Class 7

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)

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Part 2 Chapter 1 Geometric Twins MCQ improves accuracy in objective exams.

MCQ on Geometric Twins Class 7

Geometric Twins MCQ Class 7

Class 7 Maths Geometric Twins MCQ

Question 1.
In the given figures, select the option in which figures I₁ and I₂ do not appear congruent.

Solution:
(b)
The figures given in options (a), (c) and (d) have same shape and size. So, the figures are congruent.
In figure given in option (b), the shape is same, but sizes are different. So, they are not congruent.

Question 2.
If two squares are congruent, then which of the following is true?
(a) They have same length of sides.
(b) They have same length of diagonals.
(c) both (a) and (b)
(d) None of these
Solution:
(c) both (a) and (b)
If two squares are congruent, they have equal corresponding sides. Since the diagonals depend on the side length, their diagonals are also equal.

Question 3.
When checking congruence, which of the following movement is allowed?
(a) Moving (Sliding)
(b) Rotating
(c) Flipping (reflection)
(d) All of the above
Solution:
(d) All of the above
Moving, rotating or flipping the figure changes only its position or orientation. It does not change the shape or size of the figure. So, all the three movements are allowed.

Question 4.
For two line segments to be congruent, which of the following is correct?
(a) They have the same orientation (direction).
(b) They have at least one common point.
(c) They have the same length.
(d) They have the different orientation (direction).
Solution:
(c) They have the same length.
For two line segments to be congruent, only their lengths need to be equal.

Question 5.
∆ABC and ∆XYZ are two triangles such that AB = XY = 6 cm, AC = XZ = 5 cm and ∠A = ∠X = 30°. ∆ABC and ∆XYZ are congruent by:
(a) SSS congruence rule
(b) SAS congruence rule
(c) ASA congruence rule
(d) The given information is not sufficient.
Solution:
(b) SAS congruence rule

In ∆ABC and ∆XYZ, we have
AB = XY = 6 cm [Given]
AC = XZ = 5 cm [Given]
∠A = ∠X = 30° [Given]
∴ ∆ABC ≅ ∆XYZ [By SAS congruence rule]

Question 6.
If ∆ABC ≅ ∆FDE such that AB = 5 cm, ∠B = 40° and ∠A = 80°, then which of the following is true?
(a) DF = 5 cm, ∠F = 80°
(b) DF = 5 cm, ∠E = 80°
(c) DE = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40°
Solution:
(a) DF = 5 cm, ∠F = 80°
Given, ∆ABC ≅ ∆FDE,
AB = 5 cm, ∠B = 40°, ∠d = 80°

The corresponding sides are: AB and FD, BC and DE, AC and FE.
The corresponding angles are: ∠A and ∠F, ∠B and ∠D, ∠C and ∠E.
∴ ∠F = ∠A = 80° and DF = BA = 5 cm

Question 7.
In ∆PQR and ∆XYZ, PQ = XY, ∠P = ∠X and PR = XZ. ∆PQR and ∆XYZ are congruent by:
(a) SSS
(b) SAS
(c) ASA
(d) RHS
Solution:
(b) SAS
In ∆PQR and ∆XYZ, two sides and included angle are the same.

∴ ∆PQR ≅ ∆XYZ [By SAS congruence rule]

Question 8.
In the given figure, if AO = OD and BO = OC then ∆AOB ≅ ∆DOC by:

(a) SSS
(b) SAS
(c) ASA
(d) AAS
Solution:
(b) SAS
In ∆AOB and ∆DOC, we have
AO = DO [Given]
BO = CO [Given]
∠AOB = ∠DOC [Vertically opposite angles]
∴ ∆AOB ≅ ∆DOC [By SAS congruence rule]

Question 9.
Out of four given circles, identify the circles that are congruent?
(i) Circle with radius, r = 10 cm
(ii) Circle with area, A = 100π cm²
(iii) Circle with circumference, S = 30π cm
(iv) Circle with area, A = 25π cm²
Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (ii)
(d) (iii) and (iv)
Solution:
(c) (i) and (ii)
For two circles to he congruent, their radii must be the saune, and hence Perimeter (2πR) and area (πR²) will also be the same.
For (i): r = 10 cm
For (ii): A = 100π cm²
⇒ πr² = 100π cm² ⇒ r² = 100 cm²
⇒ r = 10 cm
For (iii): S = 30π cm
⇒ 2πr = 30π cm ⇒ r = \(\frac{30 \pi}{2 \pi} \mathrm{~cm}\) = 150 cm
For (iv): A = 25π cm²
⇒ πr² = 25π cm² ⇒ r² = 25 cm²
⇒ r = 5 cm
As circles given (i) and (ii) have same radii, they are congruent.

Question 10.
In the given figure, BD = CE and BD and CE are altitudes of ∆ABC.
Which of the following are correct?

(i) ∆BCD £ ∆CBE
(ii) ∠DBE = ∠DCE
(iii)) ∠DCB = ∠ECB
(iv) ∠DPE + ∠DAE = 180°
Choose the correct option from the following:
(a) (i) and (ii) only
(b) (ii), (iii) and (iv)
(c) (i), (ii) and (iv)
(d) (i) and (iv) only
Solution:
(c) (i), (ii) and (iv)
In ∆BDC and ∆CEB, we have

BC = CB [Common side (Hypotenuse)]
BD = CE [Given]
∠CDB = ∠BEC [Each 90°]
∴ ∆BDC ≅ ∆CEB [By RHS congruence rule]
⇒ ∆BCD ≅ ∆CBE
⇒ ∠DCB = ∠EBC and ∠DBC = ∠ECB
[By CPCT]
⇒ ∠DCB – ∠ECB = ∠EBC – ∠DBC
⇒ ∠DCE = ∠EBD
In quadrilateral AEPD,
∠PEA + ∠PDA + ∠DPE + ∠DAE = 360°
[Sum of all interior angles is 360°]
⇒ 90° + 90° + ∠DPE + ∠DAE = 360°
⇒ ∠DPE + ∠DAE = 360° – 180° = 180°
So, (i), (ii) and (iv) are correct.

Geometric Twins 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): If ∆ABC ≅ ∆RPQ, then BC = QR.
(R): Corresponding parts of two congruent triangles are equal.
Solution:
(d) A is false but R is true.
We have, ∆ABC ≅ ∆RPQ
Here, the corresponding vertices are: A and R, B and P, C and Question
⇒ BC – PQ [∵ Corresponding parts of two congruent triangles are equal.]
∴ Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): In ∆PQR, if PQ = PR, then ∠P = ∠R.
(R): Angles opposite to equal sides of a triangle are equal.
Solution:

(d) A is false but R is true.
In ∆PQR, we have
PQ = PR
⇒ ∠PRQ = ∠PQR
⇒ ∠R = ∠Q [Angles opposite to equal sides are equal.]
∴ Assertion (A) is false, but Reason (R) is true.

