# 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.

• #### Volume and Surface Area

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%

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%

954 124902
Q:

Insert the missing number.

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

 A) 391 B) 421 C) 481 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.

356 124725
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

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.

244 124341
Q:

Find the odd man out.

2, 5, 10, 17, 26, 37, 50, 64

 A) 50 B) 37 C) 26 D) 64

Explanation:

(1*1)+1 , (2*2)+1 , (3*3)+1 , (4*4)+1 , (5*5)+1 , (6*6)+1 , (7*7)+1 , (8*8)+1

But, 64 is out of pattern.

88 116952
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

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.

393 112159
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

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.

456 111774
Q:

Find the odd man out.

41, 43, 47, 53, 61, 71, 73, 81

 A) 41 B) 61 C) 71 D) 81

Explanation:

Each of the numbers except 81 is a prime number.

134 108116
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

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