestock	25
Metals	50
Softwares	80

(b) From the table, it can be observed that the maximum revenue was generated by the export of software, i.e. 80 crore rupees.
Also, the minimum revenue was generated by the exports of livestock, i.e. 25 crore rupees.
∴ The difference between the revenue from software and livestock = 80 – 25 = 55 crore

(c) Total revenue from exports can be calculated by summing up the revenues from all the commodities.
∴ Total revenue from exports = 30 + 70 + 25 + 50 + 80 = 255 crore rupees

Question 6.
The table shows how much money Imran’s family spends each month on different items:

Items	Expenditure (in ₹)
House rent	3,000
Food	3,400
Education	800
Electricity	400
Transport	600
Miscellaneous	1,200

Answer the following questions:
(a) Represent the given data in the form of a bar graph
(b) On which item does the Imran’s family spend the most and the second most?
(c) Is the cost of electricity about one half of the cost of education?
Solution:
(a) To represent this data in the form of a bar graph, here are the steps:

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘items’ with equal spacing between them and mark the corresponding expenditures along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length = ₹ 200, and then mark and write the corresponding values (₹ 200, ₹400, etc.) representing each unit length and calculate the heights of the bars for various items as shown below:

House rent	3000 ÷ 200	15 units
Food	3400 ÷ 200	17 units
Education	800 ÷ 200	4 units
Electricity	400 ÷ 200	2 units
Transport	600 ÷ 200	3 units
Miscellaneous	1200 ÷ 200	6 units

Here is the bar graph that we obtained based on the above steps.

(b) From the given data, we can say that on food Imran’s family spend the most and on house rent they spend the second most.

(c) Yes. As the cost of electricity is ₹ 400 and the cost of eduction is ₹ 800, we can say that the cost of electricity is about one-half the cost of education.

Question 7.
Your school is collecting data on various transportation options that students used to opt. The table below shows the data collected by you. Prepare a bar graph to be given to your school principal, also use appropriate scaling.

Transportation	Number of Students
Bus	200
Cycle	50
Walk	80
Auto	120
Car	40

Solution:
The required bar graph can be easily created using the following steps:

Step 1: Take a graph sheet and draw two mutually perpendicular lines, a horizontal line and a vertical line.

Step 2: Label the horizontal line as‘Transportation’ and write the names from the table with equal gaps between them: Bus, Cycle, Walk, Auto, Car.
Label the vertical line as ‘Number of Students’.

Step 3: (Scaling): Choose a suitable scale to show the number of students.
Here, we take:
1 unit (1 big division) = 20 students (This helps us fit large numbers like 200 easily on the graph.)

Step 4: Calculate the height of each bar:
= 200 × (\(\frac{1}{20}\)) = 10 large divisions
Height of the bar for the cycle
= 50 × (\(\frac{1}{20}\))= 2 large divisions and 5 sub-divisions
Height of the bar for the walk
= 80 × (\(\frac{1}{20}\)) = 4 large divisions
Height of the bar for the auto
= 120 × (\(\frac{1}{20}\)) = 6 large divisions
Height of the bar for the car
= 40 × (\(\frac{1}{20}\)) = 2 large divisions

Step 5: Now draw vertical bars for each mode of transportation using the heights calculated above. Make sure the bars are of equal width and there is equal spacing between them.

Question 8.
The table below displays the number of bicycles produced at a factory from 1998 to 2002.

Year	1998	1999	2000	2001	2002
No. of Bicycles	800	600	900	1100	1200

Answer the following questions:
(a) Create a bar graph to represent this information. Select your preferred scale.
(b) Which year had the greatest number of bicycles produced?
(c) Which year had the least number of bicycles produced?
Solution:
(a) To represent this data in the form of a bar

Step 1: Draw two perpendicular lines, one horizontal and one vertical.
Step 2: Along the horizontal line, mark the ‘years’ with equal spacing between them and mark the corresponding ‘No. of bicycles’ along the vertical line.
Step 3: Take bars of the same width, keeping a uniform gap between them.
Step 4: Choose a suitable scale along the vertical line. Let, 1 unit length =100 bicycles, and then mark and write the corresponding values (100 bicycles, 200 bicycles, etc.) representing each unit length and calculate the heights of the bars for various years as shown below:

1998	800 ÷ 100	8 units
1999	600 ÷ 100	6 units
2000	900 ÷ 100	9 units
2001	1100 ÷ 100	11 units
2002	1200 ÷ 100	12 units

(b) The greatest number of bicycles produced in 2002.
(c) The least number of bicycles produced in 1999.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 8 Working with Fractions MCQ improves accuracy in objective exams.

MCQ on Working with Fractions Class 7

Working with Fractions MCQ Class 7

Class 7 Maths Working with Fractions MCQ

Question 1.
The value of \(\frac{5}{6}\) of 30 is:
(a) 20
(b) 25
(c) 30
(d) 35
Solution:
(b) 25
We know, ‘of’ means multiplication.
∴ \(\frac{5}{6} \text { of } 30=\frac{5}{6} \times 30\) = 5 × 5 = 25

Question 2.
What is the value of \(\frac{5}{8}\) × 64 ?
(a) 35
(b) 40
(c) 45
(d) 50
Solution:
(b) 40
\(\frac{5}{8} \times 64\) = 5 × 8 = 40

Question 3.
The value of \(3 \frac{1}{2}+\frac{4}{3}-2 \frac{2}{5}\) is:
(a) \(\frac{89}{30}\)
(b) \(\frac{91}{30}\)
(c) \(\frac{87}{30}\)
(d) \(\frac{73}{30}\)
Solution:
(d) \(\frac{73}{30}\)
We can write, \(3 \frac{1}{2}=\frac{7}{2} \text { and } 2 \frac{2}{5}=\frac{12}{5}\)
∴ \(3 \frac{1}{2}+\frac{4}{3}-2 \frac{2}{5}=\frac{7}{2}+\frac{4}{3}-\frac{12}{5}=\frac{105}{30}+\frac{40}{30}-\frac{72}{30}\)
[∵ LCM of 2, 3 and 5 is 30.]
= \(\frac{105+40-72}{30}=\frac{73}{30}\)

Question 4.
The cost of one notebook is \(\frac{3}{4}\) rupees. What is the cost of 32 such notebooks?
(a) ₹20
(b) ₹22
(c) ₹24
(d) ₹26
Solution:
(c) ₹24
Given, the cost of one notebook is \(\frac{3}{4}\) rupees.
Therefore, the cost of 32 notebooks
= \(32 \times \frac{3}{4}\) = 8 × 3 = ₹24

Question 5.
The value of \(\frac{11}{18}+\left(\frac{7}{15} \times \frac{1}{4}\right)\) is:
(a) \(\frac{3}{180}\)
(b) \(\frac{56}{180}\)
(c) \(\frac{131}{180}\)
(d) \(\frac{9}{180}\)
Solution:
(c) \(\frac{131}{180}\)
Given, \(\frac{11}{18}+\left(\frac{7}{15} \times \frac{1}{4}\right)=\frac{11}{18}+\frac{7}{60}=\frac{110}{180}+\frac{21}{180}\)
[∵ LCM of 18 and 60 is 180]
= \(\frac{110+21}{180}=\frac{131}{180}\)

Question 6.
The reciprocal of \(\frac{8}{7}\) is:
(a) \(\frac{1}{8}\)
(b) \(\frac{7}{8}\)
(c) \(\frac{1}{7}\)
(d) \(\frac{8}{7}\)
Solution:
(b) \(\frac{7}{8}\)
The reciprocal of \(\frac{8}{7} \text { is } \frac{7}{8}\).

Question 7.
The value of \(5 \frac{2}{3} \div 1 \frac{5}{6}\) is:
(a) \(3 \frac{1}{11}\)
(b) \(2 \frac{2}{3}\)
(c) \(1 \frac{1}{6}\)
(d) \(3 \frac{2}{5}\)
Solution:
(a) \(3 \frac{1}{11}\)
We have, \(5 \frac{2}{3} \div 1 \frac{5}{6}\)
= \(\frac{17}{3} \div \frac{11}{6}=\frac{17}{8} \times \frac{6}{11}\)
[As \(5\frac{2}{3}=\frac{17}{3} \text { and } 1 \frac{5}{6}=\frac{11}{6}\)]
= \(\frac{17 \times 2}{1 \times 11}=\frac{34}{11}=3 \frac{1}{11}\)

Question 8.
\(\left(\frac{18}{6} \div \frac{3}{9}\right)+\left(\frac{21}{7} \div \frac{6}{4}\right)\) =
(a) 9
(b) 11
(c) 12
(d) 10
Solution:
(b) 11
We have, \(\left(\frac{18}{6} \div \frac{3}{9}\right)+\left(\frac{21}{7} \div \frac{6}{4}\right)\)
= \(\left(\frac{38}{6} \times \frac{3}{8}\right)+\left(\frac{21}{7} \times \frac{4}{6}\right)\)
= \(\frac{3 \times 3}{1 \times 1}+\frac{9 \times 2}{1 \times 3}\) = 9 + 2 = 11

Question 9.
The reciprocal of \(2 \frac{3}{4}\) is:
(a) \(\frac{11}{4}\)
(b) \(\frac{13}{4}\)
(c) \(\frac{4}{11}\)
(d) \(\frac{4}{13}\)
Solution:
(c) \(\frac{4}{11}\)
We have \(2\frac{3}{4}=\frac{(2 \times 4)+3}{4}=\frac{8+3}{4}=\frac{11}{4}\)
Now, reciprocal of \(\frac{11}{4} \text { is } \frac{4}{11}\)

Question 10.
The value of \(\frac{7}{8} \div 4\) is:
(a) \(\frac{7}{2}\)
(b) \(\frac{7}{32}\)
(c) \(\frac{4}{7}\)
(d) \(2\frac{1}{4}\)
Solution:
(b) \(\frac{7}{32}\)
We have, \(\frac{7}{8} \div 4=\frac{7}{8} \times \frac{1}{4}=\frac{7 \times 1}{8 \times 4}=\frac{7}{32}\)

Question 11.
The value of \(4 \frac{3}{5} \div 2 \frac{1}{4}\) is:
(a) \(2\frac{2}{45}\)
(b) \(3\frac{1}{5}\)
(c) \(1\frac{4}{9}\)
(d) \(2\frac{1}{4}\)
Solution:
(a) \(2\frac{2}{45}\)
We have, \(4 \frac{3}{5} \div 2 \frac{1}{4}=\frac{23}{5} \div \frac{9}{4}=\frac{23}{5} \times \frac{4}{9}\)
[∵ \(4 \frac{3}{5}=\frac{23}{5}, 2 \frac{1}{4}=\frac{9}{4}\)
= \(\frac{23 \times 4}{5 \times 9}=\frac{92}{45}=2 \frac{2}{45}\)