Question 3.
(A): Two congruent triangles have all corresponding angles equal.
(R): AAA criterion is sufficient to prove congruency of two triangles.
Solution:
(c) A is true but R is false.
Two triangles are said to be congruent, if they can be superimposed exactly, one over the other.
∴ Two congruent triangles have all the corresponding sides and corresponding angles equal.
But AAA criterion is not enough to prove congruency of two triangles, because size of triangles might be different.
∴ Assertion (A) is true, but Reason (R) is false.

Question 4.
(A): In ∆ABC, if AB = AC and ∠B = 50°, then ∠C = 50°.
(R): In a triangle, angles opposite to equal sides are equal.
Solution:

(a) Both A and R are true and R is the correct explanation of A.
Given, AB = AC and ∠B = 50°
We know that angles opposite to equal sides of a triangle are equal.
∴ ∠C = ∠B
⇒ ∠C = 50°
∴ Both .Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Geometric Twins Class 7 Fill in the Blanks

Question 1.
A rectangle of length 10 cm and breadth 5 cm is congruent to another rectangle with length _______ and breadth ___________ .
Solution: 10 cm, 5 cm.
For two rectangles to be congruent, their length and breadth must be same.
So, a rectangle of length 10 cm and breadth 5 cm is congruent to another rectangle with length 10 cm and breadth 5 cm.

Question 2.
If two sides and a non-including angle are given, the triangle may or may not uniquely determined. This ambiguous situation is called _______ condition, which can produce ________ different triangles.
Solution: SSA, two
If two sides and a non-including angle are given, the triangle may or may not uniquely determined. This ambiguous situation is called SSA condition, which can produce two different triangles.

Question 3.
For congruent triangles, the perimeters of both triangles are __________ .
Solution: equal
If two triangles are congruent, their corresponding sides and angles are equal.
For congruent triangles, the perimeters of both triangles are equal.

Question 4.
If ∆ABC ≅ ∆XYZ, then AB = ___________, ∠B = ______ and BC = __________ .
Solution: XY, ∠Y, YZ
If two triangles are congruent, their corresponding sides and angles are equal.
Given, ∆ABC ≅ ∆XYZ, then AB = XY, ∠B = ∠Yand BC = YZ.

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°.

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.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 4 Expressions using Letter Numbers MCQ improves accuracy in objective exams.

MCQ on Expressions using Letter Numbers Class 7

Expressions using Letter Numbers MCQ Class 7

Class 7 Maths Expressions using Letter Numbers MCQ

Question 1.
The value of the expression 2x – 7, when x = 5, is:
(a) 7
(b) 5
(c) 3
(d) -4
Solution:
(c) 3
Putting x = 5 in the expression 2x – 7, we get
2 × 5 – 7 = 10 – 7 = 3

Question 2.
The value of the expression 4p + 8q – 10, when p = 3 and q = -1, is:
(a) 8
(b) -24
(c) 14
(d) -6
Solution:
(d) -6
Putting p = 3 and q = – 1 in the expression
4p + 8q – 10,
we get 4 × 3 + 8 × (- 1) – 10 = 12 – 8 – 10
= 12 – 18 = – 6

Question 3.
The simplified form of 9x + 2y + 3x – 7y is:
(a) 12x – 5y
(b) 7xy
(c) 11x – 4y
(d) -3xy
Solution:
(a) 12x – 5y
Given, 9x + 2y + 3x – 7y
Here, 9x and 3x are like terms; 2y and -7y are like terms.
So, 9x + 2y + 3x – 7y = 9x + 3x + 2y – 7y
= 12x – 5y

Question 4.
Shown alongside is a BALANCED hanger using circular and triangular cut-outs. The weight of each circular cut-out is 1 unit and of each triangular cut-out is p units.
Which of the following will help in finding the weight of triangular cut-out?

(a) 6p = 7p
(b) 2p + 4 = 5p + 2
(c) 2(p + 4) = 5(p + 2)
(d) 4p + 2 = 2p + 5
Solution:
(b) 2p + 4 = 5p + 2
Left side: 4 circles and 2 triangles
Right side: 2 circles and 5 triangles
Thus, total weight on left side
= 4 × 1 + 2 × p = (4 + 2p) units
And total weight on right side
= 2 × 1 + 5 × p = (2 + 5p) units
Since the hanger is balanced, both sides must have equal weight.
∴ Weight on left side = Weight on right side
⇒ 4 + 2p = 2 + 5p
⇒ 2p + 4 = 5p + 2

Question 5.
Niharika is printing posters for an event using her home printer. It takes 15 seconds to set up the printer before it starts printing. Once it starts, each poster takes 6 seconds to print. Which of the following expressions shows the total time (in seconds) needed to print4p’ posters, assuming the printer is off initially?
(a) 15 + 6 + p
(b) (15 + 6)p
(c) 15 × 6p
(d) 15 + 6p
Solution:
(d) 15 + 6p
Time taken to set up the printer = 15 seconds (this happens only once).
Time taken to print 1 poster = 6 seconds
Therefore, time taken to print ‘p’ posters = 6p seconds
Thus, total time needed to print ‘p’ posters
= Time taken to set up the printer + Time taken to print ‘p’ posters = (15 + 6p) seconds

Question 6.
The simplified form of 5m – 2n – 7m + 10n + 3 is:
(a) 3
(b) 3m + 3n + 3
(c) 2m + 8n
(d) -2m + 8n + 3
Solution:
(d) -2m + 8n + 3
Given, 5m – 2n – 7m + 10n + 3
Here, 5m and -7m are like terms; -2n and 10n are like terms.
So, 5m – 2n – 7m + 10n + 3
= 5m – 7m – 2n + 10n + 3
= -2m + 8n + 3

Question 7.
Shown below is a representation of one of the properties of whole numbers.

