# 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 bag contains 50 P, 25 P and 10 P coins in the ratio 5: 9: 4, amounting to Rs. 206. Find the number of coins of each type respectively.

 A) 360, 160, 200 B) 160, 360, 200 C) 200, 360,160 D) 200,160,300

Answer & Explanation Answer: C) 200, 360,160

Explanation:

let ratio be x.

Hence no. of coins be 5x ,9x , 4x respectively

Now given total amount = Rs.206

=> (.50)(5x) + (.25)(9x) + (.10)(4x) = 206

we get x = 40

=> No. of 50p coins = 200

=> No. of 25p coins = 360

=> No. of 10p coins = 160

2020 264740
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.

562 260146
Q:

Find the odd man out?

396, 462, 572, 427, 671, 264.

 A) 671 B) 462 C) 427 D) 264

Explanation:

Here the given series is 396, 462, 572, 427, 671, 264.

In all the terms, the middle digit is the sum of first and third digit except 427.

So the Odd number in the given series is 427.

161 251633
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.

387 224237
Q:

If selling price is doubled, the profit triples. Find the profit percent ?

 A) 100% B) 200% C) 300% D) 400%

Explanation:

Let the C.P be Rs.100 and S.P be Rs.x, Then
The profit is (x-100)
Now the S.P is doubled, then the new S.P is 2x
New profit is (2x-100)
Now as per the given condition;
=> 3(x-100) = 2x-100
By solving, we get
x = 200
Then the Profit percent = (200-100)/100 = 100
Hence the profit percentage is 100%

1010 223579
Q:

A student multiplied a number by 3/5 instead of 5/3, What is the percentage error in the calculation ?

 A) 54 % B) 64 % C) 74 % D) 84 %

Answer & Explanation Answer: B) 64 %

Explanation:

Let the number be x.

Then, ideally he should have multiplied by  x by 5/3. Hence Correct result was x * (5/3)= 5x/3.

By mistake he multiplied x by 3/5 . Hence the result with error  = 3x/5

Then, error = (5x/3 - 3x/5) = 16x/15

Error %  = (error/True vaue) * 100 = [(16/15) * x/(5/3) * x] * 100 = 64 %

1002 215784
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.

458 215530
Q:

Insert the missing number.

2, 6, 12, 20, 30, 42, 56, (....)

 A) 61 B) 64 C) 72 D) 70