Class 7

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 5 Parallel and Intersecting Lines Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 5 Parallel and Intersecting Lines Solutions

Ganita Prakash Class 7 Chapter 5 Solutions

Class 7 Maths Ganita Prakash Chapter 5 Solutions Parallel and Intersecting Lines

Question 1.
List all the linear pairs and vertically opposite angles you observe in the given figure:

Solution:
We know that adjacent angles formed by two intersecting lines, are called linear pairs. Linear pairs always add up to 180°.
And opposite angles formed by two intersecting lines, are called vertically opposite angles. Vertically opposite angles are always equal to each other.

Question 2.
Using your sense of how parallel lines look, try to draw lines parallel to the line segments on this dot paper.
(a) Did you find it challenging to draw sorne of them?
(b) Which ones?
(c) How did you do ii?

Solution:

(a) Yes, some line segments are a litle more difficult to draw than others.
(b) Line segments e,f. h and g.
(c) Lines parallel to a given line segment are drawn by keeping thern equidistant from the given line segment.

Question 3.
In the figure, which line is parallel to line a — line b or line c? How do you decide this?

Solution:
Line c is parallel to line a because the corresponding dots on these lines are equidistant from each other. So, they do not intersect, no matter how far they are extended.

Question 4.
Can you draw a line parallel to l, that goes through point A? How will you do it with the tools from your geometry box? Describe your method.

Solution:
Tools needed: Ruler, Set-squares (right-angled triangle), Pencil Steps of Construction:
Step 1: Place the set square so that one side is along the line l.
Step 2: Hold the ruler against the other side of the set square (the ruler won’t move).
Step 3: Slide the set square along the ruler until one side reaches point A.
Step 4: Draw a line along the edge of the set square through point A.
Step 5: This new line is parallel to line l and passes through point A.

Question 5.
Find the angles marked below.


Solution:
(i) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain a = 48°.

(ii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain b = 52°.

(iii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain c = 81°.

(iv) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain d = 99°.

(v) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain e = 69°.

(vi) Since the sum of interior angles on the same side of a transversal intersecting a pair of parallel lines is always equal to 180°,f + 132° = 180° ⇒ f = 180°- 132° ⇒ f = 48°

(vii) Since corresponding angles formed by a transversal intersecting a pair of parallel sides are equal, we obtain g = 122°.

(viii) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain h = 75°.

(ix) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain i = 54°.

(x) Since alternate interior angles formed by a transversal intersecting a pair of parallel lines are equal, we obtain j = 97°.

Question 6.
In the figures below, what angles do x and y stand for?

Solution:
(i) The lines and angles are marked as shown in the figure.

Line m is parallel to line n and line a is a transversal.
∴ ∠2 = 65° + ∠1 [∵ Corresponding angles]
⇒ 90° = 65° + ∠1
⇒ ∠1 = 90°- 65°
⇒ ∠1 = 25°
And, ∠1 = x = 25° [∵ Vertically opposite angles]
Also, line m is parallel to line n and line b is a transversal.
∴ ∠1 +y = 180° [∵ Sum of co-interior angles = 180°]
⇒ 25° + y = 180° [∵ ∠1 = 25°]
⇒ y = 180° – 25° ⇒ y = 155°
Thus, the values of x and y are 25° and 155°, respectively.

(ii) The lines and angles are marked as shown in the figure.

Line a is parallel to line b and line d is a transversal.
∴ ∠2 = 53° [∵ Alternate interior angles]
Also, line a is parallel to line b and c is a transversal.
∴ ∠1 + ∠2 = 78° [∵ Alternate interior angles]
⇒ ∠1 + 53° = 78° [∵ ∠2 = 53°]
⇒ ∠1 = 78°- 53°
⇒ ∠1 = 25°
Therefore, ∠1 = x = 25° [∵ Vertically opposite angles]

Question 7.
What is the measure of ∠NOP in the figure?

[Hint: Draw lines parallel to LM and PQ through points N and O.]
Solution:
Lines parallel to LM and PQ through points N and O are drawn and angles are marked as shown in the figure.

Since LM is parallel to EF and MN is a transversal,
∠1 = 40° [∵ Alternate interior angles]
Now, ∠1 + ∠2 = 96°
⇒ 40° + ∠2 = 96° [∵ ∠1 = 40°]
⇒ ∠2 = 96° – 40°
⇒ ∠2 = 56°
Since EF is parallel to GH and AT is a transversal,
∠2 = ∠3 = 56° [∵ Alternate interior angles]
Also, GH is parallel to PQ and OP is a transversal.
∴ ∠4 = 52° [∵ Alternate interior angles]
So, a = ∠3 + ∠4
⇒ a = 56° + 52° ⇒ a = 108°
Thus, ∠NOP = 108°.

InText Questions

Question 1.
Can two straight lines intersect at more than one point?
Solution:
No, two straight lines cannot intersect at more than one point. If two lines intersect at more than one point, then they are coincident lines.

Question 2.
Take a plain square sheet of paper (use a newspaper for this).

(i) How would you describe the opposite edges of the sheet? They are ________ to each other.
(ii) How would you describe the adjacent edges of the sheet? The adjacent edges are ________ to each other. They meet at a point. They form right angles.
(iii) Fold the sheet horizontally in half. A new line is formed (see figure).
How many parallel lines do you see now?
(iv) Make one more horizontal fold in the folded sheet. How many parallel lines do you see now?
(v) What will happen if you do it once more? How many parallel lines will you get? Is there a pattern? Check if the pattern extends further, if you make another horizontal fold.
(vi) Make a vertical fold in the square sheet. This new vertical line is _________ to the previous horizontal lines.
(vii) Fold the sheet along a diagonal. Can you find a fold that creates a line parallel to the diagonal line?
Solution:
(i) They are parallel to each other.
(ii) The adjacent edges are perpendicular to each other.
(iii) We see three parallel horizontal lines — the top edge, the fold and the bottom edge. The new horizontal line is perpendicular to the vertical edges of paper.
(iv) On folding the paper horizontally one more time, we see five parallel horizontal lines.
(v) On folding the paper horizontally once more, we will get nine parallel lines. The number of horizontal parallel lines follow the sequence:
1^st fold → 3 lines
2^nd fold → 5 lines
3^rd fold → 9 lines and so on
So, after each fold, the number of horizontal lines increases as folding doubles the sections and add extra fold lines. The pattern continues as we fold more.

(vi) This new vertical line is perpendicular to the previous horizontal lines.

(vii) Yes, we can make a fold parallel to the diagonal by folding the sheet in the same slanting direction at equal angles or by folding a smaller triangle inside the square.

Parallel and Intersecting Lines Class 7 Extra Questions

Parallel and Intersecting Lines Class 7 Very Short Question Answer

Question 1.
In the given figure, write the vertically opposite angles of ∠AOD and ∠AOC.

Solution:
We know that when two lines intersect, the angles opposite to each other are called vertically opposite angles. They are formed without sharing a common arm and are always equal.
Thus, in the given figure, ∠BOC is vertically opposite angle of ∠AOD and ∠BOD is vertically opposite angle of ∠AOC.

Question 2.
Find the complement of each of the following angles:
(i) 28°
(ii) 66°
(iii) 75°
(iv) 80°
Solution:
We know that complement of angle x° is (90° – x°).
(i) Complement of 28° = 90° – 28° = 62°
(ii) Complement of 66° = 90° – 66° = 24°
(iii) Complement of 75° = 90° – 75° = 15°
(iv) Complement of 80° = 90° – 80° = 10°

Question 3.
Find the supplement of each of the following angles:
(i) 28°
(ii) 95°
(iii) 130°
(iv) 155°
Solution:
We know that supplement of angle x° is (180° – x°).
(i) Supplement of 28° = 180° – 28° = 152°
(ii) Supplement of 95° = 180° – 95° = 85°
(iii) Supplement of 130° = 180° – 130° = 50°
(iv) Supplement of 155° = 180° – 155° = 25°

Question 4.
Find the complement of each of the following angles:
(i) 32°
(ii) 72°
(iii) 68°
(iv) 20°
Solution:
We know that complement of angle x° is (90 – x)°.
(i) Complement of 32° = 90° – 32° = 58°
(ii) Complement of 72° = 90° – 72° = 18°
(iii) Complement of 68° = 90° – 68° = 22°
(iv) Complement of 20° = 90° – 20° = 70°

Question 5.
Find the supplement of each of the following angles:
(i) 25°
(ii) 92°
(iii) 142°
(iv) 165°
Solution:
We know that supplement of angle x° is (180 – x)°.
(i) Supplement of 25° = 180° – 25° = 155°
(ii) Supplement of 92° = 180° – 92° = 88°
(iii) Supplement of 142° = 180° – 142° = 38°
(iv) Supplement of 165° =180° – 165° = 15°

Question 6.
In the given figure, write the angles that form a linear pair with ∠AOD and with ∠BOC.

Solution:
We know that the adjacent angles formed by two lines intersecting each other, are called linear pair of angles. Linear pairs always add up to 180°.
In the given figure, ∠AOC and ∠BOD form a linear pair of angles with ∠AOD.
∠AOC and ∠BOD form a linear pair of angles with ∠BOC.

Question 7.
In the given figure, find the value of a.

Given,
∠SOR = 3a + 20° and ∠ROT = a
From the given figure, we can say that ∠SOR and ∠ROT form a linear pair.
∴ ∠SOR + ∠ROT = 180°
⇒ 3a + 20° + a = 180°
⇒ 4a = 160° ⇒ a = 40°

Question 8.
Find the value of x in the following figure, if AB is parallel to CD.

Solution:
Given, AB is parallel CD.
Since x and 3x are co-exterior angles on the same side of transversal, they add up to 180°.
∴ x + 3x = 180° ⇒ 4x = 180°
⇒ x = \(\frac{180^{\circ}}{4}\) ⇒ x = 45°

Question 9.
Find the value of x in the following figure, if AB is parallel to CD.

Solution:
Given, AB is parallel CD.
Since x and 2x are interior angles on the same side of transversal, they add up to 180°.
∴ x + 2x = 180°
⇒ 3x = 180° ⇒ x = 60°

Parallel and Intersecting Lines Class 7 Short Question Answer

Question 1.
Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 42°, 48°
(ii) 85°, 95°
(iii) 30°, 60°
(iv) 135°, 45°
Solution:
We know that if the sum of the measures of two angles is 90°, then the angles are called complementary angles and if the sum of the measures of two angles is 180°, then the angles are called supplementary angles.
(i) The given angles are 42° and 48°.
Now, sum of given angles = 42° + 48° = 90°
Thus, the given angles are complementary angles.

(ii) The given angles are 85° and 95°.
Now, sum of given angles = 85° + 95° = 180°
Thus, the given angles are supplementary angles.

(iii) The given angles are 30° and 60°.
Now, sum of given angles = 30° + 60° = 90°
Thus, the given angles are complementary angles.

(iv) The given angles are 135° and 45°.
Now, sum of given angles = 135° + 45° = 180°
Thus, the given angles are supplementary angles.

Question 2.
In the given figure, find the value of x.

Solution:
Given, ∠AOB = 47°, ∠BOC = x, ∠COD = 83°, ∠DOE = 92° and ∠EOA = 75°
Now, as ∠AOB, ∠BOC, ∠COD, ∠DOE and ∠EOA are angles at a point, they add up to 360°.
∴ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
⇒ 47° + x + 83° + 92° + 75°= 360°
⇒ 297° + x = 360°
⇒ x = 360° – 297° = 63°
Thus, the value of x is 63°.

Question 3.
In the given figure, line AB is parallel to line DG. Find the value of x + y.

Solution:
Given, line AB is parallel to line DG.
∴ ∠ACE = ∠CEG
[∵ Alternate interior angles are equal]
⇒ x = 80° [∵ ∠ACE = 80° and ∠CEF = x]
Since ∠GFH and ∠EFH form linear pair, they add up to 180°.
∴ ∠GFH + ∠EFH = 180°
⇒ 150° + y = 180° [∵ ∠GFH = 150°]
⇒ y = 180° – 150° = 30°
∴ x + y = 80° + 30° = 110°

Question 4.
In the given figure, l is parallel to m. Find the value of x and y.

Solution:
Given, l is parallel to m and n is transversal to l and m.
Since 2x and y are vertically opposite angles, they are equal.
∴ y = 2x …(i)
Since 4x and y are interior angles on the same side of the transversal, they add up to 180°.
∴ 4x + y = 180°
⇒ 4x + 2x = 180° [From (i)]
⇒ 6x = 180° ⇒ x = 30°
Substituting the value of x in (i), we get
y = 2 × 30° = 60°

Parallel and Intersecting Lines Class 7 Long Question Answer

Question 1.
Which lines appear to be perpendicular to each other in the given figure?

Solution:
When two lines intersect and the angles formed are 90° (i.e. all four angles are equal), the lines are said to be perpendicular to each other.

In the given figure, since line i is parallel to line d, perpendicular to any of these lines is also perpendicular to other. Therefore, lines b, c, h and /are perpendicular to lines i and d.

Similarly, lines d and i are perpendicular to lines b, c, h andf.

Question 2.
In the given figure, AB is parallel to CD and AD is parallel to BC. Find the values of x, y and z.

Solution:
Given, AB || CD, AD ||
BC and ∠BAD = 70°.
As AD || BC and AB is transversal, ∠ABC and ∠BAD are co-interior angles.
∴ ∠ABC + ∠BAD = 180°
⇒ x + 70° = 180° [∵ ∠BAD = 70°]
⇒ x = 180°- 70° = 110°
Now, as AB || DC and AD is transversal, ∠ADC and ∠BAD are co-interior angles.
∴ ∠ADC + ∠BAD = 180°
⇒ z + 70° = 180° [∵ ∠BAD = 70°]
⇒ z = 180°- 70° = 110°
Also, as AD || BC and DC is transversal, ∠ADC and ∠BCD are co-interior angles.
∴ ∠ADC + ∠BCD = 180°
⇒ 110°+ y = 180° [∵ ∠ADC =110°]
⇒ y = 180°- 110° = 70°

Question 3.
In the following figure, find the value of each marked angle.

Solution:
Given, line l || m || n.
Let point P lie on line l, point R lie on line p and the point of intersection of lines p and l be Q, as shown in the figure.

Since ∠PQR and 32° are vertically opposite angles, they are equal.
∴ ∠PQR = 32°
As l || m, ∠b and ∠PQR are interior angles on the same side of transversal p.
∴ ∠b + ∠PQR = 180°
⇒ ∠b + 32° = 180° [∵ ∠PQR = 32°]
⇒ ∠b = 180°- 32° = 148°
As l || n, ∠a and ∠PQR are corresponding angles.
∴ ∠a = ∠PQR
⇒ ∠a = 32° [∵ ∠PQR = 32°]

Question 4.
In the following figure, AB is parallel to CD and AD is parallel to BC. Find the values of x, y and z.

Solution:
Given, AB || CD, AD || BC, ∠DAC = 45° and ∠BAC = 30°.
As AB || CD anddC is transversal, ∠DAC and ∠ACB are alternate interior angles.
∴ ∠ACB = ∠DAC ⇒ x = 45°
[∵ ∠DAC = 45° and ∠ACB = x]
As AB || CD and AC is transversal, ∠BAC and ∠ACD are alternate interior angles.
∴ ∠ACD = ∠BAC ⇒ y = 30°
[∵ ∠BAC = 30° and ∠ACD = y]
As AD || BC and AB is transversal. ∠DAB and ∠ABC are co-interior angles.
∴ ∠DAB +∠ABC = 180°
⇒ ∠DAC + ∠CAB + ∠ABC = 180°
[∵ ∠DAB = ∠DAC + ∠CAB]
⇒ 45° + 30° + z = 180°
[∵ ∠DAC = 45°, ∠CAB = 30° and ∠ABC = z]
⇒ z = 180°- 75° = 105°

Parallel and Intersecting Lines Class 7 Case Based Questions

Question 1.
In the given figure, two straight lines PQ and RS intersect each other at O such that ∠POT = 75°.

Based on the above information, answer the following questions:
(i) Find the value of b.
(ii) Find the value of a.
(iii) Find the value of c.
Solution:
(i) Given, ∠ROP = 4b, ∠POT = 75°, ∠TOS = b
Here, ∠ROS = 180°
[Since ∠ROS is a straight angle]
⇒ ∠ROP + ∠POT + ∠TOS = 180°
⇒ 4b + 75° + b = 180°
⇒ 5b + 75° = 180°
⇒ 5b = 180° – 75° = 105°
b = \(\frac{105^{\circ}}{5}\) = 21°
Thus, the value of b is 21°.

(ii) Since ∠ROP and ∠QOS are vertically opposite angles, they are equal.
∴ ∠QOS = ∠ROP
⇒ a = 4b
⇒ a = 4 × 21° [∵ b = 21°]
⇒ a = 84°
Thus, the value of a is 84°.

(iii) Since ∠QOS and ∠QOR form a linear pair, they add up to 180°.
∴ ∠QOS + ∠QOR = 180°
⇒ a + 2c = 180°
⇒ 84° + 2c =180° [∵ a = 84°]
⇒ 2c = 180° – 84° = 96°
c = \(\frac{96^{\circ}}{2}\) = 48°
Thus, the value of c is 48°.

Question 2.
In a class, a teacher asked a student to draw three lines on the board. The student draws the lines on the board as shown.

The line l is perpendicular to line n and ∠1 = 75°.
Based on the above information answer the following questions:
(i) What is the measure of ∠6?
(ii) What is the measure of ∠5?
(iii) What is the measure of ∠3?
(iv) What is the sum of the measure of ∠2 and ∠1?
Solution:
(i) Given, line l is perpendicular to line n.
∴ ∠1 + ∠6 = 90°
⇒ 75° + ∠6 = 90° [∵ ∠1 = 75°]
⇒ ∠6 = 90° – 75° = 15°

(ii) Given, line l is perpendicular to line n.
∴ ∠5 = 90°

(iii) We have, ∠6 = 15°
Since ∠3 and ∠6 are vertically opposite angles, they are equal.
∴ ∠3 = ∠6 = 15°

(iv) Given, line l is perpendicular to line n. So, ∠2 is equal to 90°.
∴ ∠1 + ∠2 = 75° + 90° = 165°

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 4 Expressions using Letter Numbers Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 4 Expressions using Letter Numbers Solutions

Ganita Prakash Class 7 Chapter 4 Solutions

Class 7 Maths Ganita Prakash Chapter 4 Solutions Expressions using Letter Numbers

Question 1.
One plate of Jowar roti costs ₹30 and one plate of Pulao costs ₹20. If x plates of Jowar roti and y plates of pulao were ordered in a day, which expressions(s) describe the total amount in rupees earned that day?
(a) 30x + 20y
(b) (30 + 20) × (x + y)
(c) 20x + 30)
(d) (20 + 30) × (x + y)
(e) 30x – 20y
Solution:
Cost of one plate of Jowar roti = ₹30
Cost of x plates of Jowar roti = ₹30x
Cost of one plate of Pulao = ₹20
Cost of y plate of Pulao = y
So, the expression for the total amount earned that day = 30x + 20y
Hence, the correct answer is option (a).

Question 2.
Write formulas for the perimeter of:
(i) triangle with all sides equal.
(ii) a regular pentagon.
(iii) a regular hexagon.
Solution:
(i) Let side length of triangle be a. Then,
Perimeter of triangle with all sides equal = a + a + a = 3a = 3 × side

(ii) Let side length of a regular pentagon be a. Then,
Perimeter of the regular pentagon = a + a + a + a + a = 5a = 5 × side

(iii) Let side length of a regular hexagon be a. Then,
Perimeter of the regular hexagon = a + a + a + a + a + a = 6a = 6 × side

Question 3.
Munirathna has a 20 m long pipe. However, he wants a longer watering pipe for his garden. He joins another pipe of some length to this one. Give the expression for the combined length of the pipe. Use the letter- number to denote the length in metres of the other pipe.
Solution:
Length of pipe Munirathna has = 20 m
Length of another pipe Munirathna wants to join = k m
∴ Combined length of the pipe = (20 + k) m

Question 4.
What is the total amount Krithika has, if she has the following numbers of notes of ₹100, ₹20 and ₹5?
Complete the following table:

Solution:

Question 5.
In a calendar month, if any 2 × 3 grid full of dates is chosen as shown in the picture, write expressions for the dates in the blank cells if the bottom middle cell has date ‘w’.

Solution:
Date to the left to w will be 1 less than w.
Date to the right to w will be 1 more than w.
Since there are 7 clays in a week,
Date above w will be 7 less than w.
Date in the diagonally left cell to w will be 8 less than w.
Date in the diagonally right will to w will be 6 less than w.
Thus, the expressions in other five blank cells of the grid are as shown:

Question 6.
A snail is trying to climb along the wall of a deep well. During the day it climbs up ‘u’ cm and during the night it slowly slips down ‘d’ cm. This happens for 10 days and 10 nights.
(1) Write an expression describing how far away the snail is from its starting position.
(ii) What can we say about the snail’s movement if d > u?
Solution:
(i) During the day the snail climbs up ‘u’ cm.
During the night the snail slips down ‘d’ cm.
So, the net distance covered in one day is (u – d) cm.
So, in 10 days and 10 nights the net distance covered by the snail = 10(u – d) cm.
Hence, the expression describing how far away the snail is from it starting position is 10(u – d).

(ii) If d > u, snail slips down more than it climbs.
It means the snail will never reach the top.

Question 7.
Radha is preparing for a cycling race and practices daily. The first week she cycles 5 km every day. Every week she increases the daily distance cycled by ‘z’ km. How many kilometres would Radha have cycled after 3 weeks?
Solution:
In first week, Radha cycles 5 km every day.
So, she cycled 5 × 7 = 35 km in first week.
In second week, Radha cycles (5 + z) km every day.
In third week, she cycles 5 + z + z = (5 + 2z) km every day.
So, she cycled (5 + 2z) × 7 = (35 + 14z) km in third week.
Thus, number of kilometres Radha cycled in 3 weeks
= 35 + (35 + 7z) + (35 + 14z)
= (35 + 35 + 35) + (7z + 14z) = (105 + 21z)km

Question 8.
In the following figure, observe how the expression it w + 2 becomes 4w + 20 along one path. Fill in the missing blanks on the remaining paths. The ovals contain expressions and the boxes contain operations.

Solution:
[(w + 2) – 5] × 3 = [w – 3] × 3 = 3w – 9
[(w + 2) – 8] – 4 = [w – 6] – 4 = w – 10
[(w + 2) – 4] × 3 = [w – 2] × 3 = 3w – 6

Question 9.
A local train from Yahapur to Vahapur stops at three stations at equal distances along the way. The time taken in minutes to travel from one station to the next station is the same and is denoted by t. The train stops for 2 minutes at each of the three stations.
(i) What is the algebraic expression for the time taken to travel from Yahapur to Yahapur?
(ii) If t = 4, what is the time taken to travel from Yahapur to Vahapur?
Solution:
(i)
Let the time taken to travel from one station to another station = t
So, time taken to travel from Yahanpur to Vahapur = 4t
As there are three stoppages between these two stations and the train stops for 2 minutes at each stoppage, total time taken during stoppages = 2 × 3 = 6 minutes.
So, the algebraic expression for total time taken (in minutes) is (4t + 6).

(ii) From (i), the algebraic expression for total time (in minutes) taken from Yahanpur to Vahapur is (4t + 6). So, the time taken to travel from Yahapur to Vahapur = 4 × 4 + 6 = 16 + 6 = 22 minutes.

Question 10.
Simplify the following expressions:
(i) 3a + 9b – 6 + 8a – 4b – 7a + 16
(ii) 3(3a – 3b) – 8a – 4b – 16
(iii) 2(2x – 3) + 8x + 12
(iv) 8x – (2x – 3) + 12
(v) 8h – (5 + 7h) + 9
(vi) 23 + 4(6m – 3n) – 8n – 3m – 18
Solution:
(i) 3a + 9b – 6 + 8a – 4b – 7a + 16 = (3a + 8a – 7a) + (9b – 4b) + (- 6 + 16) = 4a + 5b + 10
(ii) 3(3a – 3b) – 8a – 4b – 16 = 9a – 9b – 8a – 4b – 16 = (9a – 8a) + (-9b – 4b) – 16 = a – 13b – 16
(iii) 2(2x – 3) + 8x + 12 = 4x – 6 + 8x + 12 = (4x + 8x) + (-6 + 12) = 12x + 6
(iv) 8x – (2x – 3) + 12 = 8x – 2x + 3 + 12 = 6x + 15
(v) 8h – (5 + 7h) + 9 = 8h – 5 – 7h + 9 = 8h – 7h – 5 + 9 = h + 4
(vi) 23 + 4(6m – 3n) – 8n – 3m – 18 = 23 + 24m – 12n – 8n – 3m – 18
= 24m – 3m – 12w – 8n + 23 – 18 = 21m – 20n + 5

Question 11.
Add the expressions given below:
(i) 4d – 7c + 9 and 8c – 11 +9d
(ii) – 6f + 19 – 8s and – 23 + 13f + 12s
(iii) 8d – 14c + 9 and 16c – (11 + 9d)
(iv) 6f – 20 + 8s and 23 – 13f – 12s
(v) 13m – 12n and 12n – 13m
(vi) – 26m + 24n and 26m – 24n
Solution:
(i) (4d – 7c + 9) + (8c – 11 + 9d) = (4d + 9d) + (- 7c + 8c) + (9 – 11) = 13d + c – 2
(ii) (-6f + 19 – 8s) + (-23 + 13f + 12s) = (-6f + 13f) + (-8s + 12s) + (19 – 23) = 7f + 4s – 4
(iii) (8d – 14c + 9) + [16c – (11 + 9d)] = (8d – 14c + 9) + (16c – 11 – 9d)
= (8d – 9d) + (-14c + 16c) + (9 – 11) = -d + 2c – 2
(iv) (6f – 20 + 8s) + (23 – 13f – 12s) = (6f – 13f) + (8s – 12s) + (-20 + 23) = -7f – 4s + 3
(v) (13m – 12n) + (12n – 13m) = (13m – 13m) + (- 12n + 12n) = 0
(vi) (-26m + 24n) + (26m – 24n) = (-26m + 26m) + (24n – 24n) = 0

Question 12.
Imagine a straight rope. If it is cut once as shown in the picture, we get 2 pieces. If the rope is folded once and then cut as shown, we get 3 pieces. Observe the pattern and find the number of pieces if the rope is folded 10 times and cut. What is the expression for the number of pieces when the rope is folded r times and cut?

Solution:
Step 1 (0 fold): We get 0 + 2 = 2 pieces
Step 2 (1 fold): We get 1 + 2 = 3 pieces
Step 3 (2 folds): We get 2 + 2 = 4 pieces
In the same way, if rope is folded 10 times and cut, we get 10 + 2 = 12 pieces.
In the same way, when the rope is folded r times and cut, we get r + 2 pieces.

Question 13.
Look at the matchstick pattern below. Observe and identify the pattern. How many matchsticks are required to make 10 such squares. How many are required to make w squares?

Solution:
Step 1: To make 1 square, we need 4 matchsticks.
Step 2: To make 2 squares, we need 4 + 3 = 7 matchsticks.
Step 3: To make 3 squares, we need 4 + 3 + 3 = 10 matchsticks.
So, to make w squares, we need —
4 + (w – 1) × 3 = 4 + 3 (w – 1) = 4 + 3w – 3 = (3w + 1) matchsticks.
To make 10 squares, substituting w = 10, we get
Number of required matchsticks = 3(10) + 1 = 30 + 1 = 31

Question 14.
Observe the pattern below. How many squares will be there in Step 4, Step 10, Step 50? Write a general formula. How would the formula change if we want to count the number of vertices of all the squares?

Solution:
Number of squares in step 1 = 5
Number of squares in step 2 = 5 + 4 = 9
Number of squares in step 3 = 5 + 4 + 4 = 5 + 2 × 4 = 5 + 8 = 13
Thus, number of squares in step 4 = 5 + 3 × 4 = 5 + 12 = 17
Number of squares in step 10 = 5 + 9 × 4 = 5 + 36 = 41
And, number of squares in step 50 = 5 + 49 × 4 = 5 + 196 = 201
So, the general formula for number of squares in step n = 5 + (n – 1) × 4 = 5 + 4 (n – 1) = 5 + 4n – 4 = 4n + 1
Number of vertices in step 1 = 16
Number of vertices in step 2 = 16 + 12 = 28
Number of vertices in step 3 = 16 + 12 + 12 = 16 + 2 × 12 = 16 + 24 = 40
Number of vertices in step n = 16 + (n – 1) × 12 = 16 + 12n – 12 = 12n + 4

InText Questions

Question 1.
Shabnam is 3 years older than Aftab. When Aftab’s age is 10 years, Shabnam’s age will be 13 years. Now, Aftab’s age is 18 years, what will Shabnam’s age be?
Solution:
Shabnam’s age = Aftab’s age + 3
When .Aftab’s age = 18 years,
Shabnam’s age = 18 + 3 = 21 years

Question 2.
Find the values of the following arithmetic expressions:
(i) 23 – 10 × 2
(ii) 83 + 28 – 13 + 32
(iii) 34 – 14 + 20
(iv) 42 + 15 – (8 – 7)
(v) 68 – (18 + 13)
(vi) 7 × 4 + 9 × 6
(vii) 20 + 8 × (16 – 6)
Solution:
(i) 23 – 10 × 2 = 23 – 20 = 3
(ii) 83 + 28 – 13 + 32
= (83 – 13) + (28 + 32)
= 70 + 60 = 130
(iii) 34 – 14 + 20 = (34 – 14) + 20 = 20 + 20 = 40
(vi) 42 + 15 – (8 – 7) = 42 + 15 – 1
= 42 + 14 = 56
(v) 68 – (18 + 13) = 68 – 31 = 37
(vi) 7 × 4 + 9 × 6 = 28 + 54 = 82
(vii) 20 + 8 × (16 – 6) = 20 + 8 × 10
= 20 + 80 = 100

Question 3.
Some simplifications are shown below where the letter-numbers are replaced by numbers and the value of the expression is obtained.
1. Observe each of them and identify if there is a mistake.
2. If you think there is a mistake, try to explain what might have gone wrong.
3. Then, correct it and give the value of the expression.