Question 12.
Which of the following is/are correct?
(i) \(36 \div \frac{3}{4}=48\)
(ii) \(20 \div 2 \frac{1}{2}=8\)
(iii) \(\frac{5}{6} \div \frac{1}{3}=\frac{5}{2}\)
(iv) \(3 \frac{1}{2} \div \frac{7}{4}=2\)
Choose the correct option from the following:
(a) (i), (ii) and (iii) only
(b) (i), (iii) and (iv) only
(c) (ii), (iii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Solution:
(d) (i), (ii), (iii) and (iv)

Question 13.
Which of the following is/are correct?
(i) \(36 \div \frac{3}{4}=48\)
(ii) \(45 \div 5 \frac{5}{6}=6\)
(iii) \(\frac{7}{9} \div \frac{1}{3}=\frac{7}{2}\)
(iv) \(3 \frac{1}{2} \div \frac{7}{4}=2\)
Choose the correct option from the following:
(a) (i), (ii) and (iv)
(b) (ii), (iii) and (iv)
(c) (i) and (iv) only
(d) (i), (ii) and (iii)
Solution:
(c) (i) and (iv) only
(i)
= 48 – correct

(ii)
= \(\frac{9 \times 6}{7}=\frac{54}{7} \neq 6\) = Incorrect

(iii) – Incorrect

(iv) – Correct

A Tale of Three Intersecting Lines Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The product of \(\frac{2}{5} \text { and } \frac{3}{4} \text { is } \frac{6}{20}\)
(R): To multiply two fractions, we take the LCM of the denominators and then add the numerators.
Solution:
(c) A is true but R is false.
To multiply two fractions, we do not take the LCM or add numerators. Instead, we multiply numerators and denominators directly.
∴ \(\frac{2}{5} \times \frac{3}{4}=\frac{2 \times 3}{5 \times 4}=\frac{6}{20}\)
Thus, Assertion (A) is true, but Reason (R) is false.

Question 2.
(A): \(\frac{4}{3}\) of 15 is 20.
(R): In \(\frac{4}{3}\) of 15 , ‘of’ means ‘multiplication’.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
In \(\frac{4}{3}\) of 15 , ‘of ’ means ‘multiplication’.
Thus, \(\frac{4}{3}\) of 15 = \(\frac{4}{3} \times 15\) = 4 × 5 = 20
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): The product of \(\frac{2}{3}, \frac{3}{4} \text { and } \frac{1}{2} \text { is } \frac{1}{4}\)
(R): To multiply three fractions, we take the LCM of the denominators and then add the numerators.
Solution:
(c) A is true but R is false.
To multiply three fractions, we do not take the LCM or add the numerators.
Instead, we multiply the numerators and multiply the denominators directly.
∴ \(\frac{2}{3} \times \frac{3}{4} \times \frac{1}{2}=\frac{2 \times 3 \times  1}{3 \times 4 \times 2}=\frac{1}{4}\)
Therefore, Assertion (A) is true, but Reason (R) is false.

Question 4.
(A): The product of two improper fractions is not smaller than any of the two fractions.
(R): Multiplication of two improper fractions gives a result that is greater than both the fractions or equal to either of them.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
An improper fraction has its numerator equal to or greater than the denominator (e.g. \(\frac{5}{4}, \frac{7}{3},\) 1, etc.). Its value is greater than or equal to 1.
Thus, the product of two improper fractions is equal to either of them or greater than each fraction.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

A Tale of Three Intersecting Lines Class 7 Fill in the Blanks

Question 1.
The product of two proper fractions is found by multiplying the _______ of both and the _____ of both.
Solution: numerators, denominators
The product of two proper fractions is found by multiplying the numerators of both and the
denominators of both.

Question 2.
Priya is making lemonade. She uses \(\frac{2}{3}\) of a lemon for one glass. If she makes \(\frac{3}{4}\) of a glass, she will use ______ of a lemon.
Solution: \(\frac{1}{2}\) of a glass
Given, lemon used for 1 full glass is \(\frac{2}{3}\) and glass prepared = \(\frac{3}{4}\) of a glass
∴ Required lemon = \(\frac{3}{4} \times \frac{2}{3}=\frac{3 \times 2}{4 \times 3}=\frac{6}{12}=\frac{1}{2}\) = of a lemon.

Question 3.
The product of \(\frac{5}{8}\) and 96 is _______ .
Solution: 60
Product of \(\frac{5}{8}\) and 96 is \(\frac{5}{8} \times 96\) = 5 × 12 = 60

Question 4.
Ravi is painting a wall. He uses \(\frac{3}{5}\) of a bucket of paint for one wall. If he paints \(\frac{2}{3}\) of a wall, he wil use ______ of a bucket.
Solution: \(\frac{2}{5}\)
Given, paint required to paint 1 wall = \(\frac{3}{5}\) of a bucket
Paint required to paint \(\frac{2}{3}\) of a wall
= \(\frac{2}{3} \times \frac{3}{5}=\frac{2}{5}\) of a bucket

Question 5.
Complete the following:

Solution:

Question 6.
If the product of two numbers is ________ the numbers are called reciprocal of each other.
Solution: 1
If the product of two numbers is 1, then the numbers are called reciprocal of each other.

Question 7.
Reciprocal of \(\frac{5}{11}\) is ______.
Solution: \(\frac{11}{5}\)
Reciprocal of \(\frac{5}{11} \text { is } \frac{\mathbf{1 1}}{\mathbf{5}}\)

Question 8.
\(\frac{1}{5}\) is ________ of 5.
Solution: reciprocal or multiplicative inverse
\(\frac{1}{5}\) is reciprocal or multiplicative inverse of 5.

Question 9.
\(\frac{8}{5} \div \frac{9}{}=\frac{8}{5} \times \frac{7}{9}\)
Solution: \(\frac{9}{7}\)
\(\frac{8}{5} \div \frac{9}{7}=\frac{8}{5} \times \frac{7}{9}\)

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines MCQ improves accuracy in objective exams.

MCQ on A Tale of Three Intersecting Lines Class 7

A Tale of Three Intersecting Lines MCQ Class 7

Class 7 Maths A Tale of Three Intersecting Lines MCQ

Question 1.
A triangle with two sides equal is called:
(a) Scalene triangle
(b) Equilateral triangle
(c) Isosceles triangle
(d) None of these
Solution:
(c) Isosceles triangle
We know that an isosceles triangle has exactly two equal sides and two equal angles opposite to equal sides.
Thus, a triangle with two equal sides is called isosceles triangle.

Question 2.
An equilateral triangle has angles measuring:
(a) 60°, 60°, 60°
(b) 90°, 45°, 45°
(c) 100°, 40°, 40°
(d) 120°, 30°, 30°
Solution:
(a) 60°, 60°, 60°
We know that all angles are equal in an equilateral triangle and each angle is 60°.

Question 3.
The symbol used to represent a triangle is:
(a) ∆
(b) ∠
(c) ||
(d) ⊥
Solution:
(a) ∆
The triangle is commonly denoted using the symbol ∆ as in ∆ABC. The other symbols given in the options represent angle (∠), two parallel lines (||) and two perpendicular lines (⊥).

Question 4.
Consider AB < BC < CA. Which of the following is sufficient to say ∆ABC exists?
(a) BC + CA > AB
(b) AB + CA > BC
(c) AB + BC > CA
(d) None of these
Solution:
(c) AB + BC > CA
Given, AB < BC < CA If sum of two smaller sides is greater than the third side (longest), then triangle inequality is automatically satisfied four all other combination of sides. Thus AB + BC > CA is sufficient to say ∆ABC exists.

Question 5.
A triangle with sides XY = 6.6 cm, YZ = 6 cm and XZ = 6.6 cm, is:
(a) Isosceles
(b) Scalene
(c) Equilateral
(d) None of these
Solution:
(a) Isosceles
Given, XY = 6.6 cm, YZ = 6 cm and XZ = 6.6 cm
Here, two sides have equal length.
We know that if any two sides of a triangle are equal in length, then it is called an isosceles triangle.
Thus, triangle XYZ is an isosceles triangle.

Question 6.
In a ∆ABC,which of the given condition holds?
(a) AB – BC > CA
(b) AB + BC < CA
(c) AB – BC < CA
(d) AB + CA < BC
Solution:
(c) AB – BC < CA
We know that the sum of any two sides of a triangle is greater than the third sides and the difference between the lengths of any two sides of a triangle is always smaller than the length of the third side. Therefore, AB – BC < CA.

Question 7.
Which of the following is true for an obtuse angled triangle?
(a) One angle = 90°
(b) All angles > 90°
(c) 90° < One angle < 180°
(d) Two angles > 90°
Solution:
(c) 90° < One angle < 180°
We know that if an angle of a triangle is greater than 90° (between 90° and 180°), then it is called an obtuse angled triangle.

Question 8.
If AB = 7 cm, ∠T = 50° and ∠B = 60°, then ∠C =
(a) 60°
(b) 70°
(c) 80°
(d) 100°
Solution:
(b) 70°
Given, AB = 7 cm, ∠A = 50° and ∠B = 60°
We know that the sum of all angles of a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ 50° + 60° + ∠C = 180°
⇒ 110° + ∠C = 180°
⇒ ∠C = 180°- 110° = 70°

Question 9.
If AB = 5 cm, AC = 6 cm and ∠A = 90°, what type of triangle will be constructed?
(a) Acute angled triangle
(b) Right-angled triangle
(c) Obtuse angled triangle
(d) Equilateral triangle
Solution:
(b) Right-angled triangle
Given, AB = 5 cm, AC = 6 cm and ∠A = 90°
As ∠A = 90°, the triangle is a right-angled triangle.

Question 10.
How many altitudes (maximum) can a triangle have?
(a) 1
(b) 2
(c) 3
(d) Depend on type of triangle
Solution:
(c) 3
We know that every triangle has three altitudes, each drawn from a vertex to its opposite side.

Question 11.
Which of the following sets of data is sufficient to construct a triangle?
(i) Two sides and the included angle
(ii) Three angles
(iii) Two angles and the included side
(iv) Three sides
Choose the correct option from the following:
(a) Only (i) and (Hi)
(b) Only (i) and (iv)
(c) Only (i), (ii) and (iii)
(d) (i), (ii), (iii) and (iv)
Solution:
(d) (i), (ii), (iii) and (iv)
Triangles can be constructed if we are given three sides, three angles, two sides with the included angle, or two angles with the included side.
Therefore, all the data sets are sufficient.

