Mathematics Class VII

Question 1 of 25

1. Question
1 point(s)
Find the ratio of area of rectangle of length 12 cm and breadth 10 cm to that of area of square of side 11 cm.
- $ 120 : 121 $
- $121 : 120 $
- $12 : 11 $
- $11 : 12 $
Correct

Incorrect

Hint

Area of rectangle $12 \times 10 = 120 \ cm^{2}$
Area of square $= 11 \times 11 = 121 \ cm^{2}$
Ratio = 120 : 121
Question 2 of 25

2. Question
1 point(s)
If the perimeter of a square is doubled then the ratio of original area to new area is
- $ 4 : 1 $
- $ 2: 1 $
- $ 1 : 2 $
- $ 1 : 4 $
Correct

Incorrect

Hint

Perimeter of a square = 4a
Area is $a^{2}$
Perimeter of new square $= 2 \times 4a = 4 \times 2a$
Area $= (2a)^{2} = 4a^{2}$
Ratio $= a^{2} : 4a^{2} = 1 : 4$
Question 3 of 25

3. Question
1 point(s)
The perimeter of a rectangle is 3 times the perimeter of square of side 12 cm. If the length of rectangle is 54 cm. Area of rectangle is
- $ 972 \ cm^{2} $
- $700 \ cm^{2} $
- $792 \ cm^{2} $
- $ 729 \ cm^{2} $
Correct

Incorrect

Hint

Perimeter of rectangle = 3 (perimeter of square)
$2 \times (l + b) = 3(4a) \\ 2 \times (54 + b) = 3(4 \times 12) \\ 54 + b = 72 \ \rightarrow b = 18$
Area $18 \times 54 = 972 \ cm^{2}$
Question 4 of 25

4. Question
1 point(s)
If length of a rectangle is increased 10 times and its breadth is decreased to one – fifth how will the area be changed?
- double
- half
- one – fourth
- none of these
Correct

Incorrect

Hint

Original area $= l \times b$
New area $= (10 l) \times \left(\dfrac{1}{5}{b}\right) = 2 \times (l \times b) = 2 \times (original \ area)$
Question 5 of 25

5. Question
1 point(s)
A garden is 56 m by 35 m. A strip of 2 m width is dug all around it on the outside. Find the cost of digging at the rate of RS 4.20 per $m^{2}$.
- Rs 1596
- Rs 2000
- Rs 1296
- Rs 1600
Correct

Incorrect

Hint

Area of shaded region $(60 \times 39) - (56 \times 35) = 2340 - 1960 = 380 \ m^{2}$
Cost of digging $= 4.20 \times 380 = Rs \ 1596$
Question 6 of 25

6. Question
1 point(s)
A plot of land is in the form of a right triangular region , the legs are of length 10 m and 6 m. What is its area?
- $20 \ m^{2}$
- $ 30 \ m^{2}$
- $ 50 \ m^{2}$
- $ 15 \ m^{2}$
Correct

Incorrect

Hint

Area $= \dfrac{1}{2} \times 10 \times 6 = 30 \ m^{2}$
Question 7 of 25

7. Question
1 point(s)
The base of triangle is 23 m long and its area is $ 161 \ m^{2}$. Find the length of the corresponding altitude.
- 12 m
- 10 m
- 14 m
- 11 m
Correct

Incorrect

Hint

Area $= \dfrac{1}{2} \times b \times h \ \rightarrow = \dfrac{1}{2} \times 23 \times h \ \rightarrow h = \dfrac{161 \times 2}{23} = 14 \ m$
Question 8 of 25

8. Question
1 point(s)
The base of the right angled triangle is 4 cm and hypotenuse is 5 cm long. Find its area.
- $6 \ cm^{2} $
- $10 \ cm^{2} $
- $4 \ cm^{2} $
- $15 \ cm^{2} $
Correct

Incorrect

Hint

$h^{2} + 4^{2} = 5^{2} \ \rightarrow h = \sqrt{9} = 3 \\ Area = \dfrac{1}{2} \times 3 \times 4 = 6 \ cm^{2}$
Question 9 of 25