Solution:
(1) If a = – 4, then 10 – a = 6 is wrong.

(2) If d = 6, then 3d = 36 is wrong.
As 10 – a = 10 – (-4) = 10 + 4 = 14
As 3d = 3 × d = 3 × 6 = 18
So, if a = -4, then 10 – a = 14 is correct.
So, if d = 6, then 3d = 18 is correct.

(3) Ifs = 7, then 3s – 2 = 15 is wrong.

(4) If r = 8, then 2r + 1 = 29 is wrong.
As 3s – 2 = 3 × 7 – 2 = 21 – 2 = 19
As 2r + 1 = 2 × 8 + 1 = 16 + 1 = 17
So, if s = 7, then 3s – 2 = 19 is correct.
So, if r = 8, then 2r + 1 = 17 is correct.

(5) If j = 5, then 2j = 10 is correct.

(6) If m = – 6, then 3(m + 1) = 19 is wrong.
As 2j = 2 × 5 = 10
As 3(m + 1) = 3 × (- 6 + 1) = 3 × (- 5) = – 15
So, if m = – 6, then 3(m + 1) = – 15 is correct.

(7) If f = 3, g = 1, then 2f – 2g = 2 is wrong.

(8) If t = 4, b = 3, then 2t + b = 24 is wrong.
As 2f – 2g = 2 × 3 – 2 × 1 = 6 – 2 = 4
As 2t + b = 2 × 4 + 3 = 8 + 3 = 11
So, if f = 3, g = 1, then 2f – 2g = 4 is correct.
So, if t = 4, b = 3, then 2t + b = 11 is correct,

(9) If h = 5, n = 6, then h – (3 – n) = 4 is wrong.
As h – (3 – n) = 5 – (3 – 6) = 5 – (- 3) = 5 + 3 = 8
So, if h = 5, n = 6, then h – (3 – n) = 8 is correct.

Question 4.
Here is a table showing the number of pencils and erasers sold in a shop. The price per pencil is c, and the price per eraser is d. Find the total money earned by the shopkeeper during these three days.

(i) If c = ₹50, find the total amount earned by the sale of pencils.
(ii) Write the expression for the total money earned by selling erasers. Then, simplify the expression.
Solution:
Total money earned by the shopkeeper
= Money earned on day 1 + Money earned on day 2 + Money earned on day 3
= 5c + 4 d + 3c + 6d + 10c + d = 18c + 11d

(i) Total amount earned by the sale of pencils
= 5c + 3c + 10c = 18c
= 18 × ₹50 = ₹900

(ii) Given, the price per eraser is d.
Total money earned by selling erasers
= 4d + 6d + d = 11 d

Question 5.
Fill in the blanks below by replacing the letter- numbers by numbers; an example is shown. Then compare the values that 5u and 5 + u take.

Solution:

We see that the values of 5u and 5 + u are not equal for different values of u. So, the expressions 5w and 5 + u are not equal.

Question 6.
Are the expressions 10y – 3 and 10(y – 3) equal?
After filling in the two diagrams (given below), do you think the two expressions are equal?

Solution:

Since the values of 10v – 3 and 10(y – 3) are not equal for different values of y, the expressions 10y – 3 and 10(y – 3) are not equal.

Question 7.
Find the formulas of the number machines below and write the expression for each set of inputs.

Solution:
(i) The formula for the number machine is “sum of first number and second number minus two” and the expression is a + b – 2.
The expression for each set of inputs are:
5 + 2 – 2 = 5, 8 + 1 – 2 = 7,
9 + 11 – 2 = 18, 10 + 10 – 2 = 18
and a + b – 2

(ii) The formula for the number machine is “product of first number and second number plus one” and the expression is a × b + 1.
The expression for each set of inputs are:
4 × 1 + 1= 5, 6 × 0 + 1 = 1,
3 × 2 + 1 = 7, 10 × 3 + 1 = 31
and a × b + 1 = ab + 1

Question 8.
Somjit noticed a repeating pattern along the border of a saree.

Use this to find what design appears at positions 99, 122 and 148.
Solution:
We can see the design A, B and C appear at the positions 3n – 2, 3n – 1 and 3n, respectively.
For 99, the remainder on division by 3 is 0,’i.e. it is a multiple of 3.
So, at position 99, design C will appear.
For 122, the remainder on division by 3 is 2, i.e. it is 1 less than a multiple of 3, i.e. 3n – 1.
So, at position 122, design B will appear.
For 148 , the remainder on division by 3 is 1, i.e. it is 2 less than a multiple of 3, i.e. 3n – 2.
So, at position 148, design A will appear.

Expressions using Letter Numbers Class 7 Extra Questions

Expressions using Letter Numbers Class 7 Very Short Question Answer

Question 1.
Simplify the expression 7(u – 2v) + 2v.
Solution:
Using the distributive property, this expression can be simplified as
7(u – 2v) + 2v = 7u – 7 × 2v + 2v
= 7u – 14v + 2v = 7u – 12v
[Adding like terms together]

Question 2.
What is the sum of the numbers in the given picture (unknown values are denoted by letter- numbers)?

Solution:
Adding row wise, we get Sum of numbers
= (5 × 4) + (2g + 2h) + (2g + 2h) + (5 × 4)
= 20 + 2g + 2 h + 2g + 2h + 20
= (20 + 20) + (2g + 2g) + (2h + 2 h)
= 40 + 4g + 4 h

Expressions using Letter Numbers Class 7 Short Question Answer

Question 1.
Add the numbers in each picture below. Write their corresponding expressions and simplify them.

Solution:
(i) (x + y + y + x + y + y + x + y + y + x + y + y) + (4 × 6)
= 4x + 8y + 24

(ii) 2 × 3y + 4 × 2 + 2 × (-4x) + 4 × (-8)
= 6y + 8 – 8x – 32
= 6y – 8x – 24

(iii) 4 × (- 3n) + 6 × 7m
= -12m + 42m
= 42m – 12n

Question 2.
Simplify each of the following expressions:
(i) e + e + e + f + f
(ii) m + m – (m – n) – n – n
(iii) l – 7m + l – (m – 1)
(iv) 2x – x – (x + x)
(v) (3u – 2v) – (2u – 3v)
Solution:
(i) e + e + e + f + f = 3e + 2f
(ii) m + m – (m – n) – n – n = 2m – (m – n) – 2n
= 2m – m + n – 2n
= m – n

(iii) l – m + l – (m – l) = l + l – m – (m – l)
= 2l – m – m + l
= 3l – 2m

(iv) 2x – x – (x + x) = 2x – x – 2x = -x

(v) (3u – 2v) – (2u – 3v) = 3u – 2v – 2u + 3v
= 3u – 2v – 2v + 3u
= u + v

Expressions using Letter Numbers Class 7 Long Question Answer

Question 1.
Simplify the following expressions:
(i) 2x – 3y + 7x + y – 12
(ii) 4(x – 2y) + 3x – 5y + 1
(iii) 7 + 3h – 2g – (5h – 4g)
(iv) 3g + 9h – (5 + 6h – 2g)
(v) 5(2u + 7v – 2) – 3(8u – 10v + 4)
(vi) 17 – 13a + 19b – 2(9 + 7a – 3b)
Solution:
(i) 2x – 3y + 7x + y – 12
= 9x – 2y – 12

(ii) 4(x – 2y) + 3x – 5y + 1
= 4x – 8y + 3x – 5y + 1
= 7x – 13y + 1

(iii) 7 + 3h – 2g – (5h – 4g)
= 7 + 3h – 2g – 5h + 4g
= 7 – 2h + 2g

(iv) 3g + 9h – (5 + 6h – 2g)
= 3g + 9h – 5 – 6h + 2g
= 5g + 3h – 5

(v) 5(2u + 7v – 2) – 3(8u – 10v + 4)
= 10u + 35v – 10 – 24u + 30v – 12
= – 14u + 65v – 22

(vi) 17 – 13a + 19b – 2(9 + 7a – 3b)
= 17 – 13a + 19b – 18 – 14a + 6b
= – 27a + 25b – 1

Question 2.
Add the expressions given below:
(i) 3d+ 6c – 7 and 3c – 4d + 2
(ii) -8g + 3h + 2 and 6g – 4h – 12
(iii) 11k + l + 10 and k – (11l – 10)
(iv) 21m – 7n and -17m + 10n + 13
(v) 5(x + 7y) – 3 and 2x + 3y – 2
(vi) 7a + 2b + 9 and 3a + 8b + 1
Solution:
(i) (3d + 6c – 7) + (3c – 4d + 2)
= 6c + 3c + 3d – 4d – 7 + 2 = 9c – d – 5

(ii) (-8g + 3h + 2) + (6g – 4h – 12)
= -8g + 6g + 3h – 4h + 2 – 12
= – 2g – h – 10

(iii) (11k + l + 10) + {k – (11l – 10)}
= 11k + l + 10 + k = 11l + 10
= 11k + k + l – 11l + 10 + 10
= 12k – 10l + 20

(iv) (21m – 7n) + (-17m + 10n + 13)
= 21m – 17m – 7n + 10n + 13
= 4m + 3n + 13

(v) 5(x + 7y) – 3 + (2x + 3y – 2)
= 5x + 35y – 3 + 2x + 3y – 2
= 5x + 2x + 35y + 3y – 3 – 2
= 7x + 38y – 5

(vi) (7a + 2b + 9) + (3a + 8b + 1)
= 7a + 3a + 2b + 8b + 9 + 1
= 10a + 10b + 10

Expressions using Letter Numbers Class 7 Case Based Questions

Question 1.
A movie theatre has 6 columns of seats (labelled A to F) and the seats are arranged in endless rows, starting from the front. Each seat is numbered sequentially row wise, beginning with Seat 1 in Column A of Row 1, moving left to right.

Based on the above information, answer the following questions:
(i) In the seating layout, imagine selecting any 2 × 3 block of seats (2 rows and 3 columns), like the one shown in the picture.

If the seat in the top middle of selected block is numbered ‘x’ write expressions to represent the seat numbers in the other five blank positions of the block.

(ii) Look at the group of seats arranged in the shape.

Find the sum of all the seat numbers in this shape. Then compare this total with the number
at the centre i.e. 11.
Try this again with a different set of numbers arranged in the same shape and write your observation.
Solution:
(i) Number to the left of Y will be 1 less than x.
Number to the right of V will be 1 more than x.
Since there are 6 seats in one row,
Number below x will be 6 more than x.
Number in the diagonally left cell to ‘x’ will be 5 more than x for one less than x + 6).
Number in the diagonally right cell to ‘x’ will be 7 more thain x (or one more than x + 6).
Thus, the expressions in other five blank s positions of the block are as shown:

(ii) Sum of all numbers
= 4 + 6 + 11 + 16 + 18 = 55 = 5 × 11
The sum is 5 times the number in the centre.
Now, let the number at the centre be 14, then the shape is given below:

Sum of all the numbers
= 7 + 9 + 14 + 19 + 21
= 70 = 5 × 14
Again, the sum is 5 times the number in the centre.
Now, let the number at the centre be 9, then the shape is given below:

Sum of all the numbers
= 2 + 4 + 9 + 14 + 16 = 45 = 5 × 9
Again, the sum is 5 times the number in the centre.

Question 2.
Observe the picture below. It shows a growing pattern of huts made using toothbrushes.
In Step 1, there is 1 hut.
In Step 2, there are 2 huts.
In Step 3, there are 3 huts and this pattern continues.

Based on the above information, answer the following questions:
(i) What is the general rule to find the number of toothbrushes in step n?
(ii) How many toothbrushes will be there in step 9, step 37 and step 54?
Solution:
(i) We can see that the number of toothbrushes increases by 3 at each step.

We can write general rule to find the number of toothbrushes in step n as:
4 + (n – 1) × 3 = 4 + 3n – 3 = 3n + 1,
where n = 1, 2, 3, …

(ii) Number of toothbrushes in step 9
= 3 × 9 + 1 = 27 + 1 = 28
Number of toothbrushes in step 37
= 3 × 37 + 1 = 111 + 1 = 112
Number of toothbrushes in step 54
= 3 × 54 + 1 = 162 + 1 = 163

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 6 Number Play Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 6 Number Play Solutions

Ganita Prakash Class 7 Chapter 6 Solutions

Class 7 Maths Ganita Prakash Chapter 6 Solutions Number Play

Question 1.
Arrange the stick figure cutouts in order of their heights, ensuring that each child (from left to right) states the number of children standing ahead who are taller than them.


(i) 0, 1 , 1 , 2, 4, 1 , 5
(ii) 0, 0, 0, 0, 0, 0, 0
(iii) 0, 1, 2, 3, 4, 5, 6
(iv) 0, 1,0, 1, 0, 1, 0
(v) 0, 1, 1, 1, 1, 1, 1
(vi) 0,0,0,3,3,3,3
Solution:
(i) The required arrangement is FCBGADF:

(ii) The required arrangement is AECGBDF.

(iii) The required arrangement is FDBGCEA

(iv) The required arrangement is EAGCDBF

(v) The required arrangement is FAECGBD

(vi) The required arrangement is BDFAECG

Question 2.
We know that:
(i) even + even = even
(ii) 0dd + 0dd = 0dd
(iii) even + odd = odd
Similarly, find out the parity for the below scenarios:
(iv) even – even = ___________
(v) 0dd – 0dd = _______
(vi) even – odd = ________
(vii) odd – even = ________
Solution:
(iv) even – even
Example: 6 – 2 = 4 → even
8 – 4 = 4 → even
Parity of result = even
∴ even – even = even

(v) odd – odd
Example: 7 – 3 = 4 → even
9 – 5 = 4 = 4 → even
Parity of result = even
∴ old – old = even

Question 3.
How many different magic squares can be made using numbers 1-9?
Solution:
Using the numbers 1-9, there is exactly one unique magic square (excluding rotations and reflections).

If transformations like rotations are allowed, then there are 8 variations of this magic square.

Question 4.
Create a magic square using numbers 2-10. What strategy would you use for this? Compare it with the magic squares made using 1-9.
Solution:
The numbers 2-10 are 9 consecutive numbers, just like 1-9, but increased up by 1.
Strategy: Start with the classic 1-9 magic square, and add 1 to each number.

Original: After adding 1 to each.
The magic square for numbers 2-10 would have a different magic sum (18) compared to numbers 1-9 (15). The structure remains similar, but the values are shifted up to 1.

Question 5.
Take a magic square, and
(i) increase each number by 1
(ii) double each number
In each case, is the resulting grid also a magic square?
How do the magic sums change in each case?
Solution:
Original:

(i) After increasing each number by 1:

This is still a magic square.
New magic sum = 15 + 3 × 1 = 18

(ii) After doubling each number

Still a magic square
New magic sum = 15 × 2 = 30

In case (i), adding a constant to every number → magic sum (for 3×3 grid) is increased by three times of that constant.
In case (ii), multiplying all by a constant → magic sum multiplied by that constant.

Question 6.
Using the generalised form, find a magic square if the centre number is 25.
Solution:
The generalised form of a magic square is given below.

If the centre number is 25, then m = 25.
Substituting m = 25 in generalised form of a magic square, we get

Question 7.
Write the result obtained by—
(i) adding 1 to every term in the generalised form.
(ii) doubling every term in the generalised form.
Solution:
The generalised form of a magic square is given below.

(i) Adding one to every term:

(ii) Doubling every term:

Question 8.
Create a magic square whose magic sum is 60.
Solution:
A 3 × 3 magic square’s sum is 3 × middle element.
So, for a sum is 60, the middle element should be \(\frac{60}{3}\) = 20 .
To get a magic sum of 60, we will multiply the original magic square by 4 i.e.

Question 9.
Is it possible to get a magic square by filling nine non-consecutive numbers?
Solution:
Yes, it is possible.
Justification: Let us consider the two magic squares with magic sum 45.

9 consecutive numbers
and

9 non-consecutive numbers

Question 10.
A light bulb is ON. Dorjee toggles its switch 77 times. Will the bulb be on or off? Why?
Solution:
Dorjee toggles the switch 77 times.

Each toggle changes the state of the bulb (ON to OFF or OFF to ON). Starting from ON.

An odd number of toggles will leaves the bulb OFF and an even number of toggles will leave the bulb ON. Since 77 is odd, after 77 toggles, the bulb will be OFF.

Question 11.
Here is a 2 × 3 grid. For each row and column, the parity of the sum is written in the circle; ‘e’ for even and ‘o’ for odd. Fill the 6 boxes with 3 odd numbers (‘o’) and 3 even numbers (‘e’) to satisfy the parity of the row and column sums.

Solution:
Let’s label the cells as:

So the constraints are:

• Row 1 (A, B, C): sum is odd
• Row 2 (D, E, F): sum is even
• Column 1 (A, D): sum is even
• Column 2 (B, E): sum is even
• Column 3 (C, F): sum is odd

We’ll track parities only (o or e), not actual numbers.
Row 1: A = o, B = e, C = e, then o + e + e = odd
Column 1 (A, D) – e means A must be paired with D as odd to get the sum as even.
So, if A = o, D = o, then o + o = even
Similarly, if B = e, E = e, then e + e = even
Again, if C = e, F = o, then e + o = odd
So, the 6 boxes with 3 odd numbers and 3 even numbers can be filled as follows:

Question 12.
Make a 3 × 3 magic square with 0 as the magic sum. All numbers can not be zero. Use negative numbers, as needed.
Solution:
It is given that

• The magic square is 3 × 3.
• The magic sum is 0.
• All numbers in the square cannot be zero, we can use negative numbers as needed.

So, we will use the numbers (- 4) to 4 to create a magic square whose magic sum is 0.
The required magic square is given below.

Question 13.
Two consecutive numbers in the Virahanka sequence are 987 and 1597. What are the next 2 numbers in the sequence? What are the previous 2 numbers in the sequence?
Solution:
Given numbers are 987 and 1597.
In the Virahanka sequence, each number is the sum of the two preceding numbers.
The next two numbers are:
987 + 1597 = 2584 and 1597 + 2584 = 4181
The previous two numbers are:
1597 – 987 = 610 and 987 – 610 = 377
The sequence is …, 377, 610, 987, 1597, 2584, 4181,…

Question 14.
What is the parity of the 20th term of the Virahanka sequence.
Solution:
Consider the Virahanka sequence given below:
1,2, 3, 5, 8, 13, 21,34, 55, 89, 144, 233, 377, 610, 987, …
Let us observe the pattern of odd/even in Virahanka sequence.
Here 1 → odd;
2 → even;
3 → odd
5 → odd;
8 → even;
13 → odd
21 → odd;
34 → even;
55 → odd
So parity cycle: odd, even, odd, (repeats every 3 terms)
So the parity of 20th term in Virahanka sequence is even.

Question 15.
Solve the following cryptarithm:

Solution:
Here, in TAT, letter T is a hundreds place.
So, T = 1.
⇒ A = 0 and U = 9.
So, we have U = 9, T = 1 and A = 0.

InText Questions

Question 1.
Kishor has a set of number cards and 5 boxes. If each box must contain exactly one number card, suggest an arrangement to help him distribute the cards, so that 5 cards add to 30? Is it possible?

Solution:
No, it is not possible, as the sum of 5 odd numbers is always odd and 30 is an even number.

Question 2.
In a 3 × 3 grid, there are 9 small squares, which is an odd number. Meanwhile, in a 3 × 4 grid, there are 12 small squares, which is an even number.
Given the dimensions of a grid, can you tell the parity of the number of small squares without calculating the product?

Solution:
Yes, we can determine the parity of the number of small squares in a grid without directly calculating the full product, simply by observing the parity of the dimensions.
Rule: The product of two numbers is:

• Even if at least one of the numbers is even.
• Odd if both numbers are odd.

Question 3.
We can describe how the numbers within the magic square are related to each other, i.e. the structure of the magic square.
Choose any magic square that you have made so far using consecutive numbers. If m is the letter- number of the number in the centre, express how other numbers are related to m, how much more or less than m.
Solution:
Consider the magic square

We can express it using the letter- number m for the number in the centre as:

Question 4.
Write the next 3 numbers in the sequence:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, , , , …
If you have to write one more number in the Sequence above, can you tell whether it will be an odd number or an even number (without adding the two previous numbers)?
Solution:
The next 3 terms in the sequence are:
55 + 89 = 144; 89 + 144 = 233; 144 + 233 = 377
Yes, we can determine the parity without adding the two previous number.
Since odd + odd = even
Hence, parity of next number in sequence is even.

Question 5.
Find out what each letter stands for.

Solution:
(i) YY is a two-digit number where both digits are the same. So it can be 99, 88, …
But Z is a 1-digit number and ZOO is a 3-digit number.
So, Y = 9, Z = 1 and O = 0.

\(\begin{array}{r}
99 \\
+\quad 1 \\
\hline 1 \quad 00 \\
\hline
\end{array}\)

(ii)

(iii) Here, KP is a 2-digit number and PRR is a 3-digit number. Basically, 2 × (KP) = PRR.
If P = 1, then R = 2.
Hence, K = 6, P = 1 and R = 2.

Here, C + 1 is a two-digit number i.e. 10.

Number Play Class 7 Extra Questions

Number Play Class 7 Very Short Question Answer

Question 1.
A bag holds 100 balls, each marked with an odd number. If two balls are picked at random and their numbers are added, what will be the parity of the result?
Solution:
We know that odd number + odd number = even number.
Since each ball is marked with an odd number. If two balls are picked at random and their numbers are added, the parity of the sum will be even.

Question 2.
A number has odd parity. What will be the parity of:
(i) Number + 7
(ii) Number – 2
(iii) Number + Number
Solution:
Given, a number has odd parity.
(i) Since the parity of the sum of two odd numbers is even, the parity of (Number + 7) will be even.
(ii) Since the parity of the difference of an even and an odd number is odd, parity of (N umber – 2) will be odd.
(iii) Since the parity of the sum of two odd numbers is even, the parity of (Number + Number) will be even.

Question 3.
Neha wants to climb a staircase with 7 steps. She has a simple rule: She can take either 1 step or 2 steps at a time. For example, one possible way to climb is: 2, 1, 2, 2.
In how many different ways can Neha climb to the top of the 7-step staircase?
Solution:
The number of different ways in which Neha can climb to the top of the 7-step staircase taking either 1 step or 2 steps at a time, is the 7^th element of Virahanka sequence.
Virahanka sequence: 1, 2, 3, 5, 8, 13, 21, …
7^th element of Virahanka sequence = 21
∴ Neha can climb to the top of the 7-step staircase in 21 different ways.

Number Play Class 7 Short Question Answer

Question 1.
During a maths quiz, a student says, “Two consecutive numbers add up to 98.” Is this claim correct? Justify your answer with reasoning.
Solution:
(i) The claim is not correct.
(ii) The counting numbers 1, 2, 3, 4, 5, … alternate between even and odd numbers.
In any two consecutive numbers, one will always be even and the other will always be odd.
We know that the parity of the sum of an even number and an odd number is odd.
Since 98 is an even number, his claim is incorrect.

Question 2.
Form two expressions, one that always gives odd parity and another that always gives even parity.
Solution:
We know that the parity of the product is even if at least one of them is even.
Also, in any two consecutive numbers, one is alwavs even and the other is always odd.
∴ For any number n, the expression n(n + 1) = n² + n always has even parity.
Since even number + odd number = odd number.
∴ The expression n² + n + 1 always has odd parity.

Question 3.
Solve the following cryptarithms:
(i)
\(\begin{array}{r}
\mathrm{B} 2 \\
+6 \mathrm{C} \\
\hline \mathrm{EC} 6
\end{array}\)
(ii)
\(\begin{array}{r}
\mathrm{PQ} \\
+\mathrm{QR} \\
\hline \mathrm{QRQ}
\end{array}\)
Solution:
We know that in cryptarithms:

• Each letter stands for unique digit and different letters represent different digits.
• The same letter always means the same digit.

(i) Here, 2 + C = 6 ⇒ C = 4
Now, B + 6 = EC = E4 [Since C = 4]
⇒ B = 8 and E = 1

(ii) Here, Q + R = Q ⇒ R = 0
Now, P + Q = QR
⇒ P + Q = Q0
⇒ P = 9 and Q = 1

Number Play Class 7 Long Question Answer

Question 1.
Aarav stores his stamp collection in small boxes. He has:

• an odd number of boxes with 5 stamps each,
• an odd number of boxes with 7 stamps each, and
• an even number of boxes with 10 stamps each. He counts all his stamps and says the total is 175.

Did Aarav make a mistake? Explain your reasoning.
Solution:
We know that the parity of the product of two odd numbers is odd and the parity of the product of two even numbers is even. Thus,

• An odd number of boxes with 5(odd) stamps each gives an odd number of stamps.
• An odd number of boxes with 7(odd) stamps each also gives an odd number of stamps.
• An even number of boxes with 10(even) stamps each gives an even number of stamps.

Since, odd number + odd number = even number And even number + even number = even number, the final total must be even number. But Aarav claims the total is 175, which is odd. Therefore, he must have made a mistake in his counting.

Question 2.
Two consecutive numbers in the Virahanka sequence are 377 and 610. What are the previous 2 numbers in the sequence?
Solution:
We know that in Virahanka sequence, a number is the sum of previous two numbers.
Let the previous two numbers be x and y respectively.
The sequence will be as: …, x,y, 377, 610, …
Then, y + 377 = 610
⇒ v = 610-377 = 233
Now, x + v = 377 ,
⇒ x + 233 = 377
⇒ x = 377 – 233 = 144
Thus, the previous two numbers in the Virahanka sequence are 144 and 233.

Question 3.
Solve the following cryptarithms:
(i)
\(\begin{array}{r}
\mathrm{Tl} \\
+1 \mathrm{~T} \\
\hline 66
\end{array}\)
(ii)
\(\begin{array}{r}
\mathrm{D} 3 \\
+3 \mathrm{D} \\
\hline \mathrm{PPQ}
\end{array}\)
Solution:
(i) T + 1 = 6 ⇒ T = 5
Thus,
\(\begin{array}{r}
51 \\
+\quad 15 \\
\hline 66
\end{array}\)

(ii) A three-digit number abc can be written as 100a + 10b + c.
D3 = 10D + 3, 3D = 3 × 10 + D = 30 + D and PPQ = 100P + 10P + Q = 110P + Q
Now, D3 + 3D = (10D + 3) + (30 + D)
= 11 D + 33
When D ranges from 1 to 6, the expression 11 D + 33, gives a two-digit number.
At D = 7, 11 × 7 + 33 = 77 + 33 = 110
Putting P = 1 and Q = 0, we get
110P + Q = 110 × 1 + 0 = 110
∴ P = 1, Q = 0 and D = 7

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 7 A Tale of Three Intersecting Lines Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 7 A Tale of Three Intersecting Lines Solutions

Ganita Prakash Class 7 Chapter 7 Solutions

Class 7 Maths Ganita Prakash Chapter 7 Solutions A Tale of Three Intersecting Lines

Question 1.
Use the points on the circle and/or the centre to form isosceles triangles.

Solution:
Select any two points on the circle and connect them with the centre of the circle.
Also, join these points to each other. This will form an isosceles triangle as the two radii are equal in length.

Question 2.
Can we say anything about the existence of a triangle for each of the following sets of lengths?
(i) 10 km, 10 km and 25 km
(ii) 5 mm. 10 mm and 20 mm
(iii) 12 cm. 20 cm and 40 cm
Solution:
(i) When we take direct path = 25 km.
Then roundabout path =10 km + 10 km = 20 km.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

(ii) When we take direct path = 20 mm.
Then roundabout path = 10 mm + 5 mm = 15 mm.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

(iii) When we take direct path = 40 cm.
Then roundabout path =12 cm + 20 cm = 32 cm.
Since direct path is longer than the roundabout path.
So, existence of a triangle is not possible.

Question 3.
Does there exist an equilateral triangle with sides 50, 50, 50? In general, does there exist an equilateral triangle of any sidelength? Justify your answer.
Solution:
Yes, an equilateral triangle with sides 50, 50, 50 exists because the sum of two sides is greater than the third side. For any positive number say x > 0, x + x > x. So, an equilateral triangle with all side lengths ‘x’ exists.

