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WBBSE Class 7 Math Solution Chapter 7 Areas Of Rectangles And Square Exercise 7 Solved Problems

⇒ Surface: A surface is that plane figure which has only length and breadth, but no thickness.

⇒ Rectangle: A parallelogram is called a rectangle D whose each angle is right angle.

⇒ In a rectangle opposite sides are parallel and equal in length.

⇒ In rectangle ABCD, AB = DC, AD = BC and ∠A= ∠B =∠C = ∠D = 90°

Read and Learn More WBBSE Solutions for Class 7 Maths

⇒ The greater side of a rectangle is its length and the smaller side is its breadth or width. The length and breadth are known as the dimensions of the rectangle.

⇒ Square: A rectangle whose adjacent side are equal in length is called a square.

In square PQRS, PQ = QR = RS = SP and ∠P = ∠Q= ∠R = ∠S = 90°

⇒ Perimeter: The total length of the bounding lines of a figure is known as its perimeter. The perimeter is the sum of the lengths of its all sides. [semi-perimeter is the half of perimeter]

⇒ Area: Area is the measure of the quantity of surface occupied by a figure.

“WBBSE Class 7 Maths Arithmetic Chapter 7 solved problems step-by-step”

Some important formulas:

1. Perimeter of rectangle = 2 (length + breadth)
⇒ Semi-perimeter of rectangle = length + breadth
⇒ Area of rectangle = length x breadth

2. Perimeter of square = 4 x length of each side
⇒ Area of square = (length of side)²

3. Area of 4 walls of a room = perimeter x height
= 2 x (length + breadth) x height

Points to Note:

1. 1 Arc 100 sq. mt
2. 1 Sq. mt = 10000 sq. cm

Wbbse Class 7 Maths Solutions

Question 1. Choose the correct answer 

1. The perimeter of a square of side 3.5 cm is

1. 10-5 cm
2. 14 cm
3. 12-25
4. None of these

Solution: The perimeter of a square of side 3-5 cm is (4 x 3-5) cm are 14 cm

So the correct answer is 2. 14 cm

2. The area of a rectangle whose length is 18 cm and breadth is 15 cm is

1. 66 cm
2. 66 sq. cm
3. 270 cm
4. 270 sq cm

Solution: The area of a rectangle is (18 x 15) sq. cm = 270 sq. cm

So the correct answer is 4. 270 sq cm

3. If the area of a square is 12.25 sq. cm, then its perimeter is

1. 14 cm
2. 3.5 cm
3. 28 cm
4. 4.5 cm.

Solution: The area of square is 12.25 sq. cm

Let the length of each side of the square is x cm [x > 0]
∴ Area = x² sq. cm

According to the question:

x² = 12.25

⇒ x=√12.25

⇒x= 3.5

∴ Length of each side of a square is 3.5 cm

Perimeter is (4 x 3.5) cm = 14.0 cm = 14 cm

So the correct answer is 1. 14 cm

“Areas of Rectangles and Squares Exercise 7 Class 7 WBBSE Maths full solutions”

Question 2. Write true or false

1. 1 Sq. km = 1000 Sq. m

Solution: 1 Sq. km = 1 km x 1 km

= 1000 m x 1000 m

= 1000000 sq.m

So the statement is false.

2. If the area of 3 cm square is 9 Sq. cm

Solution: The length of each side of a square is 3 cm

Area is (3 x 3) sq. cm = 9 sq. cm

So the statement is true.

3. If the length, breadth, and height of a room are 8m, 6,m, and 5m respectively, then the area of its four walls is 140 sq. m.

Solution: Area of four walls of a room
= 2 (length + breadth) x height sq. unit
= 2 (8+6) x 5 sq. m
= 2 x 14 x 5 sq. m
= 140 sq. m

So the statement is true.

Question 3. Fill in the blanks

1. 1 Sq. m = _____ Sq. Hm.

Solution: 1 Sq. m = 1 m x 1m

= \(\frac{1}{100} \mathrm{Hm} \times \frac{1}{100} \mathrm{Hm}\)

= \(\frac{1}{10000}\) Sq. Hm

= 0.0001 Sq.cm

1 Sq. m = 0.0001 Sq.cm

2. 5 Sq. cm = _____ Sq.mm.

Solution: 5 Sq. cm = 5 cm x 1 cm

= 5 x 10 mm x 10 mm

= 500 sq. mm

5 Sq. cm = 500 sq. mm

“WBBSE Class 7 Maths Chapter 7 Exercise 7 important questions and answers”

3. The area of a square piece of land is 3600 sq. m. The cost of fencing all around the land at ₹ 4.50 per metre is ₹ _____

Solution:

Given

The area of a square piece of land is 3600 sq. m

∴ The length of each side = √3600 m

= 60 m.

Perimeter = (4 x 60) m = 240 m.

The cost of fencing all around the land at ₹4.50 per metre is ₹(240 x 4.50) or 1080.

Question 4. Distinguish between 5 metres square and 5 square metres.

Solution: 5 metres square means the area of the square each of whose sides is 5 metres.

So area of the square is 5 x 5 sq. m or 25 sq. m.

Again 5 Sq. m is meant the area of a rectangle or square the product of whose length and breadth is 5 sq. mt.

(such as 5 m x 1 m, 10m x 1/2 m etc).

Question 5. The length and breadth of a rectangular plot of land are 40 m and 30 m respectively. There is path 3 m wide running all around the plot outside. Calculate the area of the path.

Solution:

Given

⇒ The length and breadth of a rectangular plot of land are 40 m and 30 m respectively. There is path 3 m wide running all around the plot outside.

⇒ The area of the rectangular plot of land of length 40 m and breadth 30 m without path is (40 x 30) sq. m or 1200 sq. m.

⇒ The length of the plot of land including the path is (40+ 3 x 2) m or 46 m and that of breadth is (30+ 3 x 2) m or 36 m.

∴ The area of the plot of land including the path is (46 x 36) sq. m = 1656 sq. m.

∴  Area of the path is (1656-1200) sq. m = 456 sq. m.

Question 6. The length of the rectangular room is thrice of its breadth. The cost of varnishing the floor of the room at ₹ 7.50 per square metre is 810. Find the perimeter of the room.

Solution:

Given

The length of the rectangular room is thrice of its breadth. The cost of varnishing the floor of the room at ₹ 7.50 per square metre is 810.

Let the breadth of the room is x m.
∴ Length is 3x m.

Area of the floor is (3x x x) Sq. m = 3x² Sq.m.

The cost of varnishing the floor of the room at ₹ 7.50 per sq. m is ₹(3x² × 7.50)

According to question, 3x² x 7.50 =₹ 810

⇒ x² = \(\frac{810}{3 \times 7 \cdot 50}\)

⇒x=√36=6

∴ Breadth of the room is 6 m and length is (6 x 3) m or 18 m.

Perimeter is 2 (18 +6) m = 48 m.

“Class 7 Maths Areas of Rectangles and Squares WBBSE solved examples”

Question 7. The cost of cultivation of a 25 m long piece of land in 150. If the breadth of the land be 10 m less, the cost would have been 75. Calculate the breadth of the land.

Solution:

Given

The cost of cultivation of a 25 m long piece of land in 150. If the breadth of the land be 10 m less, the cost would have been 75

Let the breadth of the land is x m.
Length of the land is 25 m.
Area of land is (25 x x) sq. m = 25 x sq. m

⇒ If the length of the land is 25 m and its breadth be (x 10)m. Then the area of land will be 25 (x-10)sq. m

⇒ The cost of cultivation of 25x sq. m land is ₹ 150

⇒ The cost of cultivation of 1 sq. m land is ₹  \(\frac{150}{25 x}\)

⇒ The cost of cultivation of 25(x-10) sq. m land is ₹ \(\frac{150 \times 25(x-10)}{25 x}\)

⇒ According to the question,

⇒ 150×1500 = 75x
⇒ 150x75x = 1500
⇒ 75x = 1500

⇒ x= 20

∴ The breadth of the land is 20 m.

Question 8. The length and breadth of a rectangular hall is 24 m and 15 m respectively. Find how many square tiles of side 5 dcm

Solution:

Given

⇒ The length and breadth of a rectangular hall is 24 m and 15 m respectively.

⇒ The area of floor is (24 x 15) sq. m = 360 sq. m

⇒ Area of each square tiles is (5 x 5) sq. dcm = \(\frac{25}{100}\) sq.m

‎ ∴ Number of tiles is \(\left(360 \div \frac{25}{100}\right)\)

= 1440

Question 9. The length, breadth and height of a room of my school are 8 m, 6 m and 5 m respectively.

1. Find the cost of cementing the floor at ₹ 75 per square metre.
2. Find the cost of whitewashing its ceiling at ₹ 52 per sq. m..
3. The room has 2 doors each 1-5 m wide and 1.8 m high and has 2 windows each 1-2 m wide and 1-4 m high. Calculate the cost of painting doors and windows at ₹ 260 per square metre.
4. Also calculate the total cost of plastering the walls without doors and windows at ₹ 95 per sq m and cost of painting the walls at ₹ 40 per sq. m.

Solution:

Given

The length, breadth and height of a room of my school are 8 m, 6 m and 5 m respectively.

1. Area of the floor =(8 x 6) sq. m = 48 sq. m
∴ The cost of cementing the floor at ₹ 75 per sq. m is (48 x 75) = 3600

2. Area of ceiling = Area of floor = 48 sq. m
∴ The cost of whitewashing its ceiling at ₹ 52 per sq. m is ₹ 48 x 52 = ₹  2496

3. Area of 2 doors is (2 x 1.5 x 1·8) sq. m = 5.4 sq. m
⇒ Area of 2 windows is = (2 x 12 x 1.4) sq. m = 3.36 sq. m

⇒ Total areas of doors and windows is (5.4 + 3.36) sq. m = 8.76 sq. m

⇒ The cost of painting doors and windows at ₹ 260 per sq. m is ₹ 260 x 8.76 = ₹ 2277.60

4. Area of 4 walls is 2 (8 + 6) x 5 sq. m =

∴ Area of walls without doors and windows is (140 – 8.76) sq. m = 131.24 sq. m

⇒ Cost of plastering the walls without doors and windows at 95 per sq. m is ₹(131-24 × 95) = ₹ 12467-80

⇒ Cost of painting the walls at ₹ 40 per sq.m is ₹ (131-24 x 40) = ₹ 5249-60

∴ Total cost is ₹ (12467-80 + 5249-60) = ₹ 17717-40.

“WBBSE Class 7 Maths Arithmetic Chapter 7 Exercise 7 practice problems”

Question 10. If the area of the square is 110.25 sq. cm then find its perimeter.

Solution:

Given

⇒ Area of square = 110.25 sq. cm

⇒ Length of each side = √110.25 cm = 10.5 cm

⇒ Perimeter is (10.5 x 4) cm = 42.0 cm = 42 cm

Question 11.  In the adjacent length and breadth of a rectangular land are 50m and 40m respectively. The two paths with 2m wide and are parallel to length and breadth. Find the total area of the two paths.

Solution:

Given

⇒ In the adjacent length and breadth of a rectangular land are 50m and 40m respectively. The two paths with 2m wide and are parallel to length and breadth.

⇒ The area of the path parallel to the length is (50 x 2) sq. m = 100 sq. m
⇒ The area of the path parallel to breadth is (40 x 2) sq. m = 80 sq. m

The area of common square region of side 2 m is (2 x 2) sq. m = 4 sq. m
∴ Total area of two paths is (100+ 804) sq. m = 176 sq. m

WBBSE Class 7 Math Solution Areas Of Rectangles And Squares

Areas Of Rectangles And Squares Exercise 7.1

Question 1. 1 sq. km. = (100 × 100) sq.
Solution:

1 sq. km. = (100 × 100) sq.Dm.

= 10000 sq.Dm.

1 sq. km. = 10000 sq.Dm.

Question 2. 1 sq. km. = (10 ×10) sq.Hm. = 1 00 sq.Hm.
Solution:  

1 sq. km. = (10 ×10) sq.Hm.

= 1 00 sq.Hm.

1 sq. km. = 1 00 sq.Hm.

Question 3. In the graph has an area of [3] sq.cm, approximatily sq.m.
Solution:

= \(\frac{3}{100 \times 100}\)sq.m, approximately

= 3 × 10 × 10 = 300 sq. mm, approximately.

Question 4. Area of 6 in the graph = [8] sq. cm.
Solution :

Area of 6 = 8 sq. cm.

= \(\frac{8}{10000 \times 10000}\) = sqHm.

= 0.00000008 sq.Hm.

= 8 × 100 × 100 sq.mm. = 80000 sq.mm.

Question 5. Let us write the areas of Fig No. 7 and 8 in sq.cm. sq.Dm. & sq m.
Solution :

Area of 7 = 4 sq.cm.

= \(\frac{4}{1000 \times 1000}\) sq.Dm.

= 0.000004 sq.Dm.

= \(\frac{4}{100 \times 100}\)

= 0.0004 sq m.

Area of 8 = 6 sq.cm = \(\frac{6}{1000 \times 1000}\) sq.Dm.

= 0.000006 sq.Dm.

= \(\frac{6}{100 \times 100}\)= 0.0006 sq

“How to calculate areas of rectangles and squares Class 7 WBBSE Maths Chapter 7”

Question 6. Let’s write the approximate value of areas of no 9 & 10 in ______ sq. cm _________ sq.cm.
Solution :

Area of  9 = 7 sq.cm. (Approx)

Area of 10 = 5 sq. cm. (Approx)

Class VII Math Solution WBBSE Areas Of Rectangles And Squares Exercise 7.2

Question 1. The height of the second new room of the village hospital is 7 m. Let us find the area of its 4,walls including doors and windows.
Solution :

Given

⇒ The height of the second new room of the village hospital is 7 m.

⇒ Length of the 2nd new room = 25 m

⇒ Breadth of the 2nd new room = 20 m

⇒ Height of the 2nd new room = 7m

∴ Area of 4 walls including doors & windows = 2 (L + B) x h

= 2 (25 + 20) × 7 sq m. = 630 sq m.

⇒ Area of 4 walls including doors & windows = 630 sq m.

Question 2. In the second room there are 2 doors of height 2 – 5 m. and 1 . 5 m. wide. Also, there are 2 windows which are 1 5 m. wide and 1 .8 m- high. Let’s calculate the cost of plastering its walls excluding the doors and windows at the rate of Rs. 75 per square meter.
Solution :

Given

⇒ In the second room there are 2 doors of height 2 – 5 m. and 1 . 5 m. wide. Also, there are 2 windows which are 1 5 m. wide and 1 .8 m- high.

⇒ Area of each door of the room

= 2 .5m. x 1.8 m. = 4.5sq m.

∴ Area of 2 doors of the room = 2 × 4. 50 sq m. = 9 sq m.

⇒ Area of each window of the room = 1.5m. × 1. 8 m. = 2.7 sq m.

∴ Area of 2 windows of the room = 2 × 2 . 7 sq m. = 5. 4 sq m.

Total area of 2 door & 2 window of the room

9.0 sqm. + 5.4 sqm. = 14. sqm.

∴ Total area of walls excluding 2 doors & 2 window

= (630 – 1 44) sqm = 6156 sq m.

∴ Cost of plastering wall excluding door & windows at the rate Rs. 75 per sq m.

= Rs. 6156 × 75.

= Rs. 46170.

Question 3. Let’s calculate the cost of painting the doors and windows of the second room at the rate of Rs. 300 per square meter.
Solution :

⇒ Total area of 2 doors & 2 windows = 2 × 45 + 2 × 27

= 9 + 54 = 144 sq m.

The cost of painting the doors & windows at the rate of Rs. 300 per sq m.

= Rs. 300 × 144 = Rs. 4320.

“WBBSE Class 7 Maths Chapter 7 Areas of Rectangles and Squares textbook solutions”

Question 4. Let’s calculate the cost of painting the ceiling with white color at the rate of Rs. 55 per square meter.
Solution:

⇒ Area of ceiling = L x B = 25m × 20m = 500 sq m.

⇒ The cost of painting the ceiling with white color at the rate Of Rs. 55 per sq m.

= Rs. 55 × 500 = Rs. 27,500.

Class VII Math Solution WBBSE Areas Of Rectangles And Squares Exercise 7.3

Question 1. ABCD is a rectangular plot of land. Let us find the area of a path 4m wide runs all around inside it, as seen in the figure.

Solution:

Given

⇒ ABCD is a rectangular plot of land.

⇒ AB = 54m A B

⇒ AD = BC = 28m

∴ Area of ABCD = 54m × 28m = 1 51 2 sq m.

⇒ Again, PQ = AB – 4m = (54 – 4) = 50 m

QR = (28 – 4 – 4) = 20m _

∴ Area PQRS = 50 × 20 sq m. = 1000 sq m. D’

∴ Area of the path = Area of ABCD – Area of PQRS.

= 1512 sqm – 1000 sq m.

= 512 sq m.

⇒ Area of the path = 512 sq m.

Question  2. The area of a square park PQRS is 2500 sq m. and a path 6m wide runs all around outside the park. Let us find the area of this path.

Solution:

Given

The area of a square park PQRS is 2500 sq m. and a path 6m wide runs all around outside the park.

