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Q:

An alloy of copper and bronze weight 50g. It contains 80% Copper. How much copper should be added to the alloy so that percentage of copper is increased to 90%?

A) 45 gm B) 50 gm
C) 55 gm D) 60 gm
 
Answer & Explanation Answer: B) 50 gm

Explanation:

Initial quantity of copper =80100 x 50 = 40 g 

And that of Bronze = 50 - 40 = 10 g

Let 'p' gm of copper is added to the mixture

=> 50 + p x 90100 = 40 + p

=> 45 + 0.9p = 40 + p

=> p = 50 g

Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.

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Filed Under: Alligation or Mixture
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20 7137
Q:

In the given numbers, if in each number first digit is replaced by the third digit, second digit is replaced by the first digit and third digit is replaced by the second digit then which number will be the biggest ?

653   749   872   933   541

A) 933 B) 872
C) 749 D) 653
 
Answer & Explanation Answer: C) 749

Explanation:

From the given data,

EXP.

Largest number = 974

Therefore, the required number is 749.

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Filed Under: Arithmetical Reasoning
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11 7136
Q:

In a mixture of 240 lt. water is 20% and rest is Milk. What quantity of mixture should be taken out and replaced with water so that water becomes 40%?

A) 60 lit B) 55 lit
C) 45 lit D) 50 lit
 
Answer & Explanation Answer: A) 60 lit

Explanation:

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Filed Under: Alligation or Mixture
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15 7130
Q:

Solve this Math Puzzle?

math_puzzle11533899278.jpg image

A) 36 B) 72
C) 84 D) 96
 
Answer & Explanation Answer: B) 72

Explanation:

Here in the given puzzle,

2 x (2 + 2) = 8

3 x (3 + 3) = 18

4 x (4 + 4) = 32

5 x (5 + 5) = 50

Similarly,

6 x (6 + 6) = 72.

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Filed Under: Math Puzzles
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10 7125
Q:

A person says that his son is 5 times as old as his daughter and his wife is 5 times older than his son and he is twice the age of his wife . The sum total of all the ages equals the age of the grand mother who celebrated her 81st birthday today. How old was his son?

A) 50 yrs B) 25 yrs
C) 72 yrs D) 40 yrs
 
Answer & Explanation Answer: A) 50 yrs

Explanation:

Let the daughter age be 'x'
Therefore, Son age = 5x, Mother age = 25x, Person age = 50x.

Thus we get, x+5x+25x+50x = 81
81x = 81
x = 1

Daughter's age = 1, Son's = 5, Mother's = 25, Person's age = 50.

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Filed Under: Problems on Ages
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6 7123
Q:

Rohith started a business by investing Rs. 3600. After somedays, Puneeth joined him by investing Rs. 2400. At the end of one year, the profit was divided in the ratio 2 : 1. After how many months did Puneeth join the business?

A) 2 months B) 2.5 months
C) 3 months D) 3.5 months
 
Answer & Explanation Answer: C) 3 months

Explanation:

Ratio of profit of Rohith and Puneeth = 3600 x 12 : 2400 x P

= 3600 x 12/2400P = 2/1

=> P = 9

So, Puneeth joined the business after (12 - P) = 12 - 9 = 3 months.

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Filed Under: Partnership
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12 7116
Q:

Twelve percent of Kalyan's monthly salary is equal to the sixteen per cent of Nandu's monthly salary. Sarnesh's monthly salary is half that of Nandu's monthly salary. If sarnesh's annual salary is Rs. 1.08 lacs, What is Kalyan's monthly salary ?

A) 15,000 B) 36,000
C) 24,000 D) 12,500
 
Answer & Explanation Answer: C) 24,000

Explanation:

Let K = Kalyan's monthly salary
N = Nandu's monthly salary
S = Sarnesh's monthly salary.
Then from given data,
12% of K = 16% of N
S = N/2 => N = 2S
K = 16/12 x N = 16/12 x 2S
= 16/6 x 1.08/12 = 16/6 x 0.09 = 0.24 lacs = 24,000

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Filed Under: Percentage
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4 7112
Q:

The length of a brigade which a train 130 m long and traveling at 45kmph can cross in 30 seconds, is 

A) 196 mts B) 220 mts
C) 230 mts D) 245 mts
 
Answer & Explanation Answer: D) 245 mts

Explanation:

Let the length of the brigade is x mts
Then Distance = 130 + x mts
given speed = 45 kmph = 45 x5/18 m/s
Time = 30 sec
T = D/S
=> 30 = 130+x/(45x5/18)
=> x = 245 mts.

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Filed Under: Problems on Trains
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5 7109