Question 4.
For each of the following, give at least 5 possible values for the third length so there exists a triangle having these as sidelengths (decimal values can also be chosen): ;
(i) 1, 100
(ii) 5, 5
(iii) 3, 7
Solution:
(i) 5 possible values for the third length would be 99.5. 99.8, 100, 100.5, 100.9
Since, 100 < 1 + 99.5, 100 < 1 + 99.8, 100 < 1 + 100, 100.5 < 1 + 100, and 100.9 < 1 + 100

(ii) 5 possible values for the third length would be 1, 3.5, 5, 7.5, 8.9
Since, 5 < 1 + 5, 5 < 5 + 3.5, 5 < 5 + 5, 7.5 < 5 + 5, and 8.9 < 5 + 5

(iii) 5 possible values for the third length would be 4.5, 5, 6.9, 8, 9.8
Since, 7 < 3 + 4.5, 7 < 5 + 3, 7 < 3 + 6.9, 8 < 3 + 7, 9.8 < 3 + 7

Question 5.
Construct triangles for the measurement, 3 cm, 120°, 8 cm, where the angle is included between the sides.
Solution:

Steps of construction are given below:
Step 1: Construct a side AB of length 8 cm.
Step 2: Construct ∠d = 120° by drawing the other arm of-the angle.
Step 3: Mark the point C on the other arm such that AC, = 3 cm.
Step 4: Join BC to get the required triangle.

Question 6.
Construct triangles for the measurements, 25°, 3 cm, 60° where the side is included between the angles.
Solution:

Steps of construction are given below:
Step 1: Draw the base AB of length 3 cm.
Step 2: Draw ∠d and ∠B of measure 25° and 60° respectively.
Step 3: Make the point of intersection of the two new arms of ∠d and ∠B as point C to get the required triangle.

Question 7.
Determine which of the following pairs can be the angles of a triangle and which cannot:
(i) 35°, 150°
(ii) 70°, 30°
Solution:
(i) The sum of the given angles 35° + 150° = 185°.
This is not possible because sum of the angles of the triangle exceeds 180°.

(ii) The sum of the given angles 70° + 30° = 100°.
Possible third angle 180°- 100° = 80°.
Since the possible third angle comes out positive (80°), the given angles can be angles of a triangle.

Question 8.
Find the third angle of a triangle (using a parallel line) when two of the angles are:
(i) 36°, 72°
(ii) 150°, 15°
Solution:
(i) Here ∠B = 36° and ∠C = 72°.

Since the line BC is parallel to AT.
So, ∠XAB = ∠B = 36° [Alternate angles] ………. (i)
and ∠YAC = ∠C = 72 [Alternate angles] ………. (i)
Also, ∠XAB + ∠B AC + ∠YAC = 180° [∠XAY is a straight angle]
⇒ 36° + ∠BAC + 72° = 180° [Using (i) and (ii)]
⇒ ∠BAC = 180°- 108° = 72°.

(ii) Here ∠B = 150° and ∠C = 15°.

Since the line BC is parallel to AY.
So, ∠XAB = ∠B = 150° [Alternate angles] ………. (i)
and ∠YAC = ∠C = 15° [Alternate angles] ………. (ii)
Also, ∠XAB + ∠BAC + ∠YAC = 180° [∠XAY is a straight angle]
⇒ 150° + ∠BAC + 15° = 180° [Using (i) and (ii)]
⇒ ∠BAC = 180° – 165° = 15°.

Question 9.
Here is a triangle in which we know ∠B = ∠C and ∠A = 50°, Can you find ∠B and ∠C?

Solution:
Given, ∠A = 50° and ∠B – ∠C.

Draw a line Xy which is parallel to BC.
Now, ∠XA B = ∠B and ∠YAC = ∠C [Alternate angles] ………… (i)
Also, ∠XAB + ∠BAC + ∠YAC = 180°
⇒ ∠B + 50° + ∠C = 180° [Using (i)]
⇒ ∠B + ∠C = 180° – 50° = 130°
⇒ 2 ∠B = 130° [∵ ∠B = ∠C]
⇒ ∠B = 65° = ∠C.

Question 10.
Construct a triangle TRY with RY = 4 cm, TR = 7 cm, ∠R = 140°. Construct an altitude from T to RY.
Solution:
Steps of construction are given below:

Step 1: Construct a side TR of length 7 cm.
Step 2: Construct ∠R = 140° by drawing the other arm of the angle.
Step 3: Mark the point 1 on the other arm such that RY = 4 cm.
Step 4: Join TY to get the required triangle.
Step 5: Keep the ruler aligned to RY. Place the set square along the ruler such that one of the edges of the right angle touches the ruler.
Step 6: Slide the set square along the ruler till the perpendicular edge of the set square touches the vertex T.
Step 7: Extend the line YR and then draw the altitude through T on extended YR using the perpendicular edge of the set square.

InText Questions

Question 1.
What happens when the three vertices lie on a straight line? Will these points form a triangle?
Solution:
When the three vertices lie on a straight line, they become collinear. This means they no longer form a triangle because the three points do not enclose any area.

Question 2.
Construct a triangle with sidelengths 3 cm, 4 cm, and 8 cm. What is happening? Are you able to construct the triangle?
Solution:
Let the triangle be ABC, where AB = 8 cm.

Since, the arcs from the points A and B do not meet. So, we are not able to construct the triangle with side lengths 3 cm, 4 cm, and 8 cm.

A Tale of Three Intersecting Lines Class 7 Extra Questions

A Tale of Three Intersecting Lines Class 7 Very Short Question Answer

Question 1.
In the adjoining figure:

(i) Name the vertex opposite to side PQ.
(ii) Name the side opposite to vertex Q.
(iii) Name the angle opposite to side QR.
(iv) Name the side opposite to ∠R.
Solution:
In the given figure,
(i) The vertex opposite to side PQ is R.
(ii) The side opposite to vertex Q is PR.
(iii) The angle opposite to side QR is ∠P.
(iv) The side opposite to ∠R is PQ.

Question 2.
In the given triangle ABC, find the value of x.

Solution:
Given, ∠T = 7x – 3°, ∠B = 130°, ∠C = 6x + 1°
We know that sum of all the angles in a triangle is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ 7x – 3° + 130° + 6x + 1° = 180°
⇒ 13x + 128° = 180° ⇒ 13x = 180°- 128° = 52°
⇒ x = \(\frac{52^{\circ}}{13},\) = 4°
Thus, the value of x is 4°.

Question 3.
Can a triangle be formed for the following set of angles?
(i) 80°, 70° and 50°
(ii) 56°, 64° and 60°
Solution:
We know that the sum of all angles in a triangle is 180°.
(i) Given set of angles are 80°, 70° and 50°.
Now, sum of the angles = 80° + 70° + 50° = 200° ≠ 180°
Thus, the given set of angles cannot form a triangle.

(ii) Given set of angles are 56°, 64° and 60°.
Now, sum of the angles = 56° + 64° + 60° = 180°
Thus, the given set of angles can form a triangle.

Question 4.
In the following figure, find the value of p.

Solution:
In the given figure, ∠SPR is an exterior angle, and ∠PRQ and ∠PQR are two interior opposite angles to ∠SPR.
∴ ∠SPR = ∠PRQ + ∠PQR [Exterior angle property]
⇒ p = 105° + 45° = 150°
Thus, the value of p is 150°.

Question 5.
In ∆XYZ, YX is extended to O. If ∠Y = 57° and XY = XZ, then find ∠ZXO.

Solution:
Given, ∠Y = 57° and XY = XZ
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠Z = ∠Y = 57° ……… (i)
In the given figure, ∠ZXO is an exterior angle.
And, ∠Y and ∠Z are two interior opposite angle to ∠ZXO.
We know, Exterior angle = Sum of two interior opposite angles
∴ ∠ZXO = ∠Y + ∠Z
⇒ x – 57° + 57° = 114° [Using (i)]
Thus, the value of ∠ZXO is 114°.

A Tale of Three Intersecting Lines Class 7 Short Question Answer

Question 1.
Construct a triangle PQR such that PQ =3.5 cm, QR = 6.5 cm and PR = 4 cm.
Solution:
Given, PQ = 3.5 cm, QR = 6.5 cm and PR = 4 cm

To construct the triangle PQR of given sides, we follow the steps laid down below:
Step 1: Using ruler, draw a line segment PQ, 3.5 cm long.
Step 2: With P as centre, draw an arc with radius equal to 4 cm.
Step 3: With Q as centre, draw an arc with radius equal to 6.5 cm to cut the previous arc at R.
Step 4: Join R to P and Q. Thus, ∆PQR is required triangle.

Question 2.
Construct a triangle PQR such that PQ = 7 cm, QR = 8.5 cm and PR = 9 cm.
Solution:
Given, PQ = 7 cm, QR = 8.5 cm and PR = 9 cm.

To construct the triangle PQR of given side lengths, follow the steps laid down below:
Step 1: Using ruler, draw a line segment PQ, 7 cm long.
Step 2: With P as centre, draw an arc with radius equal to 9 cm.
Step 3: With Q as centre, draw an arc with radius equal to 8.5 cm to cut the previous arc at R.
Step 4: Join R to P and Q. Thus, ∆PQR is required triangle.

Question 3.
In the given figure, PQ is parallel to RS. Find the value of y.

Solution:
In ∆PQR, we have
∠QPR = 82°, ∠PRQ = 42° and ∠PQR = x
We know that the sum of all angles in a triangle is 180°. ‘
∴ ∠QPR + ∠PRQ + ∠PQR = 180°
⇒ 82° + 42° + x = 180°
⇒ 124° + x = 180°
⇒ x = 180°-124° = 56°
Since PQ is parallel to RS and QR is a transversal, ∠PQR and ∠QRS are alternate interior angles, they are equal.
⇒ ∠QRS = ∠PQR
⇒ Y = ,x = 56°
Thus, the value of y is 56°.

Question 4.
In the given figure, find the value of x.

Solution:
In ∆DBE, ∠DEF = 150° is an exterior angle.
We know that the exterior angle of a triangle is equal to the sum of opposite interior angles.
∴ ∠DEF = ∠ERD + ∠EDB
⇒ 150° = 50° + ∠EDB
⇒ ∠EDB = 150° – 50° = 100°
Again, in ADAG, ∠EDB is an exterior angle.
∴ ∠EDB = ∠DAG + ∠AGD
⇒ 100° = ∠DAG + 70°
[∵ ∠EDB = 100°, ∠AGD = 70°]
⇒ ∠DAG = 100° – 70° = 30°
Also, ∠DAG and ∠GAH form a linear pair of angles.
∴ ∠DAG + ∠GAH = 180°
[∵ ∠DAG = 100°, ∠GAH = x]
⇒ 30° + x = 180°
⇒ x = 180°-30° = 150°

Question 5.
In ∆IJK, if IJ = IK, ∠IKJ = 50°, ∠JIK = m, find the value of m.

Solution:
Given, IJ = IK, ∠IKJ = 50°, ∠JIK = m
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠IKJ = ∠IJK = 50°
Now, sum of all the angles = 180°
⇒ ∠JIK + ∠IKJ + ∠IJK = 180°
⇒ m + 50° + 50° = 180°
⇒ m + 100° = 180°
⇒ m = 180°- 100° = 80°
Thus, the value of m is 80°.

Question 6.
In the given figure, if ∠NMO = 84° and LM = MN, then find the values of ∠N and ∠L.

Solution:
Given, ∠NMO = 84° and LM = MN
We know that in an isosceles triangle, angles opposite to equal sides are equal.
∴ ∠N = ∠L ………. (i)
In the given figure, ∠NMO = 84° is an exterior angle.
And, ∠N and ∠L are two interior opposite angles to ∠NMO.
We know,
Exterior angle = Sum of two interior opposite angles
∴ ∠NMO = ∠N + ∠L
⇒ 84° = ∠N + ∠N [Using (i)]
⇒ 2∠N = 84° ⇒ ∠N = \(\frac{84^{\circ}}{2}\) = 42°
Thus, ∠N = ∠L = 42°

Question 7.
In the given figure, find the value of y.

Solution:
In the given figure, ∠BCD is an exterior angle.
And, ∠BAC and ∠ABC are two interior opposite angles to ∠BCD.
We know, Exterior angle = Sum of two interior opposite angles
∴ ∠BCD = ∠BAC + ∠ABC
⇒ 7y + 6° = 50° + 96°
⇒ 7v + 6° = 146°
⇒ 7y = 146° – 6° = 140°
⇒ y = \(\frac{140^{\circ}}{7}\) = 20°
Thus, the value of y is 20°.

Question 8.
In ∆PQR, ZQ is thrice of ∠P and ∠R is twice of sum of ∠P and ∠Q. Find the angles.
Solution:
Let ∠P be x. Then,
∠Q = 3 x, ∠P = 3x
and ∠R = 2(∠P + ∠Q) = 2(x + 3x) = 2 × 4x = 8x
We know that sum of all the angles in a triangle is 180°.
∴ ∠P + ∠Q + ∠R = 180°
⇒ x + 3x + 8x = 180°
⇒ 12x = 180°
⇒ x = \(\frac{180^{\circ}}{12}\) = 15°
Thus, ∠P = x = 15°, ∠Q = 3x = 3 × 15° = 45° and ∠R = 8x = 8 × 15° = 120°

A Tale of Three Intersecting Lines Class 7 Long Question Answer

Question 1.
Mrs. Roy gave straws of length 21 cm to her students and asked them to cut the straw to get three pieces, which can be used to form each of the three types of triangles. The lengths of the three pieces were supposed to be whole numbers (1, 2, 3, …) only. Based on these criteria, write any one possible combination of straw lengths in Column 3 of the below table.

Solution:
We know that if a given set of three lengths satisfies the triangle inequality (each length < sum of the other two lengths), then a triangle exists having those as side lengths.
(a) If all three sides of a triangle are different in length, then it is called a scalene triangle.
Thus, the possible sets of side lengths are:
(i) 2 cm, 9 cm and 10 cm
(ii) 3 cm, 8 cm and 10 cm
(iii) 4 cm, 7 cm and 10 cm
(iv) 4 cm, 8 cm and 9 cm
(v) 5 cm, 6 cm and 10 cm
(vi) 5 cm, 7 cm and 9 cm
(vii) 6 cm, 7 cm and 8 cm

(b) If any two sides of a triangle are equal in length, then it is called an isosceles triangle.
Thus, the possible sets of side lengths are:
(i) 6 cm, 6 cm and 9 cm
(ii) 8 cm, 8 cm and 5 cm
(iii) 9 cm, 9 cm and 3 cm
(iv) 10 cm, 10 cm and 1 cm

(c) If all three sides of a triangle are equal in length, then it is called an equilateral triangle. Thus, the possible set of side lengths is 7 cm, 7 cm, 7 cm.

Question 2.
In the given figure, find ∠EAB + ∠ABC + ∠BCD + ∠CDE + ∠DEA.

Solution:
We know that the sum of angles of a triangle is 180°.
∴ In ∆ABC, we have
∠CAB + ∠ABC + ∠BCA = 180° ………… (i)
In ∆ACD, we have
∠ACD + ∠CDA + ∠DAC = 180° ………. (ii)
And in AADE, we have
∠ADE + ∠DEA + ∠EAD = 180° ……….. (iii)
Adding (i), (ii) and (iii), we get
∠CAB + ∠ABC + ∠BCA + ∠ACD + ∠CDA + ∠DAC + ∠ADE + ∠DEA + ∠EAD = 180° + 180° + 180°
⇒ (∠EAD + ∠DAC + ∠CAB) + ∠ABC + (∠BCA + ∠ACD) + (∠CDA + ∠ADE) + ∠DEA = 540°
⇒ ∠EAB + ∠ABC + ∠BCD + ∠CDE + ∠DEA = 540°
[∵ ∠EAD + ∠DAC + ∠CAB = ∠EAB, ∠BCA + ∠ACD = ∠BCD and ∠CDA + ∠ADE = ∠CDE]

Question 3.
Two line segments PS and QR intersect at O. Joining PQ and SR, we get two triangles, ∆POQ and ∆ROS as shown in the figure. Find the values of ∠P and ∠Q.

Solution:
In ∆ROS, ∠R = 35° and ∠S = 70°
We know that sum of all the angles in a triangle is 180°.
∴ ∠R + ∠ROS + ∠S = 180°
⇒ 35° + ∠ROS + 70° = 180°
⇒ ∠ROS + 105° = 180°
⇒ ∠ROS = 180° – 105° = 75°
Since ∠ROS and ∠POQ are vertically opposite angles, ∠ROS = ∠POQ = 75°.
Now, in ∆POQ, ∠P = 5x, ∠POQ =75° and ∠Q = 4x + 6°
We know that sum of all the angles in a triangle is 180°.
∴ ∠P + ∠POQ + ∠Q = 180°
⇒ 5i + 75° + 4x + 6° = 180°
⇒ 9x + 81° = 180°
⇒ 9x = 180° – 81° = 99°
⇒ x = \(\frac{99^{\circ}}{9}\) = 11°
Thus, ∠P = 5x = 5 × 11° = 55° and ∠Q = 4x + 6° = 4 × 11° + 6° = 44° + 6° = 50°

Question 4.
Construct a triangle ∆XYZ with XY = 4 cm, YZ = 7 cm and ∠Y = 108°.
Solution:
Given, in ∆XYZ, XY = 4 cm, YZ = 7 cm and ∠Y = 108°
Steps of construction of ∆XYZ are as follows:
Step 1: Using a ruler, draw a line segment XY, 4 cm long.
Step 2: Using protractor, draw the given vertex angle, ∠Y = 108°.
Step 3: On the new arm created from point F, mark a point Z such that YZ = 7 cm using a ruler and a compass.
Step 4: Using a ruler, connect points X and Z to complete the required triangle XYZ.

Question 5.
Construct a triangle IJK if IJ = 5 cm, JK = 6 cm and included angle between sides IJ and JK is 37°.
Solution:
Given, in ∆IJK, IJ = 5 cm, JK = 6 cm and included angle between sides IJ and JK is 37°.
Steps of construction of ∆IJK are as follows:
Step 1: Using a ruler, draw a line segment IJ of 5 cm length.
Step 2: Using protractor, draw the given vertex angle, ∠J = 37°.

Step 3: On the new arm created from point K mark a point K such that JK = 6 cm using a ruler anti a compass.
Step 4: Using a ruler, connect points I and K to complete the triangle IJK.

Question 6.
Construct a triangle XYZ with ∠Z = 30°, ∠Y = 115° and YZ = 5 cm.
Solution:
Given, in ∆XYZ, ∠Z = 30°, ∠Y = 115° and YZ = 5 cm
Steps of construction of ∆XYZ are as follows:
Step 1: Using a ruler, draw a line segment YZ of 5 cm length.
Step 2: Using protractor, draw the given vertex angle, ∠Y = 115°.

Step 3: Using protractor, draw the given vertex angle, ∠Z = 30°.
Step 4: Mark the intersecting point of two new lines (arms of angles) as X. Thus, the required triangle, ∆XFZ is constructed.

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 8 Working with Fractions Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 8 Working with Fractions Solutions

Ganita Prakash Class 7 Chapter 8 Solutions

Class 7 Maths Ganita Prakash Chapter 8 Solutions Working with Fractions

Question 1.
A team of workers can make 1 km of a water canal in 8 days. So, in one day, the team can make _______ km of the water canal. If they work 5 days a week, they can make _______ km of the water canal in a week.
Solution:
Water canal made by team of workers in 8 days = 1 km
So, water canal made by team of workers in 1 dav = \(\frac{1}{8}\) km
The length of the water canal made by the team of workers in 5 days = 5 × \(\frac{1}{8}\) km = \(\frac{5}{8}\) km
Hence, the team can make \(\frac{5}{8}\) km of the water canal in one week.

Question 2.
Safia saw the Moon setting on Monday at 10 pm. Her mother, who is a scientist, told her that every day the Moon sets \(\frac{5}{6}\) hour later than the previous day. How many hours after 10 pm will the moon set on Thursday?
Solution:
There are 3 days from Monday to Thursday. Since the Moon sets \(\frac{5}{6}\) hours later than previous day, the number of hours the Moon will set later on Thursday than Monday = 3 × \(\frac{5}{6}\) hours = \(\frac{15}{6}\) hours = \(\frac{5}{2}\) hours.
We know that 1 hour = 60 minutes
⇒ \(\frac{5}{2}\) hours = \(\frac{5}{2}\) × 60 minutes = \(\frac{300}{2}\) minutes = 150 minutes
Now, 150 minutes = 120 minutes + 30 minutes = 2 hours 30 minutes
Hence, on Thursday the Moon will set 2 hours 30 minutes after 10 pm.

Question 3.
Find the following products. Use a unit square as a whole for representing the fractions and carrying out the operations.
(i) \(\frac{2}{3}\) × \(\frac{4}{5}\)
(ii) \(\frac{1}{4}\) × \(\frac{2}{3}\)
Solution:
(i) Each row represents \(\frac{1}{5}\) and each column represents \(\frac{1}{3}\). The whole is divided into 5 rows and 3 columns creating 5 x 3 = 15 equal parts and 2 × 4 = 8 of the parts is double-shaded, that is \(\frac{8}{15}\) of the whole is double-shaded.
Therefore, \(\frac{2}{3}\) × \(\frac{4}{5}\) = \(\frac{8}{15}\)

(ii) Each row represents \(\frac{1}{3}\) and each column represents \(\frac{1}{4}\). The whole is divided into 3 rows and 4 columns, creating 3×4=12 equal parts and 2 of the parts are double-shaded, that is \(\frac{2}{12}\) of the whole is double-shaded.
Therefore, \(\frac{1}{4}\) × \(\frac{2}{3}\) = \(\frac{2}{12}\) = \(\frac{1}{6}\)

Question 4.
A water tank is filled from a tap. If the tap is open for 1 hour, \(\frac{7}{10}\) of the tank gets filled. How much of the tank is filled if the tap is open for
(i) \(\frac{1}{3}\) hours
(ii) \(\frac{7}{10}\) hours
Solution:
In 1 hour, part of the tank gets filled = \(\frac{7}{10}\)
(i) In \(\frac{1}{3}\) hours, part oLthe tank gels filled = \(\frac{1}{3}\) × \(\frac{7}{10}\) = \(\frac{1 \times 7}{3 \times 10}\) = \(\frac{7}{30}\)
Therefore, in \(\frac{1}{3}\) hours, \(\frac{7}{30}\) of the tank gets filled.

(ii) In \(\frac{7}{10}\) hours, part of the tank gets filled = \(\frac{7}{10}\) × \(\frac{7}{10}\) = \(\frac{7 \times 7}{10 \times 10}\) = \(\frac{49}{100}\)
Therefore, in \(\frac{7}{10}\) hours, \(\frac{49}{100}\) of the tank gets filled.

Question 5.
Tsewang plants four saplings in a row in his garden. The distance between two saplings is \(\frac{3}{4}\) m. Find the distance between the first and last sapling.
Solution:

The distance between the first and the last sapling = \(\frac{3}{4}\) + \(\frac{3}{4}\) + \(\frac{3}{4}\) = 3 × \(\frac{3}{4}\) = \(\frac{9}{4}\) m

Question 6.
Which is heavier: \(\frac{12}{15}\) of 500 grams or \(\frac{3}{20}\) of 4 kg?
Solution:
We have \(\frac{12}{15}\) of 500 g = \(\frac{12}{15}\) × 500 g = \(\frac{12 \times 500}{15}\) = 400 g
And \(\frac{3}{20}\) × 4000 g = \(\frac{3 \times 4000}{20}\) g = 600 g [∵ 1 kg = 1000 g]
∵ 600 g is heavier than 400 g.
∵ \(\frac{3}{20}\) of 4 kg is heavier than \(\frac{12}{15}\) of 500 grams.

Question 7.
(i) 3 ÷ \(\frac{7}{9}\)
(ii) \(\frac{14}{6}\) \(\frac{7}{3}\)
(iii) \(\frac{1}{6}\) ÷ \(\frac{11}{12}\)
(iv) 3\(\frac{2}{3}\) ÷ 1\(\frac{3}{8}\)
Solution:
(i) 3 ÷ \(\frac{7}{9}\) = 3 × \(\frac{9}{7}\) = \(\frac{9}{7}\) = \(\frac{3 \times 9}{7}\)
= 3\(\frac{6}{7}\)

(ii) \(\frac{14}{6}\) \(\frac{7}{3}\) = \(\frac{14}{6}\) × \(\frac{3}{7}\) = \(\frac{14 \times 3}{6 \times 7}\)
= \(\frac{42}{42}\) = 1

(iii) \(\frac{1}{6}\) ÷ \(\frac{11}{12}\) = \(\frac{1}{6}\) × \(\frac{12}{11}\) = \(\frac{1 \times 12}{6 \times 11}\)
= \(\frac{2}{11}\)

(iv) 3\(\frac{2}{3}\) ÷ 1\(\frac{3}{8}\) = \(\frac{11}{3}\) ÷ \(\frac{11}{8}\) = \(\frac{11}{3}\) × \(\frac{11}{8}\)
= \(\frac{11 \times 8}{3 \times 11}\)
= \(\frac{8}{3}\) = 2\(\frac{2}{3}\)

Question 8.
Patiganita a book written by Sridharacharya in the 9^th century CE, mentions this problem: “Friend, after thinking, what sum will be obtained by adding together 1÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) and 1 ÷ \(\frac{1}{2}\). What should the friend say?
Solution:
1÷ \(\frac{1}{6}\) = 1 × 6 = 6, 1 ÷ \(\frac{1}{10}\) = 1 × 10 = 10, 1 ÷ \(\frac{1}{13}\) = 1 × 13 = 13, 1 ÷ \(\frac{1}{9}\) = 1 × 9 = 9 and 1 ÷ \(\frac{1}{2}\) = 1 × 2 = 2
Therefore, the sum obtained by adding together
1÷ \(\frac{1}{6}\), 1 ÷ \(\frac{1}{10}\), 1 ÷ \(\frac{1}{13}\), 1 ÷ \(\frac{1}{9}\) and 1 ÷ \(\frac{1}{2}\)
= 6 + 10 + 13 + 9 + 2 = 40
Thus, the friend should say the ‘sum’ is 40.

Question 9.
Amritpal decides on a destination for his vacation. If he takes a train, it will take him 5\(\frac{1}{6}\) hours to get there. If he takes a plane, it will take him \(\frac{1}{2}\) hour. How many hours does the plane save?
Solution:
The difference between the two durations = 5\(\frac{1}{6}\) – \(\frac{1}{2}\) = \(\frac{31}{6}\) – \(\frac{1}{2}\) [∵ 5\(\frac{1}{6}\) = \(\frac{31}{6}\)]
= \(\frac{31}{6}\) – \(\frac{3}{6}\) = \(\frac{31-3}{6}\) = \(\frac{28}{6}\)
= \(\frac{14}{3}\) = 4\(\frac{2}{3}\) hours
Hence, the plane saves 4\(\frac{2}{3}\) hours.

Question 10.
What fraction of the whole square is shaded?

Solution:
In the given figure, the big square is divided into 4 identical squares. So, one small square occupies \(\frac{1}{4}\) of the area of the big square. Now, consider the smaller square.
The small square in the figure is divided into 8 identical triangles in which 3 are shaded.

So, the shaded part is \(\frac{3}{8}\) of the small square.
But the small square is \(\frac{3}{8}\) of the big square.
∴ The shaded part is \(\frac{1}{4}\) × \(\frac{3}{8}\) = \(\frac{3}{32}\) of the big square.
Hence, \(\frac{3}{32}\) of the whole square is shaded.

Question 11.
A colony of ants set out in search of food. As they search, they keep splitting equally at each point (as shown in the Fig. 8.7) and reach two food sources, one near a mango tree and another near a sugarcane field. What fraction of the original group reached each food source?