9. Question
1 point(s)
The length and breadth of a rectangular floor are 3.6 m and 2.4 m. How many square tiles will be needed to cover the floor if the side of tile is 20 cm.
- 116
- 146
- 216
- 126
Correct

Incorrect

Hint

Number of tiles $= \dfrac{360 \times 240}{20 \times 20} = 216$
Question 10 of 25

10. Question
1 point(s)
The area of the shaded region is
- $48 \ cm^{2} $
- $160 \ cm^{2} $
- $112 \ cm^{2} $
- $12 \ cm^{2} $
Correct

Incorrect

Hint

AM $= \sqrt{10^{2} - 8^{2}} = 6$
Area of the shaded region $= 2 \left[\dfrac{1}{2} \times 6 \times 8\right] = 48 \ cm^{2}$
Question 11 of 25

11. Question
1 point(s)
The filed in the form of a parallelogram has base 200 m and altitude 80 m long. The cost of watering the field at 10 paise per $m^{2}$ is
- Rs 160
- Rs 1600
- Rs 800
- Rs 1200
Correct

Incorrect

Hint

Area $= 200 \times 80 = 16000 \ m^{2} \\ Cost = 16000 \times 0.10 = Rs \ 1600$
Question 12 of 25

12. Question
1 point(s)
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and breadth is 3.6 m. Find the cost of white washing the wall. If the rate of white washing is Rs 20 per $m^{2}$.
- Rs 280
- Rs 254
- Rs 234
- Rs 284
Correct

Incorrect

Hint

Area of wall $= 4.5 \times 3.6 = 16.2 \ m^{2}$
Area of door $= 2 \times 1 = 2 \ m^{2}$
Area of remaining wall $= 16.2 - 2 = 14.2 \ m^{2}$
Cost $= 14.2 \times 20 = Rs \ 284$
Question 13 of 25

13. Question
1 point(s)
The area of a square park is same and that of a rectangular park. If the side of the square park is 60 m and length of the rectangular park is 90 m. Find the breadth of the rectangular park.
- 10 m
- 20 m
- 30 m
- 40 m
Correct

Incorrect

Hint

Breadth $= \dfrac{60 \times 60}{90} = 40 \ m$
Question 14 of 25

14. Question
1 point(s)
In order to fence a garden 72 m long and 56 m wide, metal posts are to be fixed every 5 m a part. Each post is 3 m long. Find the total length of the pipe.
- 96 m
- 32 m
- 64 m
- 36 m
Correct

Incorrect

Hint

Perimeter $2(72 + 56) = 128 \ m$
No. of metal post $= \dfrac{2 \times 128}{8} = 32$
Total length of pipe $= 32 \times 3 = 96 \ m$
Question 15 of 25

15. Question
1 point(s)
The dimensions of a photograph are $30 \ cm \times 20 \ cm $. Find the length of the wooden frame needed to frame the photograph.
- 600 cm
- 50 cm
- 100 cm
- 500 cm
Correct

Incorrect

Hint

P $= 2 \times (30 + 20)= 100 \ cm$
Question 16 of 25

16. Question
1 point(s)
A park square in shape has each side of length 62 m. If 3 layers of barbed wire has to be fixed to fence it , find the length of the wire needed.
- 447 m
- 744 m
- 600 m
- 500 m
Correct

Incorrect

Hint

Length of wire needed $= 3 \times 4 \times 62 = 744 \ m$
Question 17 of 25

17. Question
1 point(s)
If the length of a rectangle is doubled and breadth does not change, its area will be :
- double
- half
- three times
- four times
Correct

Incorrect

Hint

Area $= l times b$
New area $= 2l \times b = 2 \times (l \times b)$
Question 18 of 25

18. Question
1 point(s)
The perimeter of a rectangle is 91 cm. Its length is (2x – 1) cm and breadth is (x + 9) cm then its breadth is
- 24 cm
- 26.5 cm
- 20.5 cm
- 21.5 cm
Correct

