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- Question
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
- Correct Answer
- 20cm
ExplanationLet the breadth of the given rectangle be x then length is 2x.
thus area of the given rect isafter 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. - 1. 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
- 2. 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
- 3. 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 Discuss
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.
Options- A. 84
- B. 64
- C. 44
- D. 22 Discuss
- 9. The length of a class room floor exceeds its breadth by 25 m. The area of the floor remains unchanged when the length is decreased by 10 m but the breadth is increased by 8 m. The area of the floor is
Options- A. 5100 sq.m
- B. 4870 sq.m
- C. 4987 sq.m
- D. 4442 sq.m Discuss
- 10. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Options- A. 20%
- B. 30%
- C. 40%
- D. 50% Discuss
Area problems
Search Results
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: 4:1
Explanation:
ratio =
Correct Answer: 13 cm
Explanation:
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: 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
Correct Answer: 84
Explanation:
Let a = 13, b = 14 and c = 15. Then, =21
(s- a) = 8, (s - b) = 7 and (s - c) = 6.
Area =
=
= 84 sq.cm
Correct Answer: 5100 sq.m
Explanation:
Let the breadth of floor be 'b' m.
Then, length of the floor is 'l = (b + 25)'
Area of the rectangular floor = l x b = (b + 25) × b
According to the question,
(b + 15) (b + 8) = (b + 25) × b
2b = 120
b = 60 m.
l = b + 25 = 60 + 25 = 85 m.
Area of the floor = 85 × 60 = 5100 sq.m.
Correct Answer: 50%
Explanation:
Let original length = x and original breadth = y.
Original area = xy.
New length = x/2 and New breadth=3y
New area =
Therefore, Increase in area = New area-original area =
Therefore, Increase % = %
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