Solution:
At first point ants split in two ways. So, fraction of ants at each way is 1 ÷ \(\frac{1}{2}\) = \(\frac{1}{2}\).
At the second point. ants split in two ways.
So the fraction of ants at each war \(\frac{1}{2}\) ÷ 2 = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

At the third point, ants split in four ways.
So, fraction of ants at each way
= \(\frac{1}{4}\) ÷ 4 = \(\frac{1}{4}\) × \(\frac{1}{4}\) = \(\frac{1}{16}\)
At the fourth point, ants split in 2 wars.
So fraction of ants at each way
= \(\frac{1}{16}\) ÷ 2 = \(\frac{1}{16}\) × \(\frac{1}{2}\) = \(\frac{1}{32}\)
Hence, fraction of ants at mango tree = \(\frac{1}{2}\) + \(\frac{1}{4}\) + \(\frac{1}{16}\) + \(\frac{1}{16}\) + \(\frac{1}{32}\)
= \(\frac{16}{32}\) + \(\frac{8}{32}\) + \(\frac{2}{32}\) + \(\frac{2}{32}\) + \(\frac{1}{32}\) = \(\frac{16+8+2+2+1}{32}\)
= \(\frac{29}{32}\)
Fraction of ants near sugarcane field = \(\frac{1}{32}\) + \(\frac{1}{16}\) = \(\frac{1}{32}\) + \(\frac{2}{32}\)
= \(\frac{1+2}{32}\) = \(\frac{3}{32}\)

Question 12.
What is (1 – \(\frac{1}{2}\))?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\) ?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) ?
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\)) ?
Make a general statement and explain.
Solution:
1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\) = \(\frac{1}{2}\) × \(\frac{2}{3}\) = \(\frac{1}{3}\);
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) = \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) × \(\frac{4}{5}\) = \(\frac{1}{5}\);
(1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × (1 – \(\frac{1}{5}\)) × (1 – \(\frac{1}{6}\)) × (1 – \(\frac{1}{7}\)) × (1 – \(\frac{1}{8}\)) × (1 – \(\frac{1}{9}\)) × (1 – \(\frac{1}{10}\))
= \(\frac{1}{2}\) × \(\frac{2}{3}\) × \(\frac{3}{4}\) × \(\frac{4}{5}\) × \(\frac{5}{6}\) × \(\frac{6}{7}\) × \(\frac{7}{8}\) × \(\frac{8}{9}\) × \(\frac{9}{10}\) = \(\frac{1}{10}\)
Here, we observe that in this pattern, denominator of each term cancels the numerator of the next term, and found that the final product has the numerator of the first term and the denominator of the last term.
In general, (1 – \(\frac{1}{2}\)) × (1 – \(\frac{1}{3}\)) × (1 – \(\frac{1}{4}\)) × ………… × (1 – \(\frac{1}{n}\)) = \(\frac{1}{n}\)

InText Questions

Question 1.
In each of the figures given below, find the fraction of the big square that the shaded region occupies.

Solution:
(i) In the figure. the smaller square in the up- right corner is divided into 4 smaller squares and each square within it is further divided into 2 triangles.

∴ Total number of triangles in whole square
= (4 × 2) × 4 = 32
Out of these 32 triangles. 12 triangular regions are shaded.
∴ Fraction of shaded region = \(\frac{12}{32}\) = \(\frac{3}{8}\)
Thus, the shaded region occupies \(\frac{3}{8}\) of the area of the whole square.

(ii) In the figure, the smaller square is divided into 4 triangles and a square, which can be divided into 4 triangles having the same area. Therefore, total triangle in 1 smaller square is 8.

∴ Total number of triangles in whole square = 8 × 4 = 32
Out of these 32 triangles, 2 triangular regions are shaded.
∴ Fraction of shaded region = \(\frac{2}{32}\) = \(\frac{1}{16}\)
Thus, the shaded region occupies \(\frac{1}{16}\) of the area of the whole square.

Working with Fractions Class 7 Extra Questions

Working with Fractions Class 7 Very Short Question Answer

Question 1.
A library has 2400 books, and \(\frac{5}{8}\) of them are placed on the ground floor. The rest are kept on the first floor. How many books are on the first floor?
Solution:
Given, total number of books in library = 2400
Number of books on ground floor
= \(\frac{5}{8}\) of total books = \(\frac{5}{8}\) × 2400 = 5 × 300 = 1500
Hence, number of books on the first floor
= 2400 – 1500 = 900.

Question 2.
A water tank can hold 1250 litres of water. If 2\(\frac{1}{5}\) of such tanks are filled, how much water is used in total?
Solution:
Given, capacity of one tank = 1250 litres
And, 2\(\frac{1}{5}\) of such tanks are filled.
∴ Total water used = 2\(\frac{1}{5}\) × capacitv of one tank
= \(\frac{11}{5}\) × capacity of one tank
= \(\frac{11}{5}\) × 125o = 11 × 250
= 2750 litres

Question 3.
Divide:
(i) 25 by \(\frac{1}{3}\)
(ii) 48 by 3\(\frac{3}{4}\)
Solution:
(i) 25 ÷ \(\frac{1}{3}\) = 25 × \(\frac{3}{1}\)
= \(\frac{25 \times 3}{1}\) = 75

(ii) 48 ÷ 3\(\frac{3}{4}\) = 48 ÷ \(\frac{15}{4}\)
= 48 × \(\frac{4}{15}\) [∵ 3\(\frac{3}{4}\) = \(\frac{15}{4}\)]
= \(\frac{16 \times 4}{5}\) = \(\frac{64}{5}\) = 12\(\frac{4}{5}\)

Question 4.
Solve the following:
(i) \(\frac{3}{4}\) ÷ \(\frac{2}{5}\)
(ii) 2\(\frac{1}{2}\) ÷ \(\frac{5}{8}\)
Solution:
(i) \(\frac{3}{4}\) ÷ \(\frac{2}{5}\) = \(\frac{3}{4}\) ÷ \(\frac{5}{2}\)
= \(\frac{3 \times 5}{4 \times 2}\) = \(\frac{15}{8}\)

(ii) 2\(\frac{1}{2}\) ÷ \(\frac{5}{8}\) = \(\frac{5}{2}\) ÷ \(\frac{5}{8}\)
= \(\frac{5}{2} \times \frac{8}{5}\) = \(\frac{5 \times 8}{2 \times 5}\) = 4

Question 5.
Solve:
(i) \(\frac{5}{6}\) ÷ 3
(ii) 1\(\frac{2}{3}\) ÷ 4
Solution:
(i) \(\frac{5}{6}\) ÷ 3 = \(\frac{5}{6}\) × \(\frac{1}{3}\) = \(\frac{5 \times 1}{6 \times 3}\) = \(\frac{5}{18}\)

(ii) 1\(\frac{2}{3}\) ÷ 4 = \(\frac{5}{3}\) ÷ 4 = \(\frac{5}{3}\) × \(\frac{1}{4}\) = \(\frac{5 \times 1}{3 \times 4}\) = \(\frac{5}{12}\) [∵ 1\(\frac{2}{3}\) = \(\frac{5}{3}\)]

Question 6.
During a community health program, each participant was provided \(\frac{2}{5}\) litres of clean water per day. If 40 litres were distributed on a particular day, calculate the number of participants who received the water.
Solution:
Quantity of water for I participant = \(\frac{2}{5}\) litres
Total quantity oÍ waler = 40 litres
Now, number of participants
= \(\frac{\text { Total quantity of water }}{\text { Quantity of water for } 1 \text { participant }}\)
= 40 ÷ \(\frac{2}{5}\) = 40 × \(\frac{5}{2}\) = 100

Question 7.
What number should be multiplied by 5\(\frac{3}{4}\) to get 3\(\frac{3}{5}\)?
Solution:
Let the required number be x. Then,
x × 5\(\frac{3}{4}\) = 3\(\frac{3}{5}\)
⇒ x = \(\frac{3 \frac{3}{5}}{5 \frac{3}{4}}\)
⇒ x = \(\frac{\frac{18}{5}}{\frac{23}{4}}\) = \(\frac{15}{5}\) × \(\frac{4}{23}\) = \(\frac{18 \times 4}{5 \times 23}\)
= \(\frac{72}{115}\)

Question 8.
Each guest is served 1\(\frac{3}{5}\) litres of juice. If 40 litres of juice is available, how many guests can be served?
Solution:
Given, juice served to each guest.
= 1\(\frac{3}{5}\) litres = \(\frac{8}{5}\)litres
Total quantity of juice available = 40 litres
Now, number of guests that can be served
= \(\frac{\text { Total quantity of juice available }}{\text { Juice served to each guest }}\)
= \(\frac{40}{\frac{8}{5}}\) = 40 × \(\frac{5}{8}\) = 5 × 5 = 25

Working with Fractions Class 7 Short Question Answer

Question 1.
Multiply the following fractions and express as mixed fraction:
(i) 1\(\frac{3}{4}\) × 2\(\frac{2}{5}\)
(ii) 4\(\frac{1}{3}\) × \(\frac{3}{7}\)
(iii) 5\(\frac{2}{9}\) × 1\(\frac{1}{6}\)
(iv) 3\(\frac{4}{5}\) × 2\(\frac{3}{8}\)
Solution:
(i) 1\(\frac{3}{4}\) × 2\(\frac{2}{5}\) = \(\frac{7}{4}\) × \(\frac{12}{5}\)
= \(\frac{7 \times 3}{1 \times 5}\) = \(\frac{21}{5}\)
= 4\(\frac{1}{5}\)

(ii) 4\(\frac{1}{3}\) × \(\frac{3}{7}\) = \(\frac{13}{3}\) × \(\frac{3}{7}\)
= \(\frac{13 \times 1}{1 \times 7}\) = \(\frac{13}{7}\)
= 1\(\frac{6}{7}\)

(iii) 5\(\frac{2}{9}\) × 1\(\frac{1}{6}\) = \(\frac{47}{9}\) × \(\frac{7}{6}\)
= \(\frac{47 \times 7}{9 \times 6}\) = \(\frac{329}{54}\)
= 6\(\frac{5}{54}\)

(iv) 3\(\frac{4}{5}\) × 2\(\frac{3}{8}\) = \(\frac{19}{5}\) × \(\frac{19}{8}\)
= \(\frac{19 \times 19}{5 \times 8}\) = \(\frac{361}{40}\)
= 9\(\frac{1}{4}\)

Question 2.
Meena invited \(\frac{2}{7}\) of her students to a weekend workshop. If \(\frac{3}{5}\) of the invited students were girls, how many boys were present at the workshop if her class has 105 students?
Solution:
Given, total students in the class = 105
∴ Nunber of students invited
= \(\frac{2}{7}\) × Total students =
= \(\frac{2}{7}\) × 105 = 2 × 15 = 30
Now, girls among invited students
= \(\frac{3}{5}\) of the invited students
= \(\frac{3}{5}\) × 30 = 3 × 6 = 18
Hence, number of boys present at the workshop
= 30 – 18 = 12.

Question 3.
A library donated \(\frac{3}{8}\) of its books to a village school. If \(\frac{2}{3}\) of the donated books were story books, how many non-story books were donated to the school if the library originally had 1280 books?
Solution:
Given, total number of books in library = 1280
Number of books donated = \(\frac{3}{8}\) of total books
= \(\frac{3}{8}\) × 1280 = 3 × 160 = 480
Story books among donated books = \(\frac{2}{3}\) × of donated books = \(\frac{2}{3}\) × 480 = 2 × 160 = 320
Hence, number of non-story books among donated books = 480 – 320 = 160

Question 4.
The area of a rectangular field is 63\(\frac{3}{5}\) m². If its length is 7\(\frac{1}{2}\)m, find its breadth?
Solution:
Given area of rectangular field = 63\(\frac{3}{5}\) = \(\frac{318}{5}\) m²
length of rectangular field = 7\(\frac{1}{2}\) = \(\frac{15}{2}\) m
We know.
Area of rectangle length × Breadth
⇒ Breadth = \(\frac{\text { Area of rectangle }}{\text { Length }}\) = \(\frac{318}{5}\) ÷ \(\frac{15}{2}\)
= \(\frac{318}{5}\) × \(\frac{2}{15}\) = \(\frac{106 \times 2}{5 \times 5}\)
= \(\frac{212}{25}\) = 8\(\frac{12}{25}\) m

Question 5.
Determine the number by which 5\(\frac{4}{7}\) must be multiplied to obtain 4\(\frac{3}{5}\).
Solution:
We have. 5\(\frac{4}{7}\) = \(\frac{39}{7}\) and 4\(\frac{3}{5}\) = \(\frac{23}{5}\)
Let the number to be found b x. Then,
\(\frac{39}{7}\) × x = \(\frac{23}{5}\)
Multiplying by the reciprocal of \(\frac{39}{7}\) on both sides, we get
\(\frac{7}{39}\) × \(\frac{39}{7}\) × x = \(\frac{7}{39}\) × \(\frac{23}{5}\) [∵ Reciprocal of \(\frac{39}{7}\) is \(\frac{7}{39}\)]
⇒ x = \(\frac{7 \times 23}{39 \times 5}\) = \(\frac{161}{195}\)

Question 6.
A bookstore sells journals at ₹ 6\(\frac{1}{2}\) per copy. If the shop’s revenue from journal sales totalled ₹ 975, how many dozens of journals were sold?
Solution:
Given, total revenue from journal sales = ₹ 975
Price of 1 journal = ₹ 6\(\frac{1}{2}\) = ₹ \(\frac{13}{2}\)
Now, number of journals sold
= \(\frac{\text { Total revenue from journal sales }}{\text { Price of } 1 \text { journal }}\)
⇒ Number of journals sold = \(\frac{975}{\frac{13}{2}}\) = 975 × \(\frac{2}{13}\)
= \(\frac{75 \times 2}{1}\) = 150
Now, the number of journals sold (in dozens)
= \(\frac{150}{12}\) = \(\frac{25}{2}\) = 12\(\frac{1}{2}\) [∵ 1 dozen = 12]
Thus, 12\(\frac{1}{2}\) dozens of journals were sold.

Question 7.
Divide:
(i) 36 by \(\frac{3}{8}\)
(ii) 54 by 2\(\frac{2}{5}\)
(iii) 42 by \(\frac{7}{6}\)
Solution:
(i) 36 ÷ \(\frac{3}{8}\) = 36 × \(\frac{8}{3}\) = 12 × 8 = 96

(ii) 54 ÷ 2\(\frac{2}{5}\)
= 54 ÷ \(\frac{12}{5}\) = 54 × \(\frac{5}{12}\)
= \(\frac{9 \times 5}{2}\) = \(\frac{45}{2}\) = 22\(\frac{1}{2}\) [∵ 2\(\frac{2}{5}\) = \(\frac{12}{5}\)]

(iii) 42 ÷ \(\frac{2}{5}\) = 42 × \(\frac{6}{7}\)
= 6 × 6 = 36

Question 8.
How many square tiles with a side length of \(\frac{1}{2}\) m are required to cover an area of 12\(\frac{1}{4}\) m²?
Solution:
Given, the side length of a square tile is \(\frac{1}{2}\) m.
Therefore, area of one tile = Side × Side
= \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\) m²
Area to be covered by the tiles = 12\(\frac{1}{4}\) m² = \(\frac{49}{4}\) m²
Now, number of required tiles = Total area that need to be covered ÷ Area of one tile
= \(\frac{49}{4}\) ÷ \(\frac{1}{4}\) = \(\frac{49}{4}\) × \(\frac{4}{1}\) = 49
Thus, the number of required tiles is 49.

Question 9.
A fruit seller earns ₹ 4\(\frac{1}{2}\) per apple. He earns ₹864. How many dozens of apples did he sell?
Solution:
Given, total earnings = ₹864
Amount earned per apple = ₹4\(\frac{1}{2}\) = ₹\(\frac{9}{2}\)
Now, number of apples sold = \(=\frac{\text { Total earnings }}{\text { Amount earned per apple}}\)
= \(\frac{864}{\frac{9}{2}}\) = 864 × \(\frac{2}{9}\) = 96 × 2 = 192
So, number of apples sold (in dozens)
= \(\frac{192}{12}\) = 16 [∵ 1 dozen = 12]

Working with Fractions Class 7 Long Question Answer

Question 1.
Fill in the boxes using <, > or = without actually finding the product:
(i) \(\frac{3}{8}\) × \(\frac{2}{7}\) ☐ \(\frac{2}{7}\)
(ii) \(\frac{5}{9}\) × \(\frac{4}{11}\) ☐ \(\frac{5}{9}\)
(iii) 4\(\frac{1}{2}\) × 3\(\frac{3}{4}\) ☐ 4\(\frac{1}{2}\)
(iv) \(\frac{6}{13}\) × \(\frac{15}{4}\) ☐ \(\frac{15}{4}\)
Solution:
(i) We know, the product of two proper fractions is less than each of them.
Here, \(\frac{3}{8}\) and \(\frac{2}{7}\) are proper tractions.
∴ \(\frac{3}{8}\) × \(\frac{2}{7}\) < \(\frac{2}{7}\)

(ii) We know, the product of two proper fractions is less than each of them.
Here. \(\frac{5}{9}\) and \(\frac{4}{11}\) are proper fractions.
∴ \(\frac{5}{9}\) × \(\frac{4}{11}\) < \(\frac{5}{9}\)

(iii) We know, the product of two improper fractions is greater than both the fractions or equal to either of them.
Here, 4\(\frac{1}{2}\) and 3\(\frac{3}{4}\) are mixed tractions and mixed tractions can he converted into improper fractions.
∴ 4\(\frac{1}{2}\) × 3\(\frac{3}{4}\) > 4\(\frac{1}{2}\)

(iv) We know, the product of a proper fraction and an improper fraction (greater than 1) lies between both the fractions.
Here. \(\frac{6}{13}\) is a proper traction and \(\frac{15}{4}\) is an improper traction (greater than 1).
∴ \(\frac{6}{13}\) × \(\frac{15}{4}\) < \(\frac{15}{4}\)

Question 2.
Solve the following and write the result in simplest form:
(i) 8 × \(\frac{5}{6}\)
(ii) 4 × \(\frac{2}{5}\)
(iii) 7 × \(\frac{3}{8}\)
(iv) 5 × \(\frac{4}{3}\)
(v) \(\frac{11}{4}\) × 3
(vi) 10 × \(\frac{1}{5}\)
Solution:
(i) 8 × \(\frac{5}{6}\) = \(\frac{48 \times 5}{6}\) = \(\frac{4 \times 5}{3}\)
= \(\frac{20}{3}\)

(ii) 4 × \(\frac{2}{5}\) = \(\frac{4 \times 2}{5}\)
= \(\frac{8}{5}\)

(iii) 7 × \(\frac{3}{8}\) = \(\frac{7 \times 3}{8}\)
= \(\frac{21}{8}\)

(iv) 5 × \(\frac{4}{3}\) = \(\frac{5 \times 4}{3}\)
= \(\frac{20}{3}\)

(v) \(\frac{11}{4}\) × 3 = \(\frac{11 \times 3}{4}\)
= \(\frac{33}{4}\)

(vi) 10 × \(\frac{1}{5}\) = 2 × 1 = 2

Question 3.
Solve the following and express as a mixed fraction:
(i) 3\(\frac{5}{8}\) × 2\(\frac{1}{4}\)
(ii) 7\(\frac{2}{3}\) × 4\(\frac{3}{5}\)
(iii) 5\(\frac{3}{10}\) × 3\(\frac{4}{7}\)
(iv) \(\frac{4}{9}\) × 6\(\frac{2}{5}\)
(v) \(\frac{7}{10}\) × 9\(\frac{1}{6}\)
(vi) \(\frac{3}{8}\) × 8\(\frac{5}{9}\)
Solution:
(i) 3\(\frac{5}{8}\) × 2\(\frac{1}{4}\) = \(\frac{29}{8}\) × \(\frac{9}{4}\) = \(\frac{29 \times 9}{8 \times 4}\)
= \(\frac{261}{32}\) = 8\(\frac{5}{32}\)

(ii) 7\(\frac{2}{3}\) × 4\(\frac{3}{5}\) = \(\frac{23}{3}\) × \(\frac{23}{5}\) = \(\frac{23 \times 23}{3 \times 5}\)
= \(\frac{529}{15}\) = 35\(\frac{4}{15}\)

(iii) 5\(\frac{3}{10}\) × 3\(\frac{4}{7}\) = \(\frac{53}{10}\) × \(\frac{25}{7}\)
= \(\frac{265}{14}\) = 18\(\frac{13}{14}\)

(iv) \(\frac{4}{9}\) × 6\(\frac{2}{5}\) = \(\frac{4}{9}\) × \(\frac{32}{5}\) = \(\frac{4 \times 32}{9 \times 5}\)
= \(\frac{128}{45}\) = 2\(\frac{38}{45}\)

(v) \(\frac{7}{10}\) × 9\(\frac{1}{6}\) = \(\frac{7}{10}\) × \(\frac{55}{6}\)
= \(\frac{77}{12}\) = 6\(\frac{5}{12}\)

(vi) \(\frac{3}{8}\) × 8\(\frac{5}{9}\) = \(\frac{3}{8}\) × \(\frac{77}{9}\) = \(\frac{1 \times 77}{8 \times 3}\)
= \(\frac{77}{24}\) = 3\(\frac{5}{24}\)

Question 4.
A car travels 5\(\frac{1}{4}\) km north, then \(\frac{1}{2}\) km west and 4\(\frac{3}{8}\) km north. Find total distance travelled (in km)?
What fraction of the journey was travelled in the north direction?
Solution:
Given: Distance travelled in north direction
d₁ = 5\(\frac{1}{4}\) km = \(\frac{21}{4}\) km
Distance travelled in west direction, d₂ = \(\frac{1}{2}\) km
Distance travelled in north direction again.
d₃ = 4\(\frac{3}{8}\)km = \(\frac{35}{8}\) km
∴ Total distance travelled = d₁ + d₂ + d₃
= \(\frac{21}{4}\) + \(\frac{1}{2}\) + \(\frac{35}{8}\) = \(\frac{42}{8}\) + \(\frac{4}{8}\) + \(\frac{35}{8}\)
= \(\frac{42+4+35}{8}\) = \(\frac{81}{8}\) = 10\(\frac{1}{8}\) km
Total distance travelled in north direction
= d₁ + d₃ = \(\frac{21}{4}\) + \(\frac{35}{8}\) = \(\frac{42+35}{8}\) = \(\frac{77}{8}\) km
∴ Fraction of the journey travelled in the north direction
= \(\frac{d_1+d_3}{d_1+d_2+d_3}\) = \(\frac{\frac{77}{8}}{\frac{81}{8}}\)
= \(\frac{77}{8}\) × \(\frac{8}{81}\) = \(\frac{77}{81}\)

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 2 Arithmetic Expressions Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 2 Arithmetic Expressions Solutions

Ganita Prakash Class 7 Chapter 2 Solutions

Class 7 Maths Ganita Prakash Chapter 2 Solutions Arithmetic Expressions

Question 1.
Fill in the blanks to make the expressions equal on both sides of the ‘=’ sign:
(i) 13 + 4 = _____ + 6
(ii) 22 + = 6 × 5
(iii) 8 × = 64 4 2
(iv) 34 – _____ = 25
Solution:
(i) 13 + 4 = 11 + 6 [∵ 13 + 4 = 17 and 11 + 6 = 17]
(ii) 22 + 8 = 6 × 5 [∵ 6 × 5 = 30 and 22 + 8 = 30]
(iii) 8 × 4 = 64 ÷ 2 [∵ 64 ÷ 2 = 32 and 8 × 4 = 32]
(iv) 34 – 9 = 25

Question 2.
Arrange the following expressions in ascending (increasing) order of their values.
(i) 67 – 19
(ii) 67 – 20
(iii) 35 × 25
(iv) 5 × 11
(v) 120 ÷ 3
Solution:
(i) 67 – 19 = 48
(ii) 67 – 20 = 47
(iii) 35 + 25 = 60
(iv) 5 × 11 = 55
(v) 120 ÷ 3 = 40
Clearly, 40 < 47 < 48 < 55 < 60
∴ 120 ÷ 3 < 67 – 20 < 67 – 19 < 5 × 11 < 35 4 25
Hence, (v) < (ii) < (i) < (iv) < (iii).

Question 3.
For each of the following situations, write the expression describing the situation, identify its terms and find the value of the expression.
(i) Queen Alia gave 100 gold coins to Princess Elsa and 100 gold coins to Princess Anna last year. Princess Elsa used the coins to start a business and doubled her coins. Princess Anna bought jewellery and has only half of the coins left. Write an expression describing how many gold coins Princess Elsa and Princess Anna together have.
(ii) A metro train ticket between two stations is ₹40 for an adult and ₹20 for a child. What is the total cost of tickets:
(a) for four adults and three children?
(b) for two groups having three adults each?
(iii) Find the total height of the window by writing an expression describing the relationship among the measurements shown in the picture.

Solution:
(i) Number of gold coins Princess Elsa got = 100
Number of gold coins Princess Anna got = 100
Princess Elsa used the coins to start the business and double her coins.
So, the number of coins Princess Elsa has = 2 × 100
Princess Anna bought jewellery and has only half of the coins left.
So, the total number of coins Princess Anna has = \(\frac{100}{2}\)
Therefore, the total number of gold coins Princess Elsa and Princess Anna have together
= 2 × 100 + \(\frac{100}{2}\) = 200 + 50 = 250
Thus, the expression describing the above situation is, 2 × 100 + \(\frac{100}{2}\).
Terms: 2 × 100, \(\frac{100}{2}\)

(ii) (a) Fare of metro train ticket for an adult = ₹40
So, the fare of metro train ticket for four adults (in ₹) = 4 × 40
Fare of metro train ticket for a child = ₹20
So, the fare of metro train ticket for three children (in ₹) = 3 × 20
Therefore, the expression describing the total cost of tickets (in ₹) for four adults and three children is 4 × 40 + 3 × 20.
Total fare = 4 × ₹40 + 3 × ₹20 = ₹ 160 + ₹60 = ₹220
Terms: 4 × 40, 3 × 20

(b) Fare of metro train ticket for an adult = ₹40
So, the fare of metro train ticket for a group of three adults (in ₹) = 3 × 40
Therefore, the expression describing the total cost of tickets (in ₹) for the two groups having three adults each is 2 × (3 × 40).
Total fare = 2 × (3 × ₹ 40) = 2 × ₹120 = ₹240
Terms: 2 × (3 × 40)

(iii) By observing the given picture, the total height of the window
= Number of gaps × 5 cm + Number of grills × 2 cm + Number of borders × 3 cm
Here, total number of gaps = 7; Total number of grills = 6; Total number of borders = 2
∴ Total height of window (in cm) = 7 × 5 + 6 × 2 + 2 × 3 = 35 + 12 + 6 = 47 + 6 = 53
Terms: 7 × 5, 6 × 2, 2 ×3

Question 4.
Remove the brackets and write the expression having the same value.
(i) 14 + (12 + 10)
(ii) 14 – (12 + 10)
(iii) 14 + (12 – 10)
(iv) 14 – (12 – 10)
(v) -14 + 12 – 10
(vi) 14 – (-12 – 10)
Solution:
(i) 14 + (12 + 10) = 14 + 12 + 10 = 14 + 22 = 36
(ii) 14 – (12 + 10) = 14 – 12 – 10 = 14 – 22 = -8
(iii) 14 + (12 – 10) = 14 + 12 – 10 = 14 + 2 = 16
(iv) 14 – (12 – 10) = 14 – 12 + 10 = 14 – 2 = 12
(v) – 14 + 12 – 10 = – 14 + 2 = – 12
(vi) 14 – (-12 – 10) = 14 + 12 + 10 = 14 + 22 = 36
Here, expressions given in (i) and (vi) have same value.

Question 5.
Find the values of the following expressions. For each pair, first try to guess whether they have the same value. When are the two expressions equal?
(i) (6 + 10) – 2 and 6 + (10 – 2)
(ii) 16 – (8 – 3) and (16 – 8) – 3
(iii) 27 – (18 + 4) and 27 + (-18 – 4)
Solution:
(i) (6 + 10) – 2 = 16 – 2 = 14 and 6 + (10 – 2) = 6 + 8 = 14
Clearly, (6 + 10) – 2 = 6 + (10 – 2)
Hence, both the expressions have the same value.

(ii) 16 – (8 – 3) = 16 – 5 = 11 and (16 – 8) – 3 = 8 – 3 = 5
Clearly, 16 – (8 – 3) ≠ (16 – 8) – 3
Hence, both the expressions do not have the same value.

(iii) 27 – (18 + 4) = 27 – 22 = 5 and 27 + (-18 – 4) = 27 + (- 22) = 27 – 22 = 5
Clearly, 27 – (18 + 4) = 27 + (-18 – 4)
Hence, both the expressions have the same value.

Question 6.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(i) 34 – 9 + 12 = 13
(ii) 56 – 14 – 8 = 34
(iii) – 22 – 12 + 10 + 22 = – 22
Solution:
(i) 34 – (9 + 12) = 34 – 21 = 13
(ii) (56 – 14) – 8 = 42 – 8 = 34
(iii) – 22 – (12 + 10) + 22 = – 22 – 22 + 22 = – 22

Question 7.
Find out the sum of the numbers given in each picture below in at least two different ways. Describe how you solved it through expressions.

Solution:
For I : 5 × 4 + 4 × 8 = 20 + 32 = 52 or 4 × (4 + 8) + 4 = 4 × 12 + 4 = 52
For II : 8 × (5 + 6) = 8 × 11 = 88 or 8 × 5 + 8 × 6 = 40 + 48 = 88

Question 8.
Read the situations given below. Write appropriate expressions for each of them and find their values.
(i) The district market in Begur operates on all seven days of a week. Rahim supplies 9 kg of mangoes each day from his orchard and Shyam supplies 11 kg of mangoes each day from his orchard to this market. Find the amount of mangoes supplied by them in a week to the local district market.
(ii) Binu earns ₹20,000 per month. She spends ₹5,000 on rent, ₹5,000 on food and ₹2,000 on other expenses every month. What is the amount Binu will save by the end of a year?
(iii) During the daytime a snail climbs 3 cm up a post, and during the night while asleep, accidentally slips down by 2 cm. The post is 10 cm high, and a delicious treat is on its top. In how many days will the snail get the treat?
Solution:
(i) Amount of mangoes supplied by Rahim each day = 9 kg
Amount of mangoes supplied by Shyam each day = 11 kg
Total supplies of mangoes in the market on each day = (9 + 11) kg
∴ Total supplies of mangoes in the market in a week (7 days) = 7 × (9 + 11) = 7 × 20 = 140 kg

(ii) Binu’s per month earning = ₹20,000
Binu’s total monthly expenditures
= ₹5,000 on rent + ₹5,000 on food + ₹ 2,000 on other expenses
= ₹(5,000 + 5,000 + 2,000)
Therefore, Binu’s monthly savings = ₹20,000 – ₹(5,000 + 5,000 + 2,000) = ₹20,000 – ₹12,000 = ₹8,000
Thus, Binu’s total yearly savings = 12 × 8000 = ₹96000
Hence, Binu will save ₹96000 by the end of the year.

