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A Learning Portal from Recruitment India

The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m.What is the area of the field?

A.

77500

B.

57500

C.

37500

D.

47500

Answer with explanation

Answer: Option CExplanation

Let length = x meters, then breadth = 0.6x

Given that perimeter = 800 meters

=> 2[ x + 0.6x] = 800

=> x = 250 m

Length = 250m and breadth = 0.6 x 250 = 150m

Area = 250 x 150 = 37500 sq.m

Workspace

Water flows a tank 200 x 150m through a rectangular pipe 1.5 x 1.25m at the rate at 20kmph. In what time will be water rises by 2 meters?

A.

96min

B.

90min

C.

80min

D.

76 min

Answer with explanation

Answer: Option AExplanation

Volume of water flown in the tank

=(200 x 150 x 2) Cu.m = 60000 Cu.m

Volume flown per hour

=(3/2 x 125/100 x 20 x 1000) Cu.m =37500Cu.m

Time taken = 60000/37500 hrs = 8/5 hrs

=(8/5 x 60)min = 96 min

Workspace

The length of a rectangle is 3/5th of the side of a square. The radius of a circle is equal to side of the square. The circumference of the circle is 132 cm. What is the area of the rectangle, if the breadth of the rectangle is 15 cm?

A.

112 cm²

B.

149 cm²

C.

189 cm²

D.

Cannot be determined

Answer with explanation

Answer: Option CExplanation

Circumference of the circle = 132

2πR = 132; R = 21 cm

Side of square = 21 cm

Length of the rectangle = 3/5 * 21 = 63/5

Area of the rectangle = 63/5 * 15 = 189 cm²

Workspace

A rectangular lawn 55m by 35m has two roads each 4m wide running in the middle of it. One parallel to the length and the other parallel to breadth. The cost of graveling the roads at 75 paise per sq meter is

A.

rs.58

B.

rs.158

C.

rs.258

D.

rs.358

Answer with explanation

Answer: Option CExplanation

area of cross roads = (55 x 4) + (35 x 4)- (4 x 4) = 344sq m

cost of graveling = 344 x (75/100) = Rs. 258

Workspace

A hall 20 m long and 15 m broad is surrounded by a verandah of uniform width of 2.5 m. The cast of flooring the verandah at the rate of Rs 3.50 per sq. Meter is:

A.

Rs.500

B.

Rs. 700

C.

Rs. 600

D.

Rs. 800

Answer with explanation

Answer: Option BExplanation

Area of verandah = [(25 x 20) – (20 x 15)]Sq.m = 20 Sq.m

Cost of flooring = Rs. (200 x 3.50) = Rs. 700

Workspace

The perimeter of a square is equal to twice the perimeter of a rectangle of length 8 cm and breadth 7 cm. What is the circumference of a semicircle whose diameter is equal to the side of the square ?

A.

47.47 cm

B.

38.57 cm

C.

23.57 cm

D.

42.46 cm

Answer with explanation

Answer: Option BExplanation

Perimeter of square = 2 x Perimeter of rectangle

= 2 * 2 (8+7) = 60 cm.

Side of square = 60/4 = 15 cm = Diameter of semi-circle

Circumference of semi-circle = πd/2 + d

= (22/7) * 2 * 15 + 15 = 38.57 cm

Workspace

A.

Rs. 480

B.

Rs. 384

C.

Rs. 352

D.

Rs. 176

Answer with explanation

Answer: Option DExplanation

Workspace

The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.

A.

27 and 23

B.

24 and 23

C.

25 and 23

D.

22 and 23

Answer with explanation

Answer: Option AExplanation

Let the two parallel sides of the trapezium be a cm and b cm.

Then, a – b = 4

Workspace

The perimeter of one square is 48 cm and that of another is 20 cm. Find the perimeter and the diagonal of a square which is equal in area to these two combined?

A.

15√2 cm

B.

17√2 cm

C.

13√2 cm

D.

16√2 cm

Answer with explanation

Answer: Option CExplanation

4a = 48 4a = 20

a = 12 a = 5

a^{2} = 144 a^{2} = 25

Combined area = a^{2} = 169 => a = 13

d = 13√2

Workspace

A.

15m, 9m

B.

17m, 9m

C.

14m, 7m

D.

16m, 7m

Answer with explanation

Answer: Option BExplanation

Length = x; Breadth =y

xy – (x-5)(y+3) = 9

3x – 5y – 6 = 0 —(i)

(x+3)(y+2) – xy = 67

2x + 3y -61 = 0 —(ii)

solving (i) and (ii)

x = 17m ; y = 9m

Workspace

The rainwater from a flat roof 16 m long and 10 m wide is collected in a tank whose internal measurements are 2 m long, 2.4 m wide and 3 m deep. Before it started raining it was half-full and after the rain it was full. Find the height of rainfall.

A.

0.9 cm

B.

1.8 cm

C.

4.5 cm

D.

9 cm

Answer with explanation

Answer: Option CExplanation

Let h be the height of the rainfall = h

1.6 × 10 × h = 1/2(2 × 2.4 × 3)

h = .045m = 4.5cm

Workspace

A circular swimming pool is surrounded by a concrete wall 4ft wide. If the area of the concrete wall surrounding the pool is 11/25 that of the pool, then the radius of the pool is?

A.

10ft

B.

20ft

C.

30ft

D.

40ft

Answer with explanation

Answer: Option BExplanation

let the radius of the pool be Rft

Radius of the pool including the wall = (R+4)ft

Workspace

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