Mphasis Aptitude Questions and Answers For Freshers PDF Download

Mphasis Aptitude Questions and Answers

Mphasis Aptitude Questions and Answers with Explanation: Good News to the job aspirants who are really struggling in order to get a job in Mphasis. In this article, we have provided sample Mphasis Aptitude Questions and Answers For Freshers 2018, 2019 & 2020 Passed outs. This is a wonderful opportunity for the candidates who are preparing for placements. To gain more marks in the Mphasis Test, candidates should check out and practice these Mphasis Aptitude Questions and Answers which are given with Solutions, Explanations.



We have also provided the sectional wise Mphasis Aptitude Questions and Answers For Freshers. By practicing Mphasis Aptitude Questions and Answers with Solutions, you can know how the structure of the test paper be. So, kindly take a look at Mphasis Aptitude Questions and Answers and gain knowledge from this page. Aspirants should note that these are Mphasis model Questions but not an exact one, no one can provide the actual questions. Solve and learn the topics as well as the sub topics.

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Mphasis Aptitude Questions and Answers with Solutions

Number of Questions: 25 Questions
Time Allocated: 35 mins



TopicsQuestions asked (Approximately)
Averages2-4 Questions
Profit and Loss3-5 Questions
Time and Work2-4 Questions
Percentages2-3 Questions
Time and Distance1-3 Questions
Quadratic Equations3-5 Questions
Races and Games1-2 Questions
Mixtures and Allegations2-3 Questions
Simple Interest3-5 Questions
Compound Interest2-3 Questions
Simplification1-2 Questions
Areas2-5 Questions
Simple Equations2-4 Questions
Mensuration5-6 Questions
Boats and Streams2-5 Questions

Not only the mentioned topics as earlier, but also you should cover the remaining topics involved in quantitative aptitude section such as Surds & Indices, Pipes & Cisterns, Problems on Numbers, Problems on Trains, Ratios & Proportions, Problems on LCM & HCF etc. Because of the Aptitude section is very important in any placement exam. You can gain knowledge by learning Mphasis Aptitude Questions and Answers For Freshers (2018, 2019 & 2020 Batches) from this section.

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Mphasis Aptitude Questions and Answers For Freshers 2018, 2019 & 2020 Batches

Nowadays getting a job is not that much easy. Adn without a proper plan, you cannot get a chance to get a job in a reputed company like Mphasis. It involves a lot of hard work, you can prepare smartly by practicing the sample Mphasis Aptitude Questions and Answers with Solutions. Well, Higher authorities of Mphasis are going to recruit most talented students. So, learn Mphasis Aptitude Questions and Answers with Explanation. For the benefit of students, we have provided Mphasis Aptitude Questions and Answers in PDF format.

Mphasis Aptitude Questions and Answers PDF

1. 6, 23, 56, ?, 184, 219

A. 109
B. 107
C. 106
D. 104

Answer – A. 109

Explanation:
2^3 – 2^1
3^3 – 2^2
4^3 – 2^3
109

2. 3, 7, 12, 27, 50, ?

A. 85
B. 105
C. 100
D. 95

Answer – B. 105

Explanation:
3 * 2 + 1
7 * 2 – 2
12 * 2 + 3
27 * 2 – 4
50 * 2 + 5 = 105

3. 672, 681, 663, 672, 654, ?

A. 663
B. 687
C. 675
D. 643

Answer – A. 663

Explanation:
672+9 = 681
681-18 = 663
663+9 = 672
672-18 = 654
654+9 = 663

4. 346, 173, 259.5, 648.75, ?

A. 2270.62
B. 1450.25
C. 1560.54
D. 2125.75

Answer – A. 2270.62

Explanation:
(346÷2)*1 = 173
(173÷2)*3 = 259.5
(259.5÷2)*5 = 648.75
(648.75÷2)*7 =2270.62

5. 2, 11, 46, ?, 286

A. 136
B. 156
C. 190
D. 141

Answer – A. 141

Explanation:
2 * 5 + 1 = 11
11 * 4 + 2 = 46
46 * 3 + 3 = 141

6. 17, 17, 34, 136, ?

A. 2345
B. 3214
C. 1088
D. 1546

Answer – C. 1088

Explanation:
17*1 = 17
17*2 = 34
34*4 = 136
136*8 = 1088

7. 16, 33, 69, 126, ?