Here the square PQRS = 2500 sq m. A

∴ PQ = V2500 = 50 m

∴  AB = PQ + 6m + 6m = (50 + 6 + 6) = 62m

∴ Area of ABCD = 62m × 62m = 3844 sq m.

∴  Area of PQRS = 2500 sq m.

∴  Area of the path = Area of ABCD – Area of PQRS

= 3844 sq m. – 2500 sq m.

= 1344 sq m.

Area of the path = 1344 sq m.

Areas Of Rectangles And Squares Exercise 7.4

Question 1. Let’s try to find out the area of the figure by counting the number of squares on graph paper

Solution:

Question 2. Let us calculate mentally and write.

1. Let’s calculate the perimeter of a square of side 4 cm.
Solution:

⇒ Perimeter of a square of side 4 cm.

= 4 × 4cm = 16cm.

⇒ The perimeter of a square of side 4 cm = 16cm.

2. Let’s calculate the area of a square whose perimeter is 20 cm.
Solution :

⇒ The perimeter of a square = 20 cm.

⇒ Length of one side of the square = \(\frac{20}{4}\) = 5 cm

⇒ Area of the square of each side 5 cm = 5cm × 5cm

= 25 sq.cm.

⇒ The area of a square whose perimeter is 20 cm = 25 sq.cm.

3. Let us write the value of the area of a rectangle whose length is 8 cm. and breadth is 5cm.
Solution :

⇒ Length of the rectangle = 8cm

⇒ Breadth of the rectangle = 5cm

∴ Area of the rectangle = 8cm × 5cm

= 40 sq. cm.

Area of the rectangle = 40 sq. cm.

4. 1 sq km. =  ___________ sq Dm.
Solution :

1 sq km. = (100 × 100) sq Dm.

= 10000 sq Dm.

5. 1 sq km. = _____sq Hm. 
Solution:

1 sq km =  \(\left(\frac{1}{100} \times \frac{1}{100}\right)\)sq Hm.

= \(\frac{1}{10000}\)  sq Hm.

= 0.0001 sq Hm.

“West Bengal Board Class 7 Maths Chapter 7 solved numerical problems”

6. Let us explain, what do 5 sq m. and 5 m. square mean.
Solution:

5 sq m. means an area of a field (any shape) = 5 sq m. but, 5m sq. means an area of a square whose each side = 5m.

∴ Area of that square = 5m × 5m = 25 sq m.

7. If the area of a square is 2 cm. square then the length of its side is [Hints: Area = 2cm. square = 2 cm × 2cm.]

The area of a square is 2cm. square, then the length of each side of the square

= 2cm & its area

= 2cm × 2cm = 4sq cm.

8. Let us draw a rectangle whose area is 30 sq cm. Let us find out what are its. possible lengths and breadths. Again, if the area of a rectangle is 40 sq cm. let us find its possible lengths and breadths.
Solution:

Area of a rectangle = 30 sq cm.

It’s length & breadth will be (10cm & 3cm), (15cm, 2cm), (6cm, 5cm) etc.

Area of a rectangle = 40 sq cm.

Then its possible length & breadth will be (2cm, 20cm); (4cm> 10cm)
(8cm, 5cm), (80cm, 0 5cm) etc.

9. Mihir made a square card of cardboard whose one side is 6 cm. let’s find the area of the card.
Solution:

One side of a square cardboard = 6cm.

Area of the cardboard = 6cm x 6cm = 36 sq cm.

10. The area of a 5m. square is _____________ sqm. [Let’s put the value]
Solution:

Area of a 5m. square = 5m × 5m = 25 sq m.

Question 3. On a rectangular piece of white paper,I draw two pictures as shown in the figure.

1. Let us find how much space of white paper the No. 1 picture is occupying.
Solution:

Area of white paper in No. 1 picture

= 8cm × 8cm = 64 sq.cm.

2. Let us find how much space of white paper picture the No. 2 is occupying.
Solution :

Area of white paperiri No. 2 picture

10cm × 6cm = 60 sq.cm.

3. Let us calculate the area of white space is left in the white paper, after drawing the pictures (1) and (2).
Solution:

⇒ Total area of white paper = 32cm × 20cm = 640 sq.cm.

⇒ The total area of the two pictures = (64 + 60) sq. cm. = 12f sq. cm.

∴ Area of the white space left in the white paper after drawing the picture (1) & (2)

= 640 sq.cm – 124 sq.cm. = 516 sq.cm.

Question 4. The length and breadth of a page of my exercise book is 15 cm. and 12 cm. respectively. A margin of width 2 cm. is drawn on all sides, and wrote inside the margin. Let’s also draw a rough sketch for the problem. Let’s us find the area of the portion on which I wrote. Let’s also find the area of the portion on which I have not written.
Solution :

Given

The length and breadth of a page of my exercise book is 15 cm. and 12 cm. respectively. A margin of width 2 cm. is drawn on all sides, and wrote inside the margin.

Area of the paper ABCD= 15cm × 12 cm = 180 sq.cm.

Area of portion (Excluding margin) where I . will write

= 11cm × 8cm = 88sq.cm.

Area of the portion on which I have not written

= 180 sq.cm – 88 sq.cm.

= 92 sq.cm

Area of the portion on which I have not written = 92 sq.cm

Question 5. Rajesh has a rectangular plot of land of length 36m. and breadth 24m. There is a path 2m. wide running all around the plot outside. Let us draw a rough sketch of the problem and calculate the following 

1. Length and breadth of the plot of land including the path.
Solution :

Given

⇒ Rajesh has a rectangular plot of land of length 36m. and breadth 24m. There is a path 2m. wide running all around the plot outside.

⇒ Length of the field AB = 40m.

⇒ Breadth of Jhe field BC = 28m.

⇒ Area of the field ABCD = 40m × 28m = 1 1 20 sq m. (Ans.)

2. The area of the plot of land without the path.
Solution :

⇒ Length of the plot without a path (PQ) = 36m.

⇒ Breadth of the plot without a path (PS) = 24m

⇒ Area of the plot without path (PQRS) = 36m x 24m = 864 sq m.

3. Area of the path.
Solution:

Area of the path = Area of the plot ABCD – Area of the plot PQRS

= 1120 sq m. – 864 sq m.

= 256 sq m.

Area of the path. = 256 sq m.

Question 6. Maria’s family has a square plot of land of side 20m. A path 1m. wide runs ail around it from outside. Let’s find the area of the path.
Solution :

Given

⇒ Maria’s family has a square plot of land of side 20m. A path 1m. wide runs ail around it from outside.

⇒ Area of the square plot = 20m x 20m = 400 sq m.

⇒ Length of the plot with path = (20 + 1 + 1 )m = 22m.

⇒ Breadth of the plot with path = (20 + 1 + 1)m = 22m.

∴ Area of the plot with path = 22m × 22m = 484 sq m.

∴ Area of the path = 484 sq m – 400 sq m. = 84 sq m.

Question 7. The area of a square piece of land is 6400 sq m. Let’s find the cost of fencing all around the land at Rs. 3 50 per meter.
Solution:

Given

⇒ Area of square land – 6400 sq m.

⇒ Length of each side of the square land – 6400 m = 80 m.

⇒ The perimeter of the square land – 4 × 80m – 320m

⇒ The cost of fencing all around the land at Rs. 350/m = Rs. 350 × 320

= Rs. 1120.

“WBBSE Class 7 Maths Exercise 7 Areas of Rectangles and Squares short and long questions”

Question 8. The length of Karim’s uncle’s plot of land is twice its breadth and its area is 578 sq m. Let us find the length, breadth, and perimeter of Karim’s uncle’s land.
Solution:

Given

The length of Karim’s uncle’s plot of land is twice its breadth and its area is 578 sq m.

In this length of the land = 2 × breadth

∴ Area of the land = Length × breadth = (2 x breadth) x breadth

= 2 × (breadth)²

According to the problem

2 × (breadth)² = 578

(breadth)² = \(\frac{578}{2}\)

= 289

Breadth =  17m.

Length = 2 × 17m = 34m.

And perimeter of the land = 2 × (Length + Breadth)

= 2 × (34 + 17)m

= 2 × 51m = 102m.

perimeter of the land = 102m.

Question 9. The length of the rectangular stage of a theatre is twice its breadth. It costs Rs. 6048 to cover the whole stage with tarpaulin. If one sq m. of tarpanlin costs Rs. 21. Let’s find the length and breadth of the stage.
Solution:

Given

The length of the rectangular stage of a theatre is twice its breadth. It costs Rs. 6048 to cover the whole stage with tarpaulin. If one sq m. of tarpanlin costs Rs. 21.

Total cost to cover the whole stage (Area) with tarpaulin = Rs. 6048

Cost of 1 sq m. tarpanlin = \(\frac{\text { Rs. } 6048}{\text { Rs. } 21}\)

∴ Area of the stage = is 288 sq m.

∴  Length x breadth = 288

or, (2 x breadth) x breadth = 288

or, 2 x breadth² = 288

Area of the tarpaulin = = 288 sq m.

or, (breadth)² = \(\frac{288}{2}\) = 144

∴ Breadth = \(\sqrt{144}\) = 12 m.

∴ Length = 2 × breadth = 2 × 12m

= 24m.

Question 10. Nazreen will put Jari border on her sari of length 5 5m and breadth 1 25m. She decides to put a border of width 2 5 cm., along the breadth of the sari. Let’s find which portion of the sari will have Jari and which area of sari will be without Jari.
Solution :

Given

Nazreen will put Jari border on her sari of length 5 5m and breadth 1 25m. She decides to put a border of width 2 5 cm., along the breadth of the sari.

Area of the sari = 550cm x 125cm. = 68750 sq. cm.

Area of the sari excluding Jari = (550 – 1 0) x (1 25 – 5) = 540cm x 1 20cm

= 64800 sq. cm.

The area of sari will have Jari = (68750 – 64800)sq.cm.

= 3950 sq.cm. = 395 sq m

Now, length of the sari without Jari = (550 – 2 × 5)cm = 540cm.

Breadth of the sari without Jari = (1 .25 – 2 × 2-5)cm = 1 20 cm

Area of Sari without Jari = 540 cm × 120 cm = 64800 sq m.

= 6.48 sq m.

Question 11. As seen in the figure, two 5m. wide paths divide a rectangular garden into four equal parts. The length and breadth of the rectangular garden are 60m and 40m respectively. Let us calculate the cost of paving the path at Rs. 80 per square meter. Let’s also find the area of each part of the land. ‘
Solution :

Given

As seen in the figure, two 5m. wide paths divide a rectangular garden into four equal parts. The length and breadth of the rectangular garden are 60m and 40m respectively.

Area of the garden = 60m  × 40m = 2400 sq m.

The area of the path along the length = 60m × 5m = 300 sq m.

Area of the path along the breadth = 40m × 5m = 200 sq m.

∴ The total area of the path = is 300 sq m. + 200 sq m. – (5m × 5m)

= 500 sq m. – 25 sq m.

= 475 sq m.

Cost of paving the path at the rare of Rs. 80/sq m.

= 475 x 80 = Rs. 38000

Area of the garden without path

= 2400 sq m. – 475 sq m. = 1925 sq m.

∴ Area of each part of the land = \(\frac{1925}{4}\) = 481 25sq m.

Question 12. The path that leads to our house is 2m wide. It divides our garden into two equal squares, as shown in the figure. The path is constructed for Rs. 8000 at the rate of Rs. 500 per sq m. Let us calculate the area of each square portion of garden. If the breadth of the whole plot of rectangular land on which the house is built is 4m, let us calculate the area occupied by the house itself.

Solution :

Given

The path that leads to our house is 2m wide. It divides our garden into two equal squares, as shown in the figure. The path is constructed for Rs. 8000 at the rate of Rs. 500 per sq m.

Total construction cost = Rs. 8000.

At the rate of Rs. 500 per sq m.

∴ Area of the path = \(\frac{8000}{500}\) = 16 sq m.

Width of the path = 2m

∴ Length of the path = \(\frac{16}{2}\)  = 8m.

The area of each square portion of the garden

= 8m ×  8m = 64 sq m.

Length of the house = (8 + 2 + 8)m = 1 8m

Breadth of the house = 4m (given)

∴ Area of house = 18m × 4m = 72 sq m.

“WBBSE Maths Class 7 Areas of Rectangles and Squares full chapter solutions”

Question 13. The cost of cultivation of a 30 m long piece of land is Rs. 150. If the breadth of the land be 5m less, the cost would have been Rs. 120. Let’s calculate the breadth of the land.
Solution :

Given

The cost of cultivation of a 30 m long piece of land is Rs. 150. If the breadth of the land be 5m less, the cost would have been Rs. 120.

Cost will be decreased by (Rs. 150 – Rs. 120) = Rs. 30

If the breadth decreased by 5m.

∴ 30 : 150 : 4 :: ?

∴ Breadth = \(\frac{150 \times 5}{30}\)

= 25 cm

The breadth of the land = 25 cm

Question 14. The length and breadth of a rectangular hall is 30m and 18m respectively. Let’s find how many square tiles of side 3 dcm. will be required for flooring the hall.
Solution :

Given

The length and breadth of a rectangular hall is 30m and 18m respectively.

Length of the hall = 30 m.= 300 dcm.

Breadth of the hall = 18 m = 180 dcm.

The area of each square tile = 3 dcm. x 3 dcm.

No. of tiles = \(\frac{300 \mathrm{dcm} \times 180 \mathrm{dcm}}{3 \mathrm{dcm} \times 3 \mathrm{dcm}}\)

= 100 × 60

= 6000

No. of tiles No. of tiles

“WBBSE Class 7 Maths Chapter 7 formulas for finding areas of rectangles and squares”

Question 15. Jakir has a 18m x 14m rectangular plot of land. There is a square portion of garden of sides 3 4m. Let’s draw a rough sketch and calculate the area of the plot of land without the garden. Let’s also find how many square tiles of side 2 dcm. will be required to cover this vacant portion of land.
Solution :

Given

Jakir has a 18m x 14m rectangular plot of land. There is a square portion of garden of sides 3 4m. Let’s draw a rough sketch and calculate the area of the plot of land without the garden.

Area of Jakir’s rectangular plot of land

= 18m × 1 4m = 252 sq m.

Area of square portion of garden of sides 3 4m

= 34m × 34m = 1156 sq m.

Area of the plot of land without a garden

= (252 – 11 -56) sq m, = 24044 sq m.

Area of each square tile = 2 dcm × 2 dcm.

= 4 sq dcm. = 0 01 sq m. I_

∴ No. of tiles required = \(\frac{240 \cdot 44}{0.04}\)

= 6011

Question 16. In the figures given below, let us find the areas of the colored portions

Question 17. Let us take the values of length and breadth on our own and find the area of the colored parts.
Solution:

In 1st figure:

The area of the colored portion

= (20cm × 2cm) + (1 5cm × 2cm) – (2cm × 2cm)

= 40 sq cm. + 30 sq cm. – 4 sq cm. = (70 – 4) sq cm.

= 66 sq. cm.

In 2nd figure:

The area of the colored portion

= (23cm × 15cm) – {(23 – 4) x (15 – 4)} sq cm.

= 345 sq cm. – (19cm × 11 cm)

= 345 sq cm. – 209 sq cm. = 136 sq cm.

In 3rd figure:

⇒ Area of the whole rectangle = 18m × 12m = 216 sq m.

⇒ Are of the write portion = 9m × (18 – 4)m. = 9m  × 14m

= 126 sq m.

⇒ Area of the coloured portion = 216 sq m. – 126 sq m. = 90 sq m.

In 4th figure:

⇒ Area of two vertical coloumn = (10cm × 4cm). ×  2 = 80 sq cm

⇒ Area of middle tile = 5cm × 2cm = 10 sq cm.

⇒ Total area of coloured portion = (80 + 10) sq cm. = 90 sq cm.

In the 5th figure let the length & breadth of the rectangles be 12m & 10m respectively & the width of the shaded portion = 2m

∴ Area of the shaded portion parallel to length = 12m × 2m = 24 sqm.

Area of the shaded porition parallel to breadth = 10m × 2m = 20 sqm.

∴  Area of the middle portion = 2m × 2m = 4sq m. .

∴  Total area of the shaded portion = 24 sq m + 20 sq m. – 4 sq m.

= 40 sq m.

Total area of the shaded portion = 40 sq m.

“West Bengal Board Class 7 Maths solved problems for areas and perimeters”

Question 18. The length, breadth and height of a room of my school are 8m, 6m and 5m respectively.

1. Find the cost of cementing the floor at Rs. 75 per square metre.
Solution :

Area of the floor = Length × breadth = 8m × 6m = 48 sq m.

Cost of cementing the floor at the rate of Rs. 75 sq m.

= Rs. 75 × 48

= Rs. 3600.

The cost of cementing the floor = Rs. 3600.

2. Let’s find the cost of whitewashing its ceiling at Rs. 52 per Sq m.
Solution :

Area of the ceiling = 8m x 6m = 48 sq m.

⇒ Cost of whitewashing its ceiling at the rate of Rs. 52/sq m.

= Rs. 52 × 48

= Rs. 2496.

⇒ The cost of whitewashing its ceiling = Rs. 2496.

