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Q:

First,second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered.In how many different ways can the prizes be awarded?

A) 1736	B) 1716
C) 1216	D) 1346

Answer & Explanation Answer: B) 1716

Explanation:

$13 P_{3}$ = 13!/10! = 1716

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Q:

A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?

A) 22	B) 44
C) 66	D) 88

Answer & Explanation Answer: D) 88

Explanation:

We have: l = 20 ft and lb = 680 sq. ft.So, b = 34 ft.

Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.

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Q:

In how many different ways can five persons stand in a line for a group photograph?

A) 120	B) 240
C) 360	D) 720

Answer & Explanation Answer: A) 120

Explanation:

This is the number of permutations of five things taken all at a time.

Therefore, answer = $5 P_{5}$ = 120 ways

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Q:

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?

A) 40	B) 20
C) 30	D) none of these

Answer & Explanation Answer: D) none of these

Explanation:

Let breadth = x metres.
Then, length = (x + 20) metres.

Perimeter =5300/23.50

2[(x + 20) + x] = 200
2x + 20 = 100
2x = 80
x = 40.
Hence, length = x + 20 = 60 m.

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Q:

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:

A) 2520	B) 2420
C) 2320	D) 2620

Answer & Explanation Answer: A) 2520

Explanation:

We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
Area = (l x b) = (63 x 40) = 2520 sq.m

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Q:

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A) 1	B) 2
C) 3	D) 4

Answer & Explanation Answer: C) 3

Explanation:

Area of the park = (60 x 40) = 2400 $m^{2}$

Area of the lawn = 2109 $m^{2}$

Area of the crossroads = (2400 - 2109) = 291 $m^{2}$

Let the width of the road be x metres. Then,

$60 x + 40 x - X^{2} = 291$

$x^{2} - 100 x + 291 = 0$

(x - 97)(x - 3) = 0
x = 3.

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Q:

The percentage increase in the area of a rectangle, if each of its sides is increased by 20%

A) 22%	B) 33%
C) 44%	D) 55%

Answer & Explanation Answer: C) 44%

Explanation:

Let original length = x metres and original breadth = y metres.

Original area = xy sq.m

Increased length = $\frac{120}{100}$ and Increased breadth = $\frac{120}{100}$

New area = $\frac{120}{100} x * \frac{120}{100} y = \frac{36}{25} x y m^{2}$

The difference between the Original area and New area is:

$\frac{36}{25} x y - x y$

$\frac{11}{25} x y$ Increase % = $(\frac{\frac{11}{25} x y}{x y}) * 100$ = 44%

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Q:

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?

A) 16	B) 18
C) 20	D) 22

Answer & Explanation Answer: B) 18

Explanation:

$\frac{2 (l + b)}{b} = \frac{5}{1}$ => 2l + 2b = 5b => 3b = 2l

b=(2/3)l

Then, Area = 216 cm²

=> l x b = 216 => l x (2/3)l =216

l = 18 cm.

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