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:

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.

92 52739
Q:

The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4 ?

 A) 76 B) 79 C) 85 D) 87

Explanation:

Average = total runs / no.of innings = 32

So, total = Average x no.of innings = 32 x 10 = 320.

Now increase in avg = 4runs. So, new avg = 32+4 = 36runs

Total runs = new avg x new no. of innings = 36 x 11 = 396

Runs made in the 11th inning = 396 - 320 = 76

161 52278
Q:

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

 A) 1 B) 2 C) 3 D) 4

Explanation:

Let the numbers 13a and 13b.

Then, 13a x 13b = 2028

=>ab = 12.

Now, the co-primes with product 12 are (1, 12) and (3, 4).

[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]

So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).

Clearly, there are 2 such pairs.

125 51717
Q:

It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?

 A) Monday B) Friday C) Sunday D) Tuesday

Explanation:

On 31st December, 2005 it was Saturday.

Number of odd days from 2006 to 2009 = (1 + 1 + 2 + 1) = 5 days.

On 31st December 2009, it was Thursday.

Thus, on 1st Jan, 2010 it is Friday

183 51671
Q:

A man spends 35% of his income on food, 25% on children's education and 80% of the remaining on house rent. What percent of his income he is left with ?

 A) 6 % B) 8 % C) 10 % D) 12 %

Explanation:

Let the total income be x.

Then, income left = (100 -80)% of [100 -  (35 + 25)] % of x = 20% of 40% of x = 20/100 * 40/100 * 100) % of x = 8 % of x.

173 51662
Q:

if the price of a book is first decreased by 25% and then increased by 20%, then the net change in the price will be  :

 A) 10 B) 20 C) 30 D) 40

Explanation:

Let the original price be Rs. 100.

New final price  = 120 %  of (75 % of Rs. 100) = Rs. (120/100 * 75/100 * 100) = Rs. 90.

Decrease = 10%

191 49827
Q:

Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

 A) 1/2 B) 3/5 C) 9/20 D) 8/15

Explanation:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

P(E) = n(E)/n(S) = 9/20.

107 49646
Q:

A grocer has a sale of Rs 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs, 6500 ?

 A) 4991 B) 5467 C) 5987 D) 6453

Explanation:

Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.

Required sale = Rs.[(6500 x 6) - 34009]

= Rs. (39000 - 34009)

= Rs.  4991.