Quantitative Aptitude - Arithmetic Ability Questions

Q:

Find: S.l. on Rs 3000 at 18% per annum for the period from 4th

A) 118 B) 105
C) 108 D) 110
 
Answer & Explanation Answer: C) 108

Explanation:

(24 + 31 + 18) days = 73 days = 1/5 year .

 

P = Rs 3000 and R = 18 % p.a.

 S.I=3000*18*15100

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Filed Under: Simple Interest
Exam Prep: GRE
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22 9739
Q:

If 6 men and 3 boys working together, can do four times as much work per hour as a man and a boy together. Find the ratio of the work done by a man and that of a boy for a given time  ?

A) 1 : 2 B) 2 : 1
C) 3 : 1 D) 1 : 3
 
Answer & Explanation Answer: A) 1 : 2

Explanation:

6M + 3B = 4(1M + 1B)
6M + 3B = 4M + 4B
2M = 1B

M/B = 1/2

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Filed Under: Ratios and Proportions
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19 9739
Q:

At present, the ratio between the ages of Arun and Harish is 4:3. After 6 years, Arun's age will be 26 years. What is the age of Harish at present ?

A) 15 years B) 16 years
C) 18 years D) 21 years
 
Answer & Explanation Answer: A) 15 years

Explanation:

Let the present ages of Arun and Harish be 4x and 3x years respectively.

Then, 4x + 6 = 26 => x = 5

Harish's age = 3x = 15 years.

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Filed Under: Problems on Ages
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8 9727
Q:

In a camp, there is a meal for 120 men or 200 children. If 150 children have taken the meal, how many men will be catered to with remaining meal?

A) 50 B) 40
C) 30 D) 20
 
Answer & Explanation Answer: C) 30

Explanation:

There is a meal for 200 children. 150 children have taken the meal.

 

Remaining meal is to be catered to 50 children.

 

Now, 200 children = 120 men.

 

50 children = 12020050 =30men

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Filed Under: Chain Rule

8 9719
Q:

A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is 

A) 30 B) 36
C) 40 D) 42
 
Answer & Explanation Answer: C) 40

Explanation:

Let distance = x km and usual rate = y kmph.
Then, x/y - x/(y+3) = 40/60 --> 2y (y+3) = 9x ----- (i)
Also, x/(y-2) - x/y = 40/60 --> y(y-2) = 3x -------- (ii)
On dividing (i) by (ii), we get:

x = 40 km.

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Filed Under: Time and Distance
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12 9714
Q:

Two vessel P and Q contain milk and water in ratio 3 : 2 and 5 : 3 respectively. If 10 liters of the mixture is removed from vessels P and poured in vessel Q then ratio of milk to water in vessel Q becomes 8 : 5. Find the initial quantity of water in vessel Q.

A) 10 lit B) 6 lit
C) 16 lit D) 12 lit
 
Answer & Explanation Answer: A) 10 lit

Explanation:

Given ratio of initial mixture of milk and water in Q = 5 : 3

Let the initial quantity of mixture in vessel Q = 8x

Let quantity of Milk = 5x and

Let quantity of water = 3x

According to the question,

5x + 10 x 353x + 10 x 25  = 85

=> 25x + 30 = 24x + 32

=> x = 2

Required Initial quantity of milk = 5x = 5 x 2 = 10 lit.

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Filed Under: Ratios and Proportions
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18 9710
Q:

A trader sold an article at profit of 20%. Had he bought that article at 60% less price and sold it at Rs 90 less, then he would have gained 50%. What is the value (in Rs) of cost price?

A) 150 B) 200
C) 250 D) 300
 
Answer & Explanation Answer: A) 150

Explanation:
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Filed Under: Profit and Loss
Exam Prep: Bank Exams

4 9685
Q:

Rajan's present age is three times his daughter's and nine-thirteenth of his mother's present age. The sum of the present ages of all three of them is 125 years. What is the difference between the present ages of Rajan's daughter and Rajan's mather ?

A) 32 years B) 48 years
C) 62 years D) 50 years
 
Answer & Explanation Answer: D) 50 years

Explanation:

Let Rajan's present age be 'x' years.
=> Then his daughter's present age is = x/3 years.
His mother's present age = 13x/9 years
Now, according to the question,
x + x/3 + 13x/9 = 125
=> x = 125x9/25 = 45
Therefore, required difference = 13x/9 - x/3 = 13x - 3x/9 = 10x/9
=> 10 x 45/9 = 50 years.

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Filed Under: Problems on Ages
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10 9683