Question 12.
Which of the following statements about altitudes of a triangle are correct?
(i) An altitude alwavs lies inside the triangle.
(ii) An altitude can lie outside the triangle.
(iii) Every triangle has three altitudes.
(iv) All three altitudes are equal in length.
Choose the correct option from the following:
(a) Only (i) and (ii)
(b) Only (i) and (iii)
(c) Only (ii) and (iii)
(d) (i), (ii), (iii) and (iv)
Solution:
(c) Only (ii) and (iii)
We know that every triangle has three altitudes, each drawn from a vertex to its opposite side (or its extension).
In right-angled triangles, two altitudes lie along sides of the triangle.
In obtuse-angled triangles, 2 altitudes lie outside the triangle.
In a scalene triangle, all three altitudes are of different lengths.
Thus, statements (ii) and (iii) are true.

A Tale of Three Intersecting Lines Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): In ∆ABC, if ∠A = 60° and ∠B = 80°, then ∠C = 40°.
(R): In a triangle, the sum of all the angles is 180°.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Given, in ∆ABC, ∠d = 60° and ∠B = 80°
We know that the sum of all the angles of a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ 60° + 80° + ∠C = 180°
⇒ 140° + ∠C = 180°
⇒ ∠C = 180°- 140° = 40°
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): If two angles in a triangle are equal, the third angle must be 90°.
(R): In an isosceles triangle, the two angles are equal
Solution:
(d) A is false but R is true.
We know that if two angles of a triangle areequal, then the triangle is called an isosceles triangle.
Let ∠d and ∠B be the equal angles in ∆ABC. We know that the sum of all angles of a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠A + ∠C = 180° [∵ ∠d = ∠B]
⇒ 2∠A + ∠C = 180°
⇒ ∠C = 180° – 2∠d
As we can see that the value of ∠C will be 90° only when ∠d is 45°.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 3.
(A): If one of the interior angles of a triangle is 90°, the adjacent exterior angle is also 90°.
(R): A straight angle measures 180°.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Given, one of the interior angles of a triangle is 90°.
We know that an interior angle and its adjacent exterior angle form a straight angle.
∴ Interior angle + Adjacent exterior angle = 180°
⇒ 90° + Adjacent exterior angle = 180°
⇒ Adjacent exterior angle = 180° – 90° = 90°
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): All angles of an acute-angled triangle are less than 90°.
(R): The sum of all angles of a triangle is 180°. equal.
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
We know that if all three angles of a triangle are acute angles (between 0° and 90°), then it is called an acute angled triangle and the sum of all angles of a triangle is 180°.
Thus, both Assertion (A) and Reason (R) are true, but. Reason (R) is not the correct explanation of Assertion (A).

A Tale of Three Intersecting Lines Class 7 Fill in the Blanks

Question 1.
A triangle is a closed figure made up of three ________ line segments.
Solution: non-parallel
We know that a triangle is a closed figure made up of three non-parallel line segments.

Question 2.
Each angle in an equilateral triangle measures _________ degrees.
Solution: 60
We know that each angle in an equilateral triangle measures 60 degrees.

Question 3.
A triangle having all three sides of different lengths is called a _______ triangle.
Solution: scalene
We know that a triangle with all three sides of different lengths is called a scalene triangle.

Question 4.
An altitude of a triangle is a _________ from vertex to the opposite side.
Solution: perpendicular line segment
We know that in a triangle, an altitude is a line drawn from a corner (vertex) straight down to the opposite side, making a right angle (90°) with that side.
Thus, an altitude of a triangle is a perpendicular line segment from vertex to the opposite side.

Question 5.
The sum of the three angles in a triangle is __________ .
Solution: 180°
The sum of the three angles in a triangle is 180°.

Question 6.
In ∆ABC, if ∠A = 45° and ∠B = 65°, then the exterior angle at vertex C is _______ .
Solution: 110°
Given, ∠d = 45° and ∠B = 65°
We know that if a side of a triangle is produced or extended, then the exterior angle formed is equal to the sum of two interior opposite angles. ,
∴ Exterior angle at vertex C = Sum of interior angle at A and interior angle at B
= ∠A + ∠B = 45° + 65° = 110°
Thus, in triangle ABC, if ∠A = 45° and ∠B = 65°, then the exterior angle at vertex C is 110°.

Students can refer to BSE Odisha Class 6 Math Solution and Ganita Prakash Chapter 3 Number Play Class 6 Question Answer to understand textbook questions step by step.

Class 6 Maths Chapter 3 Number Play Solutions

Ganita Prakash Class 6 Chapter 3 Solutions

Class 6 Maths Ganita Prakash Chapter 3 Solutions Number Play

Question 1.
Fill the table below with only 4-digit numbers such that the supercells are exactly the coloured cells.

Solution:

5346

8643

1166

1258

1056

2012

8000

9635

9905

Question 2.
Fill the table below such that we get as many supercells as possible. Use numbers between 100 and 1000 without repetitions.

Solution:

999

102

909

110

918

210

928

350

958

Question 3.
Can you fill a supercell table without repeating numbers such that there are no supercells? Why or why not?
Solution:
No, we cannot fill a supercell table without repeating numbers such that there are no supercells because at least one number will always be larger than its adjacent cell unless repetition is allowed.
For example,

3

6

9

12

15

18

21

20 or 22

Here, if the last number is greater than 21, it is a supercell and if less than 21, then 21 itself is a supercell number.

Question 4.
Will the cell having the largest number in a table always be a supercell? Can the cell having the smallest number in a table be a supercell? Why or why not?
Solution:
Yes, the cell having the largest number in a table will always be a supercell, because largest number will always be greater than its neighbouring numbers.
But, the cell having smallest number in a table can never be a supercell, because it will always be smaller than its neighbouring numbers.

Question 5.
Fill a table such that the cell having the second largest number is not a supercell but the second smallest number is a supercell. Is it possible?
Solution:
Yes, it is possible.

245

147

368

313

696

758

532

590

485

In this table, the second largest number is 696, it is not a supercell number and second smallest number is 245, which is a supercell number.

Question 6.
What is the sum of the smallest and largest 5-digit palindrome? What is their difference?
Solution:
Smallest 5-digit palindrome = 10001
Largest 5-digit palindrome = 99999
Sum = 10001 + 99999 = 110000
Difference = 99999 – 10001 = 89998

Question 7.
Write an example for each of the below scenarios whenever possible.

Could you find examples for all the cases? If not, think and discuss what could be the reason. Make other such questions and challenge your classmates.
Solution:
(i) 45000 + 50000 = 95000 > 90250
(ii) 99999 + 900 = 100899
(iii) 4-digit 4- 4-digit to give a 6-digit sum. Since, the maximum sum for two 4-digit numbers (9999 + 9999) is 19998. Thus, it is not possible to get a 6 -digit number by adding two 4-digit numbers.
(iv) 50000 + 61000 = 111000
(v) The minimum sum of two 5-digit numbers (10000 + 10000) is 20000, which is greater than 18500. Therefore, it is not possible to get a sum of 18500 with two 5-digit numbers.
(vi) 70000 – 15000 = 55000 < 56503
(vii) 10000 – 999 = 9001
(viii) 12000 – 8000 = 4000
(ix) 20000 – 19500 = 500
(x) The difference between the largest 5-digit number and smallest 5-digit number (99999 – 10000) is 89999 which is less than 91500.
Therefore, it not possible to get the difference of 91500 between two 5-digit numbers.

Question 8.
There is only one supercell (number greater than all its neighbours) in this grid. If you exchange two digits of one of the numbers, there will be 4 supercells.

Figure out which digits to swap.
Solution:
Swap the digits 6 and 1 in the number 62,871. The number becomes 12,876 which will be the smallest among all the numbers.

16,200	39,344	29,765
23,609	12,876	45,306
19,381	50,319	38,408

Now, there are four supercells.

Question 9.
We are the group of 5-digit numbers between 35000 and 75000 such that all of our digits are odd. Who is the largest number in our group? Who is the smallest number in our group? Who among us is the closest to 50,000?
Solution:
The possible odd digits are: 1, 3, 5, 7 and 9.
Largest odd number is 73,999. (between 35000 and 75000)
Smallest odd number is 35,111. (between 35000 and 75000)
Closest odd number to 50,000 is 51,111.

Question 10.
Recall the sequence of powers of 2 from chapter 1. Why is the Collatz conjecture correct for all the starting numbers in this sequence?
Solution:
Powers of 2 → 2, 4, 8, 16,
For 2 (even) → 2 divide by 2 = 1
For 4 (even) → 4 divide by 2 = 2 (even) → 2 divide by 2 = 1
For 8 (even) → 8 divide by 2 = 4 (even) → 4 divide by 2 = 2 (even) → 2 divide by 2 = 1
Thus, for starting numbers that are powers of 2, the Collatz conjecture holds true because the sequence of operations simply involves a series of divisions by 2, which eventually leads to 1.

InText Questions

[Instruction for Q1 to Q7]
Some Children in a park are standing in a line. Each one says a number.

• A child says ‘1’ if there is only one taller child standing next to them.
• A child says ‘2’ if both the children standing next to them are taller.
• A child says ‘O’, if neither of the children standing next to them are taller.

That is each person says the number of taller neighbours they have.

Question 1.
Can the children rearrange themselves so that the children standing at the ends say ‘2’?
Solution:
No, the children cannot line up in a way that the ones at the ends say ‘2′ because a child only says ‘2’ when both of their neighbours are taller and children at the ends have only one neighbour, not two.

Question 2.
Can we arrange the children in a line so that all would say only 0s?
Solution:
No, it’s not possible to arrange the children in a line so that they all say only ‘0’, because a child says ‘0’ only when neither of their neighbours is taller. This would only happen if all the children are the same height.

Question 3.
Can two children standing next to each other say the same number?
Solution:
Yes, two children standing next to each other can say the same number i.e. 1 because there can be one taller and one smaller child standing next to them.

Question 4.
There are 5 children in a group, all of different heights. Can they stand such that four of them say 1 and the last one says ‘0’? Why or why not?
Solution:
Yes, because a child can say ‘0’ only if neither of the children standing next to them are taller. Arranging them in increasing or decreasing order of heights will eventually put tallest child at one of the end. So it is obvious that last child will say ‘0’.

Question 5.
For this group of 5 children, is the sequence 1, 1, 1, 1, 1 possible?
Solution:
No, the sequence 1, 1, 1, 1, 1 is not possible because there are 5 children and first four will say 1 only if they are arranged in increasing order of height. So, the last child will be tallest and will say 0 (not 1).

Question 6.
Is the sequence 0, 1,2, 1,0 possible? Why or why not?
Solution:
Yes, the sequence 0, 1, 2, 1, 0 is possible. It represents a situation where the middle child is the smallest (has two taller neighbours). The two children at the end are taller than the second last child from both the ends.

Question 7.
How would you rearrange the five children so that the maximum number of children say ‘2’?
Solution:
We can rearrange the five children so that the maximum number of children say ‘2’ as 0, 2, 0, 2, 0.
Thus, maximum number of children who can say ‘2’ is 2 . ,

Question 8.
Complete Table 2 with 5-digit numbers whose digits are ‘1’, ‘0’, ‘6’, ‘3’, and ‘9’ in some order. Only a coloured cell should have a number greater than all its neighbours.