Which of the following represents the property shown?
(a) a(b + c) = a × b + a × c
(b) a + b = b + a
(c) a × (b × c) = a × (b × c)
(d) a + (b + c) = (a + b) + c
Solution:
(a) a(b + c) = a × b + a × c
From figure, we have
Area on left side = Area on right side
⇒ 16 × 8 = 12 × 8 + 4 × 8
[∵ Area of rectangle = Length × Breadth]
⇒ 8 × 16 = 8 × 12 + 8 × 4
⇒ 8 × (12 + 4) = 8 × 12 + 8 × 4
[∵ 16 = 12 + 4]
Suppose a = 8, b = 12 and c = 4, we get
a(b + c) = a × b + a × c

Question 8.
Amit sells cold drinks and sandwiches at a school canteen. One cold drink costs ₹25 and one sandwich costs ₹40.
If m cold drinks and n sandwiches were sold in a day, which of the following expressions shows the total amount earned (in rupees) that day?
(a) 25m + 40n
(b) (25 + 40) × (m + n)
(c) 40m + 25n
(d) (25 + 40) × m + n
Solution:
(a) 25m + 40n
Cost of 1 cold drink = ₹25
Cost of 1 sandwich = ₹40
Money earned from m cold drinks
= 25 × m = 25m
Money earned from n sandwiches
= 40 × n = 40n
Total earnings = 25m + 40M

Question 9.
The formula of the given number machine is:

(a) 2a + b
(b) a × b + 5
(c) a + 5 × b
(d) None of these
Solution:
(b) a × b + 5
We can see,
4 × 1 + 5 = 445 = 9, 7 × 0 + 5 = 0 + 5 = 5,
3 × 2 + 5 = 6 + 5 = 11, 5 × 3 + 5 = 15 + 5 = 20
Here, the rule followed is “5 more than the product of two numbers”.
Thus, if two numbers are a and b, the formula for the given machine is a × b + 5.

Question 10.
Consider the following number machine:

What is the missing input?
(a) 8
(b) 10
(c) 12
(d) 15
Solution:
(a) 8
We can see,
4 × 3 + 2 × 2 = 12 + 4 = 16,
5 × 3 + 2 × 2=15 + 4=19,
7 × 3 + 3 × 2 = 21 + 6 = 27,
9 × 3 + 3 × 2 = 27 + 6 = 33
So, the rule for the given machine in algebraic expression is,
Output = a × 3 + b × 2, where a and b are inputs.
When output = 34 and b = 5,
34 = a × 3 + 5 × 2
⇒ 34 = 3a + 10 ⇒ 3a = 34 – 10 = 24
⇒ a = \(\frac{24}{3}\) = 8 3
Thus, the missing input is 8.

Question 11.
The value of expression 6m – 7n is:
(i) 6, when m = 4 and n = 2.
(ii) 9, when m = 5 and n = 3.
(iii) -9, when m = 2 and n = 3.
(iv) 10, when m = 3 and n = 1.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii) only
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii) only
When m = 4 and n = 2;
6m – 7n = 6 × 4 – 7 × 2 = 24 – 14 = 10
When m = 5 and n = 3;
6m – 7n = 6 × 5 – 7 × 3 = 30 – 21 = 9
When m = 2 and n = 3;
6m – 7n = 6 × 2 – 7 × 3 = 12 – 21 = – 9
When m = 3 and n = 1;
6m – 7n = 6 × 3 – 7 × 1 = 18 – 7 = 11
So, (ii) and (iii) are correct.

Question 12.
If p is a number, then which of the following statements are correct?
(i) 2 more than 3 times the number is 2p + 3.
(ii) 4 less than 7 times the number is 7p – 4.
(iii) 4 more than 5 times the number is 5p + 4.
(iv) 8 less than 4 times the number is 8p – 4.
Choose the correct option from the following:
(a) (i) and (ii)
(b) (ii) and (iii) only
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Solution:
(b) (ii) and (iii) only
(i) 2 more than 3 times the number is 2 + 3 × p
= 3p + 2.
(ii) 4 less than 7 times the number is 7 × p – 4
= 7p – 4.
(iii) 4 more than 5 times the number is 4 + 5 × p
= 5p + 4.
(iv) 8 less than 4 times the number is 4 × p – 8
= 4p – 8.
So, (ii) and (iii) are correct.

Question 13.
The expression 5x – 3y + 4 is equivalent to
(i) 7x – 2y – (2x + y + 2) + 6
(ii) 5x – y + 3 – 2(y + 2x) + 1
(iii) x – y – (2y – 4x) + 4
(iv) 2(2y – x – 2) + 7(1 – y + x) + 1
Choose the correct option from the following:
(a) (i) only
(b) (ii) and (iii) only
(c) (i), (iii) and (iv) only
(d) (i), (ii), (iii) and (iv)
Solution:
(c) (i), (iii) and (iv) only
(i) 7x – 2y – (2x + y + 2) + 6
= 7x – 2y – 2x – y – 2 + 6
= (7x – 2x) + (-2y – y) + 4
= 5x + (-3y) + 4 = 5x – 3y + 4

(ii) 5x – y + 3 – 2(y + 2x) + 1
= 5x – y + 3 – 2y – 4x + 1
= (5x – 4x) + (-y – 2y) + 3 + 1
= x + (-3y) + 4
= x – 3y + 4 ≠ 5x – 3y + 4

(iii) x – y – (2y – 4x) + 4
= x – y – 2y + 4x + 4
= (x + 4x) + (-y – 2y) + 4
= 5x + (-3y) + 4 = 5x – 3y + 4

(iv) 2(2y – x – 2) + 7(1 – y + x) + 1
= 4y – 2x – 4 + 7 – 7y + 7x + 1
= (-2x + 7x) + (4y – 7y) – 4 + 7 + 1
= 5x + (-3y) + 4 = 5x – 3y + 4
So, (i), (iii) and (iv) are correct.

Expressions using Letter Numbers Class 7 Fill in the Blanks

Question 1.
If n is a number, then:
(i) 7 more than the number is _____.
(ii) 3 less than the number is __.
(iii) 5 less than 7 times the number is _________.
(iv) 12 more than 9 times the number is _____.
Solution: n + 7, n – 3, 7n – 5, 9n + 12.
(i) 7 more than the number is n + 7.
(ii) 3 less than the number is n – 3.
(iii) 5 less than 7 times the number is 7n – 5.
(iv) 12 more than 9 times the number is 9n + 12.

Question 2.
If x = – 4, then 5 – x = ____.
Solution: 9
Putting x = – 4 in the expression 5 – x, we get
5 – (-4) = 5 + 4 = 9
∴ If x = – 4, then 5 – x = 9.

Question 3.
If t = 21, then 3t = _______ .
Solution: 63
Putting t = 21 in the expression 31, we get
3 × 21 = 63
∴ If t = 21, then 3t = 63.

Question 4.
If p = 7 and q = – 3, then 3p + 2q = _______.
Solution: 15
Putting p = 7 and q = – 3 in the expression 3p + 2q, we get
3 × 7 + 2 × (- 3) = 21 – 6 = 15 .
∴ If p = 7 and q = – 3, then 3p + 2q = 15.