“Class 7 WBBSE Maths Chapter 7 real-life problems on rectangle and square areas Exercise 7 solutions”

3. The room has 2 doors each 15 m wide and 18m high and has 2 windows each 12 m wide and 14 m high. Let’s calculate the cost of painting doors and windows at Rs. 260 per square metre,
Solution :

Given

⇒ The room has 2 doors each 15 m wide and 18m high and has 2 windows each 12 m wide and 14 m high.

⇒ Area of 2 doors = 2 × (1.8m  ×1.5m) = 2 × 27 sq m.

= 5.4 sq m.

⇒ Area of 2 windows = 2 x (1. 4m ×1.2m) = 2 ×1.68 sq m.

= 3 .36 sq m.

⇒ Total area of 2 doors and 2 windows

= 5.4 sq m + 3 .36 sq m. = 8 .76 sq m.

⇒ Cost of painting doors and windows at the rate of Rs 260/sq m.

= Rs 260 ×  8 .76 = Rs. 227760.

The cost of painting doors and windows = Rs. 227760.

4. Let’s also calculate the total cost ofplastering the walls, without doors and window at Rs. 95 per sq m and cost of painting the walls at Rs. 40 per sq m.
Solution :

⇒ Area of 4 walls of the room = 2 × (L + B) × H

= 2 × (8m + 6m) × 5m = 140 sq m.

⇒ Total area of 2 doors & 2 windows = 8 76 sq m.

⇒ Area of 4 walls without the area of doors & windows

= 140 sq m. – 876 sq m. = 131 -24 sq m.

⇒ Total cost of plastering and painting the walls

= Rs. 95 + Rs. 40 = Rs. 1 .35 per sq m.

⇒ Cost of plastering & painting of 131 24 sq m.

= Rs 1. 35 × 131 .24 = 1771740.

Question 19. Our local club room is square in shape of side 15m long, and 5m high. There are 4 doors in the club room each 15m wide and 2m high. Let’s calculate the cost of oil painting its 4 walls without doors at Rs. 350 per sq m.
Solution:

Given

⇒ Our local club room is square in shape of side 15m long, and 5m high. There are 4 doors in the club room each 15m wide and 2m high.

⇒ The area of 4 walls of the square room

= 2 × (L + L) × H = 2 × (15 + 15)m × 5m

= 300 sq m.

Area of 4 doors of the room

= 4 × 1.5m × 2m = 4 × 3 sq m. = 12 sq m.

Area of 4 walls without 4 doors

= (300 – 12) = 288 sqm.

Cost of oil painting = Rs. 350 per sq m.

∴  Total cost = Rs. 350 × 288 = Rs. 100800.

Question 20. Let’s draw a figure and calculate- within a rectangular plot of land, there is a square-shaped pond. There is a 3m wide paved path along its edge and on three sides of it, and on the other side there is an 18m broad garden. We calculated the area of the pond is ______________ sq m, the area of the path is _____________sq m.
Solution :

Let the length of the pond =.24m

∴ Area of the pond = 24m x 24m = 576 sq m.

Area of the path

= (30 × 3 + 24 × 3- 24 × 3) sq m.

= (90 + 72 + 72) sq m.

= 234 sq m.

Area of the path = 234 sq m.

“WBBSE Class 7 Maths Areas of Rectangles and Squares important formulas and methods”

Question 21. The length, breadth, and height of my rectangular room is _________ m, -_______ m and _________m. The areas of the 4 walls of my room including the doors and windows is ______________ sq m.
Solution :

⇒ The length of my room = (L) = 10m.

⇒ breadth of my room = (B) = 8m.

& Height of my room = (H) = 9m.

⇒ Area of 4 walls of my room = 2(L + B) × H

= 2(10 + 8) × 9 sq m.

= 324 sq m

Area of 4 walls of my room = 324 sq m

WB Class 7 Math Solution Arithmetic Chapter 6 Time And Distance Exercise 6 Solved Problems

⇒ Speed: When a moving body changes its position, it covers a certain distance at a certain time. The rate of change of distance is known as speed.

⇒ In other words, the distance covered by a moving body in a unit of time is known as speed. The distance covered is directly proportional to the time taken by the body.

1. \(\text { Speed }=\frac{\text { Distance covered }}{\text { Time required to cover the distance }}\)

2. \(\text { Time required }=\frac{\text { Distance covered }}{\text { Speed }}\)

3. Distance covered x = Speed x Required time.

Relative speed: Relative speed is the speed of one object relative to another object.

1. If two objects move in the same direction, then relative speed = difference of their speeds.
2. If two objects move in the opposite direction, their relative speed = the sum of their speeds.

Read and Learn More WBBSE Solutions for Class 7 Maths

Question 1. Choose the correct answer 

1. Ramen covered a distance of 276 km in 6 hours in a bus. The speed of the bus in km/ hr is

1. 48 km/hr
2. 46 km/hr
3. 40 km/hr
4. 42 km/hr.

Solution:

Given

⇒ Ramen covered a distance of 276 km in 6 hours in a bus.

=\(\frac{276}{6}\) km/hr

= 46 km/hr

So the correct answer is 2. 46 km/hr

“WBBSE Class 7 Maths Arithmetic Chapter 6 solved problems step-by-step”

2. A train of length 100 m. moving with a speed of 60 km / m. passes a tree. How long will it take to do so?

1. 4 sec
2. 5 sec
3. 6 sec
4. 7 sec

Solution:

Given

⇒ A train of length 100 m. moving with a speed of 60 km / m. passes a tre

⇒ To cross a tree, the train will have to travel a distance equal to its own length. So here the train will have to travel 100 m.

The speed of the train is 60 km/h.
60 km = 60 × 1000 m = 60000 m
1 hour = (60 x 60) sec = 3600 sec.

i.e., 60000 m is covered by the train in 3600 sec.

1 m is covered by the train in \(\frac{3600}{60000}\) sec

100 m is covered by the train in

= 6 sec

∴ The required time is 6 sec.

So the correct answer is 3. 6 sec

“Time and Distance Exercise 6 Class 7 WBBSE Maths full solutions”

3. A train runs at the rate of 40 km an hour. How far will it run in 21 minutes?

1. 7 km
2. 14 km
3. 18 km
4. 20 km.

Solution:

Given

⇒ A train runs at the rate of 40 km an hour.

⇒ In the mathematical language

Time (mins)
60
21

Distance (km)
40
?

⇒ Speed being constant the time and distance are in direct proportion.

∴ 60: 21 : : 40 : ? (Required distance)

∴ Required distance(4th Proportion) =

= 14 km

⇒ So the correct answer is 2. 14 km

“WBBSE Class 7 Maths Chapter 6 Exercise 6 important questions and answers”

Question 2. Write true or false 

1. A bus covers a distance of 110 km in 3 hrs 40 mins. The speed of the bus is 30 km. / h.

Solution:

Given

⇒ A bus covers a distance of 110 km in 3 hrs 40 mins.

⇒ 3 hrs 40 mins (3 x 60+40) mins. = 220 mins.

⇒ In mathematical language, the problem is

Required time (mins)
220
60

Distance covered (km)
110
?

⇒ The required time and distance covered are directly proportional.

So 220: 60 : :  110: ? (Required distance)

∴ Required distance (4th proportion) =

Speed of bus is 30 km / h

So the statement is true.

“Class 7 Maths Time and Distance WBBSE solved examples”

2. A man takes 12 hours to travel 72 km. He takes 5 hours to travel 48 km. Let he takes x hours to travel 48 km.

Solution:

Given

A man takes 12 hours to travel 72 km. He takes 5 hours to travel 48 km.

In mathematical language the problem is

Distance covered (km)
72
48

Time taken (hr)
12
x

The time taken and the distance covered is directly proportional.

So, 72: 48 : : 12: x

⇒ \(\frac{72}{48}=\frac{12}{x}\)

⇒ 72x = 12 x 48

⇒ x=8

He takes 8 hours to travel 48 km. So the statement is false.

Question 3. Fill in the blanks 

1. The distance covered in 3 hrs at 5.6 km/hr is _______ km

Solution: The distance covered in 1 hour is 5.6 km

The distance covered in 3.5 hours is (3 x 5.6) km = 16.8 km

2. The distance covered by a moving body is the product of its — and the time taken

Solution: Speed.

“WBBSE Class 7 Maths Arithmetic Chapter 6 Exercise 6 practice problems”

Question 4. To pass a bridge of length 230 m a train of length 170 m took 20 sec. Calculate to find how long will this train take to pass a 110 m long platform.

Solution:

Given

To pass a bridge of length 230 m a train of length 170 m took 20 sec.

When a train crosses a bridge, then the train has to cover a distance of its own length + the length of the bridge. That is (230+ 170) m or 400 m.

If a train crosses a platform of length 150 m, then the train has to cover a distance is (170+110) m or 280 m.

In mathematical language, the problem is

Distance (metre)
400
280

Time (sec)
20
? (x)[say]

Speed remains constant, and time and distance are in direct proportion.
∴ 400: 280 : :  20: x

⇒ \(\frac{400}{280}=\frac{20}{x}\)

⇒ 400x = 20× 280

⇒ x = 14

∴ The train would take 14 secs to cross the platform.

Question 5. A train passes two brides of lengths 210 m and 122 m in 25 sec and 17 sec respectively. Calculate the length and speed of the train.

Solution:

Given

A train passes two brides of lengths 210 m and 122 m in 25 sec and 17 sec respectively.

Let the length of the train be x m. When the train crosses a bridge of length 210 m then the train has to cover a distance of length (210 + x) m.

Similarly to cross the second bridge of length 122 m; the train has to cover a distance of length (122 + x) m.

In the mathematical language, the problem is,

Time taken (sec)
25
17

Distance covered (m)
(x + 210)
(x + 122)

Speed remains constant, time and distance are in direct proportion.
∴ 25: 17 : : (x + 210): (x + 122)

⇒ \(\frac{25}{17}=\frac{x+210}{x+122}\)

⇒ 25 (x+122) = 17 (x + 210)
⇒ 25x + 3050 = 17x+ 3570
⇒ 25x – 17x = 3570 – 3050
⇒ 8x= 520

⇒ x= 65

∴ The length of the train is 65 meter.

In 25 sec the train travels (65+ 210) m or 275 m

In 1 sec the train travels \(\frac{275}{25}\) m

In 1 hr or 3600 sec the train travels

= 39600 mt = 39.6 km

∴ The length of the train is 65 mt. and speed is 39.6 km/hr.

“How to solve time and distance problems Class 7 WBBSE Maths Chapter 6”

Question 6. Two trains of lengths 170 m and 180 m are approaching each other with speeds of 45 km/hr and 30 km/hr along two tracks. Calculate how long will they take to pass each other after meeting.

Solution:

Given

Two trains of lengths 170 m and 180 m are approaching each other with speeds of 45 km/hr and 30 km/hr along two tracks.

After the two trains meet, they would pass each other i.e., the two trains would simultaneously pass a distance equal to sum of their own length.

∴ The distance two trains will cover (170 + 180) m or 350 m.

When the two trains travel in opposite directions, in 1 hour they would cover a distance of (45+30) km or 75 km or 75000 m.

1 hour = 3600 sec.

In the mathematical language the problem is,

Distance (metre)
75000
350

Time (sec)
3600
?

Time and distance are direct proportion.
∴ 75000: 350 : : 3600 ?

∴ Time taken

= \(\frac{84}{5}\) sec. = 16.8 sec

∴ The train will pass each other in 16.8 sec.

Alternative Method:

75000 metres is covered in 3600 sec

1 metre is covered in \(\frac{3600}{75000}\) sec

350 metre is covered in

= \(\frac{84}{5}\) sec = 16.8 sec

WB Class 7 Math Solution Time And Distance

Time And Distance Exercise 6

Question 1. On Saturday, I was cycling at a speed of 13 km/hr. I covered a certain distance in 2 hrs. On Sunday cycling at the same time at a speed of 11 km/hr.I covered a certain distance. Let us find out on which day. I could travel a distance more in 2 hours and how much? Hence, keeping time constant, try to find a relation between speed and distance (direct or inverse proportion)
Solution :

Given

On Saturday, I was cycling at a speed of 13 km/hr. I covered a certain distance in 2 hrs. On Sunday cycling at the same time at a speed of 11 km/hr.I covered a certain distance.

My speed of cycling is 13 km/hr.

∴ I will cover in 2 hrs. = 13 ×  2 = 26 km.

Next day Sunday my speed is 1 1 km/hr.

∴  I will cover in 2 hrs. on Sunday = 2 ×11 =22 km.

∴ On Saturday I cover 26 – 22 = 4 km more.

∴  Speed & distance are in direct proportion.

“WBBSE Class 7 Maths Chapter 6 Time and Distance textbook solutions”

Question 2. On Monday,l went to market cycling at a speed of1 2 km/hr. But on Tuesday) went to market cycling at a speed of 15 km/hr. If the distance between my house and market is 2km, then let’s us find, on which day. I took less time to reach the market. Hence, keeping distance same, let’s form a relation between speed and time required (direct or inverse proportion).
Solution :

For Monday:

Given

On Monday,l went to market cycling at a speed of1 2 km/hr. But on Tuesday) went to market cycling at a speed of 15 km/hr. If the distance between my house and market is 2km,

Speed = 12 km/hr.

Distance = 2 km.

Time required = \(\frac{\text { Distance Covered }}{\text { Speed }}\)

= \(\frac{2 \mathrm{~km}}{12 \mathrm{~km} / \mathrm{hr} .}=\frac{2}{12} \mathrm{hr}\)

= \(\frac{1}{6} \mathrm{hr}\) hr.

= \(\frac{1}{6} \times 60\) min.

= 10 min

For Tuesday:

Speed = 15 km/hr. Distance = 2 km

Time required = \(\frac{\text { Distance Covered }}{\text { Speed }}\)

= \(\frac{2 \mathrm{~km}}{15 \mathrm{~km} / \mathrm{hr} .}=\frac{2}{15} \mathrm{hr}\)

= \(\frac{2}{15} \times 60\) min.

= 8 min

∴ On Tuesday I took less time to reach the market (10m – 8m) = 2m

Here as the distance same, when speed in increasing time is decreasing i.e. speed & time are in inverse proportion.

Question 3. Keeping speed constant, let’s find the relation between time and distance [let’s prepare a story and find the relation]
Solution :

A passenger train of length 156 meters is traveling with a speed of 24 km/hr. Find the time taken by this train to cross a goods train of length 1 64 metres,’ which .is at rest by the side of the path of the passenger train.

Total length of the two trains = (156m + 164m) = 320 meters

Speed the train = 24 km/hr. i.e. the train goes 24 km = 24000 m in 1 hr

= 60 × 60 = 3600 sec

Here distance & time are in direct proportion.

∴ 24000 : 320: : 3600:?

Time = \(\frac{320 \times 3600}{24000}\)

= 48 sec

The relation between time and distance= 48 sec

Question 4. I covered a distance of 12 km in 40 min in a bus. Let’s find the speed of the bus in km/hr.
Solution :

Given

I covered a distance of 12 km in 40 min in a bus.

Time and distance are in direct proportion

40 : 60: : 12 : ?

Distance = \(\frac{60 \times 12}{40}\)km

= 18 km

 Question 5. A train of length 100m, moving with a speed of 60 km/hr. passes a tree. How long will it take to do so, lets calculate.
Solution:

Given

⇒ A train of length 100m, moving with a speed of 60 km/hr. passes a tree.

⇒ Distance and time are in direct proportion

⇒ (60 × 1000): 100:: 60:?

⇒ Time = \(\frac{100 \times 60}{60 \times 1000}\)

= \(\frac{1}{10}\) min

= \(\frac{1}{10} \times 60\)

= 6 sec

It will take 6 sec. to do so.

“West Bengal Board Class 7 Maths Chapter 6 solved numerical problems”

Question 6. Moving with a uniform speed a taxi takes 6km 12 min to cover a distance of 217 km. Let us calculate how much time it would take to cover a distance of 273 km. (Mention the relation to calculate).
Solution :

Given

⇒ Moving with a uniform speed a taxi takes 6km 12 min to cover a distance of 217 km.

⇒ 6 hr 12 min = (6 × 60 + 12) min = 372 min

Here, Distance and time are in direct proportion.

217: 273:: 372:?

Time = \(\frac{273 \times 372}{217}\)

It would take 7 hr. 48 m. to cover the distance.

Question 7. Today, Ayanda of our locality covered a distance of 100 km in 2 hrs 5 min. on his motorbike. But Shibuya covered the same distance on his cycle in 6 km 40 min. Let’s write the ratio of their speeds.
Solution:

Given

Today, Ayanda of our locality covered a distance of 100 km in 2 hrs 5 min. on his motorbike. But Shibuya covered the same distance on his cycle in 6 km 40 min.

2 hrs 5 min = (120 + 5) min

6 hrs 40 min = (360 + 40) min = 400 min

Ayanda covered in 125 min a distance of 1 00 km. = \(\frac{125}{100}\) min

∴ He covered in 1 min a distance of km.

∴  His speed is \(\frac{125}{100}\) km/min = \(\frac{4}{5}\) km/rnln.

Shibuda covered in 400 min a distance of 100 km.

∴  He covered in 1 min a distance of \(\frac{100}{400}\) km.

∴  His speed is \(\frac{100}{400}\) km/min. = \(\frac{100}{400}\) km/min.

∴  Ratio of their speed = \(\frac{100}{400}\) km/min : \(\frac{100}{400}\) km/min.

