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Accenture Aptitude Questions for Accenture Aptitude Test 2019 | FACE Prep

Published on 13 Nov 2018

Accenture Aptitude Questions with solutions for the Quantitative Aptitude Test is given here, these topics are the basic topics that are taught in high school. It is good to know the syllabus beforehand and getting comfortable with the pattern of the exam. It is also advisable to gain practice by giving Mock Tests.

Accenture Aptitude Questions - Pattern

Accenture Online test 2019 is conducted on Wheebox platform for a lot of colleges. Check the complete pattern here. The online test consists of 60 questions and these need to be answered in 60 minutes. The level of questions ranges from easy to medium difficulty level. There are a total of 20 Accenture Aptitude Questions to be completed in 20 minutes.This is a time-consuming section, so you need to keep track of time and answer questions accordingly. To understand the Accenture Recruitment Process, read the article here.

Accenture Aptitude Questions - Syllabus

The Accenture aptitude test syllabus 2019 is given here. A total of 20 Accenture Aptitude Questions to be solved in 20 mins. Following topics will come in Aptitude Test:

  • Time Speed & Distance
  • Probability
  • Number Series
  • Time & Work
  • Profit & Loss
  • Averages
  • Mixtures & Allegations
  • Ratio & Proportion
  • Permutations & Combinations

Accenture Aptitude Questions - Previously asked

1. How many Integers are there in between 300 and 600 that are divisible by 9?

A.) 33      B.) 31        C.) 28        D.)15

Answer: Option A.

Explanation: The sequence is 306,…..,594




2. A,B,C and D are four consecutive odd numbers and their average is 42. What is the product of B and D?

A.) 1860    B.) 1890      C.) 1845       D.) 1677       E.) None of These

Answer: Option C.

Explanation: Since these are consecutive odd numbers, the differences with A, B, C and D will be the same. So, the average should possibly lie between B and C. So, B is 41 and C is 43 so D must be 45 as we have to find the product and B and D so it would be 1845.

3The difference between the ages of a man and his daughter is 28. The average of their ages now is 28. What are their present ages in years?

A.) 28, 56      B.) 14, 42       C.)7,35         D.) 35,14

Answer: Option B.

Explanation: Let the daughter’s present age be x.

Then Father’s age(y) = x+28

Average = (x+y)/2 = 28

i.e 28 = (x+x+28)/2

=> 28 = (2x+28)/2

=> 28 = x+14

=> 14 = x 

If x=14, then y=x+28= 42

4The population of mice in a market doubles everyday. Every day 20 mice are killed. How many mice are there if after every 3 days, the number of mice becomes same again?

A.) 20     B.) 10      C.) 0     D.) 40

Answer: Option A.

Explanation: Let the number of mice be x.

Then, 1st day -> x

2nd day -> (2x-20)

3rd day -> [2(2x-20)-20]

after 3rd day, value becomes equal to the first day, so (4x-40-20) = x

which gives  x = 20

5. Let C be a positive integer such that (c+7) is divisible by 5. Then the smallest integer n (>2) such that c+n^2 is divisible by 5 is 

A.) 3    B.) 5      C.) 1      D.) Does not exist

Answer: Option D.

Explanation: c + n^2 is divisible by 5 if and only if c and n^2 are both divisible by 5.

But, if c is divisible by 5 then c + 5 will not be divisible by 5.

6. There are two boxes, one containing 32 black balls and other containing 31 red balls you are allowed to move the boxes at random so that when you choose the box at random and a ball at random from the chosen box, the probability of getting the black ball is maximized. This maximum probability is

A.) 1/4   B.) 1/2    C.) 3/4    D.) 51/100

Answer: Option C.

Explanation: Move 31 balls from red ball box to other box having 31 green balls.

Now one box has one red ball and another box has 31 red balls and 31 green balls.

Then, the probability of getting red ball = 1/2 (1+31/62) = 1/2 *3/2 =3/4

7. A box contains 90 mts each of 100 gms and 100 bolts each of 150 gms. If the entire box weighs 35.5 kg., then the weight of the empty box is

A.) 10 Kg   B.) 10.5 Kg    C.) 11 Kg     D.) 11.5 Kg

Answer: Option D.

Explanation: Total weight = 90*100 + 100*150

=> 24000/1000 = 24 Kg.


empty box => 35.5 – 24 =>11.5 Kg.

8. How many times do hands of a clock coincide in 5 hours?

A.) 5     B.) 4      C.) 11      D.) 12

Answer: Option B.

Explanation: 1:05, 2:10, 3:15, 4:20

9. A certain quantity of 40% solution is replaced with 25% solution such that the new concentration is 35%. What is the fraction of the solution that was replaced?

A.)1/4     B.)1/3       C.)1/2        D.)2/3       E.)3/4

Answer: Option B.


Quantity of cheaper/Quantity of dearer =

(C.P of dearer – Mean Price)/(Mean Price – C.P of cheaper) = 


By mixture rule,

40% soln/25% soln = 10/5

Accenture Aptitude Questions

10. There are three runners Ram, Gautam, and Shyam with their respective speeds of 10kmph, 20kmph and 30kmph they are initially at P and they have to run between the two points P and Q which are 10 km apart from each other. They start their race at 6 am and end at 6 pm on the same day. If they run between P and Q without any break, then how many times they will be together either at P or Q during the given time period?

A.) 5    B.) 7     C.) 4      D.) 12 .  

Answer: Option B.

Explanation: All 3 of them will meet at point P after every 2 hours which means they will meet after every 2 hours.

So, for 12 hours they meet 6 times and including the first time, 6+1=7.

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