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Mphasis Aptitude Questions and Answers For Freshers
1. 448 ml of mixture A containing milk and water in the ratio of 9:5, respectively is mixed with ‘x’ ml of mixture B containing milk and water in the ratio of 11:10, respectively. If the ratio of milk to water in the final mixture is 3:2, then find the value of x.
- 252
- 210
- 336
- 294
Correct Choice: a
Solution
Quantity of milk in mixture A = 9/14 × 448 = 288 ml
Quantity of water in mixture A = 448 – 288 = 160 ml
Let amount of milk and water in mixture B is 11y and 10y respectively.
So, (288 + 11y)/(160 + 10y) = 3/2
Or, 576 + 22y = 480 + 30y
Or, 8y = 96
Or, y = 12
So, x = 21y = 21 × 12 = 252
Hence, option a.
2. A and B together can complete 75% of a work in 33 days while A, B and C together can complete the whole work in 26 days. If ‘C’ is 12.5% more efficient than B then find the time taken by A and C together to complete 70% of the work.
- 29.2 days
- 28.4 days
- 27.8 days
- None of these
Correct Choice: d
Solution
Total time taken by A and B together to complete the whole work = 33/0.75 = 44 days
Let total amount of work = 572 units (LCM of 44 and 26)
Efficiency of (A + B) = 572/44 = 13 units per day
Efficiency of (A + B + C) = 572/26 = 22 units per day
Efficiency of C = 22 – 13 = 9 units per day
Efficiency of B = 9/1.125 = 8 units per day
Efficiency of A = 13 – 8 = 5 units per day
Desired Time = (0.70 × 572)/14 = 28.6 days
Hence, option d.
3. A certain sum of money at a certain rate of compound interest compounded annually becomes Rs. 12500 after 2 years and Rs. 19531.25 after 4 years. Find the rate of compound interest.
- 20%
- 15%
- 17.5%
- 25%
Correct Choice: d
Solution
Let the principal amount is Rs. P and the rate of compound interest is R% p.a.
So, P(1 + R/100)2 = 12500……………………..(1)
And, P(1 + R/100)4 = 19531.25…………………..(2)
On dividing equation (2) by equation (1), we get
(1 + R/100)2 = 19531.25/12500 = 1.5625
Or, (1 + R/100) = 1.25
Or, R/100 = 0.25
Or, R = 25%
Hence, option d
4. Ratio of speed of a boat in still water to speed of stream is 9:2. The boat travels a distance of (D + 40) km in downstream and D km in upstream. If the ratio of time taken by the boat to travel in upstream and in downstream is 4:3, respectively then find the value of D.
- 220
- 240
- 212
- 224
Correct Choice: d
Solution
Let speed of boat in still water and speed of stream is 9x km/h and 2x km/h respectively.
So, Upstream speed = 9x – 2x = 7x km/h
And, downstream speed = 9x + 2x = 11x km/h
According to question;
{D/7x}/{(D + 40)/11x} = 4/3
Or, 33D = 28D + 1120
Or, 5D = 1120
Or, D = 224
Hence, option d.
5. Gunja marked an article 50% above the cost price and sold it after giving a discount of 20%. Had she bought the article for Rs. 150 less and sold it for Rs. 240 more then she would have made a profit of 60%. New selling price is how much percent more than original selling price.
- 20%
- 25%
- 15%
- None of these
Correct Choice: d
Solution
Let cost price of the article is Rs. x
Marked price of the article = 1.50 × x = Rs. 1.5x
Selling price of the article = 0.80 × 1.5x = Rs. 1.2x
According to question;
1.60 × (x – 150) = 1.2x + 240
1.6x – 240 = 1.2x + 240
Or, 0.4x = 480
Or, x = 1200
Original selling price = 1.2 × 1200 = Rs. 1440
Desired percentage = 240/1440 × 100 = 16.67%
Hence, option d.
6. Ratio of ages of A and B, 8 years ago was 5:4 respectively. If present average age of B and C is 38 years and age of C after 24 years will be 20% more than age of A after 2 years. Find the ratio of present age of B to present age of C.
- 9:10
- 10:9
- 9:8
- 8:9
Correct Choice: b
Solution
Let age of A and B, 8 years ago was 5x years and 4x years respectively.
Present age of C = ‘y’ years
So, 4x + 8 + y = 38 × 2 = 76
Or, 4x + y = 68
And, y + 24 = 1.20 × (5x + 8 + 2)
Or, y + 24 = 6x + 12
Or, 68 – 4x + 24 = 6x + 12
Or, 10x = 80
Or, x = 8
So, present age of B = 8 × 4 + 8 = 40 years
Present age of C = 68 – 4 × 8 = 36 years
Desired ratio = 40:36 = 10:9
Hence, option b.
