Area problems
- 1. The breadth of a rectangle is 25 m. The total cost of putting a grass bed on this field was ? 12375, at the rate of ? 15 per sq m. What is the length of the rectangular field?
Options- A. 27 m
- B. 30 m
- C. 33 m
- D. 32 m Discuss
Correct Answer: 33 m
Explanation:
Area to the rectangular field = 12375/15 = 825 sq m
According to the question,
(L x B) = 825 [L = length and B = breadth]
? L x 25 = 825
? L = 825/25 = 33 m
- 2. What is the area of a square having perimeter 68 cm?
Options- A. 361 sq cm
- B. 284 sq cm
- C. 269 sq cm
- D. 289 sq cm Discuss
Correct Answer: 269 sq cm
Explanation:
According to the question,
4a = 68 [ where a = side]
?a = 68/4 = 17 cm
? Required area = a2
= (17)2 = 289 sq cm
- 3. The perimeter of two squares are 68 cm. Find the perimeter of the third square whose area is equal to the different of the areas of these two squares.
Options- A. 64 cm
- B. 60 cm
- C. 32 cm
- D. 8 cm Discuss
Correct Answer: 32 cm
Explanation:
a1 = 68/4 = 17 cm
and a2 = 60/4 = 15 cm
[ where a1 and a2 are sides]
According to the question,
Area of the third square = [(17)2 - (15)2 ]
= (17 + 15) (17 - 15)
= 32 x 2
= 64 sq cm
Let a3 = Side of the third square.
According to the question, (a3)2 = 64 sq cm
? a3 = ?64 = 8 cm
? Perimeter of the third square = 4 x a3 = 4 x 8 = 32 cm.
- 4. A wheel make 4000 revolution in moving a distance of 44 km. Find the radius of the wheel.
Options- A. 15 m
- B. 27 m
- C. 25 m
- D. 28 m Discuss
Correct Answer:
Explanation:
Distance covered in 1 revolution
= (44 x 1000) / 4000
= 11 m
According to the question,
2?r = 11
? (44/7) x r = 11
? r = (11 x 7) / 44 = 1.75 m
- 5. If the area of a semi-circle be 77 sq m, find its perimeter.
Options- A. 36 m
- B. 42 m
- C. 48 m
- D. 54 m Discuss
Correct Answer: 36 m
Explanation:
According to the question,
Area of semi- circle = 77 m
(1/2) x ? x r2 = 77
? r2 = (77 x 2 x 7)/22
? r = 7m
?Circumference of semi- circle =?r + 2r
= ( ? + 2)r
= [(22/7) + 2] x 7
= 36 m
- 6. A railing of 288 m is required for fencing a semi-circular park. Find the area of the park. (? = 22/7)
Options- A. 4928 m2
- B. 9865 m2
- C. 8956 m2
- D. 9856 m2 Discuss
Correct Answer: 4928 m2
Explanation:
Let the radius of the park be r, then
?r + 2r = 288
(? + 2)r = 288
? [(22/7) + 2]r = 288
? r = (288 x 7)/36 = 56
? Area of the park = (1/2)?r2
= (1/2) x (22/7) x 56 x 56
= 4928
- 7. If the radius of a circle is increased by 6%, find the percentage increase in its area.
Options- A. 15%
- B. 12.36%
- C. 8.39%
- D. 17% Discuss
Correct Answer: 12.36%
Explanation:
Given that, a = 6
According to the formula,
Percentage increase in area
= 2a + [a2/100]%
= 2 x 6 + [36/100]%
= (12 + 0.36)%
= 12.36%
- 8. The wheel of an engine turns 350 times round its axle to cover a distance of 1.76 km. The diameter of the wheel is
Options- A. 3 cm
- B. 3?3 cm
- C. 9 cm
- D. ?3 cm Discuss
Correct Answer: ?3 cm
Explanation:
Distance covered in 1 round
= (Total Distance) / (Total round)
= (1.76 x 1000)/350
= 176/35 m
? 2?r = (1.76 x 100) / 35 cm
2? = Diameter = 17600 x 7/22 x 35 = 160 cm
- 9. The ratio of the area of the circumcircle and the incircle of a square is
Options- A. 2 : 1
- B. 1 : 2
- C. ?2 : 1
- D. 1 : ?2 Discuss
Correct Answer: 2 : 1
Explanation:
Ratio of the areas of the circumcircle and incircle of a square
= [(Diagonal)2?] / [(Side)2?]
= [(Side x ?2)2] / (Side)2 = 2/1 or 2 : 1
- 10. The radius of a circle is so increased that its circumference increased by 5%. The area of the circle, then increases by
Options- A. 12.5%
- B. 10.25%
- C. 10.5 %
- D. 11.25% Discuss
Correct Answer: 10.25%
Explanation:
Increase in circumference of circle = 5%
? Increase in radius is also 5%.
Now, increase in area of circle = 2a + (a2/100) %
Where, a = increase in radius= 2 x 5 + (5 x 5)/100 % = 10.25%
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