(iii) Since the snail climbs 3 cm up the post in daytime and slips down by 2 cm at night.
The distance climbed by the snail in a day = 3 – 2 = 1 cm
∴ The distance climbed in 7 days = 7 cm
The height of the post is 10 cm.
The distance climbed on the 8th day before slipping = 7 + 3=10 cm
So, the snail will take 8 days to reach the top of the post and get the delicious treat.

Question 9.
Find different ways of evaluating the following expressions:
(i) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10
(ii) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1
Solution:
(i) 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 + 3 + 5 + 7 + 9) + (-2 – 4 – 6 – 8 – 10) = 25 + (- 30) = – 5
OR
1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + 9 – 10 = (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + (9 – 10)
= (-1) + (-1) + (-1) + (-1) + (-1) = -5

(ii) 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 = (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1) + (1 – 1)
= 0 + 0 + 0 + 0 + 0 = 0
OR
1 – 1 + 1 – 1 + 1 – 1 + 1 – 1 + 1 – 1= (1 + 1 + 1 + 1 + 1) + (- 1 – 1 – 1 – 1 – 1)
= 5 + (- 5) = 0

Question 10.
Identify which of the following expressions are equal to the given expression without computation. You may rewrite the expressions using terms or removing brackets. There can be more than one expression which is equal to the given expression.
(i) 83 – 37 – 12
(a) 84 – 38 – 12
(b) 84 – (37 + 12)
(c) 83 – 38 – 13
(d) -37 + 83 – 12
(ii) 93 + 37 × 44 + 76
(a) 37 + 93 × 44 + 76
(b) 93 + 37 x× 76 + 44
(c) (93 + 37) × (44 + 76)
(d) 37 × 44 + 93 + 76
Solution:
(i) 83 – 37 – 12 = 83 – 37 – 12 + (1 – 1) = (83 + 1) – 37 – 1 – 12 = 84 – 38 – 12
Also, 83 – 37 – 12 = – 37 + 83 – 12
Hence, (a) and (d) are equal to the given expression 83 – 37 – 12.

(ii) 93 + 37 × 44 + 76
Rearranging the terms, we get 37 × 44 + 93 + 76, which is equal to the given expression in option (d). Hence, (d) is equal to the given expression 93 + 37 × 44 + 76.

InText Questions

Question 1.
Use ‘>’ or ‘<’ or ‘=’ in each of the following expressions to compare them. Can you do it without complicated calculations? Explain your thinking in each case.

Solution:
(i)
∴ 245 + 289 > 246 + 285

(ii)
∴ 273 – 145 = 272 – 144

(iii)
∴ 364 + 587 < 363 + 589

(iv)
∴ 142 + 245 < 129 + 245

(v)
∴ 213 – 77 < 214 – 76

Question 2.
Can you explain why subtracting a number is the same as adding its inverse, using the Token Model of integers that we saw in the Class 6 textbook of mathematics?
Solution:
In the token model:
Subtracting a positive number (e.g. subtracting 3) means removing 3 positive tokens.
Adding a negative number (e.g. adding- 3) meaning adding 3 negative tokens. These 3 negative tokens cancel out 3 existing positive tokens (by forming zero pairs), which is equivalent to removing 3 positive tokens.
Since both actions result in removing the same number of positive tokens, the final value is the same.

Question 3.
Does adding the terms of an expression in any order give the same value? Take some expressions and check. Consider expressions with more than 3 terms also.
Solution:
Yes
(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (- 5) + (-4) + (-5)
= (-9) + (-5) = -14

(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (-2) + (-7) + (-5)
= (-2) + (-12) = -14

(-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-4) + (-5)
= (-2) + (-3) + (-9)
= (-5) + (- 9) = -14
Note: Students can do on their own by taking different numbers.

Question 4.
5 × 4 + 3 ≠ 5 × (4 + 3). Can you explain why?
Is 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5?
Solution:
Expression 5 × 4 + 3 means 3 more than 5 × 4, which is equal to 23, but 5 × (4 + 3) means 5 times the sum of 3 and 4 which is equal to 35.
Hence, 5 × 4 + 3 + 5 ×(4 + 3)
Now, 5 × (4 + 3), 5 × (3 + 4), and (3 + 4) × 5 have the same meaning, which is 5 times the sum of 3 and 4 and give the same value.
Hence, 5 × (4 + 3) = 5 × (3 + 4) = (3 + 4) × 5

Question 5.
Use distributive property to find the following products:
(i) 95 × 8
(ii) 104 × 15
(iii) 49 × 50
Is this quicker than the multiplication procedure you use generally?
Solution:
(i) 95 × 8
= (100 – 5) × 8
= (100 × 8) – (5 × 8)
= 800 – 40 = 760

(ii) 104 × 15
= (100 + 4) × 15
= (100 × 15) + (4 × 15)
= 1500 + 60 = 1560

(iii) 49 × 50
= (50 – 1) × 50
= (50 × 50) – (50 × 1)
= 2500 – 50 = 2450
Yes, this procedure is quicker than the general multiplication procedure.

Arithmetic Expressions Class 7 Extra Questions

Arithmetic Expressions Class 7 Very Short Question Answer

Question 1.
Riya buys 5 notebooks per day for 4 days and 7 notebooks per day for the remaining 3 days of a week. Form an expression to represent the total number of notebooks she buys in that week.
Solution:
For 4 days: Riya buys 5 notebooks per day
For 3 days: Riya buys 7 notebooks per day
Thus, expression for total number of notebooks
= 4 × 5 + 3 × 7

Question 2.
Compare which is greater: 79 – 96 or 117 – 130.
Solution:
The value of 79 – 96 is – 17.
The value of 117 – 130 is – 13.
Clearly, – 17 < -13
Hence, 117 – 130 is greater than 79 – 96.

Question 3.
Anaya is preparing for a temple festival. She decorates 5 pillars, placing 7 marigold garlands on each. However, 2 garlands fall and cannot be used. Find the total number of garlands finally used.
Solution:
Total garlands before any fall = 5 × 7 = 35
Garlands that cannot be used = 2
Now, total garlands finally used = 35 – 2 = 33

Question 4.
Priya buys 2 packs of biscuits for ₹15 each and a juice for ₹20. What is the total cost that Priya needs to pay?
Solution:
Given, Priya buys 2 packs of biscuits for ₹15 each and a juice for ₹20.
Cost of biscuits = 2 × ₹15
Cost of juice = ₹ 20
Now, total cost = 2 × ₹ 15 + ₹20 = ₹30 + ₹20
= ₹50

Question 5.
Add brackets at appropriate places in the expressions such that they lead to the values indicated.
(i) 25 – 7 + 12 = 6
(ii) 42 – 15 – 7-9 = 29
(iii) – 32 – 18 + 14 + 32 = – 32
(iv) 35 – 18 – 5 + 10 = 32
Solution:
(i) 25 – (7 + 12) = 6
(ii) 42 – 15 – (7 – 9) = 29
(iii) – 32 – (18 + 14) + 32 = – 32
(iv) 35 – (18 – 5) + 10 = 32

Arithmetic Expressions Class 7 Short Question Answer

Question 1.
Ankit is planning a birthday party. He buys:

• 5 party hats, each costing ₹ 60
• 2 big balloons, each costing ₹90
• If the total cost exceeds ₹400, he receives a discount of ₹50.

Write the expression that shows the amount (in ₹) Ankit has to pay.
Solution:
Given, cost of five party hats = 5 × ₹ 60
Cost of two big balloons = 2 × ₹90
Discount = ₹50 [If total cost > ₹400]
Now, total cost (in ₹) = 5 × 60 + 2 × 90 = 300 + 180 = 480
As 480 > 400, Ankit gets a discount of ₹50.
So, the expression that shows the amount (in ₹) Ankit has to pay (after applying the discount) is
5 × 60 + 2 × 90 – 50.

Question 2.
Simplify:
(i) (-40) × (-1) + 28 ÷ 7
(ii) 7 – [13 – 2{4 × (- 4)}]
(iii) 81 × [59 – {7 × 8 + (13 – 2 × 5)}]
Solution:
(i) (-40) × (- 1) + 28 ÷ 7
= (40 × 1) + 28 ÷ 7
= 40 + 4 = 44 [∵ (-) × (-) = ( + )]

(ii) 7 – [13 – 2{4 × (- 4)}]
= 7 – [13 – 2 × {-16}] [∵ (+) × (-) = (-)]
= 7 – [13 + 32] = 7 – 45 = -38

(iii) 81 × [59 – {7 × 8 + (13 – 2 × 5)}]
= 81 × [59 – {7 × 8 + (13 – 10)}]
= 81 × [59 – {56 + 3}]
= 81 × [59 – 59] = 81 × 0 = 0

Question 3.
Find the following products using the distributive property.
(i) 107 × 12
(ii) 98 × 14
Solution:
(i) Let’s break 107 as 100 + 7.
Now, 107 × 12 = (100 + 7) × 12
= 100 × 12 + 7 × 12 [Using distributive property]
= 1200 + 84 = 1284
Hence, 107 × 12 = 1284

(ii) Let’s break 98 as 100 – 2.
Now, 98 × 14 = (100 – 2) × 14
= 100 × 14 – 2 × 14
= 1400 – 28 = 1372
Hence, 98 × 14 = 1372

Question 4.
Simplify: 659 – [219 – (750 — 255 ÷ 5 × 9)]
Solution:
We know, on removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
Thus, 659 – [219 – (750 – 255 ÷ 5 × 9)]
= 659 – 219 + (750 – 255 ÷ 5 × 9)
= 659 – 219 + (750 – 51 × 9) [As 255 ÷ 5 = 51]
= 659 – 219 + (750 – 459) [As 51 × 9 = 459]
= 1409-678
[As 659 + 750 = 1409 and – 219 – 459 = – 678]
= 731
Hence, 659 – [219 – (750 – 255 ÷ 5 × 9)] = 731

Question 5.
Simplify:
63 – (- 3){- 2 – 8 – 3} ÷ {5 + (- 2)(- 1)}
Solution:
63 – (-3){-2 – 8 – 3} ÷ {5 + (-2)(-1)}
= 63 – (- 3){-2 – 5} ÷ {5 + (- 2)(- 1)}
[Removal of bar]
= 63 + 3{-2 – 5} ÷ {5 + 2}
[As – (-3) = 3 and (-2) (-1) = 2]
= 63 + 3{-7} ÷ 7
= 63 – 21 ÷ 7[As 3(-7) = -21]
= 63 – 3 = 60[As 21 ÷ 7 = 3]

Question 6.
Ravi took part in a painting competition and got scores 34, 37 and 31 from three judges. Later, it was found that second judge had mistakenly given 37 instead of the correct score 35. What are Ravi’s initial and updated total scores?
Solution:
Given, Ravi got scores 34, 37 and 31 from three judges.
So, the initial total score = 34 + 37 + 31 = 102
Later, the second judge’s score was corrected from 37 to 35, which is 2 less than the initial score.
Since only the second score changed, the updated total score = 102 – 2 = 100
Hence, Ravi’s updated total score is 100.

Question 7.
Remove the brackets and write the expression having the same value.
(i) 18 + (12 + 6)
(ii) 18 + (12 – 6)
(iii) 18 – (12 + 6)
(iv) 18 – (12 – 6)
(v) 18 – (-12 – 6)
(vi) 18 – (-12 + 6)
Solution:
On removing the brackets preceded by a plus sign, the signs of all the terms inside the brackets remain same.
(i) 18 + (12 + 6) = 18 + 12 + 6
(ii) 18 + (12 – 6) = 18 + 12 – 6
On removing the brackets preceded by a minus sign, the signs of all the terms inside the brackets change.
(iii) 18 – (12 + 6) = 18 – 12 – 6
(iv) 18 – (12 – 6) = 18 – 12 + 6
(v) 18 – (-12 – 6) = 18 + 12 + 6
(vi) 18 – (-12 + 6) = 18 + 12 – 6

Question 8.
Leela is organizing candles for a festival. She places 36 candles in one box and 27 in another. She gives away 8 candles from the second box to her neighbour.
Write an expression for the number of candles Leela is left with.
Solution:
Candles in the first box = 36
Candles in the second box = 27
Candles given to neighbour = 8
Now, the number of candles left with Leela
= Candles in the first box + Candles in the second box – Candles given to neighbour
= 36 + 27 – 8, which is the required expression.

Question 9.
In three sections of a school library, 64, 93 and 81 books were recorded respectively.
Later, the librarian found that there were mistakes in two sections:

• The first section actually had 84 books, not 64.
• The second section actually had 73 books, not 93.

What is the difference between total number of books initially and after correcting the record?
Solution:
The error in the number of books in the first section was + 20 (i.e., 84 – 64 = +20).
The error in the second section was -20
(i.e., 73 – 93 = – 20).
Since these two errors cancel each other out, the overall total remains unchanged.
Therefore, there is no difference in the total number of books.

Arithmetic Expressions Class 7 Long Question Answer

Question 1.
Find the following products using the distributive property.
(i) 125 × 16
(ii) 87 × 13
(iii) 106 × 104
(iv) 107 × 91
Solution:
(i) Let’s break 125 as 100 + 25.
Now, 125 × 16 = (100 + 25) × 16
= 100 × 16 + 25 × 16 = 1600 + 400 = 2000
Hence, 125 × 16 = 2000

(ii) Let’s break 87 as 100 -13.
Now, 87 × 13 = (100 – 13) × 13
= 100 × 13 – 13 × 13 = 1300 – 169 = 1131
Hence, 87 × 13 = 1131

(iii) Let’s break 106 as 100 + 6.
Now, 106 × 104 = (100 + 6) × 104
= 100 × 104 + 6 × 104 = 10400 + 6 × (100 + 4)
[As 104 = 100 + 4]
= 10400 + 6 × 100 + 6 × 4
= 10400 + 600 + 24 = 11024
Hence, 106 × 104 = 11024

(iv) Let’s break 107 as 100 + 7.
Now, 107 × 91 = (100 + 7) × 91
= 100 × 91 + 7 × 91 = 9100 + 7 × (100 – 9)
[As 91 = 100 – 9]
= 9100 + 7 × 100 – 7 × 9
= 9100 + 700 – 63
= 9800 – 63 = 9737
Hence, 107 × 91 = 9737.

Question 2.
Simplify:
(i) 121 ÷ [17 – {15 – 3(7 – 4)}]
(ii) 32 ÷ [32 + {32 – (32 + 32 – 32 )}]
(iii) 15 – (-3) × [{4- 7 – 3} + 3 × {5 + (-3) × (-6)}]
Solution:
(i) 121 ÷ [17 – {15 – 3(7 – 4)}]
= 121 ÷ [17 – {15 – 3 × 3}]
= 121 ÷ [17 – {15 – 9}]
= 121 ÷ [17 – 6]
= 121 ÷ 11 = 11

(ii) 32 ÷ [32 + {32 – (32 + 32 – 32 )}]
= 32 ÷ [32 + {32 – (32 + 0)}]
= 32 ÷ [32 + {32 – 32}]
= 32 ÷ [32 + 0] = 32 + 32 = 1

(iii) 15 – (-3) × [{4 – 7 – 3 } ÷ 3 × {5 + (-3) × (-6)}]
= 15 + 3 × [{4 – 4} ÷ 3 × {5 + 18}]
= 15 + 3 × [0 ÷ 3 × 23]
= 15 + 3 × [0 × 23] [As 0 ÷ 3 = 0]
= 15 + 3 × 0 = 15

Arithmetic Expressions Class 7 Case Based Questions

Question 1.
At Surya Vidya Mandir, the annual Art Festival is being celebrated. Ishan is participating in the event and earning performance points based on his art scenes.
Here’s how the points are awarded:

• In the 1^st scene, he puts in a solid effort and earns 22 points.
• In the 2^nd scene, he gets more confident and scores 8 more points than in the 1^st scene.
• In the 3^rd scene, he is slightly exhausted and earns half the points he got in the 2^nd scene.

Based on the above information, answer the following questions:
(i) Write arithmetic expressions to represent the number of points earned in the 2nd and 3rd scenes respectively.
(ii) What is the total number of points earned in all three scenes?
(iii) If each point is worth ₹50, how much money can Ishan claim?
Solution:
Points earned in 1^st scene = 22
Points earned in 2^nd scene
= 8 more points than in the 1^st scene
= 22 + 8 = 30
Points earned in 3^rd scene
= \(\frac{1}{2}\) × (Points earned in 2^nd scene)
= \(\frac{1}{2}\) × 30 = 15

(i) Arithmetic expression to represent the number of points earned in 2^nd scene = 22 + 8
Arithmetic expression to represent the number of points earned in 3^rd scene = \(\frac{1}{2}\) × 30

(ii) Total number of points earned in all three scenes = 22 + 30 + 15 = 67

(iii) Given, 1 point = ₹50
Total money that Ishan can claim
= ₹50 × 67 = ₹3,350

Question 2.
Ria buys 5 packs of pencils for her art class.
Each pack has 6 coloured pencils and 4 graphite pencils. But she loses 3 coloured pencils on the way.
Based on the above information, answer the following questions

(i) How many coloured pencils does Ria have?
(ii) How many graphite pencils does she have?
(iii) What is the total number of pencils Ria finally has?
Solution:
(i) In 1 pack, there are 6 coloured pencils.
So, number of coloured pencils in 5 packs
= 5 × 6 = 30
Given, Ria lost 3 coloured pencils on the way.
Now, number of remaining coloured pencils
= 30 – 3 = 27

(ii) In 1 pack, there are 4 graphite pencils.
So, number of graphite pencils in 5 packs
= 4 × 5 = 20

(iii) Total number of pencils with Ria
= Number of coloured pencils + Number of graphite pencils
= 27 + 20 = 47

Students can refer to BSE Odisha Class 7 Math Solution and Ganita Prakash Chapter 1 Large Numbers Around Us Class 7 Question Answer to understand textbook questions step by step.

Class 7 Maths Chapter 1 Large Numbers Around Us Solutions

Ganita Prakash Class 7 Chapter 1 Solutions

Class 7 Maths Ganita Prakash Chapter 1 Solutions Large Numbers Around Us

Question 1.
Read the following numbers in Indian place value notation and write their number names in both the Indian and American systems:
(i) 4050678
(ii) 48121620
(iii) 20022002
(iv) 246813579
Solution:
(i) 4050678
Indian System: 40,50,678
Number name: Forty lakh fifty thousand six hundred seventy eight American system: 4,050,678
Number name: Four million fifty thousand six hundred seventy eight

(ii) 48121620
Indian System: 4,81,21,620
Number name: Four crore eighty one lakh twenty one thousand six hundred twenty American System: 48,121,620
Number name: Forty eight million one hundred twenty one thousand six hundred twenty

(iii) 20022002
Indian System: 2,00,22,002
Number name: Two crore twenty two thousand two American System: 20,022,002
Number name: Twenty million twenty two thousand two

(iv) 246813579
Indian System: 24,68,13,579
Number name: Twenty four crore sixty eight lakh thirteen thousand five hundred seventy nine American System: 246,813,579
Number name: Two hundred forty six million eight hundred thirteen thousand five hundred seventy nine

Question 2.
Write the following numbers in Indian place value notation:
(i) One crore one lakh one thousand ten
(ii) One billion one million one thousand one
(iii) Ten crore twenty lakh thirty thousand forty
(iv) Nine billion eighty million seven hundred thousand six hundred
Solution:
(i) 1,01,01,010
(ii) 1,001,001,001
In Indian place value notation: 1,00,10,01,001
(iii) 10,20,30,040
(iv) 9,080,700,600
In Indian place value notation: 9,08,07,00,600

Question 3.
Compare and write ‘<‘ ‘>’ or ‘ = ’:
(i) 30 thousand _________ 3 lakh
(ii) 500 lakh _______ 5 million
(iii) 800 thousand ______ 8 million
(iv) 640 crore _______ 60 billion
Solution:
(i) As 30,000 < 3,00,000
⇒ 30 thousand < 3 lakh (ii) Since 1 million = 10 lakh, 5 million = 50 lakh Clearly, 500 lakh > 50 lakh ⇒ 500 lakh > 5 million

(iii) 800 thousand = 800 × 1000 = 800,000
8 million = 8,000,000
As 800,000 < 8,000,000
⇒ 800 thousand < 8 million

(iv) Since 1 billion = 100 crore, 60 billion = 60 × 100 crores = 6,000 crores
Clearly, 640 crore < 6,000 crore
⇒ 640 crore < 60 billion

Question 4.
Find quick ways to calculate these products:
(i) 2 × 1768 × 50
(ii) 72 × 125
(iii) 125 × 40 × 8 × 25
Solution:
(i) 2 × 1768 × 50 = 2 × 1768 × \(\frac{100}{2}\) = 1768 × 100 = 1,76,800
(ii) 72 × 125 = 72 × \(\frac{1000}{8}\) = 9 × 1000 = 9,000
(iii) 125 × 40 × 8 × 25 = \(\frac{1000}{8}\) × 40 × 8 × \(\frac{100}{4}\) = 1000 × 5 × 2 × 100 = 10,00,000

Question 5.
Using all digits from 0-9 exactly once (the first cannot be 0) to create a 10-digit number, write the —
(i) Largest multiple of 5
(ii) Smallest even number
Solution:
(i) Arranging digits in descending order, we get 9, 8, 7, 6, 5, 4, 3, 2, 1, 0.
All multiple of 5 can end only in 5 or 0.
Hence, the largest multiple of 5 is 9876543210.

(ii) Arranging digits in ascending order, we get 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Since the number cannot start with 0, the smallest number = 1023456789
We know that an even number has either 2, 4, 6 or 8 at its ones place.
Thus, swapping the last two digits, we get the smallest even number = 1023456798

Question 6.
The number 10,30,285 in words is “Ten lakh thirty thousand two hundred eighty five”, which has 41 letters. Give a 7-digit number which has the maximum number of letters.
Solution:
We use 7 (seven) and 8 (eight) to make such a number since both numbers contain five letters when written in words.

One such 7-digit number is 77,77,777 (Seventy seven lakh seventy seven thousand seven hundred seventy seven).

This has 60 letters, making it one of the 7-digit numbers having maximum number of letters.

Question 7.
Write a 9-digit number where exchanging any two digits results in a bigger number. How many such numbers exist?
Solution:
To ensure that on exchanging any two digits increases the value of a 9-digit number, the digits must increase from left to right. So, the arrangement would be: 123456789.
There is only 1 number that satisfies the given condition.

Question 8.
Strike out 10 digits from the number 12345123451234512345 so that the remaining number is as large as possible.
Solution:
Given, 12345123451234512345
We have 20 digits and need to remove 10 digits, leaving us with 10 digits.
To maximise the resulting number we want the leftmost digit to be as large as possible. Looking at the original number 12345123451234512345, the first few digits are small. We can strike out the initial T234’ to get the larger digit in the first position, i.e. 5.

Now, our numbers starts with 5 as we have struck out 4 digits so far and we need to strike out 6 more.

The remaining number is 5123451234512345. We want the next digit to be as large as possible. So, we strike out ‘1234’ following the 5, leaving us with 551234512345.
We have now struck out 4 + 4 = 8 digits.
We need to strike out 2 metre digits from 551234512345.
To keep the number as large as possible we should strike out ‘ I ’ and ‘2’.

Therefore, by striking out the digits 1234, then 1234, then 1 and 2, the number is 5534512345, which is the largest possible number.

Question 9.
The words ‘zero’ and ‘one’ share letters ‘e’ and ‘o’. The words ‘one’ and ‘two’ share a letter ‘o’, and the words ‘two’ and ‘three’ also share a letter ‘t’. How far do you have to count to find two consecutive numbers which do not share an English letter in common?
Solution:
The problem involves finding two consecutive numbers whose English names share no common letters.
Here, words zero and one share e and o. Words one (1) and two (2) share o.
Words two (2) and three (3) share t. Words three (3) and four (4) share r.
Words four (4) and five (5) share/. Words five (5) and six (6) share i.
Words six (6) and seven (7) share s. Words seven (7) and eight (8) share e.
Words eight (8) and nine (9) share i and e.
Words nine (9) and ten (10) share n and e.
.
.
.
.
Words nineteen (19) and twenty (20) share t, e, n and so on.
It shows that all consecutive numbers have atleast one common letter. Hence, their is no such pair of consecutive numbers that do not share an English letter in common.

Question 10.
A calculator has only ‘+ 10,000’ and ‘+ 100’ buttons. Write an expression describing the number of button clicks to be made for the following numbers:
(i) 20,800
(ii) 92,100
(iii) 1,20,500
(iv) 65,30,000
(v) 70,25,700
Solution:
(i) 20,800 = (2 × 10,000) + (8 × 100)
Number of clicks = 2 + 8 = 10 clicks

(ii) 92,100 = (9 × 10,000) + (21 × 100)
Number of clicks = 9 + 21 =30 clicks

(iii) 1,20,500 = (12 × 10,000) + (5 × 100)
Number of clicks = 12 + 5 = 17 clicks

(iv) 65,30,000 = (653 × 10,000) + (0 × 100)
Number of clicks = 653 + 0 = 653 clicks

(v) 70,25,700 = (702 × 10,000) + (57 × 100)
Number of clicks = 702 + 57 = 759 clicks

Question 11.
You are given two sets of number cards numbered from 1-9. Place a number card in each box below to get the (i) largest possible sum (ii) smallest possible difference of the two resulting numbers.

Solution:
(i) Since each card numbered 1 – 9 is to be placed in the boxes such that no box is empty, the cards cannot be repeated.
To get the largest possible sum, both the 7-digit and 5-digit numbers need to be the largest.
Largest 7-digit number = 98,76,543; Largest 5-digit number = 98,765
Largest possible sum = 98,76,543 + 98,765 = 99,75,308

(ii) To get the smallest possible difference, the 7-digit number needs to be the smallest, and the 5-digit number needs to be the largest.
Smallest 7-digit number = 12,34,567
Largest 5-digit number = 98,765
Smallest possible difference = 12,34,567 – 98,765 = 11,35,802

Question 12.
A bar-tailed godwit holds the record for the longest recorded non-stop flight. It travelled 13,560 km from Alaska to Australia without stopping. Its journey started on 13 October 2022 and continued for about 11 days. Find out the approximate distance it covered every day. Find out the approximate distance it covered every hour.
Solution:
Given, total distance = 13,560 km and duration = 11 days
Distance covered everyday = 13,560 ÷ 11 = 1,232.72 km
Thus, the godwit covers approximately 1,233 km per day.
We know that one day has 24 hours.
Thus, distance covered every hour = 1,233 ÷ 24 = 51.375 km
Hence, the godwit covers approximately 51 km per

InText Questions

Question 1.
Observe the pattern and fill in the boxes given below.

Solution:

Question 2.
What if a person ate 3 varieties of rice every day, will he be able to taste all the lakh varieties in a 100 year lifetime? Find out.
Solution:
With 3 varieties of rice every day, he can taste 365 × 3 = 1095 varieties in a year.
To taste 1 lakh varieties, he would need 1,0,000 ÷ 1095 ≈ 91 years.
Hence, he would be able to eat all 1 lakh varieties of rice in 100 years.

Question 3.
Two of the many different ways to get 5072 are shown below:

These two ways can be expressed as:
(a) (50 × 100) + (7 × 10) + (2 × 1) = 5072
(b) (3 × 1000) + (20 × 100) + (72 × 1) = 5072
Find a different way to get 5072 and write an expression for the same.
Solution:

Expression: (5 × 1000) + (0 × 100) + (7 × 10) + (2 × 1) = 5000 + 70 + 2 = 50724.

Question 4.
The estimated population of Chintamani in the year 2024 is 1,06,000. How much more than one lakh is 1,06,000?
Solution:
Given, the estimated population of Chintamani in the year 2024 is 1,06,000.
∴ Required difference = 1,06,000 – 1,00,000
= 6,000
Thus, the population of Chintamani in 2024 is 6,000 more than one lakh.

