Quantitative Aptitude - Arithmetic Ability Questions

Q:

If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 350%, the resultant fraction is 25/51. What is the original fraction ?

A) 31/25 B) 15/17
C) 14/25 D) 11/16
 
Answer & Explanation Answer: B) 15/17

Explanation:

The original fraction is  2551 × 350+100150+100  = 15/17.

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Filed Under: Percentage
Exam Prep: GRE , GATE , CAT , Bank Exams , AIEEE
Job Role: Bank PO , Bank Clerk

61 33991
Q:

Choose the figure which is different from the rest three.

 

 

 

A) 1 B) 3
C) 2 D) 4
 
Answer & Explanation Answer: B) 3

Explanation:
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Filed Under: Odd Man Out
Exam Prep: Bank Exams

0 33965
Q:

(x-2) men can do a piece of work in x days and (x+7) men can do 75% of the same work in (x-10)days. Then in how many days can (x+10) men finish the work?

A) 27 days B) 12 days
C) 25 days D) 18 days
 
Answer & Explanation Answer: B) 12 days

Explanation:

34×(x-2)x=(x+7)(x-10)

x2-6x-280 =0 

=> x= 20   and   x=-14

 so, the acceptable values is x=20 

Therefore, Total work =(x-2)x = 18 x 20 =360 unit

 Now   360 = 30 x k         

=> k=12 days

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Filed Under: Time and Work

77 33958
Q:

A and  B can do  a piece of work in 30 days , while  B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish  the work?

A) 18 days B) 24 days
C) 30 days D) 36 days
 
Answer & Explanation Answer: A) 18 days

Explanation:

2(A+B+C)'s 1 day work = 1/30 + 1/24 + 1/20 = 1/8

=>(A+B+C)'s  1 day's work= 1/16

work done by A,B and C in 10 days=10/16 = 5/8

Remaining work= 3/8

A's 1 day's work= 116-124=148

Now, 1/48 work is done by A in 1 day.

 So, 3/8 work  wil be done by A in =48 x (3/8) = 18 days

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Filed Under: Time and Work

138 33923
Q:

The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then total quantity of the mixture in the container is:

A) 25 B) 30
C) 45 D) cannot be determined
 
Answer & Explanation Answer: B) 30

Explanation:

pool : kerosene 

   3  :  2(initially) 

   2  :  3(after replacement)

 

 Remaining QuantityInitial Quantity=1-Replaced QuantityTotal Quantity

 

(for petrol)   23=1-10k

 => K = 30

 Therefore the total quantity of the mixture in the container is 30 liters.

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Filed Under: Alligation or Mixture

83 33889
Q:

Select the option that does not belong in the following figure series.

 

 

A) A B) B
C) C D) D
 
Answer & Explanation Answer: A) A

Explanation:
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Filed Under: Odd Man Out
Exam Prep: Bank Exams

1 33854
Q:

Direction: What value should come in place of the question mark (?) in the following question?

529*6/2*?=14076

A) 204 B) 251
C) 102 D) 146
 
Answer & Explanation Answer: A) 204

Explanation:

23*3*x=14076
69*x=14076
= 69 =204

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0 33799
Q:

Two partners investede Rs. 1250 and Rs. 850 respectively in a business. They distributed 60% of the profit equally and decide to distribute the remaining 40% as the ratio of their capitals. If one partner received Rs. 30 more than the other, find the total profit?

Answer

Let the total profit be Rs.x


60% of the profit = \inline \frac{60}{100}\times x=Rs.\frac{3x}{5}


from  this part of the profit each gets = Rs.\inline \frac{3x}{10}


40% of the profit = \inline \frac{40}{100}\times x=Rs.\frac{2x}{5}


Now, this amount of Rs.\inline \frac{2x}{5} has been divided in the ratio of capitals 1250 : 850 = 25 :17


\inline \therefore Share on first capital = \inline (\frac{2x}{5}\times \frac{25}{42})=Rs.\frac{5x}{21}


Share on second capital = \inline (\frac{2x}{5}\times \frac{17}{42})=Rs.\frac{17x}{105}


Total money received by 1st investor = \inline [\frac{3x}{10}+\frac{5x}{21}]= Rs.\frac{113x}{210}


Total money received by 2nd investor = \inline [\frac{113x}{210}+\frac{97x}{210}]=Rs.\frac{97x}{210}


\inline \therefore x = 393.75


Hence total profit = Rs. 393.75

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Subject: Partnership

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