Question 5.
If x = 12, then 4x + 5 = ______.
Solution: 53
Putting x = 12 in the expression 4x + 5, we get
4 × 12 + 5 = 48 + 5 = 53
∴ If x = 12, then 4x + 5 = 53.

Question 6.
If m = – 4, then 5(m + 1) = _______.
Solution: -15
Putting m = -4 in the expression 5(m + 1), we get
5(- 4 + 1) = 5 × (- 3) = -15
∴ If = -4, then 5(m + 1) = -15.

Question 7.
If p = – 10 and q = 5, then 5p + 10q = ______.
Solution: 0
Putting p = -10 and q = 5 in the expression 5p + 10q, we get
5 × (- 10) + 10 × 5 = – 50 + 50 = 0
∴ If p = – 10 and q = 5 then 5p + 10q = 0.

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 3 A Peek Beyond the Point MCQ improves accuracy in objective exams.

MCQ on A Peek Beyond the Point Class 7

A Peek Beyond the Point MCQ Class 7

Class 7 Maths A Peek Beyond the Point MCQ

Question 1.
How many tenths make a unit?
(a) \(\frac{1}{10}\)
(b) 10
(c) 100
(d) 1
Solution:
(b) 10
We know, 1 unit = 10 one-tenths.

Question 2.
The value of \(4 \frac{4}{10}+7 \frac{3}{10}\) is:
(a) \(7 \frac{7}{10}\)
(b) \(11 \frac{7}{10}\)
(c) \(10 \frac{1}{10}\)
(d) \(7 \frac{17}{10}\)
Solution:
(b) \(11 \frac{7}{10}\)
We can write, \(4 \frac{4}{10}=\frac{44}{10} \text { and } 7 \frac{3}{10}=\frac{73}{10}\)
Now, \(4 \frac{4}{10}+7 \frac{3}{10}=\frac{44}{10}+\frac{73}{10}\)
= \(\frac{117}{10}=\frac{110}{10}+\frac{7}{10}=11+\frac{7}{10}=11 \frac{7}{10}\)

Question 3.
How is one-hundredth represented in decimal form?
(a) 0.01
(b) 0.001
(c) 1.0
(d) 0.1
Solution:
(a) 0.01
One-hundredth represents 100 parts of a unit, i.e. \(\frac{1}{100}=\frac{01}{100}\)
Here, number of zeros in denominator is 2.
So, putting decimal point in the numerator 2 places to the left, we get 0.01.
Thus, the decimal representation of one- hundredth is 0.01.

Question 4.
The place value of 5 in the decimal 11.485 is:
(a) Five-tenths
(b) Five-hundredths
(c) Five-thousandths
(d) Five-ones
Solution:
(c) Five-thousandths
The expanded form of 11.485 is:
11.485 = 10 × 1 + 1 × 1 + \(\left(4 \times \frac{1}{10}\right)\) + \(\left(8 \times \frac{1}{100}\right)+\left(5 \times \frac{1}{1000}\right)\)
Thus, the place value of 5 in the decimal 11.485 is five-thousandths, i.e. 0.005.

Question 5.
The decimal form of 729 hundredths is:
(a) 729
(b) 72.9
(c) 7.29
(d) None of these
Solution:
(c) 7.29
The decimal form of 729 hundredths is \(\frac{729}{100}\) = 7.29.

Question 6.
The decimal representation of \(\frac{3}{4}\) is:
(a) 0.75
(b) 0.85
(c) 0.65
(d) 0.95
Solution:
(a) 0.75
We have, \(\frac{3}{4}=\frac{3 \times 25}{4 \times 25}=\frac{75}{100}\) = 0.75

Question 7.
The place value of the digit 7 in the number 3.476 is:
(a) 7 ones
(b) 7 tenths
(c) 7 hundredths
(d) 7 thousandths
Solution:
(c) 7 hundredths
3.476 can he represented in the decimal place value chart as:

Tens	Ones	.	Tenths	Hundredths	Thousandths
	3	.	4	7	6

Hence, the place value of the digit 7 in the number 3.476 is 7 hundredths, i.e. 0.07.

Question 8.
If a number line is divided into 10 equal parts between 3 and 4, then the decimal number at the 7^th division is:
(a) 3.07
(b) 3.7
(c) 3.17
(d) 37
Solution:
(b) 3.7
The number line, divided into 10 equal parts between 3 and 4, is given as:

Thus, the decimal number at the 7^th division is 3.7.

Question 9.
The greatest decimal number among the following is:
(a) 1.2
(b) 1.02
(c) 1.22
(d) 1.21
Solution:
(c) 1.22
The given decimal numbers are: 1.2, 1.02, 1.22, 1.21
Converting them to like decimals, we get 1.20, 1.02, 1.22, 1.21

Now, as the whole number part of all the decimal numbers is the same, i.e. 1, we compare the digits after the decimal point from left to right.

As 1.02 has 0 at the tenths place while other decimals have 2, 1.02 is the smallest.
Now, comparing the hundredths place of 1.20, 1.22, 1.21, we get
1.22 > 1.21 > 1.20 (As 2 >1 > 0)
Thus, 1.22 is the greatest among the given numbers.

Question 10.
Which of the following sequence is decreasing?
(a) 4.5, 4.55, 4.6
(b) 2.3, 2.2, 2.1
(c) 3.1, 3.0, 3.1
(d) 1.01,2.001,3.0001
Solution:
(b) 2.3, 2.2, 2.1
A sequence is decreasing if each term is smaller
than the one before it.
2.3, 2.2, 2.1 → each term decreases by 0.1 i.e. (2.3 > 2.2 > 2.1).
Thus, 2.3, 2.2, 2.1 is the decreasing sequence.

Question 11.
How can we express 0.5 hours in minutes?
(a) 50 minutes
(b) 45 minutes
(c) 25 minutes
(d) 30 minutes
Solution:
(d) 30 minutes
We know, 1 hour = 60 minutes
Thus, 0.5 hours = 0.5 × 60 = \(\frac{5}{10}\) × 60
= 30 minutes

Question 12.
Which of the following pairs of decimal numbers are equivalent?
(i) 0.5 and 0.500
(ii) 0.05 and 0.0500
(iii) 1.06 and 1.060
(iv) 0.06 and 0.600
Choose the correct option from the following:
(a) (i) and (iv)
(b) (ii) and (iii) only
(c) (i), (ii) and (iii)
(d) (i) and (iii) only
Solution:
(c) (i), (ii) and (iii)
We know that adding trailing zeros after the last non-zero digit of a decimal does not change its value. Thus,
(i) 0.5 and 0.500 are equivalent.
(ii) 0.05 and 0.0500 are equivalent.
(iii) 1.06 and 1.060 are equivalent.
(iv) 0.06 and 0.600 are not equivalent as 0.06 represents 6-hundredths, while 0.600 represents 6-tenths.