⇒ \(\frac{4}{5}\): \(\frac{1}{4}\) = 16:5

Question 8. Moving with a uniform speed a railway wagon covers a distance of 49.5 km. in 2hrs 45 min. How long it will take to reach a station situated at a distance of 58.5 km?
Solution :

Given

Moving with a uniform speed a railway wagon covers a distance of 49.5 km. in 2hrs 45 min.

Distance & Time are in direct proportion

49.5 : 58.5 : : \(\frac{11}{4}\): ?

Time = \(\frac{58.5 \times \frac{11}{4}}{49.5}\)

= \(\frac{585}{495} \times \frac{11}{4}\)

= \(\frac{13}{4}\)

= 3 hr 15 min

∴ It will take 3 hr. 15 min. to reach the station

“WBBSE Class 7 Maths Exercise 6 Time and Distance short and long questions”

WB Class 7 Math Solution Question 9. My uncle went to Panchla on his motorbike, worked there for an hour, and returned home after 3hrs 30min. If he maintained a uniform speed of 40 km per hour, let us find the distance of Panchla from his home.
Solution :

Given

My uncle went to Panchla on his motorbike, worked there for an hour, and returned home after 3hrs 30min. If he maintained a uniform speed of 40 km per hour,

Total time taken 3 hrs 30 min but he worked there for 1 hr.

Total time for journey to go & return = 2 hr 30 min

= 2 \(\frac{1}{2}\) hr

= \(\frac{5}{2}\) hr

Time and Distance are in direct proportion,

1: \(\frac{5}{2}\) :: 40 ?

Distance = \(\frac{\frac{5}{2} \times 40}{1}\)

= 100

∴  2 ×  distance to Panchla from his home = 100 km.

∴ Distance to Panchla from his home = \(\frac{100}{2}\)

= 50 km.

Distance to Panchla from his home = 50 km.

Question 10. A bus left Kolkata at 7:30 AM and without halting anywhere on its way reached Digha at 12 noon. If the uniform speed of the bus is 45 km/ hour, let us find the distance of Digha from Kolkata.
Solution :

Given

The bus left Kolkata at 7:30 AM & reached Digha at 1 2 noon.

Total time required to go Digha from Kolkata

= (12 Noon) -(7:30 AM) = 4 \(\frac{1}{2}\) hr

= \(\frac{9}{2}\) hr

1:\(\frac{9}{2}\):: 45?

∴  Distance = \(\frac{\frac{9}{2} \times 45}{1}\)

= \(\frac{405}{2}\) km

= 202. 5km

∴  Distance from Digha to Kokata = 202.5 km.

“WBBSE Maths Class 7 Time and Distance full chapter solutions”

Question 11. A 70 m long train runs at a speed of 75 km/hr. Let us calculate how long it would take to pass a platform of length 105 metre.
Solution :

Given

A 70 m long train runs at a speed of 75 km/hr.

To pass a platform a train has to cover the length of platform + length of the train = (105 + 70) m = 175 m

Speed of the train = 75 km/hr.

∴  Distance & time are in direct proportion

75000: 175:: (60 × 60):?

Time = \(\frac{60 \times 60}{75000} \times 175\)

= \(\frac{42}{5}\) sec

= 8 \(\frac{2}{5}\)

∴  It would take 8 \(\frac{2}{5}\) sec. to pass the platform

WB Class 7 Math Solution Question 12. A 90 m long train takes 25 sec to pass a pole. Let me calculate the speed of the train in km/hr.
Solution :

Given

A 90 m long train takes 25 sec to pass a pole.

To pass the pole, a train has to cover its own length.

Time & Distance are in direct proportion.

25 :60 × 60 : : 90 : ?

∴  Distance = \(\frac{60 \times 60 \times 90}{25}\)m

= 12.96 km

∴  Speed of the train = 12.96 km/hr.

Question 13. To pass a bridge of length 250m a train of length 150m took 30 sec. Let’s calculate how long will this train take to pass a 130 m-long platform.
Solution:

Given

To pass a bridge of length 250m a train of length 150m took 30 sec.

Length of the train + length of the bridge

= (250 + 1 50) m = 400 m.

And length of platform + length of the train = (1 30 + 1 50) m = 280 m

Time & distance are in direct proportion.

400 : 280: : 30:?

Time = \(\frac{280 \times 30}{400}\)

= 21 sec

∴  This train will take 21 sec. to pass the platform

Question 14. A passenger in a train found that the train took 15 sec to pass the platform. If the speed of the train is 60km per hour, let us find the length of the platform.
Solution :

Given

A passenger in a train found that the train took 15 sec to pass the platform. If the speed of the train is 60km per hour

Here the speed of the train = 60 km/hr. = 60000 m/hr. ,

The train took 15 sec to cross a platform

Distance = \(\frac{15 \times 60000}{60 \times 60}\)

∴  Length of the platform = 250 m.

Question 15. A train takes 4 sec to pass a telegraph post and 20 sec to pass a 264 m-long bridge. Let us find the length of the train and also its speed.
Solution :

Given

A train takes 4 sec to pass a telegraph post and 20 sec to pass a 264 m-long bridge.

In 20 sec the train covered its own length + the length of the bridge (264 m) & in 4 sec the train covered its own length (Subtracting) In 16 sec the train goes 264 m

In 4 sec the train goes \(\frac{264}{16} \times 4\) = 66m

Length of the train = 66 m.

4 : 3600 : : 6 : ?

Distance = \(\frac{3600 \times 66}{4} \mathrm{~m}\)

= \(\frac{3600 \times 66}{4 \times 1000}\) km

= 59.4 km

∴  Speed of the train = 59.4 km/hr.

“WBBSE Class 7 Maths Chapter 6 speed, time, and distance problems”

WBBSE Class 7 Math Solution Question 16. A train passes two bridges of lengths 21 0m and 1 22m in 25 sec and 17 sec respectively. Let us calculate the length and speed of the train.
Solution :

Given

A train passes two bridges of lengths 21 0m and 1 22m in 25 sec and 17 sec respectively.

In 25 sec, the train covered its own length + a bridge of length 210m & in 17 sec, it covered its own. Length + a bridge of length 122m (Subtracting) we get in 8 sec, the train goes 88 m

Time & distance are in direct proportion.

8 : 3600 : : 88 : ?

Distance = \(\frac{3600 \times 88}{8}\)

= 39600m

= 39.6 km

∴  Speed of the train = 39.6 km/hr.

Now again:

Time & distance are in direct proportion.

8: 15:: 88:?

Distance = \(\frac{25 \times 88}{8}\)

∴  Length of the train = (275- 210) m = 65 m

Question 17. A 100 m long train, remaining at a speed of 48 km/hr passes a tunnel in 21 sec. Let’s calculate the length of the tunnel.
Solution :

Given

A 100 m long train, remaining at a speed of 48 km/hr passes a tunnel in 21 sec.

Speed of the train = 48 km/hr

60 × 60:21 :: 48 × 1000:?

∴ Distance = \(\frac{21 \times 48 \times 1000}{60 \times 60}\) = 280 m

∴ Length of the tunnel = Total distance – Length of train

= (280 – 100)m = 180 m

Length of the tunnel = 180 m

“West Bengal Board Class 7 Maths solved problems for speed, time, and distance”

Question 18. A train takes 1 0 sec to pass a man standing on the platform 1 50m long and passes the platform in 22 sec. Let us calculate the length and speed of the train.
Solution :

Given

A train takes 1 0 sec to pass a man standing on the platform 1 50m long and passes the platform in 22 sec.

In 22 sec. the train goes 150 m long platform + its own length

In 1 0 sec. the train goes its won length

(Subtracting) in 12 sec the train goes 150 m

12: 10:: 150:?

∴ Distance = \(\frac{10 \times 150}{12}\)= 125 m/hr

∴ Length of the train = 125 m

Again to find speed:

12 : (60×60) : : 150 : ?

∴ Distance = \(\frac{(60 \times 60) \times 150}{12}\) m/hr

= \(\frac{60 \times 5 \times 150}{1000}\) km

= 45 km

∴ Speed = 45 km/hr.

Question 19. Two trains of length 250 m and 200 m are approaching each other in two tracks side by side at speeds of 45 km/hr. and 36 km/hr. Let us find, after meeting each other, how long will they take to pass each other. [Let’s put numbers ourselves].
Solution :

Given

Two trains of length 250 m and 200 m are approaching each other in two tracks side by side at speeds of 45 km/hr. and 36 km/hr.

Total length of the two trains (250 + 200) m = 450 m.

Relative speed of the two train = (45 + 36) km/hr. = 81 km/hr

Time & distance are in direct proportion.

∴ (81 x 1000): 450:: 3600:?

∴ Required time = \(\frac{450 \times 3600}{81 \times 1000}\) = 20 sec

∴ They will take 20 sec to pass each other.

 Question 20. A goods train 250m long is running at a speed of 33 km/hr. Behind it another mail train 200m long and running at a speed of 60 km/hr. It is running in same direction on a different track & after meeting the goods train overpasses it. Let us find how long will the mail train take to overpass the goods train.
Solution :

Given

A goods train 250m long is running at a speed of 33 km/hr. Behind it another mail train 200m long and running at a speed of 60 km/hr. It is running in same direction on a different track & after meeting the goods train overpasses it.

Relative speed of the two train, when they are running in the same direction

= (60 – 33) = 27 km/hr.

Total distance covered by the mail train to over passes the other train

= (250 + 200) = 450 m.

Time & distance are in direct proportion.

∴ (27x 1 000): 450:: (60 × 60):?

∴ Time = \(\frac{450 \times 60 \times 60}{27 \times 1000}\)

= 60 sec

= 1 min

∴ The mail train will take 1 min. to overpass the goods train.

Class 7 Math Solution WBBSE Arithmetic Chapter 5 Square Root Of Fractions Exercise 5 Solved Problems

1. Square root of a fraction = \(\sqrt{\frac{\text { Numerator }}{\text { Denominator }}}\)

Example: \(\sqrt{1\frac{9}{16}}=\sqrt{\frac{25}{16}}=\frac{\sqrt{25}}{\sqrt{16}}=\frac{5}{4}=1 \frac{1}{4}\)

2. Perfect square decimal number: A decimal number whose square root is a finite decimal number is termed a perfect square decimal number.

Example: 0.8 x 0.8 (0.8)² = 0.64

Hence 0.64 is a perfect square decimal number and 0-8 is a square root.

[The perfect square of any decimal number should contain an even number of digits offer the decimal point]

3. To find the square root of the perfect square decimal number by factorization. Method

Read and Learn More WBBSE Solutions for Class 7 Maths

Method 1

By converting decimal numbers to its fractions.

Example: \(\sqrt{1 \cdot 44}=\sqrt{\frac{144}{100}}=\frac{\sqrt{144}}{\sqrt{100}}\)

= \(\frac{\sqrt{2 \times 2 \times 2 \times 2 \times 3 \times 3}}{\sqrt{2 \times 2 \times 5 \times 5}}\)

= \(\frac{\sqrt{2^2 \times 2^2 \times 3^2}}{\sqrt{2^2 \times 5^2}}=\frac{2 \times 2 \times 3}{2 \times 5}\)

= \(12\) = 1.2

∴ The square root of 1-44 is 1.2

“WBBSE Class 7 Maths Arithmetic Chapter 5 solved problems step-by-step”

Method 2

We will consider the decimal number as a whole number (by removing the decimal sign)

Example: Find the square root of 1.96

The whole number without the decimal point in 1.96 is 196.

= \(\sqrt{2^2 \times 7^2}\)

=2×7 14

Since in the square decimal number, there are 2 digits after the decimal point.

So, in the square root of 1.96, there will be 1 digit to the right after the decimal point.

∴ \(\) = 1.4

The rule for putting decimal points:

4. To find the square root of a perfect square decimal number by the method of division.

To find a square root there must be an even number of digits after the decimal point. Hence from the extreme right of the decimal point, pairs of digits are marked by an arrow sign as done on the right-hand side (if can not be paired ‘O’ is to be put to the extreme right of the decimal point)

Then we proceed by the same method as is done for finding the square root of a whole number by division.

Example: Find the values of \(\sqrt{2 \cdot 4025}\)

∴ The square root of 2.4025 is 1.55

“Square Root of Fractions Exercise 5 Class 7 WBBSE Maths full solutions”

 5. To find out the square root of numbers that are not perfect squares by the method of divisions and find their approximate values up to 3 decimal places.

[In case of division if a decimal comes we take down one ‘0’ every time after the decimal, but for finding square root by division, 2 zeros are taken down after the decimal]

Example: Find the approximate value, of √5 up to 3 decimal places.

Question 1. Choose the correct answer 

1. The area of a square root of \(1 \frac{496}{729}\) is

1. \(2 \frac{11}{27}\)

2. \(\frac{27}{65}\)

3. \(1 \frac{11}{27}\)

4. \(3 \frac{11}{27}\)

Solution:

So the correct answer is 1. \(2 \frac{11}{27}\)

“WBBSE Class 7 Maths Chapter 5 Exercise 5 important questions and answers”

2. The least positive integer which will divide the fraction \(\frac{245}{64}\) to make a perfect fraction is

1. 2
2. 3
3. 7
4. 5

Solution:
\(\frac{245}{64}=\frac{5 \times 7 \times 7}{2 \times 2 \times 2 \times 2 \times 2 \times 2}=\frac{7^2 \times 5}{2^2 \times 2^2 \times 2^2}\)

So \(\frac{245}{64}\) is to be divided by the least positive integer 5 to make it a perfect square fraction.

So the correct answer is 4. 5

Question 2. True or false

1. The square root of 0.0256 is 0.016

Solution: √0.0256=0.16

The statement is false.

2. The value of

\(\left(\sqrt{\frac{16}{25}}+\sqrt{\frac{36}{49}}\right) \sqrt{\frac{9}{16}}-\sqrt{\frac{9}{16}} \times \sqrt{\frac{16}{25}} \text { is } \frac{9}{14}\)

Solution:

The statement is true.

“Class 7 Maths Square Root of Fractions WBBSE solved examples”

Question 3. Fill up the blanks 

1. The value of \(\sqrt{240 \cdot 25}+\sqrt{2 \cdot 4025}+\sqrt{0 \cdot 024025}\) is_____

Solution: \(\sqrt{240 \cdot 25}+\sqrt{2 \cdot 4025}+\sqrt{0 \cdot 024025}\)

= 15.5+1.55+0.155 = 17.205

Question 4. The product of two positive numbers is \(\frac{6}{5}\) and their quotient is \(\frac{32}{15}\) Find the numbers.

Solution: Let the two positive numbers are x and y respectively.

According to the question, x x y= \(\frac{6}{5}\) and \(\frac{x}{y}=\frac{32}{15}\)

⇒ \(x^2=\frac{64}{25}\)

= \(x=\sqrt{\frac{64}{25}}=\frac{8}{5}\)

x x y = \(\frac{6}{5}\)

\(\frac{8}{5}\) x y = \(\frac{6}{5}\)

= \(\frac{3}{4}\)

∴ The two number are \(\frac{8}{5}\) and \(\frac{3}{4}\)

Question 5. Arrange the following in the descending order of their magnitude.

\(\sqrt{\frac{4}{25}}, \sqrt{\frac{9}{16}}, \sqrt{\frac{16}{49}}, \sqrt{\frac{1}{36}}\)

Solution: \(\sqrt{\frac{4}{25}}=\frac{2}{5}=\frac{2 \times 84}{5 \times 84}=\frac{168}{420}\)

∴ \(\frac{315}{420}>\frac{240}{420}>\frac{168}{420}>\frac{70}{420}\)

⇒ \(\sqrt{\frac{9}{16}}>\sqrt{\frac{16}{49}}>\sqrt{\frac{4}{25}}>\sqrt{\frac{1}{36}}\)

“WBBSE Class 7 Maths Arithmetic Chapter 5 Exercise 5 practice problems”

Question 6. Find which decimal number is to be added to 0.75; So that the square root of the sum will be 2.

Solution: The required number is 2² -0.75 = 4 – 0.75 = 3.25

Question 7. The area of a square is 213-16 sq. cm. Find its perimeter.

Solution:

Given

The area of a square is 213-16 sq. cm

∴ The length of each side = √213.16 cm = 14.6 cm

∴ The perimeter is (14.6 x 4) cm = 58.4 cm

Question 8. Find the approximate value √15 up to 3 decimal places.

Solution:

Given √15

∴ √15 = 3. 873

“How to find square root of fractions Class 7 WBBSE Maths Chapter 5”

Question 9. Find the least decimal number that must be subtracted from 0.000679 to make it a square decimal number.

Solution:

Given 0.000679

As (0·026)² = 0.000676

∴ The last decimal number (0.0006790-0.00676) or 0.000003 be subtracted from 0-000679 to make it a square decimal number.

Question 10. Find the values of \(\sqrt{2-(0 \cdot 01)^2}\) upto 3 decimal places.

Solution:

Given \(\sqrt{2-(0 \cdot 01)^2}\)

\(\sqrt{2-(0 \cdot 01)^2}\)= \(\sqrt{2-(0 \cdot 0001)}\)

= \(\sqrt{1.9999}\)

≈1.414

“WBBSE Class 7 Maths Chapter 5 Square Root of Fractions textbook solutions”

Question 11. A room is one and a half as long as its breadth and the area of its floor is \(98 \frac{37}{54}\) sq. m. Find its perimeter.

