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:

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A) 25200 B) 52000
C) 120 D) 24400
 
Answer & Explanation Answer: A) 25200

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3*4C2

= 210. 

 

Number of groups, each having 3 consonants and 2 vowels = 210. 

 

Each group contains 5 letters. 

 

Number of ways of arranging 5 letters among themselves = 5! = 120 

 

Required number of ways = (210 x 120) = 25200.

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257 131178
Q:

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

A) 18 litres B) 24 litres
C) 32 litres D) 42 litres
 
Answer & Explanation Answer: B) 24 litres

Explanation:

Let the quantity of the wine in the cask originally be x litres

 
Then, quantity of wine left in cask after 4 operations =x1-8x4litres

 

x1-8x4x = 1681  

 

 1-8x4=234 

 

x=24

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557 131079
Q:

A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours will be the true time when the clock indicates 1 p.m. on the following day?

A) 48 min. past 12. B) 46 min. past 12.
C) 45 min. past 12. D) 47 min. past 12.
 
Answer & Explanation Answer: A) 48 min. past 12.

Explanation:

Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours. 

24 hours 10 min. of this clock = 24 hours of the correct clock. 

1456 hrs of this clock = 24 hours of the correct clock. 

29 hours of this clock = 24*6145*29 hrs of the correct clock 

= 28 hrs 48 min of the correct clock. 

Therefore, the correct time is 28 hrs 48 min. after 8 a.m. 

This is 48 min. past 12.

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675 129458
Q:

A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:

A) No profit, no loss B) 5%
C) 8% D) 10%
 
Answer & Explanation Answer: B) 5%

Explanation:

C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.

 

S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.

 

Gain =(80/1600*100) % = 5%

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988 129446
Q:

A bag contains 6 white and 4 black balls .2 balls are drawn at random. Find the probability that they are of same colour.

A) 1/2 B) 7/15
C) 8/15 D) 1/9
 
Answer & Explanation Answer: B) 7/15

Explanation:

Let S be the sample space

 

Then n(S) = no of ways of drawing 2 balls out of (6+4) =10C2 10 =10*92*1 =45

 

Let E = event of getting both balls of same colour

 

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

 

                =6C2+4C2 = 6*52*1+4*32*1= 15+6 = 21

 

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

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722 127527
Q:

A, B and C can do a piece of work in 24 days, 30 days and 40 days respectively. They began the work together but C left 4 days before the completion of the work. In how many days was the work completed?

A) 11 days B) 12 days
C) 13 days D) 14 days
 
Answer & Explanation Answer: A) 11 days

Explanation:

One day's work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days. 

Number of days required for this initial work = 7 days. 

Thus, the total numbers of days required = 4 + 7 = 11 days.

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486 123111
Q:

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

A) 40 B) 80
C) 120 D) 200
 
Answer & Explanation Answer: A) 40

Explanation:

Let the numbers be 3x, 4x and 5x.

 

Then, their L.C.M. = 60x.

 

So, 60x = 2400 or x = 40.

 

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

 

Hence, required H.C.F. = 40.

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419 118246
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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