7. If x2 + 16x – 5 = 0, then find the value of 5x/(x2 – 9x – 5).
- 1/7
- 1/9
- -1/3
- -1/5
Correct Choice: d
Solution
5x/(x2 – 9x – 5)
= 5x/(x2 + 16x – 5 – 25x)
= 5x/-25x = -1/5
Hence, option d.
8. The average of 50 observations is 42. Later it was found that 46 was misread as 64. Find the correct average.
- 41.64
- 40.58
- 39.88
- 40.36
Correct Choice: a
Solution
Correct average = {(50 × 42) – 64 + 46}/50 = 41.64
Hence, option a.
9. Find the value of {(2744)1/3 × 25} ÷ 7.
- 20
- 30
- 40
- 50
Correct Choice: d
Solution
{(2744)1/3 × 25} ÷ 7
= (14 × 25) ÷ 7 = 50
Hence, option d.
10. Sourav invested Rs. 2500 on 30% p.a. compound interest, compounded annually for 2 years. He then gave 20% of the amount received at 40% p.a. simple interest for 3 years. Find the simple interest received.
- Rs. 1242
- Rs. 1014
- Rs. 972
- Rs. 1146
Correct Choice: b
Solution
Amount received at compound interest = 2500(1 + 30/100)2 = Rs. 4225
Interest received at simple interest = (0.20 × 4225 × 40 × 3)/100 = Rs. 1014
Hence, option b.
11. For what least value of x, the number 203×88 is divisible by 36
- 4
- 6
- 3
- 2
Correct Choice: b
Solution
Since the number is divisible by 36 therefore, it has to be divisible by 9 and 4 both. The number formed by the last two digits is 88, therefore, the whole number is divisible by 4
For the number to be divisible by 9, the sum of the numbers should be divisible by 9
(2 + 0 + 3 + x + 8 + 8 ) = (21 + x). Therefore, least number which will make the number divisible by 9 is 6. Hence, option b.
12. The ratio of the perimeters of a rectangular and squared field is 7:6. Each side of the squared field is equal to the breadth of the rectangle. Find the length of the rectangular field if the area of the rectangular field is 4800 m2.
- 60 metres
- 80 metres
- 40 metres
- 120 metres
Correct Choice: b
Solution
Let the perimeters of the rectangular and squared field be 7x metres and 6x metres respectively
Therefore, breadth of the rectangular field = 6x/4 = 1.5x metres
Or, 2(l + b) = 7x
Or, l = 3.5x – 1.5x = 2x metres
According to the question,
2x × 1.5x = 4800
Or, x2 = 1600
Or, x = 40 metres
Therefore, length of the rectangular field = 2x = 80 metres
Hence, option b.
13. The value (cos37o – sin53o) + (sec41o – cosec49o) + (tan78o – cot12o) + (tan256o – sec256o) is
1
0
-1
2
Correct Choice: c
Solution
(cos37o – sin53o) + (sec41o – cosec49o) + (tan78o – cot12o) + (tan256o – sec256o)
= {cos37o – sin(90 – 37o)} + {sec41o – cosec(90 – 41o)} + {tan78o – cot(90 – 78o) + (-1)
= (cos37o – cos37o) + (sec41o – sec41o) + (tan78o – 78o) – 1
= -1
Hence, option c.
14. An article is marked up by 120% above its cost price and then sold for Rs. 1320 after giving 20% discount. Find the cost price of the article.
- Rs. 600
- Rs. 750
- Rs. 960
- Rs. 840
Correct Choice: b
Solution
Let the cost price of the article be Rs. ‘x’
According to the question,
2.2 × 0.8x = 1320
Or, x = 1320/1.76 = Rs. 750
Hence, option b.
15. Amar is 4 times more efficient than Amish. Both working together can complete the work in 12 days. Find the number of days taken by Amar to complete the work alone.
- 16 days
- 15.8 days
- 12.5 days
- 14.4 days
Correct Choice: d
Solution
Let the efficiency of Amish be x units/day
Therefore, efficiency of Amar = 4x + x = 5x units/day
Total work = (5x + x) × 12 = 72x units
Time taken by Amar to complete the whole work alone = 72x/5x = 14.4 days
Hence, option d.
16. In a bag there are coins of Rs. 1, Rs. 2, 25 paise and 50 paise in the ratio 4:2:5:3, respectively. If the total amount in the bag is Rs. 172. Find the difference between the number of Rs. 1 coins and 50 paise coins.
- 16
- 12
- 18
- 14
Correct Choice: a
Solution
Let the number of coins of Rs. 1, Rs. 2, 25 paise and 50 paise be 4x, 2x, 5x and 3x respectively
According to the question,
4x + (2 × 2x) + (5x/4) + (3x/2) = 172
Or, 16x + 16x + 5x + 6x = 172 × 4
Or, x = (172 × 4)/43
Or, x = 16
Required difference = (4x – 3x) = x = 16
Hence, option a.
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