Question 5.
The Thoughtful Thousands only has a + 1000 button. How many times should it be pressed to show:

(i) Three thousand? 3 times
(ii) 10,000? ______
(iii) Fifty-three thousand? _________
(iv) 90,000? __________
(v) One Lakh? ________
(vi) ________? 153 times
(vii) How many thousands are required to make one lakh?
Solution:
(i) \(\frac{3000}{1000}\) = 3 ⇒ 3 times
(ii) \(\frac{10000}{1000}\) = 10 ⇒ 10 times
(iii) \(\frac{53000}{1000}\) = 53 ⇒ 53 times
(iv) \(\frac{90000}{1000}\) = 90 ⇒ 90 times
(v) \(\frac{100000}{1000}\) = 100 ⇒ 100 times
(vi) 153 × 1000 = 1,53,000
(vii) 100000 ÷ 1000 = 100. Thus, 100 thousands make 1 lakh.

Question 6.
How many zeros does a hundred thousand have?
Solution:
100 thousand = 100 × 1000 = 1,00,000
Clearly it has 5 zeros.

Question 7.
With large numbers it is useful to know the nearest thousand, lakh or crore. For example, the nearest neighbours of the number 6,72,85,183 are shown in the table below.

Write the five nearest neighbours for these numbers:
(i) 3,87,69,957
(ii) 29,05,32,481
Solution:

Question 8.
Using the meaning of multiplication and division, can you explain why multiplying by 5 is the same as dividing by 2 and multiplying by 10?
Solution:
We know that 5 × 2= 10 (multiplication fact) gives
two division facts: 10 ÷ 2 = 5 and 10 ÷ 5 = 2.
So, we can use \(\frac{10}{2}\) in place of 5 . Either we multiply a number by 5 or by \(\frac{10}{2}\), we will get the same answer.

Question 9.
Can multiplying a 3-digit number with another 3-digit number give a 4-digit number?
Solution:
The product of smallest 3-digit numbers
= 100 × 100 = 10,000 (5-digit number)
And, the product of largest 3-digit numbers
= 999 × 999 = 9,98,001 (6-digit number)
So, the product of two 3-digit numbers will have either 5 or 6 digits.
Hence, a 4-digit number cannot be obtained by multiplying two 3-digit numbers.

Question 10.
Can multiplying a 4-digit number with a 2-digit number give a 5-digit number?
Solution:
The product of smallest 4-digit number and smallest 2-digit number
= 1000 × 10 = 10,000 (5-digit number)
And, the product of largest 4-digit number and largest 2-digit number
= 9999 × 99 = 9,89,901 (6-digit number)
So, the product of 4-digit number and 2-digit number will have either 5 or 6 digits.
Hence, multiplying a 4-digit number with a 2-digit number can give a 5-digit number.

Question 11.
Roxie wondered, “If I could travel 100 kilometers every day, could I reach the Moon in 10 years?” (The distance between the Earth and the Moon is 3,84,400 km.)
(i) How far would she have travelled in a year?
(ii) How far would she have travelled in 10 years?
Solution:
(i) Distance travelled by Roxie in a day = 100 km
Thus, distance travelled by Roxie in a year = 365 × 100 = 36500 km (As 1 year = 365 days)
(ii) Distance travelled by Roxie in 10 years =100 × 365 x 10 = 36500 × 10 = 365000 km
Since 365000 < 384400, Roxie cannot reach the moon in 10 years.

Large Numbers Around Us Class 7 Extra Questions

Large Numbers Around Us Class 7 Very Short Question Answer

Question 1.
How many thousands are there in 1 million?
Solution:
Place value chart in International Number System is given below:

1 million is three places to the left of 1 thousand.
Thus, 1 million = 1,000 thousand
Hence, 1,000 thousands are there in one million.

Question 2.
How many hundreds are there in 10 lakhs?
Solution:
Place value chart in Indian Number System is given below:

10 lakh is four places to the left of 1 hundred.
Thus, 10 lakh = 10,000 hundred
Hence. 10,000 hundreds are there in 10 lakhs.

Question 3.
The annual wheat production in a region is 673400000 kilograms. Place the commas according to the International Number System format.
Solution:
In the International Number System, commas are placed in a 3-3-3-3… pattern, starting from the right. This helps to separate the digits into hundreds, thousands, millions, billions, and so on.
Number: 673,400,000
Number Name: Six hundred seventy three million four hundred thousand

Question 4.
Without multiplying, predict the number of digits in the product 113 × 98.
Solution:
Rounding 113 to nearest tens we get 110 and rounding 98 to nearest tens, we get 100.
Thus, estimated product = 110 × 100 = 11000, which is a 5-digit number.
Hence, the actual product will have 5 digits.
Verification: 113 × 98 = 11,074, which is a 5-digit number

Question 5.
Compare and write ‘>’, ‘<‘ or ‘=’’:
(i) 80 thousand ______ 8 lakh
(ii) 200 lakh ______ 2 million
Solution:
(i) 80 thousand = 80 × 1,000 = 80,000 and 8 lakh = 8 × 1,00,000 = 8,00,000
As 80,000 < 8,00,000
Thus, 80 thousand < 8 lakh

(ii) We know, 1 million = 10 lakh
So, 2 million = 20 lakh < 200 lakh Thus, 200 lakh > 2 million

Question 6.
How many ten thousands are there in the smallest 6-digit number?
Solution:
The smallest 6-digit number is 1,00,000 i.e. 1 lakh.

1 lakh is one place to the left often thousands.
Thus, 1 lakh =10 ten thousands

Question 7.
How many thousands are there in 1 lakh?
Solution:

1 lakh is 2 places to the left of thousand.
Hence, 1 lakh = 100 thousands.

Large Numbers Around Us Class 7 Short Question Answer

Question 1.
Write the number names of the following numerals in the Indian and International Numbers Systems.
(i) 437065
(ii) 42181602
(iii) 636547150
(iv) 4050607080
Solution:

Question 2.
Find the place value of underlined digits in International Number System.
(i) 3241767
(ii) 98443810
Solution:
(i)

From the table, the digit 1 is at thousands place.
Thus, the place value of 1 is 1 × 1,000 = 1,000.

(ii)

From the table, the digit 8 is at thousands place.
Thus, the place value of 8 is 8 × 1,000 = 1,000.
= 8,000,000.

Question 3.
Calculate the following products:
(i) 4 × 1522 × 50
(ii) 48 × 125
(iii) 125 × 20 × 16 × 25
Solution:
(i) 4 × 1522 × 50 = 4 × 1522 × \(\frac{100}{2}\)
= 2 × 1522 × 100 = 3044 × 100 = 304400

(ii) 48 × 125 = 48 × \(\frac{1000}{8}\) = 6 × 1000 = 6000

(iii) 125 × 20 × 16 × 25 = \(\frac{1000}{8}\) × 20 ×16 × \(\frac{100}{4}\)
= 1000 × 5 × 2 × 100 = 1000 × 10 × 100 = 1000000

Question 4.
A city’s population grew from 5,67,000 to 7.42.0 in five years. Estimate the increase using appropriate rounding.
Solution:
Given, in five years the population of a city grew from 5,67,000 to 7,42,000.
Rounding off both the numbers to nearest ten thousands, we get 5,70,000 and 7,40,000 respectively.
Thus, estimated increase in population
= 7,40,000 – 5,70,000 = 1,70,000

Question 5.
A stadium had 4,73,000 visitors in 2022 and 5.85.0 in 2023. Estimate the increase in visitors by rounding to the nearest ten thousands.
Solution:
Number of visitors in 2022 = 4,73,000
Number of visitors in 2023 = 5,85,000
Rounding both the numbers to nearest ten thousands, we get
4.73.0 → 4,70,000 and 5,85,000 → 5,90,000
Thus, estimated increase in visitors
= 5,90,000 – 4,70,000 = 1,20,000.

Large Numbers Around Us Class 7 Long Question Answer

Question 1.
Round off the following to the given nearest place.
(i) 7,065; hundreds
(ii) 55,777; thousands
(iii) 46,439; ten thousands
(iv) 30,89,732; lakhs
(v) 34,75,68,328; ten crores
Solution:
(i) In 7,065, the digit at the hundreds place is 0.
As the digit at the tens place, 6 > 5, we increase 0 by 1 and we replace the remaining digits to the right of the hundreds place by 0.
Thus, on rounding off 7065 to the nearest hundreds, we get 7100.

(ii) In 55,777, the digit at the thousands place is 5.
As the digit at the hundreds place, 7 > 5, we increase 5 by 1 and we replace the remaining digits to the right of the thousands place by 0.
Thus, on rounding off 55,777 to the nearest thousands, we get 56,000.

(iii) In 46,439, the digit at the ten thousands place is 4.
As the digit at the thousands place, 6 > 5, we increase 4 by 1 and we replace the remaining digits to the right of the ten thousands place by 0.
Thus, on rounding off 46,439 to the nearest ten thousands, we get 50,000.

(iv) In 30,89,732, the digit at the lakhs place is 0.
As the digit at the ten thousands place, 8 > 5, we increase 0 by 1 and we replace the remaining digits to the right of the lakhs place by 0.
Thus, on rounding off 30,89,732 to the nearest lakhs, we get 31,00,000.

(v) In 34,75,68,328, the digit at the ten crores place is 3.
As the digit at the crore place, 4 < 5, 3 remains unchanged and we replace the remaining digits to the right of the ten crores place by 0.
Thus, on rounding off 34,75,68,328 to the nearest ten crores, we get 30,00,00,000.

Question 2.
A clothing store earned ₹6,58,900 in January and ₹ 7,21,600 in February. Estimate the total revenue by rounding each figure to the nearest lakhs. Is your estimated total greater or smaller than the exact total?
Solution:
Revenue earned by clothing store in January = ₹6,58,900
Revenue earned by clothing store in February = ₹ 7,21,600
Rounding the numbers to nearest lakhs, we get
6,58,900 → 7,00,000
[∵ At ten thousands place, 5 = 5]
7,21,600 → 7,00,000
[∵ At ten thousands place, 2 < 5]
Now, estimated total revenue over the two months = ₹ 7,00,000 + ₹ 7,00,000 = ₹ 14,00,000
Actual total revenue = ₹6,58,900 + ₹7,21,600
= ₹ 13,80,500
Thus, the estimated total revenue for the given two months is greater than the actual revenue.

Question 3.
Say True or False.
(i) 6,78,456 is rounded off as 6,80,000 to the nearest ten thousands.
(ii) 52,25,067 is rounded off as 52,00,000 to the nearest thousands.
(iii) 2,31,48,765 is rounded off as 2,40,00,000 to the nearest ten lakhs.
Solution:
We know that, while rounding the number nearest to given place,
If the digit on the right of the given place is 5 or greater than 5, we increase the digit at that place by 1.
If the digit on the right of the given place is less than 5, we keep the digit at that place same.

(i) True
In 6,78,456, the digit at the ten thousands place is 7.
The digit to the right of 7, i.e. the digit at the thousands place is 8 and 8 > 5.
Thus, 6,78,456 is rounded off as 6,80,000 to the nearest ten thousands.

(ii) False
In 52,25,067, the digit at the thousands place is 5.
The digit to the right of 5, i.e. the digit at the hundreds place is 0 and 0 < 5.
So, 5 remains unchanged.
Thus, 52,25,067 is rounded off as 52,25,000 to the nearest thousands.

(iii) False
In 2,31,48,765, the digit at the ten lakhs place is 3.
The digit to the right of 3, i.e. the digit at the lakhs place is 1 and 1 < 5. So, 3 remains unchanged.
Thus, 2,31,48,765 is rounded off as 2,30,00,000 to the nearest ten lakhs.

Large Numbers Around Us Class 7 Case Based Questions

Question 1.
The population details of a town were published in a report.

The report is also translated for international agencies.
Based on the above information, answer the
following questions:
(i) Create the report for the international agencies using the International Number System.
(ii) What is the face value and place value of digit 3 in the population of children?
(iii) Write the number name for the population of senior citizens in the International Number System.
Solution:
i) The report in International Number System will be:

(ii) Given, the population of children = 12,35,678, which is represented in Indian Number System.

The face value of 3 is 3.
Clearly, digit 3 is at the ten-thousands place.
∴ The place value of 3 is 3 × 10,000 = 30,000.

(iii) In International Number System, population of senior citizens = 560,328
Number name: Five hundred sixty thousand three hundred twenty eight

Question 2.
The principal of Sunrise Public School is preparing a budget for renovating the school. She has received the cost from various departments:
Painting classrooms: ₹4,83,760
Replacing furniture: ₹3,27,450
Electrical work: ₹ 1,65,890
Bathroom renovations: ₹2,49,300
To present a simplified version in a meeting, she rounds off all values to the nearest lakhs.
Based on the above information, answer the following questions:
(i) What is the rounded cost of each item to the nearest lakhs?
(ii) What is the total estimated cost using the rounded values?
(iii) How much is the difference between the estimated and the total actual cost?
Solution:
(i) Rounding off all the costs to the nearest lakhs, we get

(ii) The total estimated cost is the sum of the estimated cost of each item.
Thus, total estimated cost = ₹5,00,000 + ₹ 3,00,000 + ₹ 2,00,000 + ₹ 2,00,000
= ₹ 12,00,000

(iii) The total estimated cost = ₹ 12,00,000
Now, total actual cost
= ₹4,83,760 + ₹3,27,450 + ₹ 1,65,890 + ₹2,49,300
= ₹ 12,26,400
Thus, difference between total actual cost and total estimated cost
= ₹ 12,26,400 – 12,00,000 = ₹26,400

Odisha State Board BSE Odisha 7th Class Science Solutions Chapter 17 ଆବର୍ଜନାର ପରିଚାଳନା Textbook Exercise Questions and Answers.

BSE Odisha Class 7 Science Solutions Chapter 17 ଆବର୍ଜନାର ପରିଚାଳନା

Question 1.
ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର ।
(କ) ଜଳରେ ଦୂଷିତ ପଦାର୍ଥ ମିଶିଥିଲେ ତା’କୁ ________ ଜଳ କୁହାଯାଏ ।
(ଖ) ଦୂଷିତ ଜଳକୁ ବିଶୁଦ୍ଧ କରିବାପାଇଁ ________ ଓ ________ ରାସାୟନିକ ପଦାର୍ଥ ବ୍ୟବହାର କରାଯାଏ ।
(ଗ) ହଇଜା ________ ଓ ________ ଜଳ ବାହିତ ରୋଗ କୁହାଯାଏ ।
(ଙ) ନଳାଗୁଡ଼ିକ ________ ଓ ________ ପାଇଁ ଜାମ୍ ହୋଇଯାଏ ।
(ଚ) ଘରୁ ବାହାରୁଥ‌ିବା ଦୂଷିତ ଜଳକୁ ________ କୁହାଯାଏ ।
(ଛ) ଦୂଷିତ ଜଳର ଉପଚାର ________ ରେ କରାଯାଏ ।
Solution:
(କ) ଅପଚୟିତ
(ଖ) ଚୂନ, ବ୍ଲିଚିଂପାଉଡର
(ଗ) କାମଳ, ଟାଇଫଏଡ୍
(ଙ) ପଲିଥ୍, ତୁଳା
(ଚ) ସୁଏଜ୍ ବା ନାଳ ନର୍ଦ୍ଦମା ଆବର୍ଜନା
(ଛ) ଦୂଷିତଜଳ ଉପଚାର ପ୍ଲାଣ୍ଟ

Question 2.
କାରଣ ଲେଖ :
(କ) ନଳାରେ ରନ୍ଧାତେଲ କିମ୍ବା ପୋଡ଼ା ମୋବିଲ୍ ଆଦି ପକାଇବା ଉଚିତ୍ ନୁହେଁ ।
Solution:

• ରନ୍ଧା ତେଲ କିମ୍ବା ପୋଡ଼ା ମୋବିଲ୍ ମାଟିର ଜଳ ଶୋଷଣରେ ବାଧା ସୃଷ୍ଟିକରେ ।
• ଏହା ମାଟିର ଉପର ସ୍ତରରେ ଲାଗି ରହି ଜଳକୁ ଶୋଷଣ କରିବାକୁ ନଦେବାରୁ ଜଳ ଜମିରହେ ଓ ଉପକାରୀ ଅଣୁଜୀବ ମରିଯାନ୍ତି । ଏଣୁ ନଳାରେ ରନ୍ଧାତେଲ କିମ୍ବା ପୋଡ଼ା ମୋବିଲ ପକାଇବା ଉଚିତ ନୁହେଁ ।

(ଖ) କଠିନ ଆବର୍ଜନା ଯଥା ତୁଳା, କଣ୍ଢେଇ, ପ୍ଲାଷ୍ଟିକ, ଜରି ଇତ୍ୟାଦି ନଳାରେ ନ ପକାଇ ଡଷ୍ଟବିନ୍‌ରେ ପକାଇବା ଉଚିତ ।
Solution:
ତୁଳା, କଣ୍ଢେଇ, ପ୍ଲାଷ୍ଟିକ, ଜରି ଇତ୍ୟାଦି ନଳାରେ ପକାଇଲେ ନଳା ବନ୍ଦ ବା ଜାମ୍ ହୋଇଯାଏ । ଫଳରେ ଜଳ ନିଷ୍କାସନରେ ବାଧା ସୃଷ୍ଟି ହୁଏ । ଏଣୁ ଏଗୁଡ଼ିକୁ ନଳାରେ ନ ପକାଇ ଡଷ୍ଟବିନ୍‌ରେ ପକାଇବା ଉଚିତ ।

(ଗ) ଦୂଷିତ ଜଳ ପ୍ରବାହିତ ସ୍ଥାନରେ ଇଉକାଲିପଟାସ୍ ଗଛ ଲଗାଇବା ଆବଶ୍ୟକ ।
Solution:
ଇଉକାଲିପଟାସ୍ ଗଛ ଅତି ଶୀଘ୍ର ଦୂଷିତ ଜଳ ଶୋଷଣ କରିପାରେ ଏବଂ ଉଦ୍ବେଦନ ପ୍ରକ୍ରିୟାରେ ପରିଷ୍କାର ବାଷ୍ପ ବାୟୁମଣ୍ଡଳକୁ ଛାଡ଼ିଥାଏ । ଫଳରେ ପାଣି ଜମି ରହେ ନାହିଁ ଓ ପରିବେଶକୁ ପ୍ରଦୂଷିତ ହେବାରୁ ରକ୍ଷା କରେ ।

(ପ) ଦୂଷିତ ଜଳରେ ଉପଚାର ସମୟରେ ପ୍ରଥମେ ବାରସ୍କ୍ରିନ୍ ଦେଲ ଦିଆଯାଏ ।
Solution:
ଦୂଷିତ ଜଳରେ ଉପଚାର ସମୟରେ ଏହାକୁ ପ୍ରଥମେ ବାରସ୍କ୍ରିନ୍ ମଧ୍ୟଦେଇ ପ୍ରବାହିତ ହେବାକୁ ଦିଆଯାଏ । କାରଣ ଏହାଦ୍ଵାରା ସେଥ‌ିରେ ଥ‌ିବା କାଠି, କୁଟା, ପ୍ଲାଷ୍ଟିକ, ଜରି, ତୁଳା ଚିରାକନା ଇତ୍ୟାଦି କଠିନ ଭାସମାନ ଆବର୍ଜଣା ଛାଣି ହୋଇଯାଏ ।

(ଙ) ଖୋଲା ପଡ଼ିଆରେ ଝାଡ଼ା ପରିସ୍ରା କରିବା ମହାମାରୀର କାରଣ ।

• ଖୋଲା ପଡ଼ିଆରେ ମଳତ୍ୟାଗ କଲେ ତାହା ବର୍ଷା ପାଣିରେ ଧୋଇହୋଇ ନଦୀ, ପୋଖରୀ ଓ କେନାଲ ପାଣିରେ ମିଶେ ।
• ସେଥ‌ିରେ ନାନା ପ୍ରକାର ରୋଗଜୀବାଣୁ, କୃମି, କୃମିର ଅଣ୍ଡା, ନାନା ବିଷାକ୍ତ ପଦାର୍ଥ ଜଳକୁ ଦୂଷିତ କରେ ।
• ଏହି ଦୂଷିତ ଜଳକୁ ପିଇଲେ ଏବଂ ସେଥ‌ିରେ ଗାଧୋଇଲେ ଆମେ ହଇଜା, ଆମ୍ବିକଜ୍ଵର, କାମଳ ଓ ଚର୍ମରୋଗରେ ଆକ୍ରାନ୍ତ ହୋଇଥାଉ । ଏହି ଦୂଷିତ ଜଳର ଉପଯୁକ୍ତ ନିଷ୍କାସନ ଓ ବିଶୋଧନ ନହେଲେ ଚାରିଆଡ଼େ ମହାମାରୀ

(କ) ପରିଷ୍କାର ପରିଚ୍ଛନ୍ନତା ଓ ରୋଗ ମଧ୍ୟରେ ସମ୍ପର୍କ କ’ଣ ?
Solution:

• ଅସ୍ୱାସ୍ଥ୍ୟକର ପରିବେଶରେ ବସବାସ କରିବା ଓ ଦୂଷିତ ଜଳ ବ୍ୟବହାର କରିବା ସମସ୍ତ ରୋଗର ପ୍ରଧାନ କାରଣ ର୍ଥଟେ ।
• ଏଣେତେଣେ ମଳତ୍ୟାଗ କଲେ ଯାହାଦ୍ୱାରା ବାୟୁ ଉପର ଓ ତଳ ମାଟିର ଜଳ ଦୂଷିତ ହୋଇଥାଏ । ଫଳରେ ଆମକୁ ହଇଜା, ଆମାଶୟ ଆଦି ଜଳବାହିତ ରୋଗ ସହଜରେ ବ୍ୟାପିଥାଏ ।
• ଖାଲି ପାଦରେ ଝାଡ଼ାଗଲେ ମଳରେ ଥିବା କୃମି, ପାଦର ଚର୍ମଦେଇ ଶରୀର ଭିତରକୁ ପ୍ରବେଶ କରେ । ଏଣୁ ଚପଲ ପିନ୍ଧି ମଳତ୍ୟାଗ କରିବାକୁ ଯିବା ଆବଶ୍ୟକ । ମଳ ତ୍ୟାଗକରି ସାବୁନରେ ହାତ ଧୋଇବା ଓ ନିୟମିତ ଜଳ ପାନ କଲେ ରୋଗରୁ ରକ୍ଷା ମିଳେ ।

(ଖ) ଜଣେ ନାଗରିକ ହିସାବରେ ପରିମଳ ବ୍ୟବସ୍ଥା ପାଇଁ ତୁମର କର୍ତ୍ତବ୍ୟ କ’ଣ ?
Solution:

• ଜଣେ ସଚେତନ ଓ ଉତ୍ତମ ନାଗରିକ ହିସାବରେ ପରିମଳ ବ୍ୟବସ୍ଥା ପାଇଁ ପୌର ପରିଷଦର କର୍ମକର୍ତ୍ତାଙ୍କୁ ଖବର ଦେଇ ନଳା, ରାସ୍ତା ଓ ସମ୍ପୂର୍ଣ୍ଣ ପରିବେଶକୁ ପରିଷ୍କାର ଓ ବିଶୋଧନ କରିବାପାଇଁ ଉପଯୁକ୍ତ ପଦକ୍ଷେପ ନେବା ଆବଶ୍ୟକ । କରାଇବା ଆବଶ୍ୟକ ।
• ଗ୍ରାମ ପଞ୍ଚାୟତରେ ସରପଞ୍ଚଙ୍କୁ ଅନୁରୋଧ କରି ଖୋଲା ନଳାଗୁଡ଼ିକୁ ଘୋଡ଼ଣୀ ଦେଇ ବନ୍ଦ ମଶା, ମାଛି ଏବଂ ଅନ୍ୟାନ୍ୟ ଜୀବାଣୁମାନେ ସେଠାରେ ବଂଶବିସ୍ତାର କରନ୍ତି ।

(ଗ) ଦୂଷିତ ଜଳର ପୁନଃ ବ୍ୟବହାର ପାଇଁ ତୁମେ ଦିଦ୍ୟାଳୟରେ କ’ଣ କରିପାରିବ ?
Solution:
ଦୂଷିତ ଜଳର ପୁନଃ ବ୍ୟବହାର ପାଇଁ ବିଦ୍ୟାଳୟରେ ବ୍ୟବହୃତ ଜଳକୁ ନାଳି କାଟି ବଗିଚାରେ ଥ‌ିବା ଫଳ ଫୁଲ ଗଛ ମୂଳକୁ ଛାଡ଼ି ଦେଇପାରିବୁ କିମ୍ବା ଅନ୍ୟ ଆଡ଼କୁ ନଛାଡ଼ି ସୋକ୍‌ପିଟ୍‌ରେ ନିର୍ଦ୍ଦିଷ୍ଟ ଜାଗାକୁ ଛାଡ଼ିପାରିବୁ ନଚେତ୍ ଅଧିକ ଜଳକୁ ମୁଖ୍ୟ ନାଳ ସହ ସଂଯୋଗ କରି ଛାଡ଼ି ପାରିବୁ ।

(ଘ) ବିଦ୍ୟାଳୟ ପରିବେଶରେ ଶ୍ରେଣୀ କୋଠରିରୁ ବାହାରୁଥ‌ିବା ଆବର୍ଜନାର ନିଷ୍କାସନ ପାଇଁ କ’ଣ କରିବ ?
Solution:

• ବିଦ୍ୟାଳୟ ପରିବେଶରେ ଶ୍ରେଣୀ କୋଠରିରୁ ବାହାରୁଥିବା କଠିନ ଆବର୍ଜନାର ନିଷ୍କାସନ ପାଇଁ ଦୁଇ ପ୍ରକାର ଡଷ୍ଟବିନ୍ ପ୍ରତି ଶ୍ରେଣୀରେ ରହିବା ଆବଶ୍ୟକ ।
• ଗୋଟିଏ ସବୁଲ ରଲର ଯେଉଁଥିରେ ଲେବ ନିମ୍ନାକାରଣ ଯୋଗ୍ୟ ଆଦଲ୍ୟା ଆତ୍ ମାଟିରେ ମିଶିଯାଉଥିବା ଆବର୍ଜନା ରଖାଯିବ ଓ ଅନ୍ୟଟିରେ ଜୈବ-ନିମ୍ନକରଣ ଅଯୋଗ୍ୟ ଆବର୍ଜନା ରଖାଯିବ ।
• ଏହି ସବୁଜରଙ୍ଗର ଆବର୍ଜନାକୁ ବଡ଼ ବଡ଼ ଗାତ ମଧ୍ଯରେ ପକାଇ ପଚନ ପଦ୍ଧତିରେ ଏହାକୁ କମ୍ପୋଷ୍ଟରେ ପରିଣତ କରାଯାଇପାରିବ ଏବଂ ଏହି ସାରକୁ ଫଲ ଫୁଲ ଗଛ୍ଠିରେ ଦେବତାର ବ୍ୟବହୃତ ହୋଇପାରିବ |
• ଲାଲ ରଙ୍ଗର ଆବର୍ଜନାକୁ ଅର୍ଥାତ୍ କାଚ, ପ୍ଲାଷ୍ଟିକ, ପଲିଥ୍, ଟିଣ ଓ ଲୁହା ଆଦିକୁ ବିଭିନ୍ନ କଳକାରଖାନାକୁ ପଠାଳ ପୁନଚକୃଣ ପଦତିଦ୍ୱାରା ନୂତନ ଜିନିଷ ପୃସ୍ତୁତ କରାଯାଲପାରିବ |

(ଙ) ତୁମେ ବିଦ୍ୟାଳୟର ଶୌଚାଳୟ/ପାଇଖାନାର ସଦୁପଯୋଗ ପାଇଁ କ’ଣ ସବୁ କରିବ ?
Solution:

• ବିଦ୍ୟାଳୟର ପାଇଖାନାର ସଦୁପଯୋଗ ପାଇଁ ଆମେ ପୃତି ବିଦ୍ୟାଳୟରେ ପୁଅ ଟିଥମାନକ ପାଳ ଥିଲଗା ପାଇଖାନା ଓ ପରିସ୍ରାଗାର ରଖୁବା ।
• ଚପଲ ପିନ୍ଧି ପାଇଖାନାକୁ ଯିବାକୁ ପିଲାଙ୍କୁ ଶିଖାଇବା ।
• ମଳତ୍ୟାଗ କରି ସାବୁନ୍‌ରେ ହାତଧୋଇବା ଅଭ୍ୟାସ କରାଇବା । ମଳତ୍ୟାଗ ପୂର୍ବରୁ ଓ ପରେ ପାଇଖାନାରେ ପାଣି ଢାଳିବାକୁ କହିବା ।
• ନିୟମିତ ବ୍ୟବଧାନରେ ପାଇଖାନାକୁ ପରିଷ୍କାର କରିବା ଉଚିତ ।
• ପାଇଖାନାରେ ବାଲ୍ଟି, ମଗ୍ , ସାବୁନ ଓ ପାଣି ରଖୁବା ନିହାତି ଆବଶ୍ୟକ । ପରିବେଶକୁ ସୁସ୍ଥ ଓ ପରିଷ୍କାର ରଖୁବା ଆମ ସମସ୍ତଙ୍କର କର୍ତ୍ତବ୍ୟ ।

