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Q:

A single pipe of diameter x has to be replaced by six pipes of diameters 12 cm each. The pipes are used to covey some liquid in a laboratory. If the speed/flow of the liquid is maintained the same then the value of x is ?

A) 14.69 cm	B) 29.39 cm
C) 18.65 cm	D) 22.21 cm

Answer & Explanation Answer: B) 29.39 cm

Explanation:

Volume of water flowing through 1 pipe of diameter x = Volume discharged by 6 pipes of diameter 12 cms.

As speed is same, area of cross sections should be same.

Area of bigger pipe of diameter x = Total area of 6 smaller pipes of diameter 12

i.e ${πR}^{2} = 6 πR 1^{2}$

Here R = R and R1 = 12/2 = 6

$\Rightarrow R^{2} = 6 \times 6 \times 6$

R = 14.696

=> D = X = 14.696 * 2 = 29.3938 cm.

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Filed Under: Volume and Surface Area
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Q:

The sum of one-half, one-third and one-fourth of a number exceeds the number by 22. The number is

A) 264	B) 284
C) 215	D) 302

Answer & Explanation Answer: A) 264

Explanation:

Let the number be 'x'. Then, from given data

x/2 + x/3 + x/4 = x+22
13x/12 = x+22
x = 264

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Filed Under: Problems on Numbers
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Q:

Hemavathi gets 3 marks for each right sum and loses 2 marks for each wrong sum. He attempts 35 sums and obtains 60 marks. The number of sums attempted correctly is ?

A) 23	B) 24
C) 25	D) 26

Answer & Explanation Answer: D) 26

Explanation:

Let, Hema attempted 'k' sum correctly, then

k x 3 -2 x(35-k) = 60
5k = 130
k = 26

so 26 correct sums.

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Filed Under: Simplification
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Q:

There are 15 lines in plane. How many intersections (Maximum) can be made ?

A) 55	B) 105
C) 215	D) 148

Answer & Explanation Answer: B) 105

Explanation:

First line will cut all other 14, similarly second will cut 13, and so on
Total = 14+13+12+11+10+9+8+7+6+5+4+3+2+1 = 105.

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Filed Under: Problems on Numbers
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Q:

Vinod have 20 rupees. He bought 1, 2, 5 rupee stamps. They are different in numbers by the reason of no change, the shop keeper gives 3 one rupee stamps. So how many stamps Vinod have ?

A) 10	B) 18
C) 12	D) 15

Answer & Explanation Answer: A) 10

Explanation:

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

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Filed Under: Ratios and Proportions
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Q:

There are 3 sections with 5 questions each. If four questions are selected from each section, the chance of getting different questions is ?

A) 1000	B) 625
C) 525	D) 125

Answer & Explanation Answer: D) 125

Explanation:

Methods for selecting 4 questions out of 5 in the first section = 5 x 4 x 3 x 2 x 1/4 x 3 x 2 x 1 = 5. Similarly for other 2 sections also i.e 5 and 5

So total methods = 5 x 5 x 5 = 125.

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Filed Under: Permutations and Combinations
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Q:

The ratio of present ages of X and Y is 4:5. Which of the following can't be the ratio of ages of X and Y, 20 years ago ?

A) 2 : 5	B) 8 : 15
C) 9 : 10	D) 3 : 5

Answer & Explanation Answer: C) 9 : 10

Explanation:

The ratio of X and Y ages at present is 4:5.

Then, the ratio 20 years ago will not be more than the ratio at present.

So, from the options 9:10 is not satifying => its ratio is 0.9 which is greater than presnt ratio of 0.8.

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Filed Under: Problems on Ages
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Q:

A farmer divides his herd of k cows among his 4 sons, so that one son gets one half of the herd, the second gets one-fourth, the third gets one-fifth and the fourth gets 9 cows. Then k is equal to ?

A) 160	B) 120
C) 140	D) 180

Answer & Explanation Answer: D) 180

Explanation:

Given total number of cows = k
Now, 1st son share = k/2
2nd son share = k/4
3rd son share = k/5
4th son share = 9

(k) + (k/4) + (k/5) + 9 = k
=> k - (19k/20) = 9
=> (20k-19k)/20 = 9
=> k = 180.

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Filed Under: Percentage
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