Once you have filled the table above, put commas appropriately after the thousands digit.
The biggest number in the table is _______.
The smallest even number in the table is ______.
The smallest number greater than 50,000 in the table is _______.
Solution:

96,310	96,301	36,109	36,190′
93,610.	13,609	60,319	19,306
93,106	10,639	60,193	30,196
10,369	10,963	10,396	31,906

The biggest number in the table is 96,310.
The smallest even number in the table is 10,396.
The smallest number greater than 50,000 in the table is 60,193.

Question 9.
Among the numbers 1-100, how many times will the digit 7’ occur? Among the numbers 1-1000, how many times will the digit ‘7’ occur?
Solution:
Among the numbers 1-100, we can divide the range into two parts: When ‘7’ appears in the ones place and when ‘7’ appears in the tens place.

In the ones place: The numbers that have ‘7’ in the ones place are: 7, 17, 27, 37, 47, 57, 67, 77, 87, and 97. So, there are 10 numbers where ‘7’ appears in the ones place.

In the tens place: The numbers 70 to 79 contain the digit ‘7’ in the tens place. So, there are 10 numbers where ‘7’ appears in the tens place.

Total occurrences of ‘7’ between 1 and 100:
Ones place: 10 times and Tens place: 10 times
Total = 10 + 10 = 20
So, digit ‘7’ appears 20 times among the numbers 1-100.

Among the numbers 1-1000, we can divide the range into three parts:
hundreds place, tens place and ones place
In the hundreds place: The numbers 700 to 799 contain the digit ‘7’ in the hundreds place. There are 100 numbers front 700 to 799.

In the tens place: We already know that in each set of 100 numbers (i.e. 0-99, 100-199, …, 900-999),
‘7’ will appear in the tens place 10 times (just like in the 1-100 range). Since we have 10 sets of 100 numbers, ‘7’ will appear in the tens place, 10 × 10 = 100 times.

In the ones place: Similarly, for each set of 100 numbers (i.e. 0-99, 100-199, …. 900-999), 7’ will appear in the ones place 10 times (like in the 1-100 range). So, across 10 sets of 100 numbers, 7’ will appear in the ones place 10 × 10 = 100 times.

Total occurrences of 7’ between 1 and 1000:
Hundreds place: 100 times, Tens place: 100 times, Ones place: 100 times
Total = 100 + 100 + 100 = 300
So, digit ‘7’ appears 300 times among the numbers 1-1000.

Question 10.
Write all possible 3-digit palindromes using 1, 2, 3.
Solution:
All the possible 3-digit palindromes using the digits ‘1’, ‘2’, and ‘3’ are: 111, 121, 131, 212, 222, 232, 313, 323 and 333.
There are 9 possible palindromes using the digits ‘1’, ‘2’, and ‘3’.

Question 11.
Puzzle time:

I am a 5-digit palindrome.
I am an odd number.
My ‘t’ digit is double of my ‘M’ digit.
My ‘h’ digit is double of my ‘t’ digit.
Who am I? _______
Solution:
Since palindrome is an odd number, u digit would be 1, 3, 5, 7 or 9 and tth digit would be same as u digit.
Since double of 5, 7 and 9 is not a digit, u digit would be either 1 or 3.
Double of 1 is 2 and double of 3 is 6. So, t digit would be either 2 or 6.
Since double of 6 is not a digit, u digit is 1 and t digit is 2.
Then, h digit would be 4.
So, the required 5-digit odd palindrome number is 12421.
In words: Twelve thousand four hundred twenty one

Number Play Class 6 Extra Questions

Number Play Class 6 Very Short Question Answer

Question 1.
What is the largest 4-digit number with non-zero digits such that its digits add up to 14?
Solution:
The largest 4-digit number must have largest digit ‘9’ at thousands place and the remaining digits in descending order such that the sum of all digits is 14.
Hence, the required number is 9311.

Question 2.
Write all 2-digit palindromic numbers.
Solution:
The 2-digit palindromic numbers are: I 1, 22, 33, 44, 55, 66, 77, 88 and 99.

Question 3.
Write all possible 3-digit palindromes using the digits 7, 8, 9.
Solution:
Using digits 7, 8 and 9, we get the following palindromes:
777, 787, 797, 878, 888, 898, 979, 989, 999

Question 4.
Write two 4-digit numbers such that their difference is a 2-digit number.
Solution:
Let us take 5500 and 5460 as two 4-digit numbers.
Clearly, 5500 – 5460 = 40 which is a 2-digit number.

Question 5.
Write two 5-digit numbers whose sum is 32500.
Solution:
Let us take 12500 and 20000 as two 5-digit numbers.
Clearly, 12500 + 20000 = 32500

Question 6.
Write two 4-digit numbers whose difference is 4680.
Solution:
Let us take 7500 and 2820 as two 4-digit numbers.
Clearly, 7500 – 2820 = 4680

Question 7.
Write the smallest and largest 4-digit palindromes. Find their sum and difference.
Solution:
Smallest 4-digit palindrome = 1001 and largest 4-digit palindrome = 9999
Required sum = 1001 + 9999 = 11000
Required difference = 9999 – 1001 = 8998

Question 8.
Write the second smallest and second largest 5-digit palindromes. Find their sum.
Solution:
Second smallest 5-digit palindrome = 10101
Second largest 5-digit palindrome = 99899
Required sum = 10101 + 99899 = 110000

Question 9.
Write a 6-digit number and a 4-digit number such that their difference is a 5-digit number.
Solution:
Let us take 103200 as a 6-digit number and 8200 as a 4-digit number.
Clearly, 103200 – 8200 = 95000 which is a 5-digit number.

Question 10.
Write a 6-digit number and a 5-digit number such that their difference is a 5-digit number.
Solution:
Let us take 120500 as a 6-digit number and 40500 as a 5-digit number.
Clearly, 120500 – 40500 = 80000, which is a 5-digit number.

Question 11.
Rearrange the digits of 48900125 to get the largest 8-digit number?
Solution:
In order to get the largest 8-digit number using the digits of 48900125, digits should be in descending order on moving from left to right.
Therefore, the required largest number is 98542100.

Question 12.
What is the smallest number whose digit sum is 16?
Solution:
There is no single digit number to give digit sum equal to 16.
The 2-digit numbers whose digits add up to 16 are: 79, 88 and 97.
We see that 79 is the smallest number among these numbers.
Therefore, 79 is the smallest number whose digit sum is 16.

Question 13.
Rearrange the digits of 4500731 to get the smallest 7-digit number?
Solution:
In order to get the smallest 7-digit number using the digits of 4500731, the first digit from left to right should be the smallest digit except 0. So, in this case, it is 1. Then, on moving from left to right further, digits should be in ascending order.
Therefore, the required smallest number is 1003457.

Question 14.
What is the largest 4-digit number whose digits add up to 15?
Solution:
The largest 4-digit number should have the largest digit at thousands place. Since 9 is the largest digit, it takes thousands place. Then, 6 takes the hundreds place because 9 + 6 = 15. As the sum is exhausted by thousands and hundreds place digits, rest places will take digit ‘0’.
Therefore, the required largest 4-digit number is 9600.

Number Play Class 6 Short Question Answer

Question 1.
Fill the given table such that the cell having the 2nd largest number is not a supercell but the cell having 2nd smallest number is a supercell. Use numbers between 10 and 100.

Solution:
To make sure the cell with the second largest number is not a supercell, it has to be next to the cell with the largest number. This is because the cell with the largest number is always a supercell, and two supercells cannot be next to each other.

To make the cell with the second smallest number a supercell, it should be one of the extreme end cells and the smallest number should be in the adjacent cell.

53

88

72

64

22

32

Here, 72 is the second largest number and 32 is the second smallest number
Activity: Think of another numbers by yourself.

Question 2.
Mark the supercells in the table below.

6828	670	9435	2180
3780	3708	7308	9225
8000	5583	52	5001

Solution:
6828 is a supercell because it is larger than its neighbours 670 and 3780.
9435 is a supercell because it is larger than its neighbours 670, 2180 and 7308.
9225 is a supercell because ii is larger than its neighbours 2180, 7308 and 5001.
8000 is a supercell because it is larger than its neighbours 3780 and 5583.

6828	670	9435	2180
3780	3708	7308	9225
8000	5583	52	5001

Question 3.
Fill in the table using only 3-digit numbers, making sure that the supercells line up exactly with the coloured cells.

Solution
Since 2^nd cell is a supercell, it must contain a number greater than 576. Let it he 832.

Since 3^rd and 5^th cells are not supercells and 4^th cell is a supercell, the numbers in 3^rd and 5^th cells should be smaller than the number in 4^th cell i.e., 188. Let them he 112 and 176 respectively.

576

832

112

188

176

912

The last cell is a supercell, and its neighbouring cell contains 912. So, the number in the last cell should be greater than 912. Let it be 950.

Since 7^th cell is not a supercell, it must contain number smaller than 912. Let it be 492.

Also, 6^th cell is not a supercell, it must contain a number smaller than the number present in 7^th cell, i.e. 492. Let it be 350.

Now, the complete table is as follows:

576

832

112

188

176

350

492

912

950

Activity: Think of another numbers by yourself.

Question 4.
Fill in the table using only 3-digit numbers, making sure that the supercells line up exactly with the coloured cells.

Solution:
Since 2^nd cell is a supercell, it must contain a number greater than 576. Let it be 832. Then, number in 1^st cell should be less than 432. Let it be 100.

100

432

120

312

731

512

Since 4^th cell is not supercell, it may Contain any 3-digit number less than 312. Let it be 230.

Also. 7^th cell is not supercell and second last cell is a supercell. it must contain number smaller than 731 and 512. I et it be 219.

The second last cell is a supercell, and it contains 512. So, the number in the last cell should be smaller than 512. Let it he 194.

Now, the complete table is as follows:

100

432

120

230

312

731

219

512

494

Question 5.
Calculate the digit sums of 2-digit numbers whose digits are consecutive. Do you observe a pattern?
Solution:
2-digit numbers having consecutive digits are: 12, 23, 34, 45, 56, 67, 78 and 89 .
The sums of digits of these 8 numbers arc:
1 + 2 = 3
2 + 3 = 5
3 + 4 = 7
4 + 5 = 9
5 + 6 = 11
6 + 7 = 13
7 + 8 = 15
8 + 9 = 17
We find that each sum is 2 more than the preceding sum.

Question 6.
What is the largest 5-digit number with non-zero digits whose digits add up to 17?
Solution:
Non-zero digits are 1, 2, 3, 4, 5, 6, 7, 8, 9.
The largest 5-digit number must have largest digit ‘9’ at ten thousands place (tth) and the remaining digits in descending order such that the sum of all digits is 17.

tth

th

h

t

u

Question 7.
Calculate the digit sums of 4-digit numbers whose digits are consecutive. Do you observe any pattern?
Solution:
The 4-digit numbers having consecutive digits are 1234,2345, 3456, 4567, 5678 and 6789.
The sum of digits of these 6 numbers are 10, 14, 18, 22, 26 and 30 respectively.
We find that each sum is 4 more than the preceding sum.

