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Q:

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.

A) B=900;H=300	B) B=300;H=900
C) B=600;H=700	D) B=500;H=900

Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
$13.5 * 10000 m^{2} = 135000 m^{2}$
Let altitude = x metres and base = 3x metres.
Then, $\frac{1}{2} * 3 x * x = 135000 \Rightarrow x^{2} = 90000 \Rightarrow x = 300$

Base = 900 m and Altitude = 300 m.

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Q:

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

A) b=6; l=18; H=6	B) b=5; l=6; H=18
C) l=6; b=18; H=15	D) l=5; b=18; H=18

Answer & Explanation Answer: A) b=6; l=18; H=6

Explanation:

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m

$x \times \frac{3 x}{2} = 54 \Rightarrow x^{2} = 54 \times 2 \Rightarrow x = 6$

So, breadth = 6 m and length = $\frac{3 (6)}{2}$ = 9 m.

Now, papered area = (1720/10) = 172 sq.m

Area of 1 door and 2 windows = 8 sq.m

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m

$2 \times (9 + 6) H = 180 \Rightarrow H = 6$

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Q:

In how many different ways can 3 students be associated with 4 chartered accountants,assuming that each chartered accountant can take at most one student?

A) 12	B) 36
C) 24	D) 16

Answer & Explanation Answer: C) 24

Explanation:

Number of permutations = $4 P_{3}$ = 24

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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 tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

A) rs.458	B) rs,558
C) rs.658	D) rs.758

Answer & Explanation Answer: B) rs,558

Explanation:

Area to be plastered= [2(l + b) x h] + (l x b)

= [2(25 + 12) x 6] + (25 x 12)

= (444 + 300)

= 744 sq.m

Cost of plastering = Rs.744 x (75/100)

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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:

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:

How many 3 letters words can be formed using the letters of the words hexagon?

A) 120	B) 210
C) 160	D) 200

Answer & Explanation Answer: B) 210

Explanation:

Since the word hexagon contains 7 different letters,the number of permutations is $7 P_{3}$ = 7 x 6 x 5 =210

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