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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:

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

A) 105555	B) 106665
C) 106656	D) 108333

Answer & Explanation Answer: C) 106656

Explanation:

The number of arrangements of 4 different digits taken 4 at a time is given by $4 P_{4}$ = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

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Q:

If each side of a square is increased by 25%, find the percentage change in its area.

A) 56.25%	B) 36.25%
C) 16.25%	D) 12.25%

Answer & Explanation Answer: A) 56.25%

Explanation:

Let each side of the square be a. Then, area = . $a^{2}$
New side = $\frac{125 a}{100} = \frac{5 a}{4}$ . New area = ${(\frac{5 a}{4})}^{2}$ = $(\frac{25 a^{2}}{16})$

Increase in area = $\frac{25 a^{2}}{16} - a^{2} = \frac{9 a^{2}}{16}$
Increase% = $[\frac{9 a^{2}}{16} * \frac{1}{a^{2}} * 100]$ % = 56.25%.

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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:

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 length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

A) 20%	B) 30%
C) 40%	D) 50%

Answer & Explanation Answer: D) 50%

Explanation:

Let original length = x and original breadth = y.

Original area = xy.

New length = x/2 and New breadth=3y

New area = $\frac{3}{2} x y$

Therefore, Increase in area = New area-original area = $\frac{3}{2} x y - x y = \frac{1}{2} x y$

Therefore, Increase % = $\frac{i n c r e a s e i n a r e a}{o r i g i n a l a r e a} * 100 = (\frac{\frac{1}{2} x y}{x y}) * 100 = 50$ %

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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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