Question 13.
Which of these are correct comparisons of decimals?
(i) 0.70 = 0.7
(ii) 0.701 > 0.7
(iii) 0.65 < 0.605
(iv) 0.506 < 0.56
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iv)
(d) (i), (iii) and (iv)
Solution:
(c) (i), (ii) and (iv)
(i) 0.70 and 0.7 are equal in value, i.e. 7 tenths. Thus, this comparison is correct.
(ii) 0.701 is greater than 0.7 because it has a 1 in the thousandths place, while 0.7 has 0. Thus, this comparison is correct.
(iii) 0.65 is greater than 0.605, since 0.65 = 0.650. So, this comparison is incorrect.
(iv) 0.506 < 0.56, since 0.506 = 506 thousandths, and 0.56 = 560 thousandths. Thus, this comparison is correct.

Question 14.
Which of the following statements are true?
(i) The decimal number with the greater whole number part is greater.
(ii) Milligrams are used for measuring heavier weights like bags of rice, etc. while kilograms are used for very light weights like medicine dosage, etc.
(iii) 3.5 feet means 3 feet and 5 inches.
(iv) 2.4 hours means 2 hours 24 minutes.
Choose the correct option from the following:
(a) (ii) and (iv)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (i) and (iii)
Solution:
(c) (i) and (iv)
The decimal number with greater whole number part is greater. For example, 7.1 > 6.99 because 7 > 6.
Kilograms are used for heavier things (like bags of rice) while milligrams are used for very tiny weights (like medicine).
Since 1 foot =12 inches, 0.5 feet = 6 inches. So, 3.5 feet = 3 feet 6 inches.
0.4 hours = 0.4 × 60 = 24 minutes. So, 2.4 hours = 2 hours 24 minutes.
Thus, statements (i) and (iv) are true.

A 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): Fractional part of a number is always less than 1.
(R): Decimal places represent values like
\(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000},\) and so on. These are all parts of a whole.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Decimal places represent values like \(\frac{1}{10}, \frac{1}{100}, \frac{1}{1000},\)
and so on. These are a parts of a whole, they are less than 1.
That’s why fractional part of a number is always less than 1.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 2.
(A): \(3 \frac{6}{100}\) is greater than \(3 \frac{6}{10}\).
(R): \(\frac{6}{100}\) has a larger denominator than \(\frac{6}{10}\).
Solution:
(d) A is false but R is true.
We can write \(3 \frac{6}{100}=\frac{306}{100}\) and \(3 \frac{6}{10}=\frac{36}{10}=\frac{360}{100}\)
Clearly, 360 > 306 ⇒ \(\frac{360}{100}>\frac{306}{100}\)
⇒ \(3 \frac{6}{10}>3 \frac{6}{100}\)
Thus, Assertion (A) is false.
Now, as 100 > 10, \(\frac{6}{100}\) has larger denominator than \(\).
Thus, Reason (R) is true.

Question 3.
(A): If we subtract 47.38 from 89.62, the whole number parts differ by 42, and the actual result is more than 41 and less than 43.
(R): When subtracting two decimal numbers, the result lies between one less than and one more than the difference of their whole number parts.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
Whole number part of 89.62 = 89
Whole number part of 47.38 = 47
Difference of whole number parts
= 89 – 47 = 42
Actual difference: 89.62 – 47.38 = 42.24
Now, (42 – 1) < 42.24 < (42 + 1) i.e.
41 < 42.24 < 43.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Question 4.
(A): 38 paise = ₹0.38
(R): One rupee =100 paise
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that 1 rupee = 100 paise.
So, 1 paisa = \(\frac{1}{100}\) rupee
Therefore, 38 paise = ₹ \(\frac{38}{100}\) = ₹0.38
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

A Peek Beyond the Point Class 7 Fill in the Blanks

Question 1.
The length \(2\frac{7}{10}\) cm is read as _____.
Solution: two and seven- tenths centimetres
The length \(2\frac{7}{10}\) cm is read as two and seven- tenths centimetres.

Question 2.
Kashvi cuts a 1-metre ribbon into 10 equal pieces. The length of one piece is ______ metre.
Solution: \(\frac{1}{10}\) or one-tenth
Given, length of ribbon is 1 metre and number of equal pieces is 10.
∴ Length of each piece = \(\frac{1}{10}\) metre
Thus, the length of one piece is \(\frac{1}{10}\) or one-tenth metre.

Question 3.
Fill in the blanks:
(i) __________ mm = 42.9 cm
(ii) 826 cm = ________ m
(iii) _____ g = 9.5 kg
(iv) 1834 mm = _____ cm
(v) 1749 g = ____ kg
(vi) ₹7 = _____ paise
(vii) ₹ 18.75 = ______ paise
(viii) ₹ _______ = 4250 paise
Solution: 429, 8.26, 9500, 183.4, 1.749, 700, 1875, 42.50
(i) We know that 1 cm = 10 mm.
Therefore, 42.9 cm = 42.9 × 10 mm
= \(\frac{429}{10}\) × 10 mm = 429 mm

(ii) We know that 1 cm = \(\frac{1}{100}\) m
Therefore, 826 cm = \(\frac{826}{100}\) m = 0.86 m

(iii) We know that 1 kg = 1000 g.
Therefore, 9.5 kg = 9.5 × 1000 g 95
= \(\frac{95}{10}\) × 1000 g = 9500 g

(iv) We know that 1 mm = \(\frac{1}{10}\) cm.
Therefore, 1834 mm = \(\frac{1834}{10}\) cm = 183.4 cm

(v) We know that 1 g = \(\frac{1}{1000}\) kg.
Therefore, 1749 g = \(\frac{1749}{100}\) kg = 1.749 kg

(vi) We know that 1 rupee = 100 paise.
Therefore, 7 rupees = 7 × 100 paise
= 700 paise

(vii) We know that 1 rupee =100 paise.
Therefore, ₹ 18.75 = 18.75 × 100 paise
= \(\frac{1875}{100}\) × 100 = 1875 paise

(viii) We know that 1 paise = \(\frac{1}{100}\) rupee
4250 paise = \(\frac{4250}{100}\) rupees = 42.50 rupees

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 2 Arithmetic Expressions MCQ improves accuracy in objective exams.