Question 3.
ସଂକ୍ଷିପ୍ତ ଟିପ୍‌ପଣୀ ଲେଖ :
(i) ଦୂଷିତ ଜଳ :
Solution:
(a) ତରଳ ଆବର୍ଜନାର କାରକ ଗୁଡ଼ିକ ଜଳସହିତ ମିଶି ଏହାକୁ ଦୂଷିତ କରିଥାଏ ।
(b) ବିଭିନ୍ନ କାରଣରୁ ଜଳ ପ୍ରଦୂଷିତ ହୋଇଥାଏ; ଯଥା – ଶିଳ୍ପଜନିତ, କୃଷି ସମ୍ପର୍କିତ ଓ ଗୃହକାର୍ଯ୍ୟ ଜନିତ ଆଦି । ଏହି ଆବର୍ଜନା ମିଶା ଦୂଷିତ ଜଳକୁ ଅପଚୟିତ ଜଳ କୁହାଯାଏ ।
(c) ଅପଚୟିତ ଜଳରେ କେତେକ ଦୂଷିତ ପଦାର୍ଥ ଦ୍ରବୀଭୂତ ହୋଇଥାଏ ଓ କେତେକ ଭାସୁଥାଏ । ସେଥ୍ରେ କିଛି ହାନିକାରକ ଜୈବିକ ଓ ଅଜୈବିକ ପଦାର୍ଥ ରହିଥାଏ । ରୋଗ ସୃଷ୍ଟି କରୁଥିବା କେତେକ ବୀଜାଣୁ ଏବଂ ଅନ୍ୟାନ୍ୟ
(d) ଦୂଷିତ ଜଳ ପାନକଲେ ଏବଂ ସେଥ‌ିରେ ଗାଧୋଇଲେ ଆମେ ହଇଜା, ଆମାଶୟ, କାମଳ, ଟାଇଫଏଡ୍, କାଛୁକୁଣ୍ଡିଆ ଓ ଯାଦୁ ଆଦି ଜଳବାହିତ ରୋଗରେ ଆକ୍ରାନ୍ତ ହୋଇଥାଉ ।

(ii) ଜୈବିକ ଗ୍ୟାସ :
Solution:
(a) ମନୁଷ୍ୟ ଓ ଗାଈଗୋରୁଙ୍କ ମଳରୁ ବାହାରୁଥିବା ଗ୍ୟାସ୍‌ ଜୈବିକ ଗ୍ୟାସ୍ କୁହାଯାଏ ।
(b) ବାୟୋଗ୍ୟାସ୍ ପ୍ଲାଣ୍ଟରେ ବୀଜାଣୁମାନଙ୍କ ଦ୍ବାରା ମଳ ଅପଘଟିତ ହୋଇ ସେଥୁରୁ ଗ୍ୟାସ୍ ବାହାରେ । ଏହି ଗ୍ୟାସ୍ ରନ୍ଧନ କାର୍ଯ୍ୟରେ ବ୍ୟବହୃତ ହୁଏ ।
(c) ଏହି ଗ୍ୟାସ ପ୍ଲାଣ୍ଟରୁ ଗ୍ୟାସ୍‌ ସରିଲା ପରେ ଯାହା ଅବଶିଷ୍ଟ ରହିଥାଏ, ତାହା ଶସ୍ୟକ୍ଷେତ୍ର ପାଇଁ ଉତ୍ତମ ସାର ରୂପେ ବ୍ୟବହାର କରାଯାଏ ।
(d) ଏ ଦ୍ବିବିଧ ଲାଭ ପାଇଁ ଏବେ ଚୀନ ଓ ଭାରତର ଗାଁମାନଙ୍କରେ ଏହିପରି ଜୈବିକ ଗ୍ୟାସ୍ ପ୍ଲାଣ୍ଟ୍ ହଜାର ହଜାର ସଂଖ୍ୟାରେ ସ୍ଥାପନ କରିବା ପାଇଁ ଚେଷ୍ଟା ଚାଲିଛି ।

(iii) କଲାଭ ଉପଚାର :
Solution:
(a) ଜୈବିକ, ଭୌତିକ ଓ ରାସାୟନିକ ପ୍ରକ୍ରିୟାରେ ଦୂଷିତ ଜଳକୁ ବିଭିନ୍ନ ଉପଚାର କରାଯାଇ ସେଥ‌ିରେ ଥ‌ିବା ଅଦରକାରୀ ଭୌତିକ, ରାସାୟନିକ ଓ ଜୈବିକ ପଦାର୍ଥଗୁଡ଼ିକର ପରିବର୍ତ୍ତନ କରାଯାଇଥାଏ ।
(b) ପ୍ରଥମେ ଦୂଷିତ ଜଳ ବାର୍‌ସ୍କ୍ରିନ୍ ମଧ୍ୟଦେଇ ପ୍ରବାହିତ ହେବା ସମୟରେ ଏଥ‌ିରେ ଥ‌ିବା କାଠି, କୁଟା, ଜରି, ତୁଳା ଆଦି ଭାସମାନ ପଦାର୍ଥ ଛାଣି ହୋଇଯାଏ ।
(c) ଏହାପରେ ଜଳ ଆଉ ଏକ ଟ୍ୟାକୁ ଯାଏ, ଯେଉଁଠାରେ ଜଳରେ ଥିବା ବାଲି, ଗୋଡ଼ି ସବୁ ତଳେ ବସିଯାଏ । ଏହାପରେ ଜଳ ଗୋଟେ ବଡ଼ ଟାଙ୍କିରେ ରହେ ।
(d) ଏଠାରେ ବାଲି ଓ ଗୋଡ଼ି ତଳକୁ ରହିଯାଏ ଓ ଗୋଟେ କୋରଣା ଦ୍ଵାରା ବାହାରିଯାଏ । ଏଠାରେ ଚିକ୍କଣ ଅଠାଳିଆ ପଙ୍କୁଆ ପଦାର୍ଥକୁ ପଙ୍କ (Sludge) କୁହାଯାଏ ।
(e) ଜଳରେ ଭାସୁଥିବା କଠିନ ପଦାର୍ଥ, ଗାଡ଼ିରୁ ବାହାରି ଥ‌ିବା ତେଲ ଆଦିକୁ ଅଲଗା କରି ସ୍କିମର (Skimmer) ବାହାର କରିଦିଏ । ତେଲ ମିଶା କାଦୁଅକୁ ଅନ୍ୟ ଗୋଟିଏ ଟାଙ୍କିକୁ ନିଆଯାଏ । ସେଠାରେ ଥିବା ବିଜାଣୁମାନଙ୍କ ଦ୍ୱାରା ଏହା ଅପଘଟିତ କରାଯାଏ ।
(f) ଏଠାରେ ବାୟୁ ପ୍ରବେଶ କରାଯାଇ ବୀଜାଣୁଗୁଡ଼ିକୁ ବଢ଼ିବାରେ ସାହାଯ୍ୟ କରାଯାଏ । ଏମାନେ ଜଳରେ ରହିଥ‌ିବା ମଳ, ପଚାଫଳ, ସାବୁନ ଓ ଅନ୍ୟାନ୍ୟ ଅଦରକାରୀ ଜୈବିକ ପଦାର୍ଥଗୁଡ଼ିକୁ ଅପଘଟନ କରିଦିଅନ୍ତି ।
(g) କିଛି ସମୟ ପରେ ଜଳରେ ଭାସୁଥ‌ିବା ଅଣୁଜୀବଗୁଡ଼ିକ ତଳେ ବସି ଯାଆନ୍ତି । ଏହାପରେ ଉପରୁ ଜଳକୁ ବାହାର କରିଦିଆଯାଏ । ତଳେ ବସିଥିବା ମାଟି କାଦୁଅକୁ ସାର ରୂପେ ବ୍ୟବହାର କରାଯାଏ ଓ ପରିଷ୍କାର ଜଳକୁ କ୍ଲୋରିନ୍ କିମ୍ବା ଓଜୋନ୍‌ ଦ୍ବାରା ବ୍ୟବହାର ଯୋଗ୍ୟ କରାଯାଏ ।

(iv) ଖୋଲାନାଳର ଅପକାରିତା :
Solution:

• ବର୍ଷାଦିନେ ଖୋଲାଥିବା ନଳାରେ ଅର୍ଧ ପାଣି ପଶିବା ଦ୍ବାରା ନଳାର ଆବର୍ଜନା ସବୁ ରାସ୍ତା ଉପରେ ଜମା ହୋଇ ଏକ ଅସ୍ବାସ୍ଥ୍ୟକର ପରିବେଶ ସୃଷ୍ଟି କରେ ।
• ରାସ୍ତାରେ ଚାଲିବା କଷ୍ଟକର ହୋଇପଡ଼େ । କାରଣ ମଶା, ମାଛି ଏବଂ ଅନ୍ୟାନ୍ୟ ଜୀବାଣୁ ଏଠାରେ ବଂଶବିସ୍ତାର

Question 4.
ଅପଚୟିତ ଜଳରେ ଥ‌ିବା ବିଭିନ୍ନ ଉପାଦାନରୁ ଦୁଇଟି ଲେଖାଏଁ ଉଦାହରଣ ଲେଖ ।

Question 5.
ପଲିଥନ୍ ଓ ପ୍ଲାଷ୍ଟିକ ନିର୍ମିତ ପଦାର୍ଥର ବ୍ୟବହାର ପରେ ଏଣେତେଣେ ନ ଫିଙ୍ଗି ପୋଡ଼ିବା ଦ୍ବାରା ପରିବେଶର କ’ଣ କ୍ଷତି ହୁଏ ?
Solution:

• ପଲିଥନ୍ ଓ ପ୍ଲାଷ୍ଟିକ ନିର୍ମିତ ପଦାର୍ଥଗୁଡ଼ିକୁ ଏଣେତେଣେ ଫିଙ୍ଗିଲେ ଏହା ଅଣୁ ଜୀବମାନଙ୍କ ଦ୍ବାରା ଅପଘଟିତ ହୋଇନଥାଏ କିମ୍ବା ମାଟିରେ ମିଶିନଥାଏ ।
• ଏଗୁଡ଼ିକ ପୋଡ଼ିଲେ ସେଥ‌ିରୁ ବିଷାକ୍ତ ବାଷ୍ପ ଓ ଧୂଆଁ ନିର୍ଗତ ହୋଇ ବାୟୁମଣ୍ଡଳକୁ ଦୂଷିତ କରେ ।

Question 6.
ଆଜିକାଲି ଦେଖାଯାଉଥ‌ିବା ‘‘ରାସ୍ତାରୋକା’’ବେଳେ ରାସ୍ତା ମଝିରେ ଟାଏର୍ ଜଳାଯାଉଛି । ଏହାଦ୍ୱାରା ବାୟୁମଣ୍ଡଳ ଉପରେ କି ପ୍ରଭାବ ପଡ଼େ ?
Solution:
‘ରାସ୍ତାରୋକା’ ବେଳେ ରାସ୍ତା ମଝିରେ ଟାଏର୍ ଜଳାଇଲେ ବହୁ ପରିମାଣର ଅଙ୍ଗାରକାମ୍ଳ ଗ୍ୟାସ ବାୟୁ ମଣ୍ଡଳରେ ମିଶେ ଓ ଦୂର୍ଗନ୍ଧ ବାହାରି ପରିବେଶ ପ୍ରଦୂଷିତ ହୋଇଥାଏ ।

Question 7.
ପୁରୁଣା ଓ ବ୍ୟବହାର ହୋଇସାରିଥିବା ଟାଏର୍‌ଗୁଡ଼ିକୁ ସାଧାରଣତଃ ଆମେ ଅବ୍ୟବହୃତ ଜାଗାରେ ଫୋପାଡ଼ି ଦେଇଥାଉ । ଏଥୁଗୁଁ ଯେଉଁ ସ୍ଵାସ୍ଥ୍ୟଗତ ସମସ୍ୟା ସୃଷ୍ଟି ହୁଏ ଲେଖ ।
Solution:

• ଟାଏରମ୍ମତିକ୍ ସାଧାରଣଙ କେବ-ନିମ୍ନାକରଣ ଅବଲାନା | ଏଣ୍ଡ ଅଶ୍ମଜୀବମାନଜ ଦାରା ଅପଣଟିତହୁଏ ନାହିଁ କି ମାଟିରେ ମିଶେ ନାହିଁ । ଏଣୁ ସହର ଓ ଗ୍ରାମାଞ୍ଚଳର ଆବର୍ଜନା ଦୂରୀକରଣ ଏକ ଗୁରୁତ୍ଵପୂର୍ଣ୍ଣ ସମସ୍ୟା ।

Question 8.
ଅଣୁଜୀବମାନେ ଶୁଖୁଲା ପତ୍ର ସହିତ ପ୍ରତିକ୍ରିୟା କରି କ’ଣ ସୃଷ୍ଟି କରନ୍ତି ?
(କ) ବାଲି (ଖ) ଛତୁ (ଗ) ହ୍ୟୁମସ୍ (ଘ) କାଠ
Solution:
ହ୍ୟୁମସ୍

Question 9.
କଟକ ଓ ଭୁବନେଶ୍ଵର ଭଳି ବଡ଼ ବଡ଼ ସହରରୁ ନିର୍ଗତ ବହୁ ପରିମାଣରେ ବାହିତ ମଳକୁ ସଦୁପଯୋଗ କରିବା ପାଇଁ ଦୁଇଟି ଉପାୟ ଲେଖ ।
Solution:

• କଟକର CDAରୁ ବାହିତ ମଳର ସଦୁପଯୋଗ କରାଯାଇ ସେଥୁରୁ କେବଳ ସାର ସଂଗ୍ରହ କରାଯାଉଛି, ମାତ୍ର ଗୋଟିଏ ବାୟୋଗ୍ୟାସ ପ୍ଲାଣ୍ଟ କରାଗଲେ ଏଡ୍ ସହ ଜାଲେଣି ଗ୍ୟାସ ଓ ବିଦ୍ୟୁତ୍ ଶକ୍ତି ମିଳିପାରିବ ।
• ଭୁବନେଶ୍ୱରରେ ବାହିତ ମଳକୁ ନେଇ ଗୋଟିଏ City Sewage Disposal Plant ଅର୍ଥାତ୍ ସହରର ଦୂଷିତ ଜଳ ଉପଚାର ପ୍ଳାଣ୍ଟ କରାଯାଇପାରିବ । ଫଳରେ ସେଥୁରୁ ସାର, ବିଦ୍ୟୁତ୍, ଜାଳେଣୀ ଗ୍ୟାସ୍ ଓ ପାନୀୟ ଜଳ ପରିମାଣରେ ମିଳିପାରିବ ।

ବିପଯୁବସ୍ତୁ ସପୂଜାପ ପୂଚନା ଓ ବିଶେଷଣ :

→ ଉପକ୍ରମ :

• ମନୁଷ୍ୟର ଅଦରକାରୀ ବା ପରିତ୍ୟକ୍ତ ବସ୍ତୁ ସମୂହକୁ ତ୍ୟଜ୍ୟ ବସ୍ତୁ ବା ଆବର୍ଜନା କୁହାଯାଏ ।
• ଆବର୍ଜନାଗୁଡ଼ିକର ଅବସ୍ଥାକୁ ନେଇ ଏହା କଠିନ, ତରଳ ଓ ଗ୍ୟାସୀୟ ହୋଇଥାଏ ।
• ସହରର ବିଭିନ୍ନ ସ୍ଥାନରେ ଗୃହ ଆବର୍ଜନା ସବୁ ଜମା କରାଯାଏ । ପନିପରିବା ଚୋପା, ଫଳ, ମଞ୍ଜି, ରନ୍ଧନଶାଳାର ଆବର୍ଜନା, ଗଛ ପତ୍ର ଓ ଘାସ, ଭଙ୍ଗାରୁଜା ଟିଣ, ଅଦରକାରୀ କାଚ ବୋତଲ, ପ୍ଲାଷ୍ଟିକ ଓ ପଲିଥୁନ୍ ଆଦି କଠିନ ଆବର୍ଜନା ଅଟନ୍ତି ।
• ଆଉ ଏକ ପ୍ରକାରର ଆବର୍ଜନା ହେଲା ନାଳ ଓ ନର୍ଦ୍ଦମା ପାଣି | ଶିଳ୍ପାଞ୍ଚଳଗୁଡ଼ିକରୁ ନିର୍ଗତ ତରଳ ତ୍ୟଜ୍ୟ ସବୁ
• ଗାଡ଼ି ମଟର, କଳକାରଖାନା ଆଦିରୁ ଉତ୍ପନ୍ନ ଆବର୍ଜନା, ଗ୍ୟାସୀୟ ଆବର୍ଜନା ଅଟେ । ଜାଳେଣିରୁ ମଧ୍ଯ ଗ୍ୟାସୀୟ ଆବର୍ଜନା ବାୟୁମଣ୍ଡଳରେ ମିଶିଥାଏ ।
• ତରଳ ଓ ଗ୍ୟାସୀୟ ଆବର୍ଜନା ଗୁଡ଼ିକ ରାସାୟନିକ ଆବର୍ଜନା ହୋଇଥିବାରୁ ଅତ୍ୟନ୍ତ ବିଷାକ୍ତ ଓ କ୍ଷତିକାରକ ।
• ନାଳ, ନର୍ଦ୍ଦମାର ତରଳ ଆବର୍ଜନାରେ ପ୍ରଧାନ ଅଂଶ ହେଉଛି ଜଳ ଏବଂ ଏଥିରେ ଖଦା ଅଂଶ ଅତି କମ୍ । ଜଳକୁ ଖଦାରୁ ପୃଥକ କରି କ୍ଲୋରିନ ଦ୍ଵାରା ବିଶୋଧନ କରି ନଦୀ, ନାଳ, ସମୁଦ୍ର ବା ଅନ୍ୟାନ୍ୟ ଜଳଧାର ମାନଙ୍କରେ ଛାଡ଼ିଦିଆ ଯାଇପାରେ ।

→ ବିଶୁଦ୍ଧ ଜଳ :

• ଆପେ କେବଳ ବିଶ୍ଵଦ୍ଧ କାଳକୁ ବୃ ପାନାମ୍ ଉପେ ବ୍ୟବହାର କରିଥାଇ |
• ଦୂଷିତ ଜଳକୁ ପାନୀୟ ରୂପେ ବ୍ୟବହାର କରି ଲୋକମାନେ ଜଳ ବାହିତ ରୋଗ ଯଥା – କଲେରା, ଟାଇଫଏଡ଼ ଓ କାମଳ ଆଦି ରୋଗଦ୍ୱାରା ଆକ୍ରାନ୍ତ ହୋଇଥାଆନ୍ତି ।
  • ବିଶୁଦ୍ଧ ଜଳର ଆବଶ୍ୟକତାକୁ ଅନୁଭବ କରାଇ ଜନସଚେତନତା ଜାଗ୍ରତ କରିବା ପାଇଁ 2005 ମସିହାରୁ ମାର୍ଚ୍ଚ 22 ତାରିଖକୁ ଆମେ ‘ବିଶ୍ଵ ଜଳ ବିବସ’’ ରୂପେ ପାଳନ କରି ଆସୁଛୁ ।
• ଜାତିସଂଘର ଘୋଷଣା ଅନୁଯାୟୀ ଦଶନ୍ଧି 2005 – 2015 କୁ ଆନ୍ତର୍ଜାତିକ ସ୍ତରରେ ‘‘ଜୀବନ ପାଇଁ ଜଳ’’ ଉପରେ କାର୍ଯ୍ୟାନୁଷ୍ଠାନ ଗ୍ରହଣ କରି ସମସ୍ତଙ୍କୁ ବିଶୁଦ୍ଧପାନୀୟ ଜଳ ଯୋଗାଇବାର ଏହି ପ୍ରକଳ୍ପର ଉଦ୍ଦେଶ୍ୟ ।
• ତରଳ ଆବର୍ଜନାର ବିଭିନ୍ନ କାରକଗୁଡ଼ିକ ଜଳ ସହିତ ମିଶି ଜଳକୁ ଦୂଷିତ କରିଥାଏ । ବିଭିନ୍ନ କାରଣରୁ ଜଳ ପ୍ରଦୂଷିତ ହୋଇଥାଏ; ଯଥା – ଶିଳ୍ପଜନିତ, କୃଷି ସମ୍ପର୍କିତ, ଗୃହକାର୍ଯ୍ୟ ଜନିତ ଆଦି ।
• ଅପଚଯିତ କାଳର ସମ୍ୟକୟକୃପେ ସେଥିରେ ମିଶିଥିବା ସମୟ ଆନକାନାର ଉପଚାର କୁହାଯାଏ ।

→ ଆବର୍ଜନା ମିଶ୍ରିତ ଜଳ :

• ଆବର୍ଜନା ମିଶ୍ରିତ ଜଳକୁ ଅପଚୟିତ ଜଳ କୁହାଯାଏ । ପାଇଖାନା ଓ ଗାଧୁଆ ଘରୁ ବାହାରୁଥିବା ମଇଳା ଜଳ, ନାଳରୁ ବାହାରୁଥ‌ିବା ଜଳ, ବର୍ଷାଦ୍ବାରା ଧୋଇହୋଇ ଯାଉଥିବା ଜଳ ଆଦି ଅପଚୟିତ ଜଳ ।
• ଅପଚୟିତ ଜଳରେ କେତେକ ଦୂଷିତ ପଦାର୍ଥ ଭାସମାନ ଅବସ୍ଥାରେ ରହିଥାଏ ଓ ଆଉ କେତେକ ଦ୍ରବୀଭୂତ ଅବସ୍ଥାରେ ରହିଥାଏ । ସେଥ‌ିରେ କେତେକ ହାନିକାରକ ଜୈବ ଓ ଅଜୈବ ପଦାର୍ଥ ଏବଂ ରୋଗ ସୃଷ୍ଟିକାରୀ ଜୀବାଣୁ ଓ ଅଣୁଜୀବ ରହିଥାନ୍ତି ।
• ଜଳ ନିଷ୍କାସନର ସୁବ୍ୟବସ୍ଥା କରିବା ଉଚିତ ଓ ବର୍ଷା ଜଳକୁ ନଷ୍ଟ ହେବାକୁ ନଦେଇ ତା’କୁ ସଂଚୟ କରିବା ଉଚିତ୍ ।
• ଅପଚୟିତ ଜଳରେ ଥିବା କେତେକ ପଦାର୍ଥର ଉଦାହରଣ ତଳେ ଦିଆଯାଇଛି ।

ଜୈବିକ ପଦାର୍ଥ – ମନୁଷ୍ୟର ମଳ, ପଶୁପକ୍ଷୀଙ୍କ ମଇଳା, ଅପରିଷ୍କାର ତୈଳ ଜାତୀୟ ପଦାର୍ଥ, ପଲିଥନ୍,
ଅଜୈବିକ ପଦାର୍ଥ – ମାଟି, ଗୋଡ଼ି, ପାଉଁଶ, ଭଙ୍ଗାକାଚ ଇତ୍ୟାଦି ।
ପୋଷକ – ନାଇଟ୍ରୋଜେନ୍, ଫସ୍‌ଫରସ୍, ପୋଟାସିୟମ୍ ଇତ୍ୟାଦି ଧାତୁର ଯୌଗିକରୁ ପ୍ରସ୍ତୁତ ସାର ।
ବୀଜାଣୁ – ହଇଜା ଓ ଆନ୍ତ୍ରିକ ଜ୍ଵର ସୃଷ୍ଟି କରୁଥିବା ବୀଜାଣୁ ।

→ ଦୂଷିତ ଜଳର ଉପଚାର :

• ଜଣେ ସଚେତନ ଓ ଉତ୍ତମ ନାଗରିକ ହିସାବରେ, ତୁମେ ପୌରପରିଷଦର କର୍ମକର୍ତ୍ତା ତଥା ଗ୍ରାମ ସରପଞ୍ଚଙ୍କ ସହାୟତାରେ ପରିବେଶକୁ ପରିଷ୍କାର ଓ ବିଶୋଧନ କରିବାରେ ପଦକ୍ଷେପମାନ ନେବା ଆବଶ୍ୟକ ।
• ନାଳ ନର୍ଦ୍ଦମା ପାଖରେ ଇଉକାଲିପଟାସ୍ ଗଛ ଲଗାଇଲେ ତାହା ଅପରିଷ୍କାର ଜଳ ଅଧ୍ବକ ଶୋଷଣ କରି ପରିଷ୍କାର ବାଷ୍ପ ବାୟୁମଣ୍ଡଳକୁ ଛାଡ଼ିଥାଏ ।

→ ଆବର୍ଜନା ନିଷ୍କାସନର ଉତ୍ତମ ମାର୍ଗ :

• ଅଳ୍ପ ଆବର୍ଜନା ସୃଷ୍ଟି କରିବା ଓ ଦ୍ୱିତୀୟରେ ଆବର୍ଜନାର ଉପଯୁକ୍ତ ବିନିଯୋଗକରି ସେଥୁରୁ ଉତ୍ପନ୍ନ ଉପାଦାନଗୁଡ଼ିକୁ କାମରେ ଲଗାଇବା ।
• ଅଦରକାରୀ ତୈଳଜାତୀୟ ପଦାର୍ଥ ମାଟିର ଜଳ ଶୋଷଣରେ ବାଧା ସୃଷ୍ଟି କରେ ଓ ମୋବିଲ, ପୋକମରା ଔଷଧ, ବିଭିନ୍ନ ରଙ୍ଗ, ଆଦି ନାଳରେ ପକାଇଲେ ଏହା ଉପକାରୀ ଅଣୁଜୀବଙ୍କୁ ମାରି ଅଶେଷ କ୍ଷତିସାଧନ କରିଥାଏ ।
• ପଲିଥନ୍, ଛିଣ୍ଡାକନା, ତୁଳା, ଭଙ୍ଗା କଣ୍ଢେଇ ଆଦି ନାଳରେ ପକାଇଲେ ନାଳ ବନ୍ଦ ହୋଇ ଜଳ ନିଷ୍କାନରେ ବାଧାଦିଏ ।
• ସହଜରେ ମାଟିରେ ମିଶିଯାଉଥ‌ିବା ଜୈବ ନିମ୍ବୀକରଣ ଯୋଗ୍ୟ ଆବର୍ଜନା ପାଇଁ ସବୁଜ ରଙ୍ଗର ଓ ଜୈବ – ନିମ୍ନକରଣ ଅଯୋଗ୍ୟ ଆବର୍ଜନା ପାଇଁ ଲାଲରଙ୍ଗର ଡଷ୍ଟବିନ୍ ଘରେ ଓ ବିଦ୍ୟାଳୟରେ ବ୍ୟବହାର କରିବା ଉଚିତ ।

→ ପରିମଳ ବ୍ୟବସ୍ଥାର :

• ପରିମଳ ବ୍ୟବସ୍ଥାର ଉନ୍ନତି ପାଇଁ ସମସ୍ତେ ସ୍ବଚ୍ଛ ପାଇଖାନା ବ୍ୟବହାର କରିବା ଉଚିତ ।
• ଖୋଲାଜାଗାରେ ମଳତ୍ୟାଗ କରିବା ଉଚିତ ନୁହେଁ, ଯଦି ମଳତ୍ୟାଗ ହୁଏ, ତେବେ ଏହାକୁ ବାଲି କିମ୍ବା ମାଟି ଦ୍ଵାରା ଘୋଡ଼ାଇ ଦେବା ଉଚିତ ।
• ଆଜିକାଲି ଜଳମୁଦ ପାଇଖାନାରୁ ନିଷ୍କାସିତ ଜଳକୁ ପାଇପ୍ ସାହାଯ୍ୟରେ ବାୟୋଗ୍ୟାସ ପ୍ଲାଣ୍ଟ ସହ ସଂଯୋଗ କରି ସେଥ‌ିରୁ ବିଦ୍ୟୁତ୍ ଶକ୍ତି ଓ ତାପଶକ୍ତି ଉତ୍ପନ୍ନ କରାଯାଉଛି ।

→ ଯେଉଁଠାରେ ଭୂତଳ ନିଷ୍କାସନର ସୁବିଧା ନାହିଁ :

• ପର୍ବପର୍ବାଣି, ମେଳା ଓ ମହୋତ୍ସବ ପାଳନ ବେଳେ, ହାଟ ବଜାର, ରେଳଷ୍ଟେସନ୍ , ଡାକ୍ତରଖାନା ଓ ବିମାନ ବନ୍ଦରରେ ଅନେକ ଲୋକ ପ୍ରତିଦିନ ଯାତାୟତ କରନ୍ତି । ଏଣୁ ଏଠାରେ ସୃଷ୍ଟି ହେଉଥ‌ିବା ଆବର୍ଜନାର ଉପଯୁକ୍ତ ନିଷ୍କାସନ ଓ ବିଶୋଧନ ଏକାନ୍ତ ଆବଶ୍ୟକ, ନଚେତ୍ ମହାମାରୀ ରୋଗ ବ୍ୟାପିବାର ସମ୍ଭାବନ ଅଛି ।
• ପରିମଳ ବ୍ୟବସ୍ଥା ପ୍ରତି ନିଜେ ସଚେତନ ରହିବା ସହ ଅନ୍ୟମାନଙ୍କୁ ସଚେତନ କରିବା ଆମର କର୍ତ୍ତବ୍ୟ । କାରଣ ଆମ ପରିବେଶକୁ ସୁସ୍ଥ ଓ ପରିଷ୍କାର ରଖୁବା ଆମ ସମସ୍ତଙ୍କର କର୍ତ୍ତବ୍ୟ । ମିଳିତ ଉଦ୍ୟମରେ ପରିବେଶ ପ୍ରଦୂଷଣ ମୁକ୍ତ ହୋଇ ପାରିବ ।