A. 257
B. 246
C. 206
D. 298

Answer – C. 206

Explanation:
16+(16*1+1) = 16+17 = 33
33+(17*2+2) = 33 + 36 = 69
69+(18*3+3) = 69+57 = 126
126+(19*4+4) = 126+ 80 = 206

8. 980, 147, 28, 11, 8.57, ?

A. 5.675
B. 8.224
C. 12.653
D. 10.765

Answer – B. 8.224

Explanation:
980/7 + 7 = 147
147/7 +7 = 28
28/7 + 7 = 11
11/7 + 7 = 8.57
8.57 /7 +7 = 8.224

9. 5 kg of an article is bought at Rs 480, 1/3rd of it is sold at a profit of 20%. At what loss% should the remaining article be sold so that there is an overall profit of 3 1/3%?

A. 8.5%
B. 7%
C. 6.5%
D. 5%

Answer – D. 5%

Explanation:
Let the remaining 2/3rd be sold at x% loss
1/3(20) + 2/3(-x) = 3 1/3 [x% loss, so used – sign] 1/3(20) + 2/3(-x) = 10/3
Solve, x = 5%

10. An article is sold at a 25% profit. If the CP and the SP of the article are increased by Rs 60 and Rs 30 respectively, the profit% decreases by 15%. Find the cost price of the article.

A. Rs 285
B. Rs 305
C. Rs 190
D. Rs 240

Answer – D. Rs 240

Explanation:
CP = x, then SP = (125/100)*x = 5x/4
New CP = (x+60), new SP = (5x/4 + 30), new profit% = 25-15 = 10
So (5x/4 + 30) = (110/100) * (x+60)
Solve, x = 240
Therefore, the cost price of the article = Rs. 240.

11. A boat can row at 16 km/hr in still water and the speed of river is 10 km/hr. Find the speed of boat with the river and speed of boat against the river.

A. 26 km/hr, 6 km/hr
B. 6 km/hr, 26 km/hr
C. 13 km/hr, 3 km/hr
D. 15 km/hr, 5 km/hr

Answer – A. 26 km/hr, 6 km/hr

Explanation:
Speed with the river (downstream) = 16+10 = 26 km/hr
Speed against the river (upstream) = 16-10 = 6 km/hr

12. A man goes downstream 60 km and upstream 20 km, taking 4 hrs each. What is the velocity of current?

A. 6 km/hr
B. 5 km/hr
C. 4 km/hr
D. 8 km/hr

Answer – B. 5 km/hr

Explanation:
Downstream speed = 60/4 = 15 km/hr
Upstream speed = 20/4 = 5 km/hr
Therefore, Velocity of stream = (15-5)/2 = 5 km/hr

13. The ratio between the A and B age is 7: 9. If the difference between the present ages of Q and P’s age after 4 years is 2 then what is the total of the present ages of P and Q?

A. 42
B. 44
C. 46
D. 48

Answer – D. 48

Explanation:
let the age of B is 9x and that of A is 7x. So
9x – (7x +4) = 2, x = 3
So sum will be = 27 + 21 = 48

14. Rekha married 6 years ago. Today her age is 5/4 times her age at the time of her marriage. Her daughter age is 1/5 of her age. What is the ratio of Rekha age to her daughter age after 6 years?

A. 2:1
B. 3:1
C. 4:3
D. 5:2

Answer – B. 3:1

Explanation:
R = (5/4)*(R – 6)
R = 30 years and daughter age = 30/5 = 6 years.
After 6 years ratio will be = 36/12 = 3:1

15. A laborer was appointed by a contractor on the condition he would be paid Rs 150 for each day of his work but would be, fined a the rate of Rs 30 per day for his absent. After 20 days, the contractor paid the laborer’s 2820. The number of days the laborer absented from work days:

A. 13 days
B. 19 days
C. 5 days
D. 12 days

Answer – B. 19 days

Explanation:
Let the required number of days =x days
So, 150x-(20-x)30=2820
x=19 days
Therefore, the number of days the labourer absented from work days = 19 days

16. Rahul can finish a job in 20 days. He worked for 10 days alone and completed the remaining job working with David, in 2 days. How many days would both David and Rahul together take to complete the entire job?

