Genpact Aptitude Questions and Answers For Freshers PDF Download

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Searching for the Genpact Aptitude Questions and Answers with Explanation to include in your preparation? then, refer to this article and get all the Sample Genpact Aptitude Questions and Answers For Freshers. Candidates who are 2018, 2019 & 2020 Passouts and seeking for the Genpact Placement need to solve and know the complete details with the help of this page. You can not only get the sample Genpact Aptitude Questions but also you can know the appropriate solutions.

Clearing the Genpact Placement Test is a bit difficult. And the solution is to prepare perfectly in the test. Moreover, the Aptitude Test can only be solved with logical ability. To gain such a knowledge you have to know the Test Pattern. As we have provided the Genpact Aptitude Questions and Answers with Explanations, it will be very easy for you to solve and know the suitable corrections. Enhance your time management skills as well as answering speed during the preparation. Furthermore, we have given you the direct link for the Genpact Aptitude Questions and Answers For Freshers PDF Download.

You Can Also Check: Genpact Placement Papers

Genpact Aptitude Questions and Answers For Freshers (2018, 2019 & 2020 Batches)

Number of Questions: 15 Questions

Genpact Aptitude Questions and Answers with Solutions, Explanation

Additionally, you have to prepare all the Aptitude topics. Because you can see some of the questions which are given difficult in the question paper. Furthermore, the Genpact recruiters will always seek for the candidates who have the complete skill in clearing the Genpact Placement Test. As the Genpact Placement Test comprises 15 Questions in the Aptitude section you can estimate the test paper as per the above table is given.

So, check the below sections, which are provided with the Genpact Aptitude Questions and Answers with Explanation. Time management during the preparation is a must so that you can easily reduce your test answering time and can refer to other questions. As these are given according to the Pattern wise, you can take them for the reference purpose.

Genpact Aptitude Questions and Answers in PDF

1. P, Q, and R can do a piece of work in 8 days. Q and R together do it in 24 days. Q alone can do it in 40 days. In what time will it be done by R working alone?

A. 25 days
B. 24 days
C. 60 days
D. 20 days

Answer – C. 60 days

Explanation:
P & Q do this work in 24 days.
Q alone does this work in 40 days.
Therefore, R alone will take 1/24 – 1/40 = 2/120=1/60 ⇒ 60 days

2. Ajay and Vijay undertake to do a piece of work for Rs. 480. Ajay alone can do it in 75 days while Vijay alone can do it in 40 days. With the help of Pradeep, they finish the work in 25 days. How much should Pradeep get for his work?

A. Rs. 40
B. Rs. 20
C. Rs. 360
D. Rs. 100

Answer – B. Rs. 20

Explanation:
In 24 days, they would have done 1/3 and 5/8 of the work.
The remaining work is 1 – (1/3 + 5/8) = 1/24.
This means Pradeep has done 1/24th of the work
So, he should be paid 1/24th of the amount i.e. 480 × 1/24 = Rs. 20

3. M can do a piece of work in 12 days. N can do this work in 16 days. M started work alone. After how many days should N join him, so that the work is finished in 9 days?

A. 2 days
B. 3 days
C. 4 days
D. 5 days

Answer – D. 5 days

Explanation:
M’s work in 9 days = 9/12 = 3/4. Remaining work = 1/4.
This work was done by N in 1/4 × 16 = 4 days.
Therefore, N would have joined M after 9 – 4 = 5 days.

4. The average price of 80 mobile phones is Rs.30,000. If the highest and lowest price of mobile phones is sold out then the average price of the remaining 78 mobile phones is Rs. 29,500. The cost of the highest mobile is Rs.80,000. The cost of the lowest price of mobile is?

A. Rs. 19000
B. Rs. 16000
C. Rs. 18000
D. Rs. 15000

Answer – A. Rs. 19000

Explanation:
According to the given information
The price of the costliest and cheapest mobile = (80*3000) – (78*29500) = 99000
Cheapest Mobile Price = 99000 – 80000 = 19000.