Question 8.
What is the smallest 5-digit number whose digits add up to 18?
Solution:
In order to write smallest 5-digit number, ten thousands place must be occupied by digit ‘1’. Then, sum of the rest of the digits should be 18 – 1 = 17.

The smallest 5-digit number should have the largest digit at units place. Since 9 is the largest digit, it takes units place. Then, 8 takes the tens place because 9 + 8 = 17. As the sum is exhausted by ten thousands, tens and units place digits, rest places will take digit ‘0’.

Therefore, the required smallest 5-digit number is 10089.

Question 9.
Convert the following times from 12-hour format to 24-hour format.
(i) 02:00 PM
(ii) 10:00 AM
(iii) 07:30 PM
(iv) 12:20 AM
(v) 12:15 PM
Solution:

	12-hour Format	24-hour Format
(i)	02:00 PM	14:00 hours
(ii)	10:00 AM	10:00 hours
(iii)	07:30 PM	19:30 hours
(iv)	12:20 AM	00:20 hours
(v)	12:15 PM	12:15 hours

Question 10.
What can be the previous number in the Collatz sequence to 40?
Solution:
Let x be the required number.
If x is an even number, then
\(\frac{x}{2}\) = 40 ⇒ x = 2 × 40 ⇒ x = 80
If x is an odd number, then
3x + 1 = 40 ⇒ 3x = 40 – 1
⇒ 3x = 39 ⇒ x = \(\frac{39}{3}\) = 13
Thus, either 80 or 13 can be the number preceding 40 in a Collatz sequence.

Question 11.
Observe the following numbers written in a pattern. Find their sum.

Solution:
We find that there are 3 rows each consisting of 5 boxes written with 30 and 2 rows each consisting of 6 boxes written with 70.
∴ Total sum = 3(5 × 30) + 2(6 × 70)
= 3 × 150 + 2 × 420
= 450 + 840 = 1290

Question 12.
Apply Kaprekar’s routine on the number 3562 to get Kaprekar constant.
Solution:
The number is 3562.
Round 1: A = Largest number using digits of the number 3562 = 6532
B = Smallest number using digits of the number 3562 = 2356
C = A – B = 6532 – 2356 = 4176

Round 2: A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174

Question 13.
Apply Kaprekar’s routine on the number 8825 to get Kaprekar constant.
Solution:
The number is 8825.
Round 1:
A = Largest number using digits of the number 8825 = 8852
B = Smallest number using digits of the number 8825 = 2588
C = A – B = 8852 – 2588 = 6264

Round 2:
A = Largest number using digits of the number 6264 = 6642
R = Smallest number using digits of the number 6264 = 2466
C = A – B = 6642 – 2466 = 4176

Round 3:
A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174

Question 14.
Convert the following times from 24-hour format to 12-hour format.
(i) 18:00 hours
(ii) 06:00 hours
(iii) 12:30 hours
(iv) 00:45 hours
(v) 23:15 hours
Solution:

	24-hour Format	12-hour Format
(i)	18:00 hours	06:00 PM
(ii)	06:00 hours	06:00 AM
(iii)	12:30 hours	12:30 PM
(iv)	00:45 hours	12:45 AM
(v)	23:15 hours	11:15 PM

Question 15.
Count all the suns in the following pattern.

Solution:
There are 26 boxes with 1 sun each and 26 boxes with 4 suns each.
∴ Total number of suns = (26 × 1) + (26 × 4)
= 26 + 104 = 130

Question 16.
Find the sum of the numbers in the number pattern shown in the figure.

Solution:
We note that the number 75 occurs 20 times around the outer edge, the number 40 appears 12 times along the second layer, the number 50 is present 4 times on the edge of the inner layer, and the number 1500 is located at the centre.
∴ Required sum
= (20 × 75) + (12 × 40) + (4 × 50) + 1500
= 1500 + 480 + 200 + 1500 = 3680

Question 17.
Consider the numbers in the following boxes:

Using operations addition/subtraction on these numbers, we can get
13100 = 32000 – 20000 + 700 + 200 + 200
Similarly, obtain the following numbers by using the numbers in the boxes.
(i) 25800
(ii) 28000
(iii) 7500
(iv) 14400
(v) 53000
Solution:
(i) 25800 = 20000 + 6000 – 200
(ii) 28000 = 32000 – 3500 – 700 + 200
(iii) 7500 = 3500 + 3500 + 700 – 200
(iv) 14400 = 20000 – 6000 + 200 + 200
(v) 53000 = 32000 + 20000 – 6000 + 3500 + 3500

Question 18.
Consider the numbers in the following boxes:

Using operations addition/subtraction on these numbers, we can get
23300 = 16000 + 16000 – 9000 + 300
Similarly, obtain the following numbers by using the numbers in the boxes.
(i) 60200
(ii) 2500
(iii) 7700
(iv) 64600
(v) 17300
Solution:
(i) 60200 = 45000 4 16000-800
(ii) 2500 = 9000 + 9000 – 16000 + 800 – 300
(iii) 7700 = 9000 – 800 – 800 + 300
(iv) 64600 = 45000 + 9000 + 9000 + 800 + 800
(v) 17300 = 16000 + 800 + 800 – 300

Number Play Class 6 Long Question Answer

Question 1.
Colour or mark the supercells in the given table:

6235

970

8145

4780

4708

7084

9000

160

Solution:
Clearly, 6235 is greater than the number 970 in the neighbouring cell.
Hence, the cell containing 6235 is the supercell.
970 is smaller than 6235 and 8145, so the cell containing 970 is not a supercell.
As 8145 is greater than the numbers 970 and 4780 in the neighbouring cells, the cell containing 8145 is a supercell.
As 4780 is smaller than 8145, the cell containing 4780 is not a supercell.
As 4708 is smaller than 4780 and 7084, the cell containing 4708 is not a supercell.
As 7084 is smaller than 9000, the cell containing 7084 is not a supercell.
As 9000 is greater than the numbers 7084 and 160 in the neighbouring cells, the cell containing 9000 is a supercell.
As 160 is smaller than 9000, the cell containing 160 is not a supercell.

6235

970

8145

4780

4708

7084

9000

160

Question 2.
Mark supercells in the following grid:

3462	2198	6757	5678
1001	5982	4723	2345
2723	5600	7210	2316
3298	1465	3467	4621

Solution:
The cell containing 3462 is a supercell as the numbers 2198 and 1001 in neighbouring cells are smaller than 3462.
The cell containing 6757 is a supercell as the numbers 2198, 4723 and 5678 in neighbouring cells are smaller than 6757.
The cell containing 5982 is a supercell as the numbers 2198, 4723, 5600 and 1001 in neighbouring cells are smaller than 5982.
The cell containing 7210 is a supercell as the numbers 4723, 2316, 3467 and 5600 in neighbouring cells are smaller than 7210.
The cell containing 3298 is a supercell as the numbers 2723 and 1465 in neighbouring cells are smaller than 3298.
The cell containing 4621 is a supercell as the numbers 2316 and 3467 in neighbouring cells are smaller than 4621.

3462	2198	6757	5678
1001	5982	4723	2345
2723	5600	7210	2316
3298	1465	3467	4621

Question 3.
Identify the numbers marked on the number lines below, and label the remaining positions.

Put a circle around the smallest number and a box around the largest number in each of the sequences above.
Solution:
(i) There are 3 sub-divisions between 729 and 732.
Therefore, each sub-division represents \(\frac{732-729}{3}\) = \(\frac{3}{3}\) = 1 number.

(ii) There are 7 sub-divisions between 6236 and 6250.
Therefore, each sub-division represents \(\frac{6250-6250}{7}\) = \(\frac{14}{7}\) = 2 numbers.

(iii) There are 6 sub-divisions between 12100 and 12160.
Therefore, each sub-division represents \(\frac{12100-12160}{6}\) = \(\frac{60}{6}\) = 10 numbers.

(iv) There is 1 sub-division between 53312 and 54312.
Therefore, each sub-division represents \(\frac{53312-54312}{1}\) = 1000 numbers

Question 4.
Among the numbers 100 – 1000, how many times will the digit ‘3’ occur?
Solution:
There are 901 numbers from 100 to 1000.

When digit ‘3’ is at hundreds place:
Numbers with ‘3’ in the hundreds place look like: 300 – 399
So, total count of digit ‘3’ at hundreds place = 100

When digit ‘3’ is at tens place:
To count how often ‘3’ appears in the tens place, fix the hundreds and units digit, and loop through all possibilities:
There are:

• 9 choices for hundreds digit (1 to 9)
• 1 choice for tens digit (3)
• 10 choices for units digit (0 to 9)

So, total count of digit ‘3’ at tens place = 9 × 1 × 10 = 90

When digit ‘3’ is at units place:
To count how often ‘3’ appears in the units place, fix the hundreds and tens digit, and loop through all possibilities:
There are:

• 9 choices for hundreds digit (1 to 9)
• 10 choices for tens digit (0 to 9)
• 1 choice for units digit (3)

So, total count of digit ‘3’ at units place = 9 × 1 × 10 = 90
Total count of digit ‘3’ = 100 + 90 + 90 = 280
Thus, digit ‘3’ occurs 280 times.

Question 5.
How big a number can you form having the digit sum 16? Can you make an even bigger number?
Solution:
2-digit numbers having the digit sums 16 are: 79, 88 and 97.
3-digit numbers having the digit sums 16 are: 169, 178, 187, 196, 259, 268, 277, 286, 295, 349, 358, 367, 376, 385, 394, 439, 448, 457, 466, 475, 484, 493, 529, 538, 547, 556, 565, 574, 583, 592, 619, 628, 637, 646, 655, 664, 673, 682, 691, 709, 718, 727, 736, 745, 754, 763, 772, 781, 790, 808, 817, 826, 835, 844, 853, 862, 871, 880, 907, 916, 925, 934, 943, 952, 961 and 970.
To form a 4-digit number having the digit sum 16, put one 0 in any of the above 3-digit numbers anywhere after first digit from left or two 0 in any of the above 2-digit numbers anywhere after first digit from left.

Similarly, we can form w-digit numbers by putting an appropriate number of 0 in any of the above numbers.

Thus, there will be infinitely many numbers with digit sum 16.

Yes, we can make even much bigger numbers.