MCQ on Arithmetic Expressions Class 7

Arithmetic Expressions MCQ Class 7

Class 7 Maths Arithmetic Expressions MCQ

Question 1.
The value of expression 8 + 7 + 6 is:
(a) 21
(b) 19
(c) 20
(d) 17
Solution:
(a) 21
We have, 8 + 7 + 6 = 15 + 6 = 21

Question 2.
Compare and choose the correct option to fill the box: 24 + 16 ☐ 56 – 16.
(a) >
(b) <
(c) =
(d) None of these
Solution:
(c) =
Given, 24 + 16 ☐ 56 – 16
Left hand side (LHS) = 24 + 16 = 40
Right hand side (RHS) = 56 – 16 = 40
Clearly, LHS = RHS So, 24 + 16 = 56 – 16

Question 3.
Compare and choose the correct option to fill the box: 4 × 3 ☐ 78 ÷ 6.
(a) >
(b) <
(c) =
(d) None of these
Solution:
(b) <
Left hand side (LHS): 4 × 3 = 12
Right hand side (RHS): 78 ÷ 6 = 13
Clearly, 12 < 13 ⇒ 4 × 3 < 78 ÷ 6

Question 4.
In a sweet shop, Meena packed 12 laddoos in one box and 4 in another. Kunal packed 10 laddoos in one big box and none in the second. Who packed more laddoos?
(a) Meena
(b) Kunal
(c) Both packed the same number
(d) Depends on the sweetness
Solution:
(a) Meena
Laddoos packed by Meena = 12 + 4= 16
Laddoos packed by Kunal = 10 + 0= 10
Thus, Meena packed more laddoos.

Question 5.
The number of terms in the expression 40 – 7 × 3 + 8 ÷ 2 – 4is:
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(c) 4
We have, 40 – 7 × 3 + 8 ÷ 2 – 4

So, there are 4 terms in the given expression.

Question 6.
The value of the expression 100 – {25 – (10 × 5)} is:
(a) 125
(b) 115
(c) 120
(d) 130
Solution:
(a) 125
We have, 100 – {25 – (10 × 5)}
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
Thus, 100 – {25 – (10 × 5)}
= 100 – 25 + (10 × 5)
= 100 – 25 + 50 = 150 – 25 = 125

Question 7.
The value of the expression 240 – [60 – (30 – 8 ÷ 2) + 10] is:
(a) 192
(b) 194
(c) 196
(d) 198
Solution:
(c) 196
240 – [60 – (30 – 8 ÷ 2) + 10]
= 240 – [60 – (30 – 4) + 10]
= 240 – [60 – 26 + 10]
= 240 – 44 = 196

Question 8.
Which of the following is equal to the expression ‘53 – 17 + 4’?
(a) 53 – (17 + 4)
(b) 53 + (17 – 4)
(c) 53 – (17 – 4)
(d) 4 + (17 – 53)
Solution:
(c) 53 – (17 – 4)
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
So, 53 – (17 – 4) = 53 – 17 + 4

Question 9.
Which of the following expressions are the arithmetic expressions?
(i) 12 + (2 × 4 – 6)
(ii) 7x + 2y – 3
(iii) 18 ÷ 3 + 7
(iv) 4xy – 11
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii), (iii) and (iv)
(d) None of these
Solution:
(a) (i) and (iii) only
We know, an arithmetic expression is a
mathematical sentence that contains numbers and operations like addition (+), subtraction (-), multiplication (×) or division (÷).
So, 12 + (2 × 4 – 6) and 18 ÷ 3 + 7 are arithmetic expressions.
And, 7x + 2y – 3 and 4xy – 11 are not arithmetic expressions because x and y are not numbers.

Question 10.
Which of the following are correct?
(i) The signs ‘>’, ‘=’ and ‘<’ are used to compare the values of two expressions.
(ii) 276- 19 < 275- 18 (iii) 366 – 20 = 380 – 34 (iv) 490 – 20 > 396 – 30
Choose the correct option from the following:
(a) (i) and (iii) only
(b) (ii) and (iv) only
(c) (i), (ii) and (iii)
(d) (i), (iii) and (iv)
Solution:
(d) (i), (iii) and (iv)
(i) Yes, the signs ‘>’, ‘=’ and ‘<’ are used to compare the values of two expressions.
So, (i) is correct.

(ii) Left-hand side (LHS): 276 – 19 = 257
Right-hand side (RHS): 275 – 18 = 257
Clearly, 276- 19 = 275- 18.
So, (ii) is incorrect.

(iii) Left-hand side (LHS): 366 – 20 = 346
Right-hand side (RHS): 380 – 34 = 346
Clearly, 366 – 20 = 380 – 34.
So, (iii) is correct.

(iv) Left-hand side (LHS): 490 – 20 = 470
Right-hand side (RHS): 396 – 30 = 366
Clearly, 470 > 366
⇒ 490 – 20 > 396 – 30
So, (iv) is correct.

Arithmetic Expressions 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): 110 – 27 > 112 – 27
(R): When the same number is subtracted from two different numbers, the difference is greater for the larger number.
Solution:
(d) A is false but R is true.
We know, when the same number is subtracted
from two different numbers, the difference is greater for the larger number.
Since 112 is larger than 110,
110 – 27 < 112 – 27 Therefore, Assertion (A) is false, but Reason (R) is true.

Question 2.
(A): 568 + 247 > 572 + 248
(R): If a < b and c < d, then a + c < b + d, where a, b, c and d are numbers.
Solution:
(d) A is false but R is true.
If a < b and c < d, then a + c < b + d, where a,
b, c and d are numbers.
Clearly, 568 < 572 and 247 < 248
So, 568 + 247 < 572 + 248
Therefore, Assertion (A) is false, but Reason (R) is true.

Question 3.
(A): 5 – 2 ≠ 5 + (-2)
(R): Subtracting a number is the same as adding its additive inverse.
Solution:
(d) A is false but R is true.
We know that subtracting a number is the same
as adding its additive inverse.
∴ 5 – 2 = 5 + (- 2) as – 2 is additive inverse of 2.
Therefore, Assertion (A) is false and Reason (R) is true.

Question 4.
(A) : 16 – (- 7 + 19) = 16 – 7 + 19
(R) : On removing the brackets preceded by a plus “ + ’ sign, the signs of all the terms inside the brackets remain same.
Solution:
(d) A is false but R is true.
We know, on removing the brackets preceded by a ‘+’ sign, the signs of all the terms inside the brackets remains same.
But in expression ‘16 – (-7 + 19)’, brackets are preceded by minus sign.
Therefore, 16 – (- 7 + 19) = 16 + 7 – 19
Therefore, Assertion (A) is false, but Reason (R) is true.