Solution:

Given

⇒ A room is one and a half as long as its breadth and the area of its floor is \(98 \frac{37}{54}\) sq. m.

⇒ Let the breadth of the room is x m.

⇒ Length is \(\left(1 \frac{1}{2} \times x\right)\) m or, \(\frac{3 x}{2}\) m

⇒ Area = \(\left(\frac{3 x}{2} \times x\right)\) sq.m.= \(\frac{3 x^2}{2}\) sq.m.

⇒ According to question \(\left(\frac{3 x}{2} \times x\right)\) = \(98 \frac{37}{54}\)

⇒ \(x=\sqrt{\frac{5329}{81}}\)

⇒ \(x=\frac{73}{9}\)

Breadth of room is \(\frac{73}{9}\) m and length is

= \(\frac{73}{6}\)

Perimeter = 2\(\left(\frac{73}{9}+\frac{73}{6}\right)\) m

=2 x \(\left(\frac{146+219}{18}\right)\) m

= \(\frac{365}{9} \mathrm{~m}=40 \frac{5}{9} \mathrm{~m}\)

Class VII Math Solution WBBSE Square Root Of Fractions

Square Root Of Fractions Exercise 5.1

Question 1. Let’s find the square of the following fractions.

1. \(\frac{4}{5}\)
Solution:

Square of (\(\frac{4}{5}\))²

= \(\frac{4}{5}\)

= \(\frac{16}{25}\)

The square of \(\frac{4}{5}\) = \(\frac{16}{25}\)

2. \(\frac{6}{7}\)
Solution :

Square of \(\frac{6}{7}\)

= (\(\frac{6}{7}\))²

= \(\frac{36}{49}\)

The square of \(\frac{6}{7}\) = \(\frac{36}{49}\)

3. \(\frac{8}{10}\)
Solution:

Square of \(\frac{8}{10}\)

= (\(\frac{4}{5}\))²

= \(\frac{16}{25}\)

The square of \(\frac{8}{10}\) = \(\frac{16}{25}\)

4. \(\frac{11}{12}\)
Solution:

Square of \(\frac{11}{12}\)

= (\(\frac{11}{12}\))²

= \(\frac{11 \times 11}{12 \times 12}\)

= \(\frac{121}{144}\)

The square of  \(\frac{16}{25}\) = \(\frac{121}{144}\)

Question 2. Let’s find the square root of the following

1. \(\frac{16}{25}\)
Solution:

⇒ \(\sqrt{\frac{16}{25}}=\sqrt{\frac{4 \times 4}{5 \times 5}}\)

= \(\sqrt{\frac{4^2}{5^2}}\)

= \(\frac{4}{5}\)

The square root of \(\frac{16}{25}\) = \(\frac{4}{5}\)

2.\(\frac{9}{64}\)
Solution:

⇒ \(\sqrt{\frac{9}{64}}=\sqrt{\frac{3 \times 3}{8 \times 8}}\)

= \(\sqrt{\frac{3^2}{8^2}}\)

= \(\frac{3}{8}\)

The square root of \(\frac{9}{64}\) = \(\frac{3}{8}\)

3. \(\frac{36}{121}\)
Solution:

⇒ \(\sqrt{\frac{36}{121}}=\sqrt{\frac{6 \times 6}{11 \times 11}}\)

= \(\sqrt{\frac{6^2}{11^2}}\)

= \(\frac{6}{11}\)

The square root of \(\frac{36}{121}\) =  \(\frac{36}{121}\)

4. \(\frac{144}{169}\)
Solution:

⇒ \(\sqrt{\frac{144}{169}}=\sqrt{\frac{12 \times 12}{13 \times 13}}\)

= \(\sqrt{\frac{12^2}{13^2}}\)

= \(\frac{12}{13}\)

The square root of \(\frac{144}{169}\) = \(\frac{12}{13}\)

5. \(\frac{225}{289}\)
Solution:

⇒ \(\sqrt{\frac{225}{289}}=\sqrt{\frac{15 \times 15}{17 \times 17}}\)

= \(\sqrt{\frac{15^2}{17^2}}\)

= \(\frac{15}{17}\)

The square root of \(\frac{225}{289}\) = \(\frac{15}{17}\)

Class VII Math Solution WBBSE Square Root Of Fractions Exercise 5.2

Question 1. Let’s find the least positive integer which will multiply the given fractions to make them perfect square fractions.

1. \(\frac{64}{147}\)
Solution:

Given

⇒ \(\frac{64}{147}=\frac{8^2}{7^2 \times 3}\)

∴ \(\frac{64}{147}\) Is not perfect square fraction

To make if a perfect square, we have to multiply 3 with

∴  \(\frac{64}{147}\) × 3

=  \(\frac{64}{49}\)

= \(\frac{8^2}{7^2}\)

∴ \(\frac{64}{49}\) will be a perfect square fraction.

∴ The required number is 3

2. \(\frac{25}{162}\)
Solution:

Given

⇒ \(\frac{25}{162}\)

= \(\frac{5 \times 5}{9 \times 9 \times 2}\)

∴ \(\frac{5^2}{9^2 \times 2}\) is not perfect square fraction.

To make \(\frac{25}{162}\) a perfect square fraction we have to multiply by 2. i.e

= \(\frac{25}{162}\) × 2 = \(\frac{25}{81}\)

= \(\frac{5^2}{9^2}\),  It Will be a Perfect square fraction.

∴ The required number is 2

3. \(\frac{100}{128}\)
Solution:

Given

⇒ \(\frac{100}{128}\) = \(\frac{25}{32}\)

= \(\frac{5^2}{4^2 \times 2}\) which is not perfect square fraction

To make \(\frac{25}{32}\) a perfect square fraction we have to multiply it by 2

i.e \(\frac{25}{32}\) × 2

= \(\frac{25}{16}\)

= \(\frac{5^2}{4^2}\)

∴ The required number is 2

4. \(\frac{81}{288}\)
Solution:

Given

⇒ \(\frac{81}{288}\) = \(\frac{9}{32}\)

= \(\frac{3^2}{4^2 \times 2}\) which is not a perfect of square fraction

To make it a perfect square fraction, we have to multiply it by 2

i.e \(\frac{9}{32}\) × 2

⇒ \(\frac{9}{16}\)

= \(\frac{3^2}{4^2}\)

∴ The required number is 2

Question 2. Let’s find the least positive integer which will divide the given fraction to make the perfect square fraction.

1. \(\frac{450}{625}\)
Solution :

Given

⇒ \(\frac{450}{625}\)

= \(\frac{18}{5}\)

= \(\frac{3^2 \times 2}{5^2}\) Which is not a perfect square fraction.

To make it a perfect square fraction, we have to divide it by 2.

⇒ \(\frac{450}{625}\) × \(\frac{1}{2}\)

= \(\frac{225}{625}\)

= \(\frac{9}{25}\)

= \(\frac{3^2}{5^2}\), which is a perfect square fracation.

The required number is 2.

2. \(\frac{320}{121}\)
Solution:

Given

⇒ \(\frac{320}{121}\) x \(\frac{8^2 \times 5}{5^2}\)

To make it a perfect square fraction, we have to divide it by 5.

⇒ \(\frac{320}{625}\) × \(\frac{1}{5}\)

⇒ \(\frac{64}{121}\)

= \(\frac{8^2}{11^2}\), Which is perfect square fraction.

∴ The required number is 5.

3. \(\frac{245}{64}\)
Solution:

Given

⇒ \(\frac{245}{64}\) = \(\frac{49 \times 5}{64}\)

= \(\frac{7^2 \times 5}{8^2}\), Which is not a perfect square fraction

To make it a perfect square fraction we have to divide it with 5

⇒ \(\frac{245}{64} \times \frac{1}{5}\)

=  \(\frac{49}{64}\)

= \(\frac{7^2}{8^2}\) , which is the Perfect square fraction.

∴ The required number is 5.

4. \(\frac{243}{14}\)
Solution:

Given

⇒ \(\frac{243}{144}\)= \(\frac{27}{16}\)

=  \(\frac{3^2 \times 3}{4^2}\) , Which is not a Perfect square fraction.

⇒ To make it a perfect square fraction we have to divide it by 3

∴ \(\frac{3^2 \times 3}{4^2} \times \frac{1}{3}\)

= \(\frac{3^2}{4^2}\), Which is a perfect square fraction..

∴ The required number is 3.

Class VII Math Solution WBBSE Square Root Of Fractions Exercise 5.3

Question 1. The area of a square is \(\frac{1089}{625}\) sq.cm. Let’s find the length of its side.
Solution :

Given

⇒ The area of a square = \(\frac{1089}{625}\) sq.cm.

⇒ Length of its side = \(\sqrt{\frac{1089}{625}}\)

= \(\sqrt{\frac{3 \times 3 \times 11 \times 11}{25 \times 25}}\) cm

= \(\frac{3 \times 11}{25}\) cm.

= \(\frac{33}{25}\) cm.

Length of its side = \(\frac{33}{25}\) cm.

“West Bengal Board Class 7 Maths Chapter 5 solved numerical problems”

Question 2. Let us find the square root of the following fractions

1.  3\(\frac{22}{49}\)
Solution: 

Given

3\(\frac{22}{49}\)

⇒ 3 \(\frac{22}{49}\) = \(\frac{169}{49}\)

= \(\frac{13 \times 13}{7 \times 7}\)

= \(\frac{13^2}{7^2}\)

⇒ \(\sqrt{3 \frac{22}{49}}=\sqrt{\frac{13^2}{7^2}}\)

= \(\frac{13}{7}\)

= \(1 \frac{6}{7}\)

The square root of 3\(\frac{22}{49}\) = \(1 \frac{6}{7}\)

2. \(\frac{375}{1215}\)
Solution:

Given

⇒   \(\frac{375}{1215}\)

= \(\frac{25}{81}\)

= \(\frac{5^2}{9^2}\)

⇒ \(\sqrt{\frac{375}{1215}}=\sqrt{\frac{5^2}{9^2}}\)

= \(\frac{5}{9}\)

The square root of  \(\frac{375}{1215}\) = \(\frac{5}{9}\)

3. 6 \(\frac{433}{676}\)
Solution:

Given

6 \(\frac{433}{676}\)

⇒ \(\frac{433}{676}\)

=  \(\sqrt{\frac{4489}{676}}=\sqrt{\frac{67^2}{26^2}}\)

= \(\frac{67}{26}\)

= \(2 \frac{15}{26}\)

The square root of  6 \(\frac{433}{676}\) = \(2 \frac{15}{26}\)

4. 1\(\frac{496}{729}\)
Solution:

Given

1\(\frac{496}{729}\)

⇒ \(1 \frac{496}{729}=\frac{1225}{729}\)

= \(\frac{5 \times 5 \times 7 \times 7}{3 \times 3 \times 3 \times 3 \times 3 \times 3}\)

= \(\frac{5^2 \times 7^2}{3^2 \times 3^2 \times 3^2}\)

= \(\sqrt{1 \frac{496}{729}}=\sqrt{\frac{5^2 \times 7^2}{3^2 \times 3^2 \times 3^2}}\)

= \(\frac{5 \times 7}{3 \times 3 \times 3}\)

= \(\frac{35}{27}\)

The square root of  1\(\frac{496}{729}\) = \(\frac{35}{27}\)

5. \(\frac{324}{576}\)
Solution:

Given

⇒  \(\frac{324}{576}\)

= \(\frac{9}{16}\)

= \(\frac{3^2}{4^2}\)

The required square = \(\sqrt{\frac{3^2}{4^2}}\)= \(\frac{3}{4}\)

Question 3. With what should the square root of \(\frac{121}{169}\)be multiplied to give 1, let’s find.
Solution :

Given

⇒ \(\sqrt{\frac{121}{169}}=\sqrt{\frac{11^2}{13^2}}\)

= \(\frac{11}{13}\)

= \(\frac{11}{13}\) × 1

∴ The required number = 1 × \(\frac{13}{11}\)

= \(\frac{13}{11}\)

The square root of \(\frac{121}{169}\) = \(\frac{13}{11}\)

Question 4. Two positive numbers are such that one is twice the other. The product of these two numbers is 1 \(\frac{17}{32}\), let’s find the numbers

Given

Two positive numbers are such that one is twice the other. The product of these two numbers is 1 \(\frac{17}{32}\),

The product of two number = 1\(\frac{17}{32}\)

As one is twice the other

∴ Square of 1st number \(\frac{49}{32}\) × \(\frac{1}{2}\)

= \(\frac{49}{64}\)

= \(\frac{7^2}{8^2}\)

∴ 1st number = \(\sqrt{\frac{7^2}{8^2}}=\frac{7}{8}\)

∴  2nd number = 2 × \(\frac{7}{8}\) = \(\frac{7}{8}\)

Question 5. Let’s find a fraction, which when multiplied by itself gives \(6 \frac{145}{256}=\frac{1681}{256}\)
Solution :

⇒ \(6 \frac{145}{256}=\frac{1681}{256}\)

= \(\frac{41^2}{16^2}\)

The required no. = \(\sqrt{\frac{1681}{256}}\)

= \(\sqrt{\frac{41^2}{16^2}}=\frac{41}{16}\)

Question 6. By which fraction should \(\frac{49}{91}\) be multiplied, so that the square root of the product is 1 Let’s find.
Solution :

⇒ \(\frac{49}{91}\)×  _________ = 1²

∴ The required number = 1 × \(\frac{91}{49}\)

= 1\(\frac{42}{49}\)

= 1 \(\frac{6}{7}\)

Question 7. Let’s find, by which fraction should \(\frac{35}{42}\) be multiplied so that the square root of the product is 2.
Solution :

⇒ \(\frac{35}{42}\)×  _________ = 2²

∴ The required number = \(\frac{4 \times 42}{35}\)

= \(\frac{24}{5}\)

= 4 \(\frac{4}{5}\)

WBBSE Class 7 Math Solution Question 8. Let’s find the least positive integer which when multiplied — makes it a perfect square.
Solution :

⇒ \(\frac{9}{50}=\frac{3^2}{5^2 \times 2}\) is not a Perfect Square

If we multiply 2 with it, it will be \(\frac{3^2}{5^2 \times 2} \times 2=\frac{3^2}{5^2}\)= a perfect

∴ The required number = 2.

Question 9. The product of two positive numbers is \(\frac{14}{15}\) and their quotient is \(\frac{14}{15}\), let’s find the numbers.
Solution :

Given

The product of two positive numbers is \(\frac{14}{15}\) and their quotient is \(\frac{14}{15}\)

Let one number = x & the other number = y

∴ x × y = \(\frac{14}{15} \& \frac{x}{y}=\frac{35}{24}\)

∴  (x+y) × = \(\frac{x}{y}=\frac{14}{15} \times \frac{35}{24}\)

= \(\frac{49}{36}=\frac{7^2}{6^2}\)

x = \(\left(\frac{7}{6}\right)^2\)

And y = \(\frac{14}{15} \times \frac{6}{7}\)

= \(\frac{4}{5}\)

One number = \(\frac{7}{6}\) & the other number = \(\frac{4}{5}\)

Question 10. The product of two positive numbers is \(\frac{16}{50}\) and their quotient is \(\frac{1}{2}\) let’s find the numbers.
Solution :

Given

The product of two positive numbers is \(\frac{16}{50}\) and their quotient is \(\frac{1}{2}\)

Let one number be x & the other number is y.

