Deloitte **Aptitude Test 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 to get comfortable with the pattern of the exam – **Check complete Deloitte Aptitude Test pattern & Syllabus here**

**Deloitte Aptitude Test Pattern**

**Total Questions:**25**Time Allotted:**35 minutes**Cut-Off:**70%**Minimum questions to attempt:**18-20

The aptitude test of Deloitte Online Test consists of questions based on basic aptitude topics such as profit, loss and percentages, Time & Work, Speed & Distance, AP, GP, Geometry, Mensuration etc. The section difficulty level varies from easy to moderate. **The total time allotted for this section is 35 minutes**. Given below are the topics to be covered for the test:

**Number Systems****Averages****Percentage****Simple Interest & Compound Interest****Time, Speed & Distance****Geometry****Coordinate Geometry****Logarithms****Quadratic Equations****Probability****Permutation & Combination****Time and Work****Alligation**- Few Miscellaneous Questions
- Questions based on general logic.

**Deloitte Aptitude Test Questions with Answers**

**1) Find the HCF of 18, 108, and 264**

A. 4 B. 3 C. 6 D. 8

**Answer:** Option C

We must first dissolve the numbers into their prime factors. The product of the prime factors that are common to all of the number will be the HCF.

18 = 2 x 3^2

108 = 2^2 x 3^3

264 = 2^3 x 3 x 11

HCF = 2 x 3 = 6

**2) Find the smallest number 4 digit number, when divided by 7, 9 and 11 leaves a remainder of 3, 5 and 7 in each case.**

A. 1355 B. 1366 C. 1377 D. 1388

**Answer:** Option D

Since the difference between the divisor and remainder is the same in each case, 7-3 = 9-5 = 11-7 = 4, the answer will be the smallest 4 digit LCM of 7, 9 and 11, -4.

LCM of 7, 9 and 11 = 7 x 9 x 11 as they are prime to each other = 696.

Smallest 4 digit multiple of the LCM = 696 x 2 = 1392.

Number = 1392 – 4 = 1388.

**3) If 43322K is exactly divisible by 72, what is K?**

A. 4 B. 2 C. 6 D. 8

**Answer:** Option A

If 43322K is divisible by 72, then it will be divisible by both 8 and 9

To check divisibility of 8, we have to check the last three digits (as all 000s are divisible by 8)

224 is the only one divisible by 8.

**4) Find the remainder if 2^39 is divided by 7.**

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

**Answer:** Option C

Using the pattern method;

2^1 / 7 = 2 (REM)

2^2 / 7 = 4

2^3 / 7 = 1

2^4 / 7 = 2

2^5 / 7 = 4

And so on

Every third power of 2 when divided by 7 gives a remainder of 4. Therefore remainder = 4. (As 2^39 is the 13th 3rd power of 2)

**5) Find the last digit of 587^1997**

A. 7 B. 9 C. 3 D. 1

**Answer:** Option A

We can use the power cycles of 7 to solve this question

Last digits of powers of 7 —> 7, 9, 3, 1, 7, 9, 3, 1..

Cycle repeats after every 4 numbers, I.e multiple of 4 will end with 1

587^2000 will end with 1, therefore 587^1997 will end with 7

**6) Find the last 2 digits of 2^265**

A. 16 B. 32 C. 64 D. 08

**Answer:** Option B

For 2, the formula is as follows;

Last 2 digits of (2^10)^n = 76, if n is even, and = 24, if n is odd

2^265 = (2^10)^26 x 2^5

= 76 x 32

= 2432 , last two digits = 32.

**7) In how many ways can the word ‘MANGOES’ be arranged, in which A and N are at the end positions?**

A. 200 B. 240 C. 400 D. 480

**Answer:** Option D

When A is at the start, and N is at the end, the 5 other letters can be arranged in 5! ways.

When N is at the start, and A is at the end, the 5 other letters can be arranged in 5! ways.

Answer = 5! + 5! = 480

**8) How many 7 digit numbers can be formed using the digits 1, 3, 0, 3, 5, 3, 5?**

A. 120 B. 360 C. 240 D. 420

**Answer:** Option B

In the first place, we have 6 options as 0 can’t be used. The other places can be filled normally.

6 x 6 x 5 x 4 x 3 x 2 x 1 = 6! x 6

However, we have to account for the repetition of the numbers;

6! x 6 / 2!x3! = 360

**9) How many terms of the A.P. 1,4,7,… are needed to give the sum 1001?**

A. 25 B. 26 C. 27 D. 28

**Answer:** Option B

Sn = 1001 = n/2 x {2×1 + (n-1)x3}

2002 = 3n^2 – n

3n^2 – n – 2002 = 0

n = 26

**10) A single soul can merge with another soul in every second, and thus two souls get reduced to a single soul. If a soul cannot merge with another soul, then it becomes a ghost. Every soul tries not to become a ghost. On a particular day, there were (2^42000 – 1) souls. After how many seconds will all souls become ghosts.**

A. 42000 seconds B. 21000 seconds C. 41999 seconds D. 20999 seconds

**Answer:** Option A

2^1 -1 soul become ghosts in 1 second.