Arithmetic Expressions Class 7 Fill in the Blanks

Question 1.
Fill in the blanks with the correct sign: ‘<’ , ‘>’ or ‘=’:
(i) 111 – 28 _____ 85
(ii) 55 ÷ 11 ____ 5
Solution: <, =
(i) LHS: 111 – 28 = 83
RHS: 85
Clearly, 83 < 85 ⇒ 111 – 28 < 85

(ii) LHS: 55 ÷ 11 = 5
RHS: 5
Clearly, LHS = RHS ⇒ 55 – 11 = 5

Question 2.
Fill in the blanks to make the expressions equal on both sides of the ‘ = ’ sign:
(i) 19 + 6 = ____ + 9
(ii) 9 × _____ = 72 ÷ 2
Solution: 16, 4
(i) 19 + 6 = 16 + 9
[∵ 19 + 6 = 25 and 16 + 9 = 25]
(ii) 9 × 4 = 72 ÷ 2
[∵ 72 ÷ 2 = 36 and 9 × 4 = 36]

Question 3.
In the blanks below, fill ‘<’, ‘>’ or ‘=’ after analysing the expressions on the LHS and RHS. Use reasoning and understanding of terms and brackets to figure this out, and not by evaluating the expressions.
(i) (12 – 7) × 36 _____ (7 – 12) × 36
(ii) 20 + 6 × 15 _____ (20 + 6) × 15
Solution: <, <
(i) LHS : (12 – 7) × 36 = Positive number × 36 → Positive number
RHS : (7 – 12) × 36 = Negative number × 36 → Negative number
We know, a positive number is always greater than a negative number.
Thus, LHS > RHS
Hence, (12 – 7) × 36 > (7 – 12) × 36

(ii) RHS : (20 + 6) × 15

Thus, LHS < RHS
Hence, 20 + 6 × 15 < (20 + 6) × 15

Question 4.
Fill in the blanks with numbers and boxes with operation signs such that the expressions on both sides are equal.
(i) 21 + (___ ☐ ____) = 21 + 12 – 5
(ii) 32 – (3 + 8) = 32 ☐ 3 – _____
(iii) 25 – (11 ☐ 7) = 25 – 11 + 7
(iv) 19 – (17 – 12) = 19 ☐ 17 ☐ 12
Solution: 12 – 5, 8, -, -, +
(i) 21 + ( 12 – 5) = 21 + 12 – 5
(ii) 32 – (3 + 8) = 32 – 3 – 8
(iii) 25 – (11 – 7) = 25 – 11 + 7
(iv) 19 – (17 – 12) = 19 – 17 + 12

Regular revision with Class 7 Maths MCQ with Answers and Ganita Prakash Class 7 Maths Chapter 1 Large Numbers Around Us MCQ improves accuracy in objective exams.

MCQ on Large Numbers Around Us Class 7

Large Numbers Around Us MCQ Class 7

Class 7 Maths Large Numbers Around Us MCQ

Question 1.
Which of the following numbers has 7 at the lakhs place?
(a) 7,50,812
(b) 70,15,902
(c) 5,74,210
(d) 52,08,701
Solution:
(a) 7,50,812
In 7,50,812, the digit 7 is at lakhs place.
In 70,15,902, the digit 7 is at ten lakhs place.
In 5,74,210, the digit 7 is at ten thousands place.
In 52,08,701, the digit 7 is at hundreds place.

Question 2.
(20 × 10,000) + (53 × 1,000) + (1 × 100) + (2 × 1) represents the number:
(a) 53,20,102
(b) 20,53,102
(c) 2,53,102
(d) 5,20,301
Solution:
(c) 2,53,102
(20 × 10,000) + (53 × 1,000) + (1 × 100) + (2 × 1) = 2,00,000 + 53,000 + 100 + 2
= 2,53,102

Question 3.
The correct way of writing ‘sixty five lakh nine thousand four hundred twenty two’ in numbers is:
(a) 65,90,422
(b) 65,09,422
(c) 6,59,422
(d) 65,40,922
Solution:
(b) 65,09,422
Sixty five lakh nine thousand four hundred twenty two = 65,09,422

Question 4.
10 million is equal to:
(a) 100 crore
(b) 1 lakh
(c) 100 lakh
(d) 10,000 lakh
Solution:
(c) 100 lakh
We know,
1 million = 10 lakh
Thus, 10 million =10 × 10 lakh = 100 lakh

Question 5.
The estimated difference of 1,76,346 and 3,865 rounded off to the nearest thousand is:
(a) 1,70,000
(b) 1,72,000
(c) 1,90,000
(d) 1,54,000
Solution:
(b) 1,72,000
1,76,346 is rounded off to nearest thousands as
1,76,000. [∵ At hundreds place, 3 < 5] 3,865 is rounded off to nearest thousands as 4.000. [∵ At hundreds place, 8 > 5]
Thus, estimated difference = 1,76,000 – 4,000
= 1,72,000

Question 6.
The product 974 × 95 will likely have:
(a) 3 digits
(b) 4 digits
(c) 6 digits
(d) 5 digits
Solution:
(d) 5 digits
Rounding 974 to nearest hundreds we get
1,0 and rounding 95 to nearest tens we get 100.
Thus, estimated product = 1,000 × 100
= 1,00,000.
So, the product will likely have five digits and it will be less than 1,00,000 as the numbers are rounded up and 1,00,000 is the smallest 6-digit number.
Verification: 974 × 95 = 92,530, which has five digits.

Question 7.
The estimated sum of 5,47,86,326 and 45,56,767 rounded off to the nearest lakhs is:
(a) 5,94,00,000
(b) 59,40,000
(c) 54,90,000
(d) 9,47,000
Solution:
(a) 5,94,00,000
5,47,86,326 is rounded off to nearest lakhs as
5,48,0,000. [∵ At ten thousands place, 8 > 5]
45,56,767 is rounded off to nearest lakhs as
46,0,000. [∵ At ten thousands place, 5 = 5]
Thus, estimated sum
= 5,48,00,000 + 46,00,000 = 5,94,00,000

Question 8.
The product 105 × 93 will likely have:
(a) 3 digits
(b) 4 digits
(c) 6 digits
(d) 5 digits
Solution:
(b) 4 digits
Rounding off 105 to nearest hundreds, we get 100 and rounding off 93 to nearest tens we get 90,
Thus, estimated product = 100 × 90 = 9,000
So, the product will likely have four digits and it will be close to 9,000.
Verification: 105 × 93 = 9,765, which has four digits.