∴ x ×  y = \(\frac{16}{50}\) × \(\frac{x}{y}\)

= \(\frac{1}{2}\)

∴ (x ×  y) ×  (\(\frac{x}{y}\)) = \(\frac{16}{50} \times \frac{1}{2}\)

= \(\frac{4}{25}\)

= \(\frac{2^2}{5^2}\)

(x)² = \(\left(\frac{2}{5}\right)^2\)

∴ x = \(\frac{2}{5}\)

∴ y = \(\frac{16}{50}\) × \(\frac{5}{2}\)

= \(\frac{4}{5}\)

= One number = \(\frac{2}{5}\) and other number \(\frac{4}{5}\)

Question 11. \(\sqrt{\sqrt{\frac{9}{64}}+\sqrt{\frac{25}{64}}}\) let’s find its value
Solution:

Given

⇒ \(\sqrt{\sqrt{\frac{9}{64}}+\sqrt{\frac{25}{64}}}=\sqrt{\sqrt{\frac{3^2}{8^2}}+\sqrt{\frac{5^2}{8^2}}}\)

= \(\sqrt{\frac{3}{8}+\frac{5}{8}}\)

= \(\sqrt{\frac{3+5}{8}}\)

= \(\sqrt{\frac{8}{8}}\)

=\(\sqrt{1}\)

= 1

\(\sqrt{\sqrt{\frac{9}{64}}+\sqrt{\frac{25}{64}}}\) = 1

Question 12. \(\sqrt{\frac{1}{4}}+\sqrt{\frac{1}{9}}-\sqrt{\frac{1}{16}}-\sqrt{\frac{1}{25}}\) let s find the value
Solution:

⇒ \(\sqrt{\frac{1}{4}}+\sqrt{\frac{1}{9}}-\sqrt{\frac{1}{16}}-\sqrt{\frac{1}{25}}c\)

= \(\sqrt{\frac{1^2}{2^2}}+\sqrt{\frac{1^2}{3^2}}-\sqrt{\frac{1^2}{4^2}}-\sqrt{\frac{1^2}{5^2}}\)

= \(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\)

= \(\frac{30+20-15-12}{60}\)

= \(\frac{50-27}{60}\)

= \(\frac{23}{60}\)

\(\sqrt{\frac{1}{4}}+\sqrt{\frac{1}{9}}-\sqrt{\frac{1}{16}}-\sqrt{\frac{1}{25}}\) = \(\frac{23}{60}\)

WBBSE Class 7 Math Solution Question 13. Arrange the following in the descending order of their magnitude
\(\)
Solution:

⇒ \(\sqrt{\frac{1}{16}}, \sqrt{\frac{1}{25}}, \sqrt{\frac{1}{36}}, \sqrt{\frac{1}{49}}\)

⇒  \(\sqrt{\frac{1}{16}}=\sqrt{\frac{1^2}{4^2}}=\frac{1}{4}\)

⇒  \(\sqrt{\frac{1}{25}}=\sqrt{\frac{1^2}{5^2}} \quad=\frac{1}{5}\)

⇒  \(\sqrt{\frac{1}{36}}=\sqrt{\frac{1^2}{6^2}} \quad=\frac{1}{6}\)

⇒  \(\sqrt{\frac{1}{49}}=\sqrt{\frac{1^2}{7^2}}, \quad=\frac{1}{7}\)

⇒  \(\frac{1}{4}>\frac{1}{5}>\frac{1}{6}>\frac{1}{7}\)

∴   \(\frac{1}{16}>\frac{1}{25}>\frac{1}{36}>\frac{1}{49}\)

“WBBSE Class 7 Maths Exercise 5 Square Root of Fractions short and long questions”

Question 14. Let’s find, by what magnitude is \((\sqrt{25}+\sqrt{81})\) more than \((\sqrt{16}+\sqrt{36})\)
Solution:

⇒ \((\sqrt{25}+\sqrt{81})\) – \((\sqrt{16}+\sqrt{36})\)

= (5 + 9) -(4 + 6) = 14 -10

= 4

Question 15. Let’s find the square roots of the following fractions

1. \(3 \frac{22}{49}\)
Solution:

⇒ \(\sqrt{3 \frac{22}{49}}\)

=  \(\sqrt{\frac{169}{49}}=\sqrt{\frac{13^2}{7^2}}\)

=  \(\frac{13}{7}\)

= 1 \(\frac{6}{7}\)

The square root of \(3 \frac{22}{49}\) = 1 \(\frac{6}{7}\)

2. \(7 \frac{57}{256}\)
Solution:

⇒ \(\sqrt{7 \frac{57}{256}}\)

= \(\sqrt{\frac{1849}{256}}=\sqrt{\frac{43^2}{16^2}}\)

= \(\frac{43}{16}\)

= 2 \(\frac{11}{16}\)

The square root of  \(7 \frac{57}{256}\) = 2 \(\frac{11}{16}\)

3. \(\frac{1089}{2025}\)
Solution:

⇒ \(\sqrt{\frac{1089}{2025}}\)

=  \(\sqrt{\frac{33^2}{45^2}}\)

= \(\frac{33}{45}\)

= \(\frac{11}{15}\)

The square root of \(\frac{1089}{2025}\) = \(\frac{11}{15}\)

4. \(3 \frac{814}{1225}\)
Solution:

⇒ \(\sqrt{3 \frac{814}{1225}}\)

= \(\sqrt{\frac{4489}{1225}}=\sqrt{\frac{67^2}{35^2}}\)

= \(\frac{67}{35}\)

= 1 \(\frac{32}{35}\)

The square root of \(3 \frac{814}{1225}\) = 1 \(\frac{32}{35}\)

We found that in perfect square decimal numbers, there are even numbers of digits after the decimal point

WBBSE Class 7 Math Solution Square Root Of Fractions Exercise 5.4

Question 1. Let’s find the value of the square of the following decimal numbers 

1. 0.7
Solution :

0.7 × 0.7 = 0.49

(0.7)² = 0.49

2. 0 .16
Solution :

= 0.16 × 0.16 = 0.0256

(0.16)² = 0 . 0256

3. 0. 08
Solution :

= 0 . 08 × 0 . 08 = 0. 0064

(0. 08) = 0 .0064

4. 0.25
Solution :

0 . 25 × 0 . 25 = 0 . 0625

(0-25)² = 0. 0625

Question 2. By counting the number of digits after the decimal point, let’s identify the square decimal numbers from the following decimal numbers 

1. 22.5
Solution:

= 22.5 – Here one digit after the decimal point, so it is not a square decimal number.

2. 1.44
Solution :

1 .44 – Here after the decimal point there is two digits, so it is a square decimal number.

3. 62.5.
Solution:

62.5 – Here after the decimal point there i§ one digit, so it is not a square decimal number.

4. 12.1
Solution:

12.1 – Here after the decimal point there is one digit, so it is not a square decimal number.

Question 3. Let’s find the square root of the following decimal numbers.

1.  4. 41
Solution:

⇒ \(\sqrt{4 \cdot 41}=\sqrt{\frac{441}{100}}\)

= \(\sqrt{\frac{7 \times 7 \times 3 \times 3}{10 \times 10}}\)

= \(\sqrt{\frac{7^2 \times 3^2}{10^2}}\)

= \(\frac{7 \times 3}{10}\)

= \(\frac{21}{10}\)

= 2.1

4. 41 = 2.1

2. 4. 41
Solution:

⇒  \(\sqrt{2 \cdot 25}=\sqrt{\frac{225}{100}}\)

= \(\sqrt{\frac{5 \times 5 \times 3 \times 3}{10 \times 10}}\)

= \(\sqrt{\frac{5^2 \times 3^2}{10^2}}\)

= \(\frac{5 \times 3}{10}\)

= \(\frac{15}{10}\)

= 0.16

4. 41 = 0.16

3. 0.0484
Solution:

⇒  \(\sqrt{0 \cdot 0.0484}=\sqrt{\frac{484}{10000}}\)

= \(\sqrt{\frac{2 \times 2 \times 11 \times 11}{100 \times 100}}\)

= \(\sqrt{\frac{2^2 \times 11^2}{100^2}}\)

= \(\frac{2 \times 11}{100}\)

= \(\frac{22}{100}\)

= 0.22

0.0484 = 0.22

WBBSE Class 7 Math Solution Square Root Of Fractions Exercise 5.5

Question 1.  Let’s find the square roots of the following decimal numbers by the method of division.

1. 0.000256
Solution :

⇒ \(\sqrt{0.000256}\)

∴ \(\sqrt{0.000256}\) = 0.016

2. 0. 045369
Solution :

⇒  \(\sqrt{0. 045369}\)

∴ \(\sqrt{0. 045369}\)= 0.213

3.1.0609
Solution :

⇒  \(\sqrt{1.0609}\)

∴ \(\sqrt{1.0609}\) = 1.03

4. 75.609
Solution :

⇒  \(\sqrt{75.609}\)

∴ \(\sqrt{75.609}\)

Question 2. The area of a square is 32.49 sq. cm. Let’s find the length of one side of the square.
Solution :

The area of a square = 32.49 sq. cm.

∴ The length of one side of the square = \(\sqrt{32.49}\) cm

= 5.7m

The length of one side of the square = 5.7m

Question 3. Let’s find the length of one side of a square whose area is equal to the sum of the areas of rectangles of areas 2.1214 sq. cm, and 2.9411 sq. cm.
Solution :

The area of the square = (2.1214 + 2.9411) sq. cm.

= 5.0625 sq. cm.

∴ The length of one side of the square =

= 2.25

The length of one side of the square = 2.25

Question 4. Let’s calculate what must be added to 0.28 so that the square root of the sum is 1.
Solution :

Square of 1 = 1

∴ The required number = 1 – 0.28 = 0.72

Question 5. Let’s find the square root of the product of 0.162 and 0.2
Solution:

= 0.18

The square root of the product of 0.162 and 0.2 = 0.18

Question 6. Let’s calculate the value of \(\sqrt{240.25}+\sqrt{2.4025}+\sqrt{0.024025}\)
Solution:

⇒ \(\sqrt{240.25}+\sqrt{2.4025}+\sqrt{0.024025}\)

15.5 + 1. 55 + 0.155 =17.205

Question  7. Of the two squares of areas 1.4641 sq.m, and 1.0609 sq.m, Let’s Find which one has a bigger side and by how much it is big.
Solution:

Area of the 1st square = 1 .4641 sq.m.

Area of the 2nd square = 1 .0609 sq.m.

∴ Each, side of the 1st square = \(\sqrt{1.4641}\)

= 1.21

The each side of the 2nd square = \(\sqrt{1.0609}\)

∴ Side of 1st square is bigger than the 2nd square by (1.21 – 1 . 03 ) m = 0 . 18 m

WB Class 7 Math Solution Question  8. The sum of the squares of 0.4 and 0.3 is the squares of which number, let’s find.
Solution :

(0.4)² + (0.3)² = 0.16 + 0.09 = 0.25

The required number = \(\sqrt{0.25}\)

= \(\sqrt{\frac{25}{100}}\)

= \(\frac{5}{10}\)

= 0.5

The sum of the squares of 0.4 and 0.3 is the squares of  0.5

Question  9. Let’s find the square root of the following by the method of division.

1. 2.56
Solution :

⇒ \(\sqrt{0.25}\)

2. 4.84
Solution :

⇒ \(\sqrt{4.84}\)

= 2.2

3. 5.76
Solution :

⇒ \(\sqrt{5.76}\)

= 2.4

4. 6.76
Solution :

⇒ \(\sqrt{6.76}\)

= 2.6

5. 0.045369
Solution :

⇒ \(\sqrt{0.045369}\)

= 0.213

6. 0.000169
Solution :

⇒ \(\sqrt{0.000169}\)

= 0.013

7. 76.195441
Solution :

⇒ \(\sqrt{76.195441}\)

= 8.729

8. 170.485249
Solution :

⇒ \(\sqrt{76.195441}\)

= 13.0.57

9. 5505.64
Solution :

⇒ \(\sqrt{5505.64}\)

= 74.2

“WBBSE Maths Class 7 Square Root of Fractions full chapter solutions”

Question 10. Let’s find the decimal number which when multiplied by itself gives the product as 1.1025.
Solution :

The required number \(\sqrt{1.1025}\)

= 1.05

Question 11. Let’s find, which decimal number is to be added to 0.75 so that square root of the sum will be 2.
Solution :

Square of 2 = 2² = 4

∴ The required number = 4 – 0.75 = 3.25

11. Let’s find, which decimal number is to be subracted from 48.09 so that the square root of the result is 5.7.
Solution :

Square of 5.7 = (5.7)² = 32.49

The required number = 48.09 – 32.49

=15.6

15.6 is to be subracted from 48.09 so that the square root of the result is 5.7

Question 12. Let’s find the least decimal number that must be subtracted from 0.000328 to make it a square decimal number (up to 6 decimal place
Solution :

⇒ \(\sqrt{0.000328}\)

If we subtract 0.000004 from 0.000328

It will be = 0.000324 = (0.01 8)²

The required number = 0.000004

Question 13. Let’s find the approximate value of the following 

1. \(\sqrt{6}\) (upto 2 decimal places)
Solution:

⇒ \(\sqrt{6}\)

= 2.45

2. \(\sqrt{8}\) (upto 2 decimal places)
Solution:

⇒ \(\sqrt{8}\)

= 2.828

= 2.83 (upto 2 decimal places)

3. \(\sqrt{11}\) (upto 3 decimal places)
Solution:

⇒  \(\sqrt{11}\)

= 3.3166

= 3.317 (up to 3 decimal places)

“WBBSE Class 7 Maths Chapter 5 square root of decimal fractions problems”

4. \(\sqrt{12}\) (upto 3 decimal places)
Solution:

⇒ \(\sqrt{11}\)

= 3.4641

= 3.464 (up to 3 decimal places)

Question 14. Let’s find the approximate value of 7T5 upto 2 – places of decimal. Let’s then square this approximate value to find how big or less it is from 15.
Solution :

= 3.872

= 3.87 (up to 2 decimal places)

Now (3.87)² = 14.9761

Difference

= 15.9761

= 0.0231

WBBSE Class 7 Math Solution Arithmetic Chapter 4 Approximation Of Values Exercise 4 Solved Problems

Approximate value: In our daily transactions it is not always possible to find the exact value of a quantity or to ascertain accurate values, lengths, weights, etc. of a thing.

⇒ So in calculations involving values, length, weights, etc, we have to take the approximate values just a little above or a little below the accurate values. This is called the approximate value of a quantity or the nearest value or measurement of a thing.

⇒ To find the approximate value up to certain places of the digit, if that digit is any number between 5 to 9, then 1 is added to the digit at that certain place, or else the same digit is kept.

⇒ The sign for approximate value is “≈” (almost equal)

Read and Learn More WBBSE Solutions for Class 7 Maths

Question 1. Determine the approximate value of 3746 up to tens, hundreds, and thousands of places.
Solution:

Given

3746

⇒ The number at the tens place is 4 and on its immediate right side is 6 which is between 5 to 9.

⇒ So up to tens place, the approximate value is (374 + 1) ten = 3750.

⇒ The digit on the hundreds place is 7 and on its immediate right side is 4. So the approximate value upto hundreds place is 3700.

⇒ The digit on the thousands place is 3 and on its right side we have 7, so the approximate value upto thousands place is 4000.

We can write:

3746 ≈ 3750 (up to tens place).
3746 ≈ 3700 (upto hundreds place)
3746 ≈ 4000 (upto thousands of places)

Question 2. Find the approximate value of 3-853672 up to 1, 2, 3, 4, and 5 decimal places and whole numbers.

Solution:

Given 3-853672

⇒ 3.853672 ≈ 3.8537
[6th place after decimal is 2]

⇒ 3.853672 ≈ 3.853672
[5th place after decimal is 7 hence I is added to 4th decimal place value 6, 6 + 1 = 7]

⇒ 3.853672 ≈ 3.854
[4th place after decimal is 6 hence 1 is added to 3rd decimal place value 3, 3+1 = 4]

⇒ 3.853672 ≈ 3.85
[3rd place after decimal is 3]

⇒ 3.853672 ≈ 3.9
[2nd place after decimal is 5 hence 1 is added to 1 decimal place value 4, 8+ 1 = 9]

⇒ 3.853672 ≈ 4
[1 place after decimal is 8 hence 1 is added to the whole number 3, 3+1 = 4]

“WBBSE Class 7 Maths Arithmetic Chapter 4 solved problems step-by-step”

Question 3. Find the approximate value of \(\frac{22}{7}\) upto 2, 3, 4, and 5 decimal places.

Solution:

Given \(\frac{22}{7}\)

\(\frac{22}{7}\) ≈ 3.142857……

\(\frac{22}{7}\) ≈ 3.14 [two decimal places]

\(\frac{22}{7}\) ≈ 3.143 [3 decimal places]

\(\frac{22}{7}\) ≈ 3.1429 [4 decimal places]

\(\frac{22}{7}\) ≈3-14286 [5 decimal places]

“Approximation of Values Exercise 4 Class 7 WBBSE Maths full solutions”

Question 4. Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km. Find one distance traveled by light in 1 second in kilometers approximately.

Solution:

Given

Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km.

1 mile = 1.6093 km

186000 miles = (186000 x 1.6093) km

= 299,329.8 km

≈ 299,330 km [1 place after the decimal is 8. Hence 1 is added to the whole number at the unit place i.e., 9 + 1 = 10]

Question 5. Divide 22 among 8 boys and 7 girls equally. Find how much each would get (approximated up to 2 places of the decimal). Also, find the total money received by 8 boys and that received by 7 girls. Then find the total money received by 8 boys and 7 girls and how much this total amount is more or less than ₹ 22.

Solution:

⇒ Total number of boys and girls is (8 + 7) or 15.

⇒ ₹ 22 is divided among 15 boys and girls.

⇒ Each gets = ₹ \(\frac{22}{15}\)

= \(\frac{2200}{15}\) paise

≈ 146.67 paise

⇒ The total money received by 8 boys is (146-67 x 8) paise
= 1173.36 paise
≈ ₹ 11.73.

⇒ The total money received by 7 girls is (146-67 x 7) paise
= 1026.69 paise
≈ ₹ 10.27

⇒ Total money received by 8 boys and 7 girls is (11.73 + 10.27) = ₹ 22.00 = ₹ 22

⇒ This amount is equal to 22.

“WBBSE Class 7 Maths Chapter 4 Exercise 4 important questions and answers”

Question 6. If \(\frac{2}{7}\) radian = 180°, find the value of 1 radian in degrees upto 2 places of decimals.