Question 6.
Solve the puzzle:
My father’s salary is a 6-digit palindrome.
It is an even number.
Its tens digit is double of units digit and hundreds digit is double of tens digit.
What is my father’s salary?
Solution:
Units digit of an even number is 0 or 2 or 4 or 6 or 8. Since it is a palindrome, 0 cannot be the units digit. It is given that tens digit is double of units digit. So, 6 and 8 cannot be the units digit as 6 × 2 = 12 and 8 × 2 = 16 but 12 and 16 are not digits. If 4 is the units digit, then 8 would be the tens digit. But it is given that hundreds digit is double of tens digit, then hundreds digit would be 16, which is not possible.

So, units digit must be 2 only. Then, tens digit would be 4 and hundreds digit would be 8.

2

4

8

8

4

2

Thus, the palindromic number is 248842.
Hence, father’s salary is 248842.

Question 7.
List all 3-digit palindromic numbers and highlight those whose digit sum is greater than 15.
Solution:
3-digit palindromic numbers are given in the below table:

101	111	121	131	141	151	161	171	181	191
202	212	222	232	242	252	262	272	282	292
303	313	323	333	343	353	363	373	383	393
404	414	424	434	444	454	464	474	484	494
505	515	525	535	545	555	565	575	585	595
606	616	626	636	646	656	666	676	686	696
707	717	727	737	747	757	767	777	787	797
808	818	828	838	848	858	868	878	888	898
909	919	929	939	949	959	969	979	989	999

Highlighted (bold) palindromes have their digit sum greater than 15.

Question 8.
Write a 5-digit number, a 4-digit number and a 3-digit number such that their sum is 24680.
Solution:
Let us take 20000 as a 5-digit number. Then,
20000 + Sum of a 4-digit number and a 3-digit number = 24680
⇒ Sum of a 4-digit number and a 3-digit number
= 24680 – 20000 = 4680
Let us take 4500 as a 4-digit number. Then,
4500 + a 3-digit number = 4680
⇒ 3-digit number = 4680 – 4500 = 180
Hence, 20000 is a 5-digit number, 4500 is a 4-digit number and 180 is a 3-digit number such that their sum is 24680.
There can also be other numbers like
19000 + 5100 + 580 = 24680,
17500 + 7000 4 180 = 24680 etc.
Think of another numbers by yourself.

Question 9.
State whether the following statements are True/ False.
(i) The sum of two 5-digit numbers is always a 5-digit number.
(ii) The sum of a 5-digit number and a 3-digit number is always a 5-digit number.
(iii) 4’be sum of a 5-digit number and a 3-digit number is always a 6-digit number.
(iv) The difference of two 4-digit numbers is always a 4-digit number.
(ii) The difference between a 5-digit number and a 3-digit number is always a 4-digit number.
Solution:
(i) False; 26000 + 42000 = 68000 → a 5-digit number
76000 + 84000 = 160000 + not → a 5-digit number
Thus, the sum of two 5-digit numbers may or may not be a 5-digit number.

(ii) False; 10000 + 100 = 10100 → a 5-digit number
99900 + 900 = 100800 → not a 5-digit number
Thus, the sum of a 5-digit number and a 3-digit number may or may not be a 5-digit number.

(iii) False; 10000 + 100 = 10100 → not a 6-digit number
99900 + 900 = 100800 → a 6-digit number
Thus, the sum of a 5-digit number and a 3-digit number may or may not be a 6-digit number.

(iv) False; 8040 – 6080 = 1960 → a 4-digit number
1980 – 1200 = 780 → not a 4-digit number
Thus, the difference of two 4-digit numbers may or may not be a 4-digit number.

(v) False; 10250 – 750 = 9500 → a 4-digit number
15500 – 900 = 14600 → not a 4-digit number
Thus, the difference between a 5-digit number and a 3-digit number may or may not be a 4-digit number.

Question 10.
Apply Kaprekar’s routine on a 3-digit number. What number will start repeating?
Solution:
Let the number be 729.
Round 1:
A = Largest number using digits of the number 729 = 972
B = Smallest number using digits of the number 729 = 279
C = A – B = 972 – 279 = 693

Round 2:
A = Largest number using digits of the number 693 = 963
B = Smallest number using digits of the number 693 = 369
C = A – B = 963 – 369 = 594

Round 3:
A = Largest number using digits of the number 594 = 954
B = Smallest number using digits of the number 594 = 459
C = A – B = 954 – 459 = 495

Round 4:
A = Largest number using digits of the number 495 = 954
B = Smallest number using digits of the number 495 = 459
C = A – B = 954 – 459 = 495
Clearly, 495 repeats after 3 iterations.

Question 11.
How many rounds does the number 7443 take to reach the Kaprekar constant.
Solution:
The number is 7443.
Round 1:
A = Largest number using digits of the number 7443 = 7443
B = Smallest number using digits of the number 7443 = 3447
C = A – B = 7443 – 3447 = 3996

Round 2:
A = Largest number using digits of the number 3996 = 9963
B = Smallest number using digits of the number 3996 = 3699
C = A – B = 9963 – 3699 = 6264

Round 3:
A = Largest number using digits of the number 6264 = 6642
B = Smallest number using digits of the number 6264 = 2466
C = A – B = 6642 – 2466 = 4176

Round 4:
A = Largest number using digits of the number 4176 = 7641
B = Smallest number using digits of the number 4176 = 1467
C = A – B = 7641 – 1467 = 6174, which is the Kaprekar constant
Thus, we reach at the Kaprekar constant in four rounds.

Question 12.
Make the Collatz sequence by starting with number 24.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
First term: 24
Second term: \(\frac{24}{2}\) = 12 [As 24 is an even number.]
Third term: \(\frac{12}{2}\) = 6 [As 12 is an even number.] 0
Fourth term: \(\frac{6}{2}\) = 3 [As 6 is an even number.]
Fifth term: 3 × 3 + 1 = 10 [As 3 is an odd number.]
Sixth term: \(\frac{10}{2}\) = 5 [As 10 is an even number.]
Seventh term: 3 × 5 + 1 = 16 [As 5 is an odd number.]
Eighth term: \(\frac{16}{2}\) = 8 [As 16 is an even number.]
Ninth term: \(\frac{8}{2}\) = 4 [As 8 is an even number.]
Tenth term: \(\frac{4}{2}\) = 2 [As 4 is an even number.]
Eleventh term: \(\frac{2}{2}\) = 1 [As 2 is an even number.]
Hence, the Collatz sequence starting with 24 is:
24, 12, 6, 3, 10, 5, 16, 8, 4, 2, 1

Question 13.
Make the Collatz sequence by starting with number 17.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till I is obtained.
1^st term : 17
2^nd term : 3 × 17 + 1 = 52 [As 17 is an odd number]
3^rd term : \(\frac{52}{2}\) = 26 [As 52 is an even number]
4^th term : \(\frac{26}{2}\) = 13 [As 26 is an even number]
5^th term : 3 × 13 + 1 = 40 [As 13 is an odd number]
6^th term : \(\frac{40}{2}\) = 20 [As 40 is an even number]
7^th term : \(\frac{20}{2}\) = 10 [As 20 is an even number]
8^th term : \(\frac{10}{2}\) = 5 [As 10 is an even number]
9^th term : 3 × 5 + 1 = 16 [As 5 is an odd number]
10^th term : \(\frac{16}{2}\) = 8 [As 16 is an even number]
11^th term : \(\frac{8}{2}\) = 4 [As 8 is an even number]
12^th term : \(\frac{4}{2}\) = 2 [As 4 is an even number]
13^th term : \(\frac{2}{2}\) = 1 [As 2 is an even number]
Hence, the Collatz sequency starting with 17 is: 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.

Question 14.
Find the Collatz sequence by starting with number 11.
Solution:
The rule followed by a Collatz sequence is: Start with any number; if the number is even, take half of it; if the number is odd, multiply it by 3 and add 1; repeat till 1 is obtained.
1^st term : 11
2^nd term : 3 × 11 + 1 = 34 [As 11 is an odd number]
3^rd term : \(\frac{34}{2}\) = 17 [As 34 is an even number]
4^th term : 3 × 17 + 1 = 52 [As 1 7 is an odd number]
5^th term : \(\frac{52}{2}\) = 26 [As 52 is an even number]
6^th term : \(\frac{26}{2}\) = 13 [As 26 is an even number]
7^th term : 3 × 13 + 1 = 40 [As 13 is an odd number]
8^th term : \(\frac{40}{2}\) = 20 [As 40 is an even number]
9^th term : \(\frac{20}{2}\) = 10 [As 20 is an even number]
10^th term : \(\frac{10}{2}\) = 5 [As 10 is an even number]
11^th term : 3 × 5 + 1 = 16 [As 5 is an odd number]
12^th term : \(\frac{16}{2}\) = 8 [As 16 is an even number]
13^th term : \(\frac{8}{2}\) = 4 [As 8 is an even number]
14^th term : \(\frac{4}{2}\) = 2 [As 4 is an even number]
15^th term : \(\frac{2}{2}\) = 1 [As 2 is an even number]

Question 15.
Make a quick estimate. Take about 30 seconds. Then compare your answer with your friends.
(i) Number of tiles on your classroom floor:
(a) More than 200
(b) Less than 200
(Hint: Count how many tiles there are*in one row and one column.)

(ii) Time taken to write your full name:
(a) More than 10 seconds
(b) Less than 10 seconds

(iii) Estimate the number of books in your school library.
(a) More than 2000
(b) Less than 2000

(iv) Weight of a fully packed school bag:
(a) More than 4 kg
(b) Less than 4 kg

(v) Estimate the number of leaves on a big tree near your school.
(a) Around 1000
(b) Around 10,000
(c) More than 50,000
Solution:
(i) Estimate the number of tiles in one row and one column. Multiply the two to get the total. If it is 20 rows and 15 columns, then 20 × 15 = 300 tiles.
(ii) Say your name aloud and imagine writing each letter. Most students take 5-12 seconds depending on length.
(iii) Think of how many shelves there are and how many books per shelf. For example, 50 shelves × 50 books = 2500 books.
(iv) Lift your bag and guess based on how heavy it feels. Most packed school bags weigh around 3-5 kg.
(v) A large tree has thousands of branches and leaves. It’s very hard to count, but an estimate can be made.

Number Play Class 6 Case Based Questions

Question 1.
During a school trip, Class 6 students noticed a unique design featuring a 9-box row on a wall. Inspired, they decided to fill the boxes with numbers between 300 and 700, using each number only once.
They were excited to experiment with different number arrangements and see what interesting patterns might emerge.

Based on the above information, answer the following:
(ii) Fill the table such that we get maximum number of supercells.
(ii) How many supercells are there in the table.
Solution:
(i) If there are n cells in a row, then maximum number of supercells = \(\left\{\begin{array}{c}
\frac{n}{2}, \text { if } n \text { is even } \\
\frac{n+1}{2}, \text { if } n \text { is odd }
\end{array}\right.\)
Since total number of cells is 9, maximum 9 +1
number of supercell = \(\frac{9+1}{2}\) = 5
We know that two adjacent cells can never be supercells. So, the maximum number of supercells would be obtained when every alternate cell is a supercell.

