Quantitative Aptitude - Arithmetic Ability Questions


What is Quantitative Aptitude - Arithmetic Ability?

 

Quantitative Aptitude - Arithmetic Ability test helps measure one's numerical ability, problem solving and mathematical skills. Quantitative aptitude - arithmetic ability is found in almost all the entrance exams, competitive exams and placement exams. Quantitative aptitude questions includes questions ranging from pure numeric calculations to critical arithmetic reasoning. Questions on graph and table reading, percentage analysis, categorization, simple interests and compound interests, clocks, calendars, Areas and volumes, permutations and combinations, logarithms, numbers, percentages, partnerships, odd series, problems on ages, profit and loss, ratio & proportions, stocks &shares, time & distance, time & work and more .

 

Every aspirant giving Quantitative Aptitude Aptitude test tries to solve maximum number of problems with maximum accuracy and speed. In order to solve maximum problems in time one should be thorough with formulas, theorems, squares and cubes, tables and many short cut techniques and most important is to practice as many problems as possible to find yourself some tips and tricks in solving quantitative aptitude - arithmetic ability questions.

 

Wide range of Quantitative Aptitude - Arithmetic Ability questions given here are useful for all kinds of competitive exams like Common Aptitude Test(CAT), MAT, GMAT, IBPS and all bank competitive exams, CSAT, CLAT, SSC Exams, ICET, UPSC, SNAP Test, KPSC, XAT, GRE, Defence, LIC/G IC, Railway exams,TNPSC, University Grants Commission (UGC), Career Aptitude test (IT companies), Government Exams and etc.


Q:

In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If he attempts all 60 questions and secures 130 marks, the no of questions he attempts correctly is :

A) 35 B) 38
C) 40 D) 42
 
Answer & Explanation Answer: B) 38

Explanation:

Let the number of correct answers be X.

Number of incorrect answers = (60 – X).

4x – (60 – x) = 130

=> 5x = 190 

=> x = 38

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183 63036
Q:

A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. If they get Rs. 3000 as wages for the whole work, what are the daily wages of A, B and C respectively (in Rs):

A) 200, 250, 300 B) 300, 200, 250
C) 200, 300, 400 D) None of these
 
Answer & Explanation Answer: B) 300, 200, 250

Explanation:

A's 5 days work = 50%  

B's 5 days work = 33.33% 

C's 2 days work = 16.66%     [100- (50+33.33)] 

Ratio of contribution of work of A, B and C = 50 : 3313 : 1623 = 3 : 2 : 1   

A's total share = Rs. 1500 

B's total share = Rs. 1000 

C's total share = Rs. 500  

 

A's one day's earning = Rs.300  

B's one day's earning = Rs.200  

C's one day's earning = Rs.250

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346 61416
Q:

What was the day of the week on, 16th July, 1776?

A) Tuesday B) Wednesday
C) Monday D) Saturday
 
Answer & Explanation Answer: A) Tuesday

Explanation:

16th July, 1776 = (1775 years + Period from 1st Jan, 1776 to 16th July, 1776)

Counting of odd days :

1600 years have 0 odd day.

100 years have 5 odd days.

75 years = (18 leap years + 57 ordinary years) =  [(18 x 2) + (57 x 1)] = 93 (13 weeks + 2 days) = 2 odd days


1775 years have (0 + 5 + 2) odd days = 7 odd days = 0 odd day. 

 

Jan   Feb   Mar   Apr   May   Jun   Jul  

31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days= (28 weeks + 2 days) 

 

Total number of odd days = (0 + 2) = 2.  

Required day was 'Tuesday'.

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282 61005
Q:

A student has to obtain 33% of the total marks to pass. He got 125 marks and failed by 40 marks. The maximum marks are :

A) 500 B) 600
C) 800 D) 1000
 
Answer & Explanation Answer: A) 500

Explanation:

Given that the student got 125 marks and still he failed by 40 marks

 

=> The minimum pass mark = 125 + 40 = 165

 

Given that minimum pass mark = 33% of the total mark

 

=> total mark =33/100 =165

 

=> total mark = 16500/33 = 500

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262 60657
Q:

Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are black or both are queen?

A) 52/221 B) 55/190
C) 55/221 D) 19/221
 
Answer & Explanation Answer: C) 55/221

Explanation:

We have n(s) =52C2 52 = 52*51/2*1= 1326. 

Let A = event of getting both black cards 

     B = event of getting both queens 

A∩B = event of getting queen of black cards 

n(A) = 52*512*1 = 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6  and  n(A∩B) = 4C2 = 1 

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and 

P(A∩B) = n(A∩B)/n(S) = 1/1326 

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

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185 60135
Q:

Tea worth of Rs. 135/kg & Rs. 126/kg are mixed with a third variety in the ratio 1: 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be____?

A) Rs. 169.50 B) Rs.1700
C) Rs. 175.50 D) Rs. 180
 
Answer & Explanation Answer: C) Rs. 175.50

Explanation:

Since first second varieties are mixed in equal proportions, so their average price = Rs.(126+135)/2= Rs.130.50

So, Now the mixture is formed by mixing two varieties, one at Rs. 130.50 per kg and the other at say Rs. 'x' per kg in the ratio 2 : 2, i.e., 1 : 1. We have to find 'x'.

 

Cost of 1 kg tea of 1st kind         Cost of 1 kg tea of 2nd kind 

 

 

  

x-153/22.50 = 1  => x - 153 = 22.50  => x=175.50.

 Hence, price of the third variety = Rs.175.50 per kg.

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377 59908
Q:

A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning?

Answer

Let the average after 7th inning = x


Then average after 16th inning = x - 3


16(x-3)+87 = 17x 


 x = 87 - 48 = 39

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418 59846
Q:

Insert the missing number.

7, 26, 63, 124, 215, 342, (....)

A) 391 B) 421
C) 481 D) 511
 
Answer & Explanation Answer: D) 511

Explanation:

Numbers are (23 - 1), (33 - 1), (43 - 1), (53 - 1), (63 - 1), (73 - 1) etc.

So, the next number is (83 - 1) = (512 - 1) = 511.

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218 59562