2^2 -1 soul become ghosts in 2 seconds. (one merge, then ghost)

2^3 -1 soul becomes a ghost in 3 seconds. (two merges, then ghost)

Therefore, 2^42000 – 1 become ghosts in 42000 seconds.

**11) A can do a piece of work in 4 days, B can do it in 20 days. With the help of C, they finish the work in 2 days. In how many days C alone can do the whole work?**

A. 4 days B**.** 5 days C. 6 days D. 7 days

**Answer:** Option B

Efficiency of A = 25% (100/4)

Efficiency of B = 5% (100/20)

Efficiency of A+B+C = 50% (100/2)

Efficiency of C = Efficiency of A+B+C – (Efficiency of A + Efficiency of B)

= 50% – 30%

= 20%

100/20 = 5 days.

**12) A can do a piece of work in 20 days. If B is 80% more efficient than A, then the number of days required by B to do the same piece of work is:**

A.11 days B.12 days C. 8 days D. 12 days

**Answer:** Option A

Efficiency of A = 100/20 = 5%

Efficiency of B = 5 x 1.8 = 9%

Thus number of days required by B = 100/9 = 11.11 days

**13) In a village the average age of n people is 39 years. But after the verification, it was found that one person was 50 years older than his age considered in the average. So the new average, after correction increased by 1. The value of n is?**

A. 40 B. 45 C. 50 D. 55

**Answer:** Option C

It is the same as a person with 50 years more age replacing an existing person in the village.

Since the total age of the village having n persons, is being increased by 50 years, and the average age is being increased by 1 year, there are 50 people in the village.

**14) A train normally covers a certain distance at a speed of 100 km/hr. However, if it were to halt for a fixed time interval in each hour its average speed reduced to 80 km/hr. What is the average time interval for which the train halts in each hour?**

A. 10 minutes B. 12 minutes C. 15 minutes D. 20 minutes

**Answer:** Option B

Let us find the LCM of the speeds, and consider that as the distance.

Distance = 400km.

Time taken at 100 km/hr = 4 hours

Time taken at 80 km/hr = 5 hours

Instead of 4, the train takes 5 hours. So, it takes 1 hour extra over a journey of 5 hours. So it would take 12 minutes extra on each hour. Thus, the train halts for 12 minutes each hour.

**15) Monthly incomes of A and B are in the ratio of 5:4 and their savings are in the ratio of 4:3. If the expenditure of each will be 600, then the monthly incomes of each are?**

A. 3000, 2400 B. 2400, 1920 C. 5000, 4000 D. 3500, 2800

**Answer:** Option A

Income = Exp + savings

A —> 5x = 4y + 600

B —> 4x = 3y + 600

5x – 4y = 600 = 4x – 3y

5x – 4y = 4x – 3y

x = y

x = 600, y = 600

Income = A —> 600×5 = 3000, B —> 600×4 = 2400

**16) A traveled from B to C covering a total distance of 300 km in 9 hrs. He traveled by car at 25 km/h and the rest by train at 40 km/h. The distance traveled by car is:**

A. 80 B. 85 C. 95 D. 100

**Answer:** Option D

Let the distance be D

D / 25 + 300-D / 40 = 9

D = 100.

**17) The price of an item is increased by 40%. By what percent should the new price be decreased to bring it back to the original price?**

A. 40 B. 30 C. 28.57 D. 29.3

**Answer:** Option C

Let the initial price be 100.

After increase = 140.

So, to a decrease of 40 on 140 is 40/140 x 100%

= 28.57%

**18) The Indian cricket team played 20 one-day matches in a particular season of a year and won 25% of the matches they played. If they wanted a minimum success rate of 75%, what is the minimum number of matches they would have to play more?**

A. 30 B. 35 C**.** 40 D. 45

**Answer:** Option C

Required Win to loss ratio = 3:1 ( 75% : 25% )

Current losses = 15 ( 20 x (100-25)% )

Therefore, wins required = 15 x 3 = 45.

More wins required = 45 – 5 (original wins) = 40

**19) I went to buy a Maruti at Nurav Motors. The dealer offered me three discount options on the list price of 200000.**

1) Two successive discounts of 10%

2) Three successive discounts of 7%

3) Four successive discounts of 5.5%

Which option is the best for me?

A. 1 B. 2 C. 3 D. All are equal

**Answer:** Option B

We don’t need to calculate the absolute profit difference at all, and simply work with numbers and find the cheapest option.

0.9 x 0.9 = 0.81

0.93 x 0.93 x 0.93 = 0.804

0.945 x 0.945 x 0.945 x 0.945 = 0.797

Cheapest is option 2

**20) A and B start a business with some investments. A as a working partner received 20% of the annual profits as salary and the remaining was equally divided among A and B. If the entire profit was divided among A and B in the ratio of their investments, A would have received 1000rs less than what he actually got. B’s share of the profit is 6000. If B’s investment is 21000, what is A’s investment.**

A. 18000 B. 20000 C. 25000 D. 24000

**Answer:** Option D

Let the profit be P.

B’s profit = (P – 0.2P) / 2 = 0.4P = 6000

P = 15000, 0.6P = 9000

If the profit was divided equally, A would have received 9000-1000 = 8000, and B would have received 7000.

Therefore, ratio of investments = 8:7

Therefore A’s investment = 21000 x 8/7 = 24000.