In order to get the maximum number of supercells, always consider the 1st cell to be the supercell.
One such table with numbers between 300 and 700 is as follows:

699

301

698

302

697

303

696

304

695

(ii) Clearly, there are 5 supercells.

Question 2.
One day, Raghu bought a packet of pencils and found a lottery scratch coupon tucked inside. Curious, he scratched it off and revealed six two-digit numbers. The first number was 34, and to his surprise, the rest of the numbers followed the mysterious pattern of the Collatz sequence.

Based on the above information, answer the following questions:
(i) What number is written in second box?
(ii) What number is written in last box?
(iii) How many boxes are supercells?
Solution:
(i) As the numbers on the lottery ticket follow the Collatz sequence, and the first number is 34, which is even, the number in second box = \(\frac{34}{2}\) = 17

(ii) In the given Collatz sequence
First term = 34
Second term = \(\frac{34}{2}\) = 17 [As 34 is an even number]
Third term = 3 × 17 + 1 = 52 [As 17 is an odd number]
Fourth term = \(\frac{52}{2}\) = 26 [As 52 is an even number]
Fifth term = \(\frac{26}{2}\) = 13 [As 26 is an even number]
Sixth term = 3 × 13 + 1 = 40 [As 13 is an odd number]
Thus, the number in the last i.e. sixth box is 40.
The list of numbers obtained on the lottery ticket is given below.

(iii) The list of numbers obtained on the lottery ticket is given below.

34

17

52

26

13

40

Clearly, 3 boxes are supercells.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 6 Number Play MCQ improves accuracy in objective exams.

MCQ on Number Play Class 7

Number Play MCQ Class 7

Class 7 Maths Number Play MCQ

Question 1.
Which of the following expressions has odd parity?
(a) 2 + 6
(b) 8 – 4
(c) 7 + 3
(d) 4 + 5
Solution:
(d) 4 + 5
The word parity is used to denote the property
of being even or odd.
Here, 2 + 6 = 8 (Even)
8 – 4 = 4 (Even)
7 + 3 = 10 (Even)
4 + 5 = 9 (Odd)

Question 2.
Which of the following statements is not true?
(a) The parity of the sum of any count of even numbers is even.
(b) The parity of the sum of even count of odd numbers is even.
(c) The parity of the sum of odd count of odd numbers is odd.
(d) If the parity of n is odd, then the parity of n² is even.
Solution:
(d) If the parity of n is odd, then the parity of n² is even.
If n is an odd number, then n1 is also an odd number.
For example, 5 is an odd number, and 5² = 25 is also an odd number.
∴ If the parity of n is odd, then the parity of n² is also odd.
Statement (d) is not true.

Question 3.
Which of the following expressions has odd parity at n = 4?
(a) n²
(b) 2n + 1
(c) 3n
(d) n – 2
Solution:
(b) 2n + 1
At n = 4, n² = 4² = 16 (Even)
At n = 4, 2n + 1 = 2 × 4 + 1 = 9 (Odd)
At n = 4, 3n = 3 × 4 = 12 (Even)
At n = 4, n – 2 = 4 – 2 = 2 (Even)

Question 4.
Which of the following has an even parity?
(a) 3 + 8
(b) 7 – 2
(c) 5²
(d) 5 + 7
Solution:
(d) 5 + 7
The word parity is used to denote the property of being even or odd.
Here, 3 + 8 = 11 (Odd)
7 – 2 = 5 (Odd)
5² = 25 (Odd)
5 + 7 = 12 (Even)

Question 5.
Which of the following has an odd parity?
(a) 8 × 88
(b) 7 × 72
(c) 9²
(d) 82
Solution:
(c) 9²
We know that the parity of the product is odd
if both numbers are odd and the parity of the product of two numbers is even if at least one of them is even.
∴ The parity of both 8 × 88 and 7 × 72 is even.
Also, the parity of n² is the same as the parity of n.
∴ The parity of 9² is odd and the parity of 8² is even.

Question 6.
At what value of n, the expression 3n + 8 has an odd parity?
(a) 9
(b) 8
(c) 4
(d) 0
Solution:
(a) 9
Given expression, 3n + 8
At n = 9, 3 n + 8 = 3 × 9 + 8 = 27 + 8 = 35 (Odd)
At n = 8, 3 n + 8 = 3 × 8 + 8 = 24 + 8 = 32 (Even)
At w = 4, 3n + 8 = 3 × 4 + 8 = 12 + 8 = 20 (Even)
At n = 0, 3n + 8 = 3 × 0 + 8 = 0 + 8 = 8 (Even)
So, the expression has odd parity when n = 9.

Question 7.
Which of the following months has days with odd parity?
(a) April
(b) June
(c) August
(d) November
Solution:
(c) August
Months that have an even number of days show even parity, while those with an odd number of days show odd parity.

April, June, and November each have 30 days, showing even parity, while August, with 31 days, shows odd parity.

Question 8.
If each number in a 3 × 3 magic square is increased by 1, the magic sum will increase by:
(a) 2
(b) 4
(c) 3
(d) 5
Solution:
(c) 3
We know that in a 3 × 3 magic square (using 1 – 9) if we increase each number by n the magic sum increases by 3n.
∴ If each number of a magic square is increased by 1, the magic sum will increase by 3.

Question 9.
What will be the magic sum of a magic square if the central number is 15?
(a) 45
(b) 40
(c) 30
(d) 35
Solution:
(a) 45
Here, central number = 15 = 3 × 5
We know that in a 3 × 3 magic square (using 1-9) if we multiply each number by n the new magic sum becomes 15 × n.
Here, new magic sum =15 × 3 = 45

Question 10.
In how many different ways, number 6 can be written as a sum of Is and 2s?
(a) 12
(b) 13
(c) 14
(d) 15
Solution:
(b) 13
We know that the number of different ways in which number 6 can be written as the sum of Is and 2s is the 6^th element of the Virahanka sequence.
Virahanka sequence is: 1, 2, 3, 5, 8, 13, …
Thus, 6 can be written as sum of Is and 2s in 13 different ways.

Question 11.
Two consecutive numbers in the Virahanka sequence are 144 and 233. The next number in the sequence is:
(a) 297
(b) 377
(c) 89
(d) 367
Solution:
(b) 377
We know that in Virahanka sequence, a number is the sum of previous two numbers.
∴ Required number = 144 + 233 = 377

Question 12.
If each number of a 3 × 3 magic square (with numbers 1 – 9) is multiplied by 4, the magic sum will increase by:
(a) 8
(b) 60
(c) 45
(d) 30
Solution:
(c) 45
We know that the magic sum of a 3 × 3 magic square filled using numbers 1-9 is 15.
If each number of a magic square is multiplied by 4, the new magic sum will be 15 × 4 = 60.
∴ The magic sum will increase by:
60 – 15 = 45.

Question 13.
What will be the magic sum of a 3 × 3 magic square if the central number is 0?
(a) 3
(b) 2
(c) 1
(d) 0
Solution:
(d) 0
Generalised form of 3 × 3 magic square is:

m + 1	m – 4	m +3
m + 2	m	m – 2
m – 3	m + 4	m – 1

Given, m = 0. The magic square with central number 0 is obtained as:

1	-4	3
2	0	-2
-3	4	-1

Here, magic sum = 1 – 4 + 3 = 0

Question 14.
How many rhythms are there with 7 beats consisting of short syllables and long syllables?
(a) 13
(b) 21
(c) 34
(d) 55
Solution:
(b) 21
The number of rhythms with 7 beats consisting of short syllables and long syllables is the 7^th element of the Virahanka sequence.
Now, Virahanka sequence is: 1, 2, 3, 5, 8, 13, 21,…
As, the 7^th element is 21, there are 21 rythms with 7 beats.

Question 15.
The number of different magic squares which can be formed with numbers 1-9, including reflection and rotation, are:
(a) 3
(b) 4
(c) 7
(d) 8
Solution:
(d) 8
Total 8 magic squares can be formed with numbers 1-9, including reflection and rotation.

Question 16.
The expression n² – 1 has even parity when:
(i) n = 2
(ii) n = 3
(iii) n = 4
(iv) n = 5
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (in)
(c) (iii) and (iv)
(d) (ii) and (iv)
Solution:
(d) (ii) and (iv)
Given expression: n² – 1
When n = 2, n² – 1 = 2² – 1 = 4 – 1 = 3 (odd)
When n = 3, n² – 1 = 3² – 1 = 9 – 1 = 8 (even)
When n = 4, n² – 1 = 4² – 1 = 16 – 1 = 15 (odd)
When n = 5, n² – 1 = 5² – 1 = 25 – 1 = 24 (even
Thus, the expression n² – 1 has even parity when n is 3 and 5.

Question 17.
Which of the following 3 × 3 grid is a magic square:
(i)

8	1	6
3	5	7
4	9	2

(ii)

8	1	2
3	5	7
6	9	4

(iii)

6	1	8
7	5	3
2	9	4

(iv)

8	9	6
5	3	7
4	1	2

Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (ii) and (iii)
Solution:
(a) (i) and (iii)
We know that a 3 × 3 square grid of numbers is called a magic square if each row, each column and each diagonal, add up to the same number. This number is called the magic sum.

In grid (i), the sum of each row, each column and each diagonal is 15.
Thus, it is a magic square.
In grid (ii), the sum of first row (i.e. 8+1+2 = 11) is different from the sum of second row (i.e. 3 + 5 + 7 = 15).
Thus, it is not a magic square.

The grid (iii) is the vertical reflection of the grid (i).
Thus, it is also a magic square as sum of each row, each column and each diagonal is 15.

In grid (iv), the sum of first row (i.e. 8 + 9 + 6 = 23) is different from the sum of second row (i.e. 5 + 3 + 7 = 15).
Thus, it is not a magic square.

Number Play Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): The parity of 100 × 101 is even.
(R): The parity of the product of two numbers is even when at least one of the numbers is even.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that the parity of the product of two numbers is even when both numbers are even, or when one is even and the other is odd.
Thus, the parity of 100 × 101 is even as 100 is even.
Therefore, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): The expression 2n + 7 always has an odd parity.
(R): The parity of the sum of an even number and an odd number is odd.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that 2n is an even number.
Also, even number + odd number = odd number
∴ The expression 2n + 7 always has odd parity.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): In a 3 × 3 magic square with numbers 1-9, the first row is [8 5 2],
(R): The magic sum of a 3 × 3 magic square is 15.
Solution:
(d) A is false but R is true.
We know that the magic sum of a 3 × 3 magic square is 15.
Since the number occurring at the centre of a magic square filled using 1-9 must he 5. Hence, the first row cannot be [8 5 2].
∴ Assertion (A) is false, but Reason (R) is true.