Incorrect

Hint

$2[2x - 1 + x + 9] = 91 \\ \rightarrow 6x = 91 - 16 = 75 \ \rightarrow x = \dfrac{75}{6}$
Breadth $= \dfrac{75}{6} + 9 = \dfrac{129}{6} = 21.5$
Question 19 of 25

19. Question
1 point(s)
A square lawn has a 2 m wide path surrounding it. If the area of the path is 136 $m^{2}$ then the area of the lawn is
- $250 \ m^{2}$
- $300 \ m^{2}$
- $1800 \ m^{2}$
- $225 \ m^{2}$
Correct

Incorrect

Hint

$(x + 4)^{2} - x^{2} = 136 \\ x^{2} + 16 + 8x - x^{2} = 136 \ \rightarrow x = 15 \ m \\ Area = (15)^{2} = 225 \ m^{2}$
Question 20 of 25

20. Question
1 point(s)
The area of isosceles triangle having the base 6 cm and length of each equal side 5 cm is
- $9 \ cm^{2} $
- $ 15 \ cm^{2}$
- $ 12 \ cm^{2}$
- $ 30 \ cm^{2}$
Correct

Incorrect

Hint

AD $= \sqrt{5^{2} - 3^{2}} = 4 \ cm$
Area $= \dfrac{1}{2} \times 6 \times 4 = 12 \ cm^{2}$
Question 21 of 25

21. Question
1 point(s)
The rectangular lawn is 30 m by 20 m. It has two roads each 2 m wide running in the middle of it, one parallel to the length and other parallel to the breadth then the area of the road is
- $96 \ m^{2}$
- $ 100 \ m^{2}$
- $ 104 \ m^{2}$
- $ 116 \ m^{2}$
Correct

Incorrect

Hint

Area of the road $= (30 \times 2) + (20 \times 2) - (2 \times 2) = 96 \ m^{2}$
Question 22 of 25

22. Question
1 point(s)
A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. The area of the path is
- $400 \ m^{2}$
- $ 350 \ m^{2}$
- $ 380 \ m^{2}$
- $ 425 \ m^{2}$
Correct

Incorrect

Hint

Area of the path $= (45 + 5) \times (30 + 5) - (45 \times 30) = 1750 - 1350 = 400 \ m^{2}$
Question 23 of 25

23. Question
1 point(s)
A circle has area which is 100 times the area of another circle. What is the ratio of their circumference ?
- $100 : 1 $
- $1 : 1 $
- $10 : 1 $
- $1 : 2 $
Correct

Incorrect

Hint

$\dfrac{\pi r _1^2}{\pi r _2^2} = \dfrac{100}{1} \\ \dfrac{2 \pi r_1}{2 \pi r_2} = \dfrac{r_1}{r_2} = \dfrac{10}{1} = r_1 : r_2 = 10 : 1$
Question 24 of 25

24. Question
1 point(s)
From a circular sheet of radius is 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining circle.
- $ 21.98 \ cm^{2}$
- $ 29.18 \ cm^{2}$
- $ 28.19 \ cm^{2}$
- $ 12.89 \ cm^{2}$
Correct

Incorrect

Hint

Area of remaining circle $= \pi (4)^{2} - \pi (3)^{2} = \pi (16 - 9) = 7 \pi = 21.98 \ cm^{2}$
Question 25 of 25

25. Question
1 point(s)
Find the area of a triangle whose sides are 9 cm, 12 cm , 15 cm.
- $51 \ cm^{2} $
- $52 \ cm^{2} $
- $53 \ cm^{2} $
- $54 \ cm^{2} $
Correct

Incorrect

Hint

$S = \dfrac{9 + 12 + 5}{2} = \dfrac{36}{2} = 18$
Area $= \sqrt{s(s -a)(s-b)(s-c)} = \sqrt{18(18 - 9)(18 - 12)(18 - 15)} = \sqrt{18 \times 9 \times 6 \times 3} = 54 \ cm^{2}$

Mathematics Class VII

Olympiad Questions – Perimeter and Area – Class VII

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