Quantitative Aptitude - Arithmetic Ability Questions

Q:

In a 100 m race, A can beat B by 25 m and B can beat C by 4 m. In the same race, A can beat C by:

A) 21 m B) 26 m
C) 28m D) 29m
 
Answer & Explanation Answer: C) 28m

Explanation:

A : B = 100 : 75

 

B : C = 100 : 96.

 

A : C = (A/B*B/C) = (100/75*100/96) = 100/72  =100:72     

 

   A beats C by (100 - 72) m = 28 m.

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Filed Under: Races and Games

24 7237
Q:

A teacher multiplies 987 by a certain number and obtains 556781 as her answer. If in the answer, both 6 and 7 are wrong but the other digits are correct, then the correct answer will be

A) 553681 B) 555181
C) 556581 D) 555681
 
Answer & Explanation Answer: D) 555681

Explanation:

Given number is 987 = 3 x 7 x 47.

So, required number must be divisible by each one of 3, 7, 47.

None of the numbers in 553681 and 555181 are divisible by 3. While 556581 is not divisible by 7.
Correct answer is 555681.

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Filed Under: Numbers
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7 7236
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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4 7229
Q:

Two bottles contains mixture of milk and water. First bottle contains 64% milk and second bottle contains 26% water. In what ratio these two mixtures are mixed so that new mixture contains 68% milk?

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

Explanation:

% of milk in first bottle = 64%

% of milk in second bottle = 100 - 26 = 74%

Now, ATQ

 

64%            74%
         

         68%

 

6                    4

 

Hence, by using allegation method,

 

Required ratio = 3 : 2

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Filed Under: Alligation or Mixture
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11 7227
Q:

A man has only 20-paise and 25-paise coins in a bag. If he has 50 coins in all totaling to Rs.10.25, then the number of 20-paise coins is 

A) 42 B) 45
C) 38 D) 36
 
Answer & Explanation Answer: B) 45

Explanation:

Let number of 20 ps coins = x and

number of 25 ps coins = y

Given total coins in the bag = 50

x + y = 50.......(1)

But the total money in the bag = Rs. 10.25

0.20x + 0.25y = 10.25

20x + 25y = 1025.........(2)

Now multiplying (1) by 25 we get

25x+25y=1250.............(3)

By solving (2) and (3)

20x + 25y = 1025;

=> x = 45;

Then, the no. of 20 ps coins are 45.

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Filed Under: Simplification
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5 7217
Q:

Two persons Shyam and Rahim can do a job in 32 days together. Rahim can do the same job in 48 days alone. They started working together and after working 8 days Rahim is replaced by a third person Ram whose efficiency is half that of Rahim. Find in how many days the remaining work will be completed by both Shyam and Ram together?

A) 16 days B) 72/5 days
C) 15 days D) 96/5 days
 
Answer & Explanation Answer: B) 72/5 days

Explanation:

Work done by Shyam and Rahim in 8 days = 8/32 = 1/4

Remaining work to be done by Shyam and Ram = 1 - 1/4 = 3/4

Given efficieny of Ram is half of Rahim i.e, as Rahim can do the work in 48 days, Ram can do the work in 24 days.

One day work of Ram and Shyam = (1/32 - 1/48) + 1/24 = 5/96

Hence, the total work can be done by Shyam and Ram together in 96/5 days.

 

Therefore, remaining work 3/4 can be done by them in 3/4 x 96/5 = 72/5 = 14.4 days.

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Filed Under: Time and Work
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8 7213
Q:

A shop keeper sells an article at a loss of 8%, but when he increases the selling price of the article by Rs. 164 he earns a profit of 2.25% on the cost price. If he sells the same article at Rs. 1760, what is his profit percentage?

A) 2.5% B) 5%
C) 10% D) 7.5%
 
Answer & Explanation Answer: C) 10%

Explanation:

According to the given data,

Let Cost price of the article be 'cp'

Then,

102.25 cp - 92 cp = 164 x 100

10.25 cp = 16400

cp = 1600

Now, if he sells at Rs. 1760

Profit = 1760 - 1600 = 160

 

Profit% = 160/1600 x 100 = 10%.

 

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10 7204
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 7202