5. Ramesh walked 6 km to reach the station from his house, then he boarded a train whose average speed was 60 kmph and thus he reached his destination. In this way, he took a total time of 3 hours. If the average speed of the entire journey was 32 kmph then the average speed of walking is:

A. 2 kmph
B. 4 kmph
C. 5 kmph
D. 7 kmph

Answer – B. 4 kmph

Explanation:
By analyzing the given data
Total Distance = 32 * 3 = 6 + 60 * x
x = 1.5 hour;
Walking Speed = 6/1.5 = 4 kmph

6. Uday and Mahesh are partners in a business. Uday invests Rs.35,000 for 6months and Mahesh invests Rs.40,000 for 8 months. Out of a profit of Rs.28,750, Uday’s share is

A. Rs.12,765
B. Rs.10,450
C. Rs.11,300
D. Rs.11,392

Answer – D. Rs.11,392

Explanation:
According to the given information
35,000*6: 40000*8
2100:3200
21:32
Uday’s share = 21*28750/53 = 11391.5 = 11392

7. M and N invested in a business. They earned some profit which they divided in the ratio of 2:3. If M invested Rs.30000, the amount invested by N is

A. Rs.40,000
B. Rs.35,000
C. Rs.45,000
D. Rs.64,000

Answer – C. Rs.45,000

Explanation:
Given that
30,000: N = 2:3
N = 90,000/2 = 45,000

8. A sum of money is divided among A, B, C & D in the ratio 5: 4: 7: 9 respectively. If the share of A is Rs.2300 more than that of B, then what is the total amount of C & D together?

A. Rs.36000
B. Rs.35000
C. Rs.37200
D. Rs.36800

Answer – D. Rs.36800

Explanation:
Given
5x – 4x = 2300
X = 2300
C = 7*2300 = 16100
D = 9*2300 = 20700
C+D = 36800

9. X invested Rs 60000 in a business. After a few months, Y joined him with Rs 24000. The total profit was divided between them in the ratio 3: 1 at the end of the year. After how many months did Y join?

A. 4 months
B. 2 months
C. 6 months
D. 5 months

Answer – B. 2 months

Explanation:
X: Y
60*12 : 24a = 3:1
60*12/24a = 3/1 = 3
a = 60*12/24*3
= 60*4/24 = 10
12 – 10 = 2
Hence, After 2 months B joined

10. Out of the three annual examinations, each with a total of 500 marks, a student secured average marks of 45% and 55% in the first and second annual examinations. To have an overall average of 60%, how many marks does the student need to secure in the third annual examination?

A. 300
B. 350
C. 400
D. 450

Answer – C. 400
Explanation :
Total marks for three examinations = 3x 500 = 1500
Total required marks in three examinations = 60% of 1500
= 900
Marks secured in first examination = 45 % of 500
=225
Marks secured in third examination = 55 % of 500
=275
Thus, the required marks in the third examination
=900 – ( 225 + 275 )
= 900 – 500
= 400

11. The perimeter of a square is twice the perimeter of a rectangle. If the perimeter of a square is 48cms and the length of the rectangle is 7cm. Find the breadth of the rectangle?

A. 4 cm
B. 5 cm
C. 6 cm
D. 7 cm

Answer – B. 5 cm

Explanation:
P of Square = 4a = 48
A = 48/4 = 12cm
P of rectangle = 48/2 = 24cm =2(l+b)
2(7+b) = 24
B = 12-7 = 5

12. How many marbles of 10cm length and 7cm width are required to pave the floor of room 7m length and 4m breadth ?

A. 4000
B. 5100
C. 2800
D. 3200

Answer – A. 4000

Explanation:
Area of floor = 700*400 = 280000
Area of marble = 10*7 = 70
N = 280000/70 = 4000

13. Veena bought a watch costing Rs. 1404 including sales tax at 8%. She asked the shopkeeper to reduce the price of the watch so that she can save the amount equal to the tax. The reduction in the price of the watch is?

A. Rs.112
B. Rs.120
C. Rs.108
D. Rs.104

Answer – D. Rs.104

Explanation:
1.08x = 1404
x = 1300
The reduction of the price of the watch = 104

14. 80% of a small number is 4 less than 40% of a larger number. The larger number is 125 greater than the smaller one. The sum of these two numbers is

A. 325
B. 345
C. 355
D. 365

Answer – C. 355

Explanation:
smaller number = x; larger number = y
0.8x + 4 = 0.4y
4y – 8x = 40
y – x = 125
x = 115; y = 240
x + y = 355

15. 500 kg of ore contained a certain amount of iron. After the first blast furnace process, 200 kg of slag containing 12.5% of iron was removed. The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore. How many kgs of iron was there in the original 500 kg ore?

A. 54.2
B. 58.5
C. 46.3
D. 89.2

Answer – D. 89.2.

Explanation:
Let us assume that ‘x’ kg of iron in 500 kg ore.
Iron in the 200 kg of removed =200*12.5/100= 25 kg.
The percentage of iron in the remaining ore was found to be 20% more than the percentage in the original ore
So (x-25)/300 = (120/100)*x/500
=> x – 25 = 18x/25
=> 7x = 625
=> x = 89.2

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