→ ପରିଷ୍କାର ପରିଚ୍ଛନତା ଓ ରୋଗ :

• ଅସ୍ବାସ୍ଥ୍ୟକର ପରିବେଶରେ ବାସ କରିବା ଓ ଦୂଷିତ ଜଳ ପାନ କରିବା ସମସ୍ତ ରୋଗର ପ୍ରଧାନ କାରଣ ।
• ଏଣେତେଣେ ମଳତ୍ୟାଗ କରିବା ଫଳରେ ବାୟୁ, ମାଟି ଉପର ଓ ମାଟି ତଳ ଜଳ ଦୂଷିତ ହୋଇଥାଏ ।
• ଖୋଲା ପଡ଼ିଆରେ ଝାଡ଼ା ନ ଯାଇ ବରପାଲି ବା ସେପ୍‌ଟିକ୍ ଟ୍ୟାକ୍ ବ୍ୟବହାର କରିବା ଆବଶ୍ୟକ ।
• ଖାଲି ପାଦରେ ଝାଡ଼ାଗଲେ ମଳରେ ଥିବା କୃମି ପାଦର ଚର୍ମଦେଇ ଶରୀର ଭିତରକୁ ପ୍ରବେଶ କରନ୍ତି ।
  • ଦୂଷିତ ଜଳ ପାନ କଲେ ଏବଂ ସେଥ‌ିରେ ଗାଧୋଇଲେ ହଇଜା, କାମଳ,
  • ଟାଇଫଏଡ଼, କାଛୁକୁଣ୍ଡିଆ ଓ ଯାଦୁ ଆଦି ରୋଗ ହୋଇଥାଏ ।

→ ବ୍ୟବହୃତ ଜଳର ନିଷ୍କାସନ ଓ ପୁନଃ ବ୍ୟବହାର :

• ବ୍ୟବହୃତ ଜଳକୁ ଗୋଟିଏ ଜାଗାରେ ଜମିବାକୁ ନ ଦେଇ ଏହାକୁ ସୋପିଟ ମଧ୍ୟକୁ ବା ବଗିଚାର ଗଛ ମୂଳକୁ ଛାଡ଼ିଦେବା ଉଚିତ ।
• ସହରାଞ୍ଚଳରେ ବ୍ୟବହୃତ ଜଳକୁ ଏଣେତେଣେ ନ ଛାଡ଼ି ମୁଖ୍ୟନାଳକୁ ଛାଡ଼ିବା ଉଚିତ ।
• କୂଅ ଓ ନଳକୂଅ ପାଖରେ ସିମେଣ୍ଟ ଚଟାଣ କରି ଉଚ୍ଚା କଲେ ବ୍ୟବହୃତ ପାଣି ଏହା ମଧ୍ୟକୁ ନଯାଇ ଜଳକୁ ଦୂଷିତ କରି ନ ଥାଏ ।
• ନାଳ ନର୍ଦ୍ଦମାର ଆବର୍ଜନାକୁ ସଂଗ୍ରହ କରି ବିଭିନ୍ନ ପ୍ରଣାଳୀରେ ସେଥୁରୁ ଜଳ ଓ ଖଦା ଅଂଶକୁ ପୃଥକ କରି ଓ ଜଳକୁ ବିଶୋଧନ କରି ଚାଷପାଇଁ ଏହି ଜଳକୁ ବିନିଯୋଗ କରାଇପାରିବ । ଏପରିକି ମାଛ ଚାଷ ପାଇଁ ମଧ୍ୟ ଲାଭଜନକ ଭାବରେ ଏହାର ବ୍ୟବହାର କରାଯାଇପାରିବ ।

→ ଅପଚୟିତ ଜଳର ପୁନଃ ଚକ୍ରଣ ଏବଂ ପୁନଃ ବିନିଯୋଗ :

• ପ୍ରତ୍ୟେକ ଆବର୍ଜନାର ପୁନଃଚକ୍ରଣ କରି ବ୍ୟବହାର ଯୋଗ୍ୟ କରାଯାଇ ପାରିବ । କଠିନ ଆବର୍ଜନାକୁ ସହଜରେ ପୁନଃଚକ୍ରଣ କରି ବ୍ୟବହାର କରାଯାଏ; ଯଥା –ଭଙ୍ଗା ଟିଣ, କାଚ, ପ୍ଲାଷ୍ଟିକ ଆଦି ଆବର୍ଜନାକୁ ଆବଶ୍ୟକୀୟ କାରଖାନାକୁ ପଠାଯାଇ ଏହାକୁ ତରଳାଇ ସେଥୁରୁ ନୂତନ ଜିନିଷ ପ୍ରସ୍ତୁତ କରାଯାଇଥାଏ ।
• ଅପଚୟିତ ଜଳକୁ ସିଧାସଳଖ ନଦୀ ବା ସମୁଦ୍ରକୁ ଛାଡ଼ିବା ଫଳରେ ଅଶୋଧ ଜଳ ଓ ଏଥ‌ିରେ ଥିବା ମାରାତ୍ମକ ପଦାର୍ଥମାନ ଜଳଜ ଜୀବମାନଙ୍କର ଅଶେଷ କ୍ଷତି ଘଟାଇଥାଏ ।
• ଆଜିକାଲି ଅପଚୟିତ ଜଳକୁ ବିଭିନ୍ନ ପ୍ରକ୍ରିୟାଦ୍ୱାରା ପୁନଃ ବ୍ୟବହାର ଉପଯୋଗୀ କରାଯାଇ ପାରୁଛି । ଉଦାହରଣ ସ୍ଵରୂପ, ଭୁବନେଶ୍ୱର ବିଡ଼ିଏ ନିକୋ ପାର୍କକୁ ବିଭିନ୍ନ ନଳାରେ ଆସୁଥ‌ିବା ଆବର୍ଜନା ପୂର୍ଣ୍ଣ ନର୍ଦ୍ଦମା ଜଳକୁ ପରିଷ୍କାର କରି ଏକ କୃତ୍ରିମ ହ୍ରଦ କରାଯାଇଛି । ଏହି ହ୍ରଦରେ ନୌକା ବିହାର ଓ ଅନ୍ୟାନ୍ୟ ଜଳକ୍ରୀଡ଼ା କରାଯାଇ ପାରୁଛି ।
• ଭୌତିକ, ରାସାୟନିକ ଓ ଜୈବିକ ପ୍ରକ୍ରିୟାରେ ଦୂଷିତ ଜଳର ଉପଚାର କରାଯାଇ ଏହାକୁ ପରିଷ୍କାର ଓ ବ୍ୟବହାର
• ଏଥୁରୁ ଶେଷରେ ସାର ମିଳିବା ସହ ଜାଳେଣି ଗ୍ୟାସ୍ ମଧ୍ୟ ମିଳେ ଏବଂ ପରିଷ୍କାର ବିଶୋଧ ଜଳକୁ ପାନୀୟ ଜଳରୂପେ ବ୍ୟବହାର କରାଯାଏ ।

→ ଆସ, ଜାଣିବା :

• ଘରୁ, କଳକାରଖାନାରୁ, ଚାଷଜମିରୁ ନିର୍ଗତ ଓ ମନୁଷ୍ୟର ବିଭିନ୍ନ ବ୍ୟବହାରରୁ ନିର୍ଗତ ବା ନିଷ୍କାସିତ ଜଳକୁ ଆମେ ପାରୁଛି ।
• ଅପଚୟିତ ଏକ ପ୍ରକାର ତରଳ ଦୂଷିତ ପଦାର୍ଥ ଯାହା, ମାଟି ଜଳ ଏବଂ ଜମିକୁ ପ୍ରଦୂଷିତ କରିଥାଏ ।
• ଦୂଷିତ ଜଳକୁ ବିଭିନ୍ନ ପ୍ରକାର ପ୍ରକ୍ରିୟାରେ ଉପଚାର କରାଯାଏ ।
• ଯେଉଁଠାରେ ଭୂତଳ ନିଷ୍କାସନର ସୁବିଧା ନାହିଁ ସେଠାରେ କମ୍ ଖର୍ଚ୍ଚରେ ସୁଲଭ ପରିମଳ ବ୍ୟବସ୍ଥା କରାଯାଇଥାଏ ।
• ଦୂଷିତ ଜଳର ଉପଚାର ଫଳରେ ପଟୁମାଟି ଓ ଜୈବିକ ଗ୍ୟାସ୍ ଉପଲବ୍ଧ ହୋଇଥାଏ ।
• ଖୋଲା ନାଳରେ ମାଛି, ମଶା ଏବଂ ବିଭିନ୍ନ ଅଣୁଜୀବ ସୃଷ୍ଟି ହୋଇଥାଆନ୍ତି । ସେଗୁଡ଼ିକ ବିଭିନ୍ନ ରୋଗର କାରଣ ।
• ଖୋଲା ପଡ଼ିଆରେ ଝାଡ଼ା ନଯାଇ, ପକ୍କା ପାଇଖାନା ବା ବରପାଲି ପାଇଖାନା ବ୍ୟବହାର କରିବା ଆବଶ୍ୟକ ।

Odisha State Board BSE Odisha 7th Class English Solutions Test-1 Textbook Exercise Questions and Answers.

BSE Odisha Class 7 English Solutions Test-1

BSE Odisha 7th Class English Test-1 Text Book Questions and Answers

The figures in the right hand margin indicates the marks for each question.
1. Write the following Odia names of the persons in English. [10]
(Teacher will give names of ten persons in Odia)

ମହାତ୍ମା ଗାନ୍ଧୀ – Mahatma Gandhi
ମଧୁସୂଦନ ଦାସ – Madhusudan Das
ବିଜୁ ପଟ୍ଟନାୟକ – Biju Pattnaik
ବିନାପାନି ମହନ୍ତ – Binapani Mohantv
ଗୋପାବନ୍ଧୁ ଦାଶ – Gopabandhu Dash
ଗୋପୀନାଥ ମହନ୍ତ – Gopinath Mohantv
ଫାକିରମୋହନ ସେନାପତି – Fakirmohan Senapati
ରାଧନାଥ ରୋଭ – Radhanath Rov
ରାଜେନ୍ଦ୍ର ପ୍ରସାଦ – Rajendra Prasad
ଚିତ୍ତରଂଜନ ଦାସ – Chittaranjan Das

2. Write the following names of places in English. [10]
(Teacher will give names of ten places in Odia)

Answer:
ଭୁବନେଶ୍ୱର – Bhubaneswar
ଦିଲ୍ଲୀ – Delhi
ଆହ୍ଲାବାଦ – Allahabad
ବାଲାସୋର – Balasore
ଆନନ୍ଦପୁର – Anandpur
ବରିପଡା – Baripada
ଗୌହାଟୀ – Gauhati
ଚେନ୍ନାଇ – Chennai
ଫିରୋଜାବାଦ – Ferozabad
ଚମ୍ପୁଆ – Champua

3. Your teacher will give dictation of ten English words. Write them in the space given below. [10]

Answer:
rooster
monkey
money
protect
desert
teacher
wealth
singer
bottle
driver

4. Given below are some words. Your teacher will read aloud ten of them. Tick those which s/he reads aloud.
city, country, tree, crown, rooster, close, monkey, jackal, animal, camel, thirsty, climb, zoo, trumpet [10]
Answer:
(Student tick the words which their teacher reads aloud.)

5. Your teacher will read aloud a paragraph (The Jackal and the Rooster). You listen to him/her and fill in the gaps
There was a very ___________ and ___________ rooster. It looked like a __________ with a beautiful red __________. It also felt like a __________ in the morning. It sat on a ___________ place and ____________ non-stop cock-koo-doodle-doo. He knew his _____________ was much better than the songs of ____________. [15]
Answer:
There was a very big and handsome rooster. It looked like a king with a beautiful read crown. It also felt like a king. It gets up very early in the morning. It sat on a high place and sang non-stop cock-koo-doodle-doo. He knew his song was much better than the songs of other roosters of his locality.

6. Match the pairs of words which sound alike at the end part. Write the serial numbers in boxes. [15]

Answer:

7. Read the poem and answer the questions that follow. [10]
(କବିତାଟି ପଢ଼ ଏବଂ ପରବର୍ତ୍ତୀ ପ୍ରଶ୍ନଗୁଡିକର ଉତ୍ତର ଦିଅ ।)

A fox was moving on one day,
And just above his head,
He saw a vine with lovely grapes,
Rich ripe and purple-red.

Question (i).
What is the poem about?
(କବିତାଟି କେଉଁ ବିଷୟରେ ?)
Answer:
The poem is about a fox and vine with lovely grapes.

Question (ii).
What was the fox doing one day?
(ଦିନେ କୋକିଶିଆଳ କ’ଣ କରୁଥିଲା ?)
Answer:
The fox was moving here and there one day.

Question (iii).
What did he see above his head?
(ତା’ ମୁଣ୍ଡ ଉପରେ ସେ କ’ଣ ଦେଖୁଲା ?)
Answer:
He saw a vine with lovely grapes above his head.

Question (iv).
What were the grapes like? (ଅଙ୍ଗୁରଗୁଡ଼ିକ କିପରି ଥିଲା ?)
Answer:
The grapes were rich, ripe and purple red.

Question (v).
‘He’ in the third line is used for___________.
(ତୃତୀୟ ଧାଡ଼ି ‘He’ (ସେ) _________________ ପାଇଁ ବ୍ୟବହାର ହୋଇଛି ।)
Answer:
In the third line ‘He’ is used for the fox.

8. Read the following paragraph and write the answers to the questions that follow in complete sentences.
Once there lived a jackal in a forest. One day, he did not get any food. He was very hungry. In the evening, the hungry jackal came into a village. He moved here and there in the village. He was looking for food. By chance he fell into a washer-man’s tub which was full of blue water. He became blue and other jackals could not know him. [20]

Question (i).
What is the paragraph about?
(ଅନୁଚ୍ଛେଦଟି କେଉଁ ବିଷୟରେ ?)
Answer:
The paragraph is about a hungry jackal.

Question (ii).
Where did the jackal live?
(ବିଲୁଆଟି କେଉଁଠାରେ ରହୁଥିଲା ?)
Answer:
The jackal lived in a forest.

Question (iii).
Why did he become very hungry one day?
(ଦିନେ ବିଲୁଆ କାହିଁକି ଅଧିକ ଭୋକିଲା ଥିଲା ?)
Answer:
One day the jackal did not get any food anywhere, so he became hungry

Question (iv).
When did the jackal come into the village?
(ବିଲୁଆଟି କେତେବେଳେ ଗାଁ ଭିତରକୁ ଆସିଲା ?)
Answer:
The jackal came into the village in the evening.

Question (v).
What was he searching for in the village?
(ଗାଁରେ ସେ କ’ଣ ଖୋଜୁଥୁଲା ?)
Answer:
He was searching for food in the village.

Question (vi).
Where did he move in the village?
(ସେ ଗାଁର କେଉଁ ଜାଗାରେ ବୁଲୁଥିଲା ?)
Answer:
He moved here and there in the village.

Question (vii).
What did he fall into?
(କେଉଁଠାରେ ସେ ପଡ଼ିଗଲା ?)
Answer:
He fell into a washer-man’s tub which was full of blue water.

Question (viii).
What happened to him when he fell into the tub?
(ଟବ୍‌ରେ ପଡ଼ିଗଲା ପରେ ତା’ର କ’ଣ ହେଲା ?)
Answer:
When he fell into the tub, he became blue.

Question (ix).
Why couldn’t the other jackals know him?
(ଅନ୍ୟ ବିଲୁଆମାନେ ତାହାକୁ କାହିଁକି ଜାଣିପାରିଲେ ନାହିଁ ?)
Answer:
Other jackals could not know him because he was looking blue quite different from them.

Question (x).
Who is ‘He’ in the last line?
(ଶେଷ ଧାଡିରେ ‘He’ କିଏ ?)
Answer:
In the last lines ‘He’ refers to the jackal.

Odisha State Board BSE Odisha 7th Class Sanskrit Solutions Chapter 6 ପୂରଣବାଚକଶବ୍ଦାଃ Textbook Exercise Questions and Answers.

BSE Odisha Class 7 Sanskrit Solutions Chapter 6 ପୂରଣବାଚକଶବ୍ଦାଃ

४. शून्यस्थानं पूरयतु
(ଶୂନ୍ୟସ୍ଥାନଂ ପୂରୟତୁ) ଶୂନ୍ୟସ୍ଥାନ ପୂରଣ କର :

(क) ______ बालकः ।

(କ) _______ବାଳକଃ ।

उत्तर :
(क) चतुर्थः बालकः ।
(କ) ଚତୁର୍ଥୀ ବାଳକଃ ।

(ख) ________ शुकः ।

(ଖ) _______ ଶୁକଃ ।

उत्तर :
(ख) नवमः शुकः ।
(ଖ) ନବମଃ ଶୁକଃ ।

(ग) ________ बालिका: ।

(ଗ) _____ ବାଳିକା ।

उत्तर :
(ग) सप्तमी बालिका ।
(ଗ) ସପ୍ତମୀ ବାଳିକା ।

(घ) _______ माला ।

(ଘ) _________ ମାଳା

उत्तर :
(घ) तृतीया माला ।
(ଘ) ତୃତୀୟା ମାଳା।

२. किं समीचीनम् (✓) किम् असमीचीनम् (✗)
କିଂ ସମୀଚୀନମ୍ (✓) ଅସମୀଚୀନମ୍ (✗) କେଉଁଟି ଠିକ୍ କେଉଁଟି ଭୁଲ୍

यथा
(क) प्रथमा फलम् [ ]
(ख) पञ्चमी बालिका [ ]
(ग) पञ्चमः बालिका [ ]
(घ) द्वितीयं पुष्पम् [ ]
(ङ) द्वितीय: पुष्पम् [ ]
(च) चतुर्थः मार्जार: [ ]
(छ) तृतीया लता [ ]
(ज) चतुर्थं मार्जारः [ ]
(झ) तृतीयः लता [ ]
(ज) प्रथमं फलम् [ ]
उत्तर :
(क) प्रथमा फलम् [✗]
(ख) पञ्चमी बालिका [✓]
(ग) पञ्चमः बालिका [✗]
(घ) द्वितीयं पुष्पम् [✓]
(ङ) द्वितीय: पुष्पम् [✗]
(च) चतुर्थः मार्जार: [✓]
(छ) तृतीया लता [✓]
(ज) चतुर्थं मार्जारः [✗]
(झ) तृतीयः लता [✗]
(ज) प्रथमं फलम् [✓]

३. स्तम्भं योजयतु
(ସ୍ତମଂ ଯୋଜୟତୁ) ସ୍ତମ୍ଭ ମିଳନ କର:

उत्तर :

अतिरिक्त प्रश्नोत्तरम्
(ଅତିରିକ୍ତ ପ୍ରଶ୍ନୋତ୍ତରମ୍)

१. मातृभाषया अनुवादं कुरुत ।

(क) रवि : प्रथमः सोमः द्वितीयः ।
उत्तर :
ରବିବାର ପ୍ରଥମ ସୋମବାର ଦ୍ଵିତୀୟ।

(ख) स एव छात्रोत्तमः ।
उत्तर :
ସେ ହିଁ ଉତ୍ତମ ଛାତ୍ର ।

(ग) एवं क्रमेण सप्ताहः वारक्रमः ।
उत्तर :
ଏହିପରି ସପ୍ତାହର ବାରକ୍ରମ।

(घ) छात्र सुखी परिश्रमी ।
उत्तर :
ଛାତ୍ର ସୁଖୀ ପରିଶ୍ରମୀ।

(ङ) विभक्तिगणनं जानाति क्रमशः ।
उत्तर :
କ୍ରମେ ବିଭକ୍ତି ଗଣନା ଜାଣେ ।

२. एकपदेन उत्तरं लिखत ।

(क) प्रथमं श्रवणं द्वितीयं किम् ?
उत्तर :
ମନନମ୍

(ख) द्वितीयं मननं तृतीयं किम् ?
उत्तर :
ସ୍ମରଣମ୍

(ग) तृतीयं स्मरणं चतुर्थं किम् ?
उत्तर :
ଭାଷଣମ୍

(घ) पंचमं पठनं षष्ठं किम् ?
उत्तर :
ଲେଖନମ୍

(ङ) षष्ठं लेखनं सप्तमं किम् ?
उत्तर :
ଗାୟନମ୍

३. शून्यस्थानं पूरयत ।

(क) नवमं ______ दशमं कीर्त्तनम् ।
उत्तर :
ନର୍ଭନମ୍

(ख) भाषाप्रशिक्षणं सामान्यं _____।
उत्तर :
ସୋପାନମ୍

(ग) प्रथमं ______ द्वितीयं मननम् ।
उत्तर :
ଶ୍ରବଣମ୍

(घ) अष्टमं ______
उत्तर :
ବୋଧନମ୍

ङ ______ सामान्यं सोपानम्।
उत्तर :
ଭାଷା ପ୍ରଶିକ୍ଷଣମ୍

४. विक्षिप्तवर्णाना मेलनेन शुद्धपदं रचयत ।

(क) ष्ट अम्न।
उत्तर :
ଅଷ୍ଟମମ୍

(ख) प्रमः थ।
उत्तर :
ପ୍ରଥମଃ

(ग) पासो म्म ।
उत्तर :
ସୋପାନମ୍

(घ) या द्विती।
उत्तर :
ଦ୍ଵିତୀୟା

(ङ) मी प
उत्तर :
ପଞ୍ଚମୀ

५. स्तम्भमेलनं’ कुरूत ।

उत्तर :

६. मातृभाषया अर्थं लिखत ।

(क) मननम्
उत्तर :
ମନନକରିବା

(ख) सोपानम्
उत्तर :
ପାହାଚ

(ग) वोधनम्
उत्तर :
ବୋଧନ

(घ) माला
उत्तर :
ମାଳ

(ङ) शुकः
उत्तर :
ଶୁଆ

७. उत्तर चयनं कुरूत ।

(क) भौमवार:
(a) सोमवार :
(b) मंगलवार
(c) बुधवार
(d) गुरुवारः
उत्तर :
(b) मंगलवार

(ख) कः षष्ठवार ?
(a) वुधवार :
(b ) भौमवार
(c) शुक्रवार :
(d) रविवार :
उत्तर :
(c) शुक्रवार :

(ग) कः विभक्तिगणना जानाति ?
(a) परिश्रमी
(b) दृष्ट:
(c) सुखी
(d) सुखी परिश्रमी
उत्तर :
(d) ସୁଖୀ ପରିଶ୍ରମୀ

(घ) प्रथमं किम् ?
(a) पठनम्
(b) श्रवणम्
(c) लेखनम्
(d) मननम्
उत्तर :
(b) श्रवणम्

(ङ) भाषा प्रशिक्षणश्च नवमं स्थानं किम् ?
(a) कीर्त्तनम्
(b) नर्त्तनम्
(c) गायनम्
(d ) श्रवणम्-
उत्तर :
(b) नर्त्तनम्

८. संशोधन कुरूत ।

(क) तृतीय स्थानम् ।
उत्तर :
ତୃତୀୟଂ ସ୍ଥାନମ୍ ।

(ख) चतुर्थं मार्जारः ।
उत्तर :
ଚତୁର୍ଥୀ ମାର୍ଜାରଃ ।

(ग) तृतीय : लता ।
उत्तर :
ତୃତୀୟା ଲତା।

(घ) सप्तमी वालकः ।
उत्तर :
ସପ୍ତମଃ ବାଳକଃ ।

(ङ) प्रथमा फलम् ।
उत्तर :
ପ୍ରଥମଂ ଫଳମ୍ ।

पूरणवाचकशब्दाः
ପୂରଣବାଚକଶବ୍ଦ (ପୂରଣବାଚକ ଶବ୍ଦ)

१. प्रथमः बालक:

ପ୍ରଥମଃ ବାଳକଃ ( ପ୍ରଥମ ବାଳକ)

द्वितीय: बालक:

ଦ୍ଵିତୀୟ ବାଳକଃ (ଦ୍ଵିତୀୟ ବାଳକ)

तृतीय: बालक:

ତୃତୀୟ ବାଳକ (ତୃତୀୟ ବାଳକ)

चतुर्थः बालक:

ଚତୁର୍ଥୀ ବାଳକଃ (ଚତୁର୍ଥ ବାଳକ)

पश्च्रम: बालक:

ପଞ୍ଚମଃ ବାଳକଃ (ପଞ୍ଚମ ବାଳକ)

षष्ठः बालक:

ଷଷ୍ଠୀ ବାଳକଃ (ଷଷ୍ଠୀ ବାଳକ)

सप्तमः बालक:

ସପ୍ତମଃ ବାଳକଃ (ସପ୍ତମ ବାଳକ)

अष्टम: बालक :

ଅଷ୍ଟମୀ ବାଳକଃ (ଅଷ୍ଟମ ବାଳକ)

नवम: बालक:

ନବମଃ ବାଳକଃ (ନବମ ବାଳକ)

दशम: बालक:

ଦଶମଃ ବାଳକଃ (ଦଶମ ବାଳକ)

प्रथमा बालिका

ପ୍ରଥମା ବାଳିକା (ପ୍ରଥମ ଝିଅ)

द्वितीया बालिका

ଦ୍ଵିତୀୟା ବାଳିକା (ଦ୍ଵିତୀୟ ଝିଅ)

तृतीया बालिका

ତୃତୀୟା ବାଳିକା (ତୃତୀୟ ଝିଅ)

चतुर्थी बालिका

ଚତୁର୍ଥୀ ବାଳିକା (ଚତୁର୍ଥ ଝିଅ)

पश्चमी बालिका

ପଞ୍ଚମୀ ବାଳିକା (ପଞ୍ଚମ ଝିଅ)

षष्ठी बालिका

ଷଷ୍ଠୀ ବାଳିକା (ଷଷ୍ଠ ଝିଅ)

सप्तमी बालिका

ସପ୍ତମୀ ବାଳିକା (ସପ୍ତମ ଝିଅ)

अष्टमी बालिका

ଅଷ୍ଟମୀ ବାଳିକା (ଅଷ୍ଟମ ଝିଅ)

नवमी बालिका

ନବମୀ ବାଳିକା (ନବମ ଝିଅ)

दशमी बालिका

ଦଶମୀ ବାଳିକା (ଦଶମ ଝିଅ)

प्रथमं फलम्

ପ୍ରଥମ ଫଳମ୍ (ପ୍ରଥମ ଫଳ)

द्वितीयं फलम्

ଦ୍ଵିତୀୟଂ ଫଳମ୍ (ଦ୍ଵିତୀୟ ଫଳ)

तृतीयं फलम्

ତୃତୀୟଂ ଫଳମ୍ (ତୃତୀୟ ଫଳ)

चतुर्थं फलम्

ଚତୁର୍ବିଂ ଫଳମ୍ (ଚତୁର୍ଥ ଫଳ)

पश्च्रमं फलम्

ପଞ୍ଚମ ଫଳମ୍ (ପଞ୍ଚମ ଫଳ)

षष्ठं फलम्

ଷବଂ ଫଳମ୍ (ଷଷ୍ଠ ଫଳ)

सप्तमं फलम्

ସପ୍ତମଂ ଫଳମ୍ (ସପ୍ତମ ଫଳ)

अष्टमं फलम्

ଅଷ୍ଟମ ଫଳମ୍ (ଅଷ୍ଟମ ଫଳ)

नवमं फलम्

ନବମଂ ଫଳମ୍ (ନବମ ଫଳ)

दशमं फलम्

ଦଶମାଂ ଫଳମ୍ (ଚତୁର୍ଥ ଫଳ)

प्रथमं श्रवणं द्वितीयं मननम्
तृतीयं स्मरणं चतुर्थं भाषणम् ।
पश्चमं पठनं षष्ठं च लेखनम्
सप्तमं गायनम् अष्टमं बोधनम् ।
नवमं नर्त्तनं दशमं कीर्त्तनम्
भाषाप्रशिक्षणं सामान्यं सोपानम् ।

ପ୍ରଥମଂ ଶ୍ରବଣଂ ଦ୍ଵିତୀୟଂ ମନନମ୍
ତୃତୀୟଂ ସ୍ମରଣଂ ଚତୁର୍ଥୀ ଭାଷଣମ୍ ।
ପଞ୍ଚମାଂ ପଠନଂ ଷଶଂ ଚ ଲେଖନମ୍
ସପ୍ତମଂ ଗାୟନମ୍ ଅଷ୍ଟମଂ ବୋଧନମ୍ ।
ନବମଂ ନର୍ତ୍ତନଂ ଦଶମ କୀର୍ତ୍ତନମ୍
ଭାଷାପ୍ରଶିକ୍ଷଣଂ ସାମାନ୍ୟ ସୋପାନମ୍ ।