Solution:

Given

\(\frac{2}{7}\) radian = 180°

1 radian = \(\left(\frac{7}{22} \times 180\right)^{\circ}\)

= \(\left(\frac{1260}{22}\right)^{\circ}\)

= 57.27°

The value of 1 radian in degree upto 2 places of decimals = 57.27°

Question 7. Simplify the following up to 4 decimal places. \(0 \cdot 004872+13 \cdot \dot{725}+0 \cdot \dot{3} 6\)

Solution: \(0 \cdot 004872+13 \cdot \dot{725}+0 \cdot \dot{3} 6\)

=0.004872+13.725725………..+0.363636……

= 14.094233……

= 14.0942

WBBSE Class 7 Math Solution Approximation Of Values

Approximation Of Values Exercise 4.1

Question 1. Let’s write the approximate values of the following fractions correct up to two, three and four places of decimal.

1. \(\frac{13}{17}\)
Solution:

Given  \(\frac{13}{17}\)

⇒ \(\frac{13}{17}\) = 0.76470588.

≈ 0. 76 (Two decimal places)

≈  0.765 (Three decimal places)

≈  0.7647 (Four decimal places)

2. \(\frac{19}{29}\)
Solutions:

Given \(\frac{19}{29}\)

⇒  \(\frac{19}{29}\) = 0.6551724

≈  0.66 (Two decimal places)

≈  0.655 (Three decimal places)

≈  0. 6552 (Four decimal places)

Approximation Of Values Exercise 4.2

Question 1. Let’s write the approximate values of the following numbers up to 1 0’s, 1000s, and 1 0,000’s places.

Question 2. Let us divide Rs. 3 among 7 children and calculate how much Paise each wou,d get (approximately upto 2 places of decimal). Let’s then find total money received by 7 children and then try to find how more or less it will be than Rs. 3.
Solution:

Total amount- Rs. 3 = 300p

Total children = 7

Each will receive = \(\frac{300}{7}\) p

= 42.857≈ 42.86 p

∴ Total money received by 7 children = 42.86 x 7 p = 300.02 p

∴ 300.02 p – 300p = 02p more.

Question 3. 1 divided Rs. 22 among 8 boys and 7 girls equally. Let’s find how much each would get. (approximated up to 2 places of decimal). Also let’s find the total money received by 8 boys and that received by 7 girls. Then let’s find total money received by 8 boys and 7 girls and how much this total amount is more or less than Rs. 22.
Solution :

Total amount = Rs. 22 = 2200p.

Total number of boys & girls = 8 + 7=15

Each will get = \(\frac{2200}{15}\)

p = 1 46.666p

= 146.67 p (approx)

∴ 8 boys will receive = 146.67 x 8 = 1 173.36 p = Rs. 11.73

7 girls will received = 146.67 x 7 = 1026.69 p = Rs. 10.27p

Total amount = Rs. (11.73 + 10.27) = Rs. 22.

It is equal.

Question 4. Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km, let’s find the distance travelled by light in 1 second in Kilometres approximately (up to 3 decimal places of approximation).
Solution :

Given

Light travels 186000 miles in 1 second. Again 1 mile = 1.6093 km,

Distance travelled by light in 1 second

= 1 86000 mile = 1 86000 x 1 .6093 km.

= 299329.8 km = 299330 km.

“Class 7 Maths Approximation of Values WBBSE solved examples”

Question 5. Let’s write the approximate value up to 2 decimal places of 0.997.
Solution :

0.997 = 1.(up to 2 decimal) (approx).

Question 6. Let us fill in the gaps
Solution :

WBBSE Class 7 Math Solution Question 7. Let us write the approximate values of the following fractions, up to 2, 3 & 4 decimal places 

1. \(\frac{22}{7}\)
Solution :

Given \(\frac{22}{7}\)

⇒ \(\frac{22}{7}\) = 3.14 (upto 2 decimal places)

= 3.143 (up to 3 decimal places)

= 3.1429 (up to 4 decimal places)

\(\frac{22}{7}\) = 3.1429 (up to 4 decimal places)

2. \(\frac{3}{14}\)
Solution :

Given \(\frac{3}{14}\)

⇒ \(\frac{3}{14}\) = 0.21 (up to 2 decimal places)

= 0.214 (up to 3 decimal places)

= 0.2143 (4 decimal places)

\(\frac{3}{14}\) = 0.2143 (4 decimal places)

3. \(\frac{1}{5}\)
Solution :

Given \(\frac{1}{5}\)

⇒ \(\frac{1}{5}\) = 0.20 (upto 2 decimal places)

= 0.200 (up to 3 decimal places)

= 0.2000(upto 4 decimal places)

\(\frac{1}{5}\) = 0.2000(upto 4 decimal places)

 4. \(\frac{47}{57}\)
Solution :

Given

⇒  \(\frac{47}{57}\) = 0.82 (up to 2 decimal places)

= 0.825 (up to 3 decimal places)

= 0.8246 (up to 4 decimal places)

\(\frac{47}{57}\) = 0.8246 (up to 4 decimal places)

Question 8. Let’s write the approximate values of the following numbers up to lacs, thousands and hundreds place. 
Solution:

Question 9. Practical application of approximation.

1. 11 hours 9 min 40 sec – approximately the value up to min.
Solution: 11 hours 10 mins.

2. If the price of the shoe is written as Rs. 99.99, let us find what will be
its approximate value.
Solution: Rs. 100.

“WBBSE Class 7 Maths Arithmetic Chapter 4 Exercise 4 practice problems”

3. The length of a line segment is 1.59 cm, Let us find what is the
approximate value of the length.
Solution: 1. 6cm.

4. The weight Of poppy seed (post) is 102gm. Let’s find
approximately the weight for which money is charged.
Solution: 100

Class 7 Math Solution WBBSE Arithmetic Chapter 3 Proportion Exercise 3 Solved Problems

1. When the values of two ratios are equal, they are said to be in proportion and one is called proportional to the other.

2. Four quantities a, b, c and d are said to be in proportion or proportional when a: b : : c : d or a:b = c d [The symbol ‘: :’ stand for ‘as’]
a is called the 1st term of the proportions.

Similarly, b, c, and d are called the 2nd term, 3rd term and 4th term respectively.

So, \(\frac{1 \text { st term }}{2 \text { nd term }}=\frac{3 \text { rd term }}{4 \text { th term }}\)

⇒ 1st term x 4th term = 2nd term x 3rd term

i.e., The product of the extreme terms = The product of the two mean terms.

Example: 6 18 15: 45 are in proportional.

Read and Learn More WBBSE Solutions for Class 7 Maths

Here 6 x 45 18 x 15 = 270

Continued Proportion: Three quantities of the same kind are said to be in continued proportion when the ratio of the first to the second is equal to the ratio of the second to the third.

The second quantity is called a mean proportional between the 1st and 3rd.
If three quantities a, b and c are in continued proportion then a b : : b: c

⇒ \(\frac{a}{b}=\frac{b}{c} \Rightarrow b^2=a c\)

where b is called the mean proportional of a and c.

“WBBSE Class 7 Maths Arithmetic Chapter 3 proportion solutions”

Example: 3:6 : : 6: 12

Here 3 x 12 = 6 x 6 = 36

Types of Proportions: There are two types of proportion

1.  Simple proportion or Direct proportion
2. Inverse proportion

1. Simple proportion or Direct proportion: Simple proportion constitutes two ratios formed with simultaneous increase or decrease of two correlated quantities.

If two ratios a:b and c:d indicate the simultaneous increase or decrease, then the simple or direct proportion will be a: b: : c:d

2. Inverse proportion: If two ratios are such that one is equal to the inverse ratio of the other, then the proportion thus formed is an Inverse proportion.

Example: If two ratios a: b and c d then they form an inverse proportion.

i.e. a:b: : d:c [inverse of c:d]

or c: d: : b: a [inverse of a:b]

Question 1: Find if the pair of ratios given below are equal and also find if four members are in proportion

1. 6: 4 and 15: 10
2. 4-2: 1-4 and 21: 7
3. 5x: 6x and 3y: 4y [y ≠0, x ≠ 0]

Solution:

1.  6: 4 = 3: 2

15: 10  = 3: 2

∴ 64 and 15: 10 are equal.

Hence 6, 4, 15 and 10 are in proportion.
We can write 6: 4 : : 15: 10

2. 4.2: 1.4 = 3: 1

21: 7 = 3: 1

∴ 4.2: 14 and 21: 7 are equal.

Hence 4.2, 1.4, 21 and 7 are in proportion.
We can write 4.2: 1.4 : :  21: 7

3. 5x: 6x = 5: 6
3y: 4y = 3: 4

∴ 5x: 6x and 3y: 4y are not equal.

Hence 5x, 6x, 3y, 4y are not in proportion.

“WBBSE solutions for Class 7 Maths proportion exercise 3 solved problems”

Question 2. Verify whether the following numbers are in proportion

1. 3, 4, 6, 8
2. 12, 18, 10, 15
3. 18, 8, 15, 6

Solution:

1. 3, 4, 6, 8
Here, product of extreme = 3 x 8 24 and product of means = 4 x 6 = 24

∴ Product of extreme = product of means
Hence the four members are in proportion.

2. 12, 18, 10, 15
Here, product of extreme = 12 x 15 = 180 and product of means = 18 x 10 = 180

∴ Product of extreme = product of means
Hence the four members are in proportion.

3. 18, 8, 15, 6
Here, product of extreme = 18 x 6 = 108 and product of means = 8 × 15 = 120

∴ Product of extreme product of means
So, the four members are not in proportion.

Question 3. Form different proportionality with 6, 8, 12, 16:

Solution:

Given Numbers 6, 8, 12, 16

Question 4. Find the missing term from the given proportion 

1. *: 45 : 20
2. 6: * : : 10: 15
3. 8: 12 : : *: 24
4. 5: 10 : : 15: *

Solution:

1. *: 4 : : 5: 20

Let 1st term of the given proportion be x

So, x: 4 : : 5: 20

⇒ \(\frac{x}{4}=\frac{5}{20}\)

∴ 1st term is 1.

2. 6: * : : 10: 15

Let 2nd term is x

∴ 6: x : : 10:

⇒ \(\frac{6}{x}=\frac{10}{15}\)

⇒ 10x = 6 x 15

∴ 2nd term is 9.

3. 8: 12 : : *: 24

Let 3rd term is x
8: 12 : :  x: 24

⇒ \(\frac{8}{12}=\frac{x}{24}\)

⇒ 12x = 8 x 24

∴ 3rd term is 16.

4. 5: 10 : : 15: *

Let 4th term is x

⇒ \(\frac{5}{10}=\frac{15}{x}\)

⇒ 5x = 15 x 10

⇒ \(x=\frac{15 \times 10^2}{8}\)

⇒ x = 30

∴ 4th term is 30.

Question 5. Find whether the following sets of numbers are in continued proportion and write the proportionality

1. 6, 8, 12
2. 5, 25, 125
3. 4, 10, 25

Solution:

1. 6, 8, 12

6 × 12 = 72 ≠(8)²

i.e., 1st term × 3rd term ≠ (mean term)²

So, 6, 8, 12 are not in continued proportion.

2. 5, 25, 125

5 × 125 = 625 = (25)²

i.e., 1st term x 3rd term = (mean)²

So, 5, 25 and 125 are in continued proportion.
The proportionality is 5 25 25: 125

3. 4, 10, 25

4 x 25 = 100 = 10²

1st term x 3rd term = (2nd term)²

So, 4, 10 and 25 are in continued proportion.
The proportionality is 4: 10 : : 10 : 25

Question 6: Of the three numbers in continued proportion, 1st and 2nd numbers are 16 and 20 respectively. Find the third number of this proportion.

Solution: Let the 3rd term of the continued proportion be x.
So, 16, 20 and x. are in continued proportion.

∴ 16: 20 : : 20: x

⇒ \(\frac{16}{20}=\frac{20}{x}\)

⇒ 16x = 20 x 20 = 400

⇒ \(x=\frac{400}{16}=25\)

∴ 3rd term is 25.

“Class 7 WBBSE Maths Chapter 3 proportion exercise 3 step-by-step solutions”

Question 7. Find the mean proportion of

1. 9 and 16
2. a and b
3. 1.5 and 13.5

Solution:
1. Let the positive mean proportion of 9 and 16 is x

∴ 9: x : : x: 16

⇒ \(\frac{9}{x}=\frac{x}{16}\)

⇒ x² =144
⇒x= √144 = 12

∴ Mean proportion 9 and 16 is 12

2. Mean proportion of a and b is ±√ab

3. Mean proportion of 1.5 and 13.5 is

= \(\sqrt{\frac{15}{10} \times \frac{135}{10}}\)

= \(\sqrt{\frac{15 \times 15 \times 3 \times 3}{10 \times 10}}\)

= \(\frac{15 \times 3}{10}=\frac{45}{10}\)

= 4.5

Question 8. If 6 men can do a piece of work in 20 days. Find in how many days will 12 men finish the same work.

Solution:

Given 6 men can do a piece of work in 20 days.

In the mathematical language, the problem is,

Men (number)
6
12

Time (days)
20
?

For a particular work, if the number of persons increases, less time is required to finish the work.

But if the number of people decreases, the time required will be increased.

∴ So the number of men and time are in inverse proportion.

In proportionality, we take the inverse of the relation.
∴ 12: 6 : : 20: ?

Product of extreme = product of means
∴ 12 x 4th term = 6 x 20

⇒ 4 th term =

Hence 12 men will take 10 days to finish the work.

Question 9: Ramesh babu used 24 ploughs to cultivate his whole land in 15 days. Find if he wanted to cultivate the same land in 10 days, how many ploughs he would have needed?

Solution:

Given

Ramesh babu used 24 ploughs to cultivate his whole land in 15 days.

In the mathematical language the problem is,

Time (days)
15
10

Number of ploughs
24
? (x) [say]

To cultivate the whole land, as time decreases, the required number of ploughs will be increased. So the time and number of ploughs are in inverse proportion.

So, 10: 15 : : 24: x

⇒ \(\frac{10}{15}=\frac{24}{x}\)

⇒ 10 x = 24 x 15

∴ 36 ploughs would have been needed.

Wbbse Class 7 Maths Solutions

Question 10. In a flood relief camp, there is the provision of food for 4000 people for 190 days. After 30 days, 800 people went away elsewhere. Find for how many days the remaining food lasts for the remaining people in this camp.

Solution:

Given

In a flood relief camp, there is the provision of food for 4000 people for 190 days. After 30 days, 800 people went away elsewhere.

Let the required number of days be x.

In the mathematical language, the problem is,

No of people (by heads)
4000
4000 -800 = 3200

Time (days)
190 – 30 =160
x

As the number of people decreases therefore the certain quantity of food will cover more number of days for the same heads. So the number of people and the number of days are inversely proportional so,

3200: 4000 : : 160: x

⇒ \(\frac{3200}{4000}=\frac{160}{x}\)

∴ That food will go on for 200 days for 3200 people.

“Solved examples of proportion problems WBBSE Class 7 Maths”

Question11. When water freezes to ice, its volume increases by 10%. Find the ratio of a certain volume of water and its corresponding volume of ice.

Solution:

Given

When water freezes to ice, its volume increases by 10%

100 cubic units of water when freezes to ice, its volume becomes (100 + 10) cubic units or 110 units.

So the ratio of certain volume of water and its corresponding volume of ice is, 100: 110 or 10: 11

Question 12. The cost of 5 tables is 8000. Calculate the price of 8 tables.

Solution:

Given  The cost of 5 tables is 8000.

In the mathematical language the problem is,

No of Tables
5
8

Price of Tables (in rupees)
8000
?

If the number of tables increases the price will increases. If the number of tables decreases price will decrease.

So the number of tables and the price of the tables are directly proportional.

∴ 5: 8 : : 8000: ?

∴ Cost of tables (?) =

= ₹12800

The price of 8 tables = ₹12800

Question 13. The ratio of the ages of father and son is 5: 2. After 10 years the ratio of their ages will be 2: 1. Find the present ages of father and son.
years.

Solution:

Given

The ratio of the ages of father and son is 5: 2.

Let the father’s age is 5x years and son’s age is 2x years.

[where x is the common multiple and x > 0]

After 10 years the father’s age will be (5x + 10) years and son’s age will (2x + 10)

According to the conditions,

⇒ 5x+10 = 2(2x+10)

⇒ 5x+10 = 4x = 20

⇒ 5x – 4x = 20 – 10

⇒ x = 10

∴ The present age of the father is (5 x 10) years or 50 years and the age of the son is (2 x 10) years or 20 years.

Class 7 Math Solution WBBSE Proportion

Proportion Exercise 3.1

Question 1. Let’s find if the pair of ratios given below are equal and also find if four members are in proportion.

1. 7: 2 and 28: 8

Solution:

Here 7: 2 = (7 × 4) : ( 2× 4) (Multiplying both by 4)

= 28: 8

∴ 7 : 2:: 28: 8

∴ They are proportional. ,

2. 9: 7 and 18: 14
Solution:

Here 9 : 7 = (9 × 2) : (7 × 2) (Multiplying both by 2)

= 18 : 14

∴  9 : 7: : 18: 14

∴  They are proportional.

“WBBSE Class 7 Maths Chapter 3 important questions and answers”

3. 1.5 : 3 and 4.5: 9.
Solution :

Here 1 .5 : 3 = (1 .5 x 3) : (3 x 3) (Multiplying both by 3)

= 4.5 : 9

∴  1 .5 : 3 : : 4.5 : 9

∴  They are proportional.

4. 7 : 3 and 5 : 2.
Solution:

Here 7:3≠ 5:2

∴  They are not in proportional.

