Discussion
Home ‣ Aptitude ‣ Area Comments
- Question
- The difference between the circumference and the radius of a circle is 37 cm. The area of a circle is?
Options- A. 148 sq. cm
- B. 111 sq. cm
- C. 154 sq. cm
- D. 259 sq. cm
- Correct Answer
- 154 sq. cm
Explanation? 2?r - r = 37
? [(2 x 22/7) -1]r= 37
? 37r / 7= 37
? r = 7
So, area of the circle =(22/7) x 7 x 7 cm2
= 154 cm2 - 1. The radius of a circle has been reduced from 9 cm to 7 cm . the appropriate percentage decrease in area is?
Options- A. 31.5 %
- B. 39.5 %
- C. 34.5 %
- D. 65.5 % Discuss
- 2. If the radius of a circle be reduced by 50%. Its area is reduced by?
Options- A. 31.5%
- B. 39.5%
- C. 34.5%
- D. 65.5% Discuss
- 3. The area of a rectangle is thrice that of square. Length of the rectangle is 40 cm and the breadth of the rectangle is ( 3 / 2 ) times that of the side of the square. The side of the square in cm is?
Options- A. 15
- B. 20
- C. 30
- D. 60 Discuss
- 4. The length of a rectangular field is increased by 60 % . By what percent would the width have to be decreased to maintain the same area?
Options- A. 371/2%
- B. 60%
- C. 75%
- D. 120% Discuss
- 5. If the ratio of the area of two square is 9 : 1 the ratio of their perimeters is?
Options- A. 9:1
- B. 3:1
- C. 3:4
- D. 1:3 Discuss
- 6. The area of a circular field is 13.86 hectares . The cost of fencing it at the rate of 20 paisa per meter is?
Options- A. Rs. 277.20
- B. Rs. 264
- C. Rs. 324
- D. Rs. 198 Discuss
- 7. The area of circle is 38.5 sq. cm. Its circumference is?
Options- A. 6.20 cm
- B. 11 cm
- C. 22 cm
- D. 121 cm Discuss
- 8. If the circumference of a circle is the 352 meters, then its area in m square is?
Options- A. 9856
- B. 8956
- C. 6589
- D. 5986 Discuss
- 9. Area of a square with side x is equal to the area of a triangle with base x. The altitude of the triangle is?
Options- A. x/2
- B. x
- C. 2x
- D. 4x Discuss
- 10. In a triangle ABC, BC = 5 cm AC = 12 cm and AB = 13 cm. The length of a altitude drawn from B on AC is?
Options- A. 4 cm
- B. 5 cm
- C. 6 cm
- D. 7 cm Discuss
Area problems
Search Results
Correct Answer: 39.5 %
Explanation:
Original area = (22/7) x 9 x 9 cm2
New area = (22/7) x 7 x 7 cm2
? Decrease = 22/7 x [(9)2 -(7)2] cm2
=(22/7) x 16 x 2 cm2
Decrease percent = [(22/7 x 16 x 2) /( 7/22 x 9 x 9)] x 100 %
= 39.5 %
Correct Answer: 34.5%
Explanation:
Original area = ? x (r/2)2 = ?r2/4
Reduction in area = ? r2 - 3? r2/4
? Reduction per cent = [ 3?r2/4 x 4/(?r2) x 100 ] %
= 75%
Correct Answer: 20
Explanation:
Let the side of the square = y cm
Then, breadth of the rectangle = 3y/2 cm
? Area of rectangle = (40 x 3y/2) cm2
= 60y cm2
? 60y = 3y2
? y = 20
Hence, the side of the square = 20 cm
Correct Answer: 371/2%
Explanation:
Let length = L and breadth = B
Let , New breadth = Z
Then, New length = ( 160 / 100) L.
= 8L / 5
? 8L / 5 x Z = LB
or Z = 5B/8
Decrease in breadth = (B-5B/8)
= 3B/8
? Decrease in percent = (3B/8 x1/B ) x 100 %
= 371/2%
Correct Answer: 3:1
Explanation:
Let the area of square be (9x)2 m2 and (x2) m2
Then, their sides are (3x) m and x metres respectively
? Ratio of their perimeters = 12x / 4x
=3:1
Correct Answer: Rs. 264
Explanation:
? 22/7 x r2 = 13.86 x 10000
? r2 = (13.86 x 10000 x 7) / 22
? r = 210 m
? Circumference = [2 x (22/7) x 210 ] m
= 1320 m
Cost of fencing = Rs. (1320 x 20)/100
= Rs . 264
Correct Answer: 22 cm
Explanation:
? ( 22 x r2) / 7 = 38.5
? r2 =(38.5 x 7) /22
? r = 3.5 cm
? Circumference = 2 x (22/7) x 3.5 cm
= 22 cm
Correct Answer: 9856
Explanation:
circumference = 2?r = 2 x (22 / 7) x r = 352
? r = (352 x 7) / (22 x 2) =56 cm
Now area = ?r2 = (22 / 7) x 56 x 56 m2
= 9856 m2
Correct Answer: 2x
Explanation:
Let h be the altitude of triangle.
So area of triangle = (1/2)xh
area of square = x2
From question area of square = area of triangle
x2 = (1/2)xh
? h = (2x2)/x = 2x
Correct Answer: 5 cm
Explanation:
? s= (13 + 5 + 12) / 2 cm
= 15cm
s-a = 2 cm,
s-b = 10 cm and
s-c = 3 cm
? Area = ? (15 x 2 x 10 x 3 ) cm2
= 30 cm2
? (12 x h) / 2 = 30
? h = 5 cm
Comments
There are no comments.More in Aptitude:
Programming
Copyright ©CuriousTab. All rights reserved.