Question 9.
1 billion is equal to:
(i) 100 crore
(ii) 100 million
(iii) 100 lakh
(iv) 10,000 lakh
Choose the correct option from the following:
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Solution:
(c) (i) and (iv)
We know, 1 million = 10 lakh
And, 1 billion = 1,000 million
Thus, 1 billion = 1,000 × 10 lakh
= 10,000 lakh =100 crore [∵ 1 crore = 100 lakh]

Question 10.
In which of the following numbers the comma is placed incorrectly according to the Indian Number System?
(i) 23,56,76,145
(ii) 123,98,34,67
(iii) 780,444,112
(iv) 56,12,34,889
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)
We know that while writing a number in Indian Number System, the commas are placed in 3-2-2-2… pattern from right to left.

In 23,56,76,145 and 56,12,34,889, the commas are placed in 3-2-2-2… pattern from right to left. They are written correctly according to Indian Number System.

However, in 123,98,34,67 the commas are not placed in 3-2-2-2… pattern from right to left.
Also, in 780,444,112 the commas are placed according to the International Number System,
i. e. in 3-3-3… pattern from right to left.

Thus, the commas are not placed correctly according to the Indian Number System in 123,98,34,67 and 780,444,112.

Large Numbers Around Us 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): 1,000 notes of ₹100 are needed to make ₹ 1,00,000.
(R): Place value helps us break large numbers into smaller, meaningful parts.
Solution:
(b) Both A and R are true but R is not the correct explanation of A.
As ₹1 ,00,000 ÷ 100 = 1,000,
1,000 notes of ₹100 are needed to make ₹1,00,000.
Thus, Assertion (A) is true.
And we know that place value helps us to express larger numbers into smaller, meaningful parts.
Thus, Reason (R) is true, but Reason (R) is not the correct explanation of Assertion (A).

Question 2.
(A): The digit 6 in the number 9,645,123 has a place value of 600,000 in the International Number System.
(R): The place value of a digit is the product of the digit and its position value.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
We know that the place value of a digit in the number system is found by multiplying the
digit with the value of the position it occupies.

Thus, in 9,645, 123, the place value of 6 is 6 × 100,000 = 600,000.
Hence, both Assertion (A) and Reason (R) are true, and Reason (R) is tile correct explanation of Assertion (A).

Question 3.
(A): When rounding 3,42,765 to the nearest lakhs, we get 4,00,000.
(R): Rounding to the nearest lakhs depends on the digit in the ten thousands place.
Solution:
(d) A is false but R is true.
When rounding 3,42,765 to nearest lakhs, we check the digit on the just right of lakhs place, i.e. the digit at the ten thousands place.
In 3,42,765, 4 is at ten thousands place and 4 < 5. So, the digit at the lakhs place, i.e. 3 remains unchanged.
Hence, if 3,42,765 is rounded off to nearest lakhs, we get 3,00,000.
Thus, Assertion (A) is false, but Reason (R) is true.

Question 4.
(A): If 98 × 102 is calculated, then the product will be very close to 10,000.
(R): Both numbers 98 and 102 are close to 100.
Solution:
(a) Both A and R are true and R is the correct explanation of A.
98 is rounded off to nearest tens as 100 and 102 is rounded off to nearest hundreds as 100. Thus, the product is estimated as 100 × 100 = 10,000.
This means that the actual product will be close to 10,000.
Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

Large Numbers Around Us Class 7 Fill in the Blanks

Question 1.
In one lakh, 1 is follwed by _____________________ zeros.
Solution: five
We know, 1 lakh = 1,00,000, where the digit 1 is followed by five zeros.
∴ In one lakh, 1 is followed by five zeros.

Question 2.
In the Indian Number Svswm, the correct way to write the number 9625084 is __________.
Solution: 96,25,084
In the Indian Number System, commas are placed in a 3-2-2-2… pattern from the right to left. This helps to separate the digits into hundreds, thousands, lakhs, crores, and so on.
Therefore, in the Indian Number System, the correct way to write the number 9625084 is 96,25,084.

Question 3.
The smallest 7—digit number is _______ lakh.
Solution: ten
The smallest 7-digit number is 10,00,000 i.e. ten lakh.

Question 4.
The largest 8—digit number is _________
Solution: 9,99,99,999
The largest 8-digit number is 9,99,99,999.

Question 5.
The place value of 5 in 7,005,380 is _________.
Solution: 5000
Given number is 7,005,380, which is represented in International Number System.

Periods	Millions			Thousands			Ones
Place Name	HM	TM	M	HTh	TTh	Th	H	T	O
7,005,380			7	0	0	5	3	8	0

Thus, the place value of 5 in 7,005,380 is 5 thousand = 5 × 1000 = 5000.

Question 6.
The expanded form of the number 76,70,905 is _________.
Solution: (7 × 10,00,000) + (6 × 1,00,000) + (7 × 10,000) + (9 × 100) + (5 × 1)
The expanded form of the number 76,70,905 is
(7 × 10,00,000) + (6 × 1,00,000) + (7 × 10,000) + (9 × 100) + (5 × 1).

Question 7.
In a crore, 1 is followed by ________ zeros.
Solution: seven
We know that 1 crore = 1,00,00,000, where 1 is followed by seven zeros.
∴ In a crore, 1 is followed by seven zeros.

Question 8.
If 728 is rounded off to the nearest hundreds, we get ________.
Solution: 700
In 728, the tens digit is 2 and 2 < 5. So, 7 at the hundreds place remains unchanged while rounded off to the nearest hundreds.
Thus, if 728 is rounded off to the nearest hundreds, we get 700.

MCQ Questions for Class 7 Maths with Answers

Class 7 Maths MCQ Chapter Wise

Class 7 Maths MCQ Questions Part 1

• Chapter 1 Large Numbers Around Us Class 7 MCQ
• Chapter 2 Arithmetic Expressions Class 7 MCQ
• Chapter 3 A Peek Beyond the Point Class 7 MCQ
• Chapter 4 Expressions using Letter Numbers Class 7 MCQ
• Chapter 5 Parallel and Intersecting Lines Class 7 MCQ
• Chapter 6 Number Play Class 7 MCQ
• Chapter 7 A Tale of Three Intersecting Lines Class 7 MCQ
• Chapter 8 Working with Fractions Class 7 MCQ

Ganita Prakash Class 7 Part 2 MCQ

• Chapter 1 Geometric Twins Class 7 MCQ
• Chapter 2 Operations with Integers Class 7 MCQ
• Chapter 3 Finding Common Ground Class 7 MCQ
• Chapter 4 Another Peek Beyond the Point Class 7 MCQ
• Chapter 5 Connecting the Dots Class 7 MCQ
• Chapter 6 Constructions and Tilings Class 7 MCQ
• Chapter 7 Finding the Unknown Class 7 MCQ

Also Read Ganita Prakash Class 7 Solutions