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- Question
The base of a parallelogram is twice its height. If the area of the parallelogram is 338 sq.cm. Find its height?
Options- A. 13 cm
- B. 14 cm
- C. 16 cm
- D. 12 cm
- Correct Answer
- 13 cm
ExplanationLet height =x Then, base =2x2x * x = 338=>x=13 - 1. A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq m is
Options- A. Rs. 258
- B. Rs. 358
- C. Rs. 458
- D. Rs. 558 Discuss
- 2. The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.
Options- A. B=900;H=300
- B. B=300;H=900
- C. B=600;H=700
- D. B=500;H=900 Discuss
- 3. The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?
Options- A. 50
- B. 60
- C. 70
- D. 80 Discuss
- 4. The area of the largest circle that can be drawn inside a rectangle with sides 18cm by 14cm is
Options- A. 49
- B. 154
- C. 378
- D. 1078 Discuss
- 5. The ratio of the area of a square to that of the square drawn on its diagonal is
Options- A. 1:2
- B. 2;3
- C. 3:1
- D. 4:1 Discuss
- 6. The ratios of areas of two squares, one having its diagonal double than the other is
Options- A. 2:1
- B. 2:3
- C. 3:1
- D. 4:1 Discuss
- 7. A room 5m 55cm long and 3m 74cm broad is to be paved with square tiles.Find the least number of square tiles required to cover the floor?
Options- A. 156
- B. 166
- C. 176
- D. 186 Discuss
- 8. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle?
Options- A. 10cm
- B. 20cm
- C. 30cm
- D. 40cm Discuss
- 9. The diameter of the driving wheel of a bus is 140 cm. How many revolution, per minute must the wheel make in order to keep a speed of 66 kmph ?
Options- A. 150
- B. 250
- C. 350
- D. 550 Discuss
- 10. A room of 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor.
Options- A. 136
- B. 146
- C. 166
- D. 176 Discuss
Area problems
Search Results
Correct Answer: Rs. 558
Explanation:
Area to be plastered =
=
Cost of plastering =
Correct Answer: B=900;H=300
Explanation:
Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
Let altitude = x metres and base = 3x metres.
Then,
Base = 900 m and Altitude = 300 m.
Correct Answer: 60
Explanation:
let ABC be the isosceles triangle, the AD be the altitude
Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base]
since in an isoceles triange the altitude bisects the base so
BD = DC = 16-x
In a triangle ADC,
BC = 32-2x = 32-20 = 12 cm
Hence, required area = = = 60 sq cm
Correct Answer: 154
Explanation:
The diameter is equal to the shortest side of the rectangle.
So radius= 14/2 = 7cm.
Therefore,
Correct Answer: 1:2
Explanation:
ratio =
Correct Answer: 4:1
Explanation:
ratio =
Correct Answer: 176
Explanation:
Area of the room=(544 * 374)
size of largest square tile= H.C.F of 544 & 374 = 34 cm
Area of 1 tile = (34 x 34)
Number of tiles required== [(544 x 374) / (34 x 34)] = 176
Correct Answer: 20cm
Explanation:
Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is
after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=
since new area is 75 units greater than original area thus
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.
Correct Answer: 250
Explanation:
Circumference = No.of revolutions * Distance covered
Distance to be covered in 1 min. = (66 X1000)/60 m = 1100 m.
Circumference of the wheel = 2 x (22/7) x 0.70 m = 4.4 m.
Number of revolutions per min. =(1100/4.4) = 250.
Correct Answer: 176
Explanation:
area of the room = 544 x 374 sq.cm
size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm
area of 1 tile = 34x34 sq cm
no. of tiles required = (544 x 374) / (34 x 34) = 176
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