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Q:

The base of a triangle of 15cm and height is 12cm. The height of another triangle of double the area having the base 20cm is

A) 16	B) 17
C) 18	D) 20

Answer & Explanation Answer: C) 18

Explanation:

a = $\frac{1}{2} \times 15 \times 12$ = 90 sq.cm

b = 2a = $\frac{1}{2} \times 20 \times h$

h= 18cm

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Q:

The dimensions of a room are 10m x 7m x 5m. There are 2 doors and 3 windows in the room. The dimensions of the doors are 1m x 3m. One window is of size 2m x 1.5m and the other 2 windows are of size 1m x 1.5m. The cost of painting the walls at Rs. 3 per sq m is

A) Rs.174	B) Rs.274
C) Rs.374	D) Rs.474

Answer & Explanation Answer: D) Rs.474

Explanation:

Area of 4 walls = 2(l+b)h
=2(10+7) x 5 = 170 sq m
Area of 2 doors and 3 windows = 2(1x3)+(2x1.5)+2(1x1.5) = 12 sq m
area to be planted = 170 -12 = 158 sq m

Cost of painting = Rs. 158 x 3 = Rs. 474

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Q:

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

A) Rs. 258	B) Rs. 358
C) Rs. 458	D) Rs. 558

Answer & Explanation Answer: D) Rs. 558

Explanation:

Area to be plastered = $[2 (l + b) \times h] + (l \times b)$

= $[2 (25 + 12) \times 6] + (25 \times 12) = 744 s q . m$

Cost of plastering = $744 \times \frac{75}{100} = R s . 558$

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Q:

The cost of fencing a square field @ Rs. 20 per metre is Rs.10.080.How much will it cost to lay a three meter wide pavement along the fencing inside the field @ Rs. 50 per sq m

A) 53800	B) 43800
C) 83800	D) 73800

Answer & Explanation Answer: D) 73800

Explanation:

perimeter = total cost / cost per m = 10080 /20 = 504m
side of the square = 504/4 = 126m
breadth of the pavement = 3m
side of inner square = 126 - 6 = 120m
area of the pavement = (126 x126) - (120 x 120) = 246 x 6 sq m
cost of pavement = 246*6*50 = Rs. 73800

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Q:

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 & Explanation Answer: C) rs.258

Explanation:

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

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

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Q:

A took 15 seconds to cross a rectangular field diagonally walking at the rate of 52 m/min and B took the same time to cross the same field along its sides walking at the rate of 68m/min. The area of the field is?

A) 50	B) 60
C) 70	D) 80

Answer & Explanation Answer: B) 60

Explanation:

length of the diagonal = 52 x (15/60) =13m

sum of length and breadth = 68 x (15/60) = 17m

$(\sqrt{l^{2} + b^{2}}) = 13 o r l + b = 17$

area =lb = 1/2[(2lb)] = 1/2[(l+b)² - (l²+b²)] = 1/2[17² -169]

=1/2 x 120 = 60 sq.m

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Q:

Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is

A) 1.967	B) 1.867
C) 1.767	D) 1.567

Answer & Explanation Answer: A) 1.967

Explanation:

required area = (area of an equilateral triangle of side 7 cm)- (3 * area of sector with Ã = 60 degrees and r = 3.5cm)
$(\frac{\sqrt{3}}{4} * 7 * 7) - (3 * \frac{22}{7} * 3.5 * 3.5 * \frac{60}{360})$ sq cm

= $(\frac{\sqrt{3}}{4} * 49) - (11 * 0.5 * 3.5)$ sq cm

= 1.967 sq cm

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Q:

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 & Explanation Answer: B) 20ft

Explanation:

let the radius of the pool be Rft

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

Area of the concrete wall = $π [{(R + 4)}^{2} - R^{2}]$ = $π [(R + 4 + R) (R + 4 - R)] = 8 π (R + 2)$ sq feet

=> 8 $π (R + 2) = \frac{11}{25} {πR}^{2} = > 11 R^{2} = 200 (R + 2)$

Radius of the pool R = 20ft

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