Question 4.
(A): The next 3 numbers in the Virahanka sequence: 1, 2, 3, 5, 8, 13,… are 21, 34 and 45.
(R): In Virahanka sequence, a number is the sum of previous two numbers.
Solution:
(d) A is false but R is true.
We know that in Virahanka sequence, a number is the sum of previous two numbers.
∴ The next 3 numbers in the Virahanka sequence: 1, 2, 3, 5, 8, 13, … are 8 + 13 = 21, 21 + 13 = 34 and 34 + 21 = 55.
∴ Assertion (A) is false, but Reason (R) is true.

Number Play Class 7 Fill in the Blanks

Question 1.
The parity of the sum of 99 even numbers is _______.
Solution: even
We know that the parity of sum of any count of even numbers is even.
Therefore, the parity of the sum of 99 even numbers is even.

Question 2.
The parity of the expression n² + n is always ________.
Solution: even
We know that the parity of n is the same as the parity of n.
Also, even number + even number = even number and odd number + odd number = even number.
Therefore, no matter if n is even or odd, the parity of the expression n² + n is always even.

Question 3.
The parity of the sum of numbers from 1 to 20 is ______ .
Solution: even
From 1 to 20, there are 10 even numbers and 10 odd numbers.
We know that the parity of the sum of any count of even numbers is even. So, the sum of 10 even numbers is even.

Also, the parity of the sum of even count of odd numbers is even. So, the sum of 10 odd numbers is even.

Since even number + even number = even number, the parity of the sum of numbers from 1 to 20 is even.

Question 4.
If n is odd, then the parity of n × n × n is __________ .
Solution: odd
We know that,
odd number × odd number = odd number.
∴ If n is odd, then the parity of n ×n × n is odd.

Question 5.
The sum of first 6 numbers in a Virahanka sequence is ______.
Solution: 32
Virahanka sequence is: 1, 2, 3, 5, 8, 13, …
Sum of first 6 numbers = 1 + 2 + 3 + 5 + 8 +13 = 32
Thus, the sum of first 6 numbers in a Virahanka sequence is 32.

Question 6.
The magic sum of a 4 × 4 magic square using numbers 1-16 is _______.
Solution: 34
The sum of numbers from 1 to 16 = 1 + 2 + 3 + …. + 14 + 15 + 16 = 136

We know that in a magic square, each row, each column and each diagonal add up to the same number called the magic sum.
∴ Magic sum = \(\frac{136}{4}\) = 34 [As there are 4 rows and 4 columns in a 4 × 4 magic square]
Thus, the magic sum of a 4 × 4 magic square using numbers 1-16 is 34.

Question 7.
The parity of 6^th element in Virahanka sequence is _______ .
Solution: odd
Virahanka sequence is: 1, 2, 3, 5, 8, 13, …
The 6^th element of Virahanka sequence is 13, which is an odd number.
∴ The parity of 6^th element in Virahanka sequence is odd.

Question 8.
The magic sum of a 4 × 4 magic square using numbers 1-16 is 34. The total of row sum is _______ .
Solution: 136
Given, the magic sum of a 4 × 4 magic square using numbers 1-16 is 34.
Since there are 4 rows, the total of row sum is 4 × 34 = 136.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 5 Parallel and Intersecting Lines MCQ improves accuracy in objective exams.

MCQ on Parallel and Intersecting Lines Class 7

Parallel and Intersecting Lines MCQ Class 7

Class 7 Maths Parallel and Intersecting Lines MCQ

Question 1.
How many angles are formed when two lines intersect?
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
(c) 4
When two lines intersect, they form four angles at the point of intersection.

Question 2.
The sum of a linear pair of angles formed by intersecting lines is:
(a) 360°
(b) 180°
(c) 90°
(d) 45°
Solution:
(b) 180°
A linear pair of angles is formed when two angles are adjacent and their non-common arms form a straight line. This means the two angles lie on a straight line.
So, their sum is always 180°.

∠AOC + ∠BOC = 180°
So, they form linear pairs.

Question 3.
Which of the following is true about perpendicular lines?
(a) They form acute angles.
(b) They form obtuse angles.
(c) They do not intersect.
(d) They form four right angles.
Solution:
(d) They form four right angles.
When two lines are perpendicular, they form four right angles at the point of intersection .

Question 4.
If the complement of an angle is 42°, then the angle is:
(a) 28°
(b) 48°
(c) 138°
(d) 148°
Solution:
(b) 48°
We know that complement of angle x° is (90° – x°).
∴ Complement of 42° = 90° – 42° = 48°
Thus, the angle is 48°.

Question 5.
If the supplement of an angle is 42°, then the angle is:
(a) 28°
(b) 48°
(c) 138°
(d) 148°
Solution:
(c) 138°
We know that supplement ot’x0 is (180° – x°).
∴ Supplement of 42° = 180° – 42° = 138°.

Question 6.
When a transversal cuts two lines, how many angles are formed in total?
(a) 4
(b) 6
(c) 8
(d) 10
Solution:
(c) 8
When a transversal cuts across two lines (whether parallel or not), it intersects each line and forms 4 angles with first line and 4 angles with second line.
So, total of 8 angles are formed when a transversal cuts two lines.

Question 7.
In the given figure, the parallel lines l and m are intersected by a transversal t. The value of ∠a is:

(a) 50°
(b) 80°
(c) 120°
(d) 130°
Solution:
(d) 130°
In the given figure, ∠a and 130° are alternate exterior angles.
We know that alternate exterior angles formed by a transversal intersecting a pair of parallel lines are always equal to each other.
So, ∠a = 130°

Question 8.
In the given figure, which pair of angles forms alternate interior angles?

(a) ∠1 and ∠5
(b) ∠2 and ∠6
(c) ∠4 and ∠5
(d) ∠3 and ∠5
Solution:
(d) ∠3 and ∠5
We know that alternate interior angles lie between the two lines and on opposite sides of the transversal.
From the figure, as ∠3 and ∠5 are on opposite sides of the transversal t and between the two lines l and m, they form a pair of alternate interior angles.

Question 9.
Which of the following is always true when a transversal intersects two lines?
(a) All angles are right angles.
(b) All vertically opposite angles are equal.
(c) All angles are different.
(d) All corresponding angles are equal.
Solution:
(b) All vertically opposite angles are equal.

When a transversal intersects two lines (whether parallel or not), vertically opposite angles are always equal at each point of intersection. This is a basic geometric property and does not depend on whether the lines are parallel or not.

Question 10.
Two parallel lines are cut by a transversal. If ∠x = 65° and it corresponds to ∠y, what is the measure of ∠y?
(a) 115°
(b) 65°
(c) 125°
(d) cannot be determined
Solution:
(b) 65°
We know that when two parallel lines are cut by a transversal, the corresponding angles are always equal.
∴ ∠y = ∠x = 65°

Question 11.
In the figure, if two lines l and m are parallel and ∠c = 110°, what is the measure of its alternate angle?

(a) 110°
(b) 70°
(c) 90°
(d) 100°
Solution:
(a) 110°
We know that when two parallel lines l and m are cut by a transversal t, the alternate interior angles are always equal. Here, ∠f is the alternate interior angle of ∠c.
Therefore, ∠f – ∠c = 110°.

Question 12.
Which of the following are true for adjacent angles?
(i) They share a common vertex.
(ii) They lie on different planes.
(iii) They have a common arm.
(iv) Their non-common arms lie on the same side of the common arm.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)
Solution:
(c) (i) and (iii)
We know that two angles in a plane are said to be adjacent angles if:

(1) they have a common vertex,
(2) they have a common arm, and
(3) their other arms lie on the opposite side of the common arm.
Thus, (i) and (iii) are true.

Question 13.
Which of the following statements about supplementary angles are correct?
(i) Two angles are supplementary if their measures add up to 180°.
(ii) Two obtuse angles can be supplementary.
(iii) A pair of right angles is always supplementary.
(iv) Supplementary angles must always be adjacent.
(v) If one angle is 65°, its supplement is 115°.
Choose the correct option from the following:
(a) Only (i) and (iii)
(b) (i), (ii) and (iv)
(c) (i), (ii) and (v)
(d) (i), (iii) and (v)
Solution:
(d) (i), (iii) and (v)
We know that if the sum of the measures of two angles is 180°, then the angles are called supplementary angles.

Two obtuse angles cannot be supplementary angles as the sum of these angles will be more than 180°.

A pair of right angles, i.e. two right angles are always supplementary angles as the sum of these angles is 180° and supplementary angles may or may not be adjacent.

As 65° + 115° = 180°, 65° and 115° are supplement of each other.
Thus, (i), (iii) and (v) are correct.

Parallel and Intersecting Lines Class 7 Assertion and Reason Questions

The following questions are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question 1.
(A): Two right angles can be complementary.
(R): Complementary angles are those angles whose sum is 90°.
Solution:
(d) A is false but R is true.
We know that if the sum of the measures of two angles is 90°, then the angles are called complementary angles.
A right angle measures exactly 90°, so sum of two right angles = 90° + 90° = 180°.
Thus, two right angles cannot be complementary.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): Two obtuse angles can never be supplementary.
(R): Supplementary angles are those angles whose sum is 180°.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that if the sum of the measures of two angles is 180°, then the angles are called supplementary angles.

An obtuse angle is more than 90°. So, two obtuse angles will always have a sum greater than 180°, and hence they cannot be supplementary.

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 3.
(A): Linear pair of angles are always supplementary.
(R): Linear pair angles are adjacent and lie on a straight line.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that linear pair of angles are adjacent angles formed on a straight line.
Since a straight line forms 180° angle, the two adjacent angles always add up to 180°.
Hence, they are supplementary.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): Parallel lines never intersect even if extended infinitely.
(R): Parallel lines are always equidistant from each other.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that parallel lines stay the same distance apart and lie on the same plane.
So, they never intersect, no matter how far extended.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Parallel and Intersecting Lines Class 7 Fill in the Blanks

Question 1.
When two lines on a plane do not meet even when extended infinitely, they are called __________ lines.
Solution: parallel
When two lines on a plane do not meet even when extended infinitely, they are called parallel lines.

Question 2.
When two lines intersect and all four angles formed are equal, each angle measures ________ degrees.
Solution: 90
We know that when two lines intersect, they form four angles.

Given, all four angles formed are equal.
Let the four angles formed be ∠a, ∠b, ∠c and ∠d. Then,
∠a + ∠b + ∠c + ∠d = 360°
⇒ ∠a + ∠a + ∠a + ∠a = 360°
[∵ ∠a = ∠b = ∠c = ∠d]
⇒ 4∠a = 360° => ∠a = 90°
Thus, when two lines intersect and all four angles formed are equal, each angle measures 90 degrees.

Question 3.
Vertically opposite angles are always ________ .
Solution: equal
Vertically opposite angles are always equal.

Question 4.
________ lines always stay the same distance apart from each other.
Solution: Parallel
Parallel lines always stay the same distance apart from each other.