5. 3ab : 4aq and 6b : 8q.
Solution :

Here \(\left(3 a b \times \frac{2}{a}\right)\) : \(\left(4 a b \times \frac{2}{b}\right)\) = 6b : 8q

∴ 3ab ; 4aq : : 6b :.8q

∴ They are in proportion

6. 5.2: 6.5 and 4: 5
Solution:

Here \(\frac{5.2}{1.3}: \frac{6.5}{1.3}\) = 4:5

∴ 5.2: 6.5 :: 4 : 5

∴ They are in proportion.

7. 3y : 7y and 12 p : 28p.
Solution :

Here \(\left(3 y \times \frac{4 p}{y}\right):\left(7 y \times \frac{4 p}{y}\right)\)

= 12p :28p

∴ 3y : 7y : : 12p : 28y

∴ They are in proportion.

8. 5pq : 7pr and 15s : 21 q [ here a, q, y, p, r are not zeros ]
Solution :

Here 5pq : 7pr ≠ 15s : 21 q

∴ They are not in proportion.

Question 2. The length and breadth of a rectangular figure are 10 cm and 6 cm respectively. The length and breadth of figure are increased by 2 cm. Let’s find if the ratios of length to breadth remain same.
Solution :

Given

The length and breadth of a rectangular figure are 10 cm and 6 cm respectively. The length and breadth of figure are increased by 2 cm.

Length : Breadth = 10 cm : 6 cm = 5 : 3

Now, New length =.(10 + 2) cm = 12 cm.

New breadth (6 + 2) cm = 8 cm.

∴ New length : New breadth = 12 cm : 8 cm = 3 : 2

Here the length: Breadth are not the same.

Question 3. Paranbabu bought 500 gm of sugar for Rs. 17.50 and Deepenbabu bought 2 kg of sugar for Rs. 70. Let’s find if the 12p ; 28py amounts of sugar and their prices are in proportion.
Solution :

Given

Paranbabu bought 500 gm of sugar for Rs. 17.50 and Deepenbabu bought 2 kg of sugar for Rs. 70.

Paranbabu bought 500 gm sugar for Rs. 17.50

∴ Here the cost price = Rs. 35 per kg.

Deepenbabu bought 2 kg sugar for Rs. 70.

∴ Here the cost price = Rs. 35 per kg.

∴ They bought equal quantity of sugar at same price

Question 4. Fill in the blank squares :

1. 5 : 7 : : 25 : 35
Solution:

⇒ 5 : 7 : : 25 : 35

As 5 × 5 = 25

7× 5 = 35

2. 6 : 7 : : 30 : 35
Solution:

⇒ 6 : 7 : : 30 : 35

As 6 × 5 = 30

7 × 5 = 35

3. 21 : 28 : : 3 : 4
Solution:

21 : 28 : : 3 : 4

⇒  As \(\frac{21}{7}\) = 3

= \(\frac{28}{7}\) = 4

4. 9 : 24 : : 3 : 8
Solution:

9 : 24 : : 3 : 8 A

As \(\frac{9}{3}\) = 3

⇒ \(\frac{24}{3}\) = 8

Class 7 Math Solution WBBSE Proportion Exercise 3.2

Question 1. Let’s verify whether 7, 5, 14 and 10 are in proportion.
Solution :

Given

7, 5, 14 and 10

Here product of extreems = 7 × 10 = 70

And product of means = 5 × 14 = 70

∴ Product of extremes = product of means

Hence, the four numbers are in proportion.

Question 2. Let’s verify whether 10, 5, 14 and 7 are in proportion.
Solution :

Given

10, 5, 14 and 7

Here the product of extreme = 10 × 7 = 70

And product of means = 5 × 14 = 70

∴  Product of extremes = Product of means

Hence, the four numbers are in proportion.

Question 3. Let’s verify whether 14, 5, 10 and 7 are in proportion.
Solution :

Given

14, 5, 10 and 7

Here product of extremes = 1 4 × 7 = 98

And product of means = 5 × 10 = 50

∴ Product of extremes

Hence, the four numbers are Product Not in proportion.means

Question 4. Let’s prepare table as above and form different proportion with 5, 15, 10 and 30.

Question 5. Let’s prepare table as above and form different proportions with 7, 8, 14, 16.
Solution:

Question 6. Let’s form different proportions with 9, 11, 27 and 33
Solution:

Class VII Math Solution WBBSE Proportion Exercise 3.3

Question 1. Let’s find whether the following sets of numbers are in continued proportion, and let’s write the proportionality :

1. 5, 10, 20
Solution:

⇒ 5, 10, 20

Here 5 ×  20 = 100= (10)²

Hence 1 st term x 3rd term = (2nd term)² or (mean)²

∴ 5, 10 and 20 are in continued proportion.

∴ 5: 10: : 10: 20

2. 8, 4, 2
Solution: 

⇒ 8, 4, 2

Here 8 × 2 = 16= (4)²

i.e. 1st term x 3rd term = (2nd term)² or (mean)²

∴ 8, 4, and 2 are in continued proportion.

∴ 8: 4 :: 4 : 2

“Step-by-step solutions for proportion problems Class 7 WBBSE”

3. 7, 14, 28
Solution:

⇒ 7, 14, 28

Here 7× 28 = 196= (14)²

Hence, 1st term x 3rd term = (2nd term)² or (mean)²

∴ 7, 14, and 28 are in continued proportion.

∴  7: 14 : : 14: 28

4. 81, 9, 18
Solution :

⇒ 81, 9, 18

Here 81 x 18 = 1458

Henc# 1st term x 3rd term= (2nd term)²

∴ 81, 9 and 18 are not in continued proportion.

5. 4, 6, 12
Solution :

⇒ 4, 6, 12

Here 4 x 12 = 48 ≠ (6)²

Hence, 1st term x 3rd term ≠ (2nd term)²

∴ 4, 6 and 12 are not in continued proportion.

6. 4, 10, 25
Solution :

⇒ 4, 10, 25

Here 4 × 25 = 100 = (10)²

Hence, 1st term x 3rd term = (2nd term)² or (mean)²

∴ 4, 10, and 25 are in continued proportion

∴ 4: 10 :: 10 : 25

Class VII Math Solution WBBSE Proportion Exercise 3.4

Proportion Exercise 3.5

Question 1. Sumit bought 2 exercise books for Rs. 1 4. Let’s find how much he will pay for 7 such exercise books.
Solution :

Given

Sumit bought 2 exercise books for Rs. 1 4.

Here no. of exercise books and cost price are in direct proportion

∴  2 : 7 : : 14: ?

Or, 2 × 4th term = 7 × 14

∴ 4th term = \(\frac{7 \times 14}{2}\) = 49

∴ Price of 7 exercise book = Rs. 49.

Question 2. A jeep covers a distance of 320 km in 8 hours, keeping the same speed, let’s find how long the jeep will take to cover a distance of 120 km.
Solution :

Given

A jeep covers a distance of 320 km in 8 hours,

Here distance and time are in direct proportion.

∴ 320 : 120: : 8 : ?

∴ 320 × 4th term = 120 × 8

∴ 4th term = \(\frac{120 \times 8}{320}\) = 3

∴ Time required to cover 120 km = 3 hour

Question 3. 720g of chromium is needed to make 6kg of stainless steel. Let’s find how much cromium will be needed to make 11 kg of stainless steel.
Solution :

Given

720g of chromium is needed to make 6kg of stainless steel.

Quantity of stainless steel (kg) Quantity of Cromium (gm) .

Here quantity of stainless steel & quantity of cromium are in direct proportion.

∴ 6:11 : : 720 : ?

∴  6 × 4th term = 11 x 720

∴  4th term = \(\frac{11 \times 720}{6}\)

= 1320 gm = 1.32 kg.

∴  1.32 kg of cromium will be needed.

Question 4. There are 3 litres of syrup in 10 litre of a soft drink (sherbat). Let’s find how much syrup will be required to make 5 litres of soft drink.
Solution :

Given

There are 3 litres of syrup in 10 litre of a soft drink (sherbat).

Volume of Soft drink (litre) Volume of syrup (litre)

Here volume of soft drink & volume of syrup are in direct proportion.

∴ 10 : 5 : : 3 : ?

∴  10 × 4th term = 5 × 3

∴  4th term = \(\frac{5 \times 3}{10}=\frac{3}{2}\)

= 1.5 litre of syrup is required.

 Question 5. Let me form a practical problem on direct proportion and try to solve it.

A car can travel 100 km on the cost of petrol of Rs. 120

1. How much would the petrol cost for a journey of 350 km?
2. How many would they travel for the cost of Rs. 300 petrol?

1. How much would the petrol cost for a journey of 350 km?
Solution:

Here the distance travelled & the cost of petrol are in direct proportion.

100 : 350: : 120: ?

∴ 350 × 120

∴ 100 x 4th term = 350 x 1 00

∴ 4th term = \(\frac{350 \times 120}{100}\)  = 420

∴ To travel 350 km. cost of petrol will be Rs. 420.

Question 6. How many would the can travel for the cost of Rs. 300 petrol?
Solution:

Here the cost of petrol & the distance travelled are in direct proportion.

∴ 120 : 300: :100: ?

∴ 120 ×  4th term = 300 × -1 00

∴ 4th term = \(\frac{300 \times 100}{120}\) = 250

With cost of Rs. 300, petrol, one can go = 250 km

WBBSE Class 7 Math Solution Proportion Exercise 3.6

Question 1. Let’s complete the table given below

Question 2. 8 men can do a piece of work in 15 days. Let’s find in how many days will 10 men finish the same work.
Solution :

Given

8 men can do a piece of work in 15 days.

For doing a work if the number of men were increased the time required would decrease.

∴ Number of men and time (no. of days) are in inverse proportion.

In proportionality let’s take the inverse of relation.

∴ 10: 8 = 15: ?

∴ Product of extremes = Product of means

10 ×  4th term = 8 ×15

∴  4th term = \(\frac{8 \times 15}{10}\)

Hence, 10 men will take 12 days to finish the work.

“Best guide for Class 7 Maths WBBSE proportion exercise 3 problems”

Question 3. 1 2 men had provision of food for 20 days. Let’s find, if there are 40 men, how long will the provision last?
Solution:

Given

12 men had provision of food for 20 days.

Here if the number of men increased, no. of days decreased.

A number of men and a number of days are in inverse proportion.

Let us take the inverse relation.

12 : 40 = 7 × 20

Product of extreems = Product of means

12 × 20 = 40  × ?

∴ 4th term = \(\frac{12 \times 20}{40}\) = 6

∴  For 40 men provision last for 6 days.

“WBBSE Class 7 Maths proportion exercise 3 problem-solving techniques”

Question 4. Arunbabu used 1 6 ploughs to cultivate his whole land in 1 0 days. Let’s find, if he wanted to cultivate the same land in 8 days, how many ploughs he would have needed.
Solution :

Given

Arunbabu used 16 ploughs to cultivate his whole land in 1 0 days.

No. of days No. of Ploughs

Here no. of days & no. of ploughs are in inverse proportion.

10 : 8 = ? : 16

∴ Product of extrems = Product of means

10 × 16 = 8 × ?

∴ 4th term = \(\frac{10 \times \cdot 16}{8}\) = 20

∴ 20 ploughs are required to cultivate the same land.

Question 5. In a flood relief camp, there is a provision of food for 4000 people for 1 90 days. After 30 days, 800 people went away elsewhere. Let’s find for how many days the remaining food lasts for the remaining people in this camp.
Solution:

Given

In a flood relief camp, there is a provision of food for 4000 people for 1 90 days. After 30 days, 800 people went away elsewhere.

Here No. of men decreased, so no. of days will increase.

No. of men and no. of days are in inverse proportion.

3200 : 4000= 160 : ?

.;. Product of extremes- Product of mean

3200 x? = 4000 × 160

= \(\frac{4000 \times 160}{3200}\)

= 200

∴ Remaining food last for 200 days.

“Understanding proportion concepts with solved problems Class 7 WBBSE”

Question 6. The cost of 3 umbrellas or 1 chair is Rs. 600. Let’s calculate the price of 2 umbrellas and 2 chairs.
Solution :

Given

The cost of 3 umbrellas or 1 chair is Rs. 600.

Cost of 1 chair = Cost of 3 umbrellas

Cost of 2 chair = Cost of 3 × 2 = 6 umbrellas.

Cost of 2 umbrellas + 2 chairs =-2 + 6 = 8 umbrellas.

Here no. of umbrella & the price are in direct proportion.

∴ 3 : 8 = 600 : ?

∴  i.e Product of extremes = Product of means

3 × ? = 8 ×  600

∴ Price = \(\frac{8 \times 600}{3}\) = 1600

∴ Required price = Rs. 1600.

“WBBSE Class 7 Maths proportion exercise 3 worksheet with answers”

WBBSE Class 7 Math Solution Question 7. Let’s find the ratio of the number of students present and number of students absent today in our class. We also find the ratio of number of students present and the number of students absent today in class six. Let’s find in the ratios are equal and also find if the four numbers are in proportion.
Solution :

Let in class 7 no. of student present & absent are 42 & 8.

Ratio of number of students absent today in class 7

= 42 : 8 = 21 : 4

Again in class 6 no. of student present & absent are 40 & 1 0

Ratio of number of student present and number of student absent to day in Class 6 = 40: 1 0 = 4 : 1

These two ratios are not equal,

And the four numbers (42, 8, 40, 10) are not in proportional.

Question 8. Let us count the number of squares of different colours and answer the questions given below.

1. Ratio of red squares and blue squares.
Solution: 9:2.

2. What is the ratio of brown & violet squares?
Solution: 12:13.

What is the ratio of red & green squares?
Solution: 3 :2.

Ratio of brown and yellow squares?
Solution: 3 :2.

Which four of the coloured squares are in proportion?
Solution: Red, green, brown & yellow.

“Basic proportion formulas and examples Class 7 WBBSE solutions”

Question 9. In two types of ‘sherbet’ the ratios of syrup and water are 2: 5 and 6:10. Let’s find, which one is sweeter.
Solution :

Given

In two types of ‘sherbet’ the ratios of syrup and water are 2: 5 and 6:10.

In 1st type of Sherbat, Syrup : Water = 2:5= \(\frac{2}{5}\)

In 2nd type of Sherbat, Syrup : Water = 6:10 = 3:5= \(\frac{3}{5}\)

∴ 2nd one is more sweeter.

Question 10. When water freezes to ice, its volume inceases by 1 0%. Let’s find the ratio of a certain volume of water and its corresponding volume of ice.
Solution :

Given

When water freezes to ice, its volume inceases by 1 0%.

Set the volume of water = 100 cc.

∴ The volume of ice = \(\left(100+\frac{10}{100} \times 100\right) \mathrm{cc}\)

= (100 + 10) cc. = 110 cc.

∴ Ratio of volume of water and corresponding volume of ice

= 1 00 cc : 1 1 0 cc =10:11.

Ratio of volume of water and corresponding volume of ice =10:11.

WBBSE Class 7 Math Solution Question 11. My age is 12 years and my father’s age is 42 years. Let’s find the ratio of our ages.
Solution :

Given

My age 12 years

My father’s age = 42 years

∴The ratio of my age and my father’s age

= 12 year : 42 years = 2:7.

“Tips and tricks to solve proportion problems Class 7 WBBSE”

Question 12. The ratio of storybooks and textbooks of Preetam is 2: 5. If Preetam has 4 story books, then let’s find how many textbooks he has.
Solution :

Given

The ratio of story books and textbooks of Preetam = 2:5

⇒  \(\frac{\text { No. of story books }}{\text { No. of textbooks }}=\frac{2}{5}\)

⇒ \(\frac{4}{\text { No. of text books }}=\frac{2}{5}\)

Or, 2 x number of textbooks = 4 × 5

Number of text books\(\frac{4 \times 5}{2}\) = 10

Question 13. For making a flower garland of china roses (Jaba) and Indian marigold (gada), 105 of these flowers were collected. If the ratio of chinaroses and marigold is 3 : 4, let’s find how many china roses and marigolds are there. Let’s also find how many more chinaroses must be collected so that the ratio of these two types of flowers become equal.
Solution :

Given

For making a flower garland of china roses (Jaba) and Indian marigold (gada), 105 of these flowers were collected. If the ratio of chinaroses and marigold is 3 : 4,

Total No. of flowers = 105

The ratio of no. of Jaba flowers and no. of gada flowers = 3:4 =

∴ No. of Chinaroses (Jaba) flowers = \(\frac{3}{7}\) × 105 = 45

Number of Marigold (gada) flowers= \(\frac{4}{7}\) × = 60

To make the ratio of two types of flowers equal,

Number of  chinaroses (Jaba) flowers required = 60 – 45 = 1 5.

“WBBSE Class 7 Maths Chapter 3 complete guide with solved exercises”

Question 14. Let the small squares in the square given below, are coloured according to your own will with 5 different colours from those 5 types of coloured squares, let’s choose any pair of colours and find the ratio between their number of squares. Of these ratios, also identity the ratios of greater inequality, ratio of lesser inequality and ratio of equality. Out of these ratios, if the numbers of four different coloured squares are found in proportion, then note that proportionality too.
Solution :

Given

Let the small squares in the square given below, are coloured according to your own will with 5 different colours from those 5 types of coloured squares,