A. 10
B. 12
C. 4
D. 5

Answer – C. 4

Explanation:
Rahul alone finished 1/2 of the work in 10 days.
Remaining 1/2 was finished by Rahul and David together in 2 days.
Therefore, they both together can finish the complete job in 4 days.
Hence, it is proved

17. Two pipes P and Q can fill a tank in 10 min and 12 min respectively and a waste pipe can carry off 12 liters of water per minute. If all the pipes are opened when the tank is full and it takes one hour to empty the tank. Find the capacity of the tank.

A. 60
B. 75
C. 30
D. 45

Answer – A. 60

Explanation:
Let the waste pipe take ‘T’ time to empty the tank.
(1/10 + 1/12 – 1/T)*60 = -1
We will get T = 5 min
Therefore, the capacity of the tank is 5*12 = 60 liters

18. Two pipes can separately fill the tank in 15hrs and 30hrs respectively. Both the pipes are opened and when the tank is 1/3 full a leak is developed due to which 1/3 water supplied by the pipe leaks out. What is the total time to fill the tank?

A. 40/3 hr
B. 50/3 hr
C. 20/3 hr
D. 35/3 hr

Answer – A. 40/3 hr

Explanation:
(1/15 + 1/30)*T1 = 1/3, T1 = 10/3 hr
now after leak is developed, [(1/15 + 1/30) – (1/3)*(1/15 + 1/30)]*T2 = 2/3
T2 = 10 hr. So total time to fill the tank = 10 + 10/3 = 40/3 hr

19. There are 2 containers of equal capacity. The ratio of milk to water in the first container is 4:5 and in the second container is 3:7. If they are mixed up then the ratio of milk to water in the mixture will be

A. 34: 75
B. 67: 113
C. 17: 63
D. 65: 96

Answer – B. 67: 113

Explanation:
4+5 = 9=> 40: 50
3+7 = 10=> 27:63
40+27: 50: 63 = 67: 113
the ratio of milk to water in the mixture is 67: 113

20. A bag contains 10p,25p and Rs50p coins in the ratio of 5: 2: 1 respectively. If the total money in the bag is Rs.120.Find the number of 25p coins in that bag?

A. 110
B. 90
C. 160
D. 130

Answer – C. 160

Explanation:
10*5 : 25*2 : 50*1 = 50:50:50 = 1:1:1
120/3 = Rs.40
Rs. 1 = 4
Rs.40 = 4*40 = 160 coins

22. Ram and Gopal have money in the ratio 5: 12 and Gopal and Krishna also have money in the same ratio 5: 12. If Ram has Rs. 500, Krishna has

A. Rs.2500
B.Rs.2880
C. Rs.1850
D. Rs.3100

Answer – B. Rs.2880

Explanation:
Ram: Krishna = 5/12* 5/12 = 25/144
Hence, Krishna has 144*500/25 = 2880

23. The ratio of students of three classes is 2:3:4. If 12 students are increased in each class then their ratio turns into 13:18:23. What was the total number of students in all the three classes originally ?

A. 225
B. 190
C. 250
D. 215

Answer – A. 225

Explanation:
50:75:100
15 students increased
65:90:115 => 13:18 :23
Total no of students in all the three classes = 50+75+100 = 225

24. A circular wire of diameter 84 cm is bent into a rectangle with sides ratio 6: 5. What are the respective sides of the rectangle?

A. 72 cm, 60 cm
B. 90 cm, 75 cm
C. 60 cm, 72 cm
D. 78 cm, 65 cm

Answer – A. 72 cm, 60 cm

Explanation:
Length of wire = 2ᴨr = 2(22/7)*42 = 264 cm
Perimeter of rectangle = 2(6x+5x) = 264
Solve, x = 12
So dimensions – 12*6, 12*5 = 72, 60

25. The total cost of painting the walls of a room is Rs 475. Find the cost of painting the walls of another room whose length, breadth and height each are double than the dimensions of the previous room.

A. Rs 1846
B. Rs 1960
C. Rs 1780
D. Rs 1900

Answer – D. Rs 1900

Explanation:
Area of first room = 2(l+b)*h
After all dimensions doubled, new area = 2(2l+2b)*2h = 4[2(l+b)*h ] = 4 times previous area.
Then, cost of painting = 4*475

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