Sutherland Aptitude Questions and Answers For Freshers PDF Download

Sutherland Aptitude Questions and Answers

Sutherland Aptitude Questions and Answers with Explanation For Freshers: The current article helpful for those who are looking for the Sample Sutherland Aptitude Questions and Answers For Freshers 2018, 2019 & 2020 passed out batches. Well, By practicing Sutherland Aptitude Questions and Answers with Solutions you can find the difficulty level of the exam according to that you can guess the questions may come in the test. Therefore, kindly learn Sutherland model Aptitude Questions. Candidates should keep in mind that these are sample Sutherland Aptitude Questions and Answers with Explanation but not the exact one. For the benefit of job aspirants, we have listed them. You cannot find the exact questions, because they may vary from test to test.



You Can Also Check: Sutherland Placement Papers 

Sutherland Aptitude Questions and Answers (Topics)

Number of Questions: 20 Questions

TopicsQuestions asked( Approximately)
Averages3-4 Questions
Profit and Loss3-5 Questions
Time and Work2-4 Questions
Percentages1-3 Questions
Time and Distance2-3 Questions
Quadratic Equations3-5 Questions
Races and Games1-2 Questions
Mixtures and Allegations2-3 Questions
Simple Interest1-3 Questions
Compound Interest2-3 Questions
Simplification1-2 Questions
Areas2-5 Questions
Simple Equations2-4 Questions
Mensuration3-4 Questions
Boats and Streams4-5 Questions

 



In addition to that, you should cover the remaining topics in the aptitude section such as Odd man out, problems on LCM and HCF, Probability, Permutation, Ratios & Proportions etc. Sutherland is a most reputed company not only IT but also service and Business. Therefore, the higher authorities of Sutherland is going to recruit the most talented, skillful and dynamic candidates. This is a wonderful opportunity for job aspirants in order to gain knowledge by preparing Sutherland Aptitude Questions and Answers with Explanation. By this kind of preparation, applicants can aware of exam pattern and can save time.

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

Nowadays there is very tough in order to secure a job in MNC like Sutherland. Students can fulfill their goal by learning Sutherland Aptitude Questions and Answers. For every employee, problem-solving skills are very important by practicing the listed questions you are able to increase that as well as the questions may come in the exam. Well, we have arranged Sutherland Aptitude Questions and Answers in PDF format with free of cost in the below section. Therefore, students can download it and prepare each and every question without fail, which may come in your exam.

Sutherland Aptitude Questions and Answers with Solutions

1. Rice worth Rs. 110 per kg and Rs. 95 per kg are mixed with a third variety in ratio 1:1:2. If the mixture is worth Rs. 115 per kg, the price of the third variety per kg will be

A. 137.5
B. 147.5
C. 117.5
D. 127.5

Answer – D.127.5

Explanation:
First, two types of rice are mixed in 1:1 so the total cost for 2 kg of rice is 205, so average price = 102.5
So, x – 115 = 12.5, x = 127.5

2. There are three vessels each of 20-liter capacity is filled with a mixture of milk and water. The ratio of milk and water are 2:3, 3:4 and 4:5 respectively. All the vessels are emptied into the fourth vessel, then find the ratio of milk and water in the final mixture.

A. 401/544
B. 401/545
C. 401/546
D. 401/543

Answer – A. 401/544

Explanation:
Milk = 2/5 + 3/7 + 4/9 and water = 3/5 + 4/7 + 5/9
so the ratio will be 401/544

3. Two vessels contain milk and water in the ratio of 7:3 and 2:3 respectively. Find the ratio in which the contents of both the vessels must be mixed to get a new mixture containing milk and water in the ratio 3:2.

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

Answer – C. 2:1

Explanation:
let us assume that, the ratio be k:1
then in the first mixture, milk = 7k/10 and water = 3k/10
and in the second mixture, milk = 2/5 and water = 3/5
[7k/10 + 2/5]/[3k/10 3/5] = 3/2
K = 2, so the ratio will be 2:1

4. In 80 liter mixture of milk and water, the water content is 40 percent. The trader gives 20 liters of the mixture to the customer and adds 20 liters of water to the mixture. What is the final ratio of milk and water in the mixture?

A. 11:9
B. 12:7
C. 9:13
D. 9:11

Answer – D. 9:11

Explanation:
milk = 48 and water = 32 litre initially
then milk = 48 – 20*3/5 = 36 and water = 32 – 20*2/5 + 20 = 44
so ratio = 9:11

5. 16, 9, 8, 13, 25, ?

A. 63.5
B. 68.5
C. 64.5
D. 59.5

Answer – A. 63.5

Explanation:
16 * 0.5 + 1 = 9
9 * 1 – 1 = 8
8 * 1.5 + 1 = 13
13 * 2 – 1 = 25
25 * 2.5 + 1 = 63.5

6. 3 8 27 76 ? 696 2099

A. 275
B. 255
C. 245
D. 235

Answer – D. 235

Explanation:
3 * 3 – 1 = 8
8 * 3 + 3 = 27
27 * 3 – 5 = 76
76 * 3 + 7 = 235.

7. 972 488 242 123 ? 31.75 13.875

A. 87.5
B. 59.5
C. 54.5
D. 69.5

Answer – B. 59.5

Explanation:
972 / 2 + 2 = 488
488 / 2 – 2 = 242
242 / 2 + 2 = 123
123/2 – 2 = 59.5

8. 52 54 28 30 ? 18 10

A. 18
B. 15
C. 14
D. 16

Answer – D. 16

Explanation:
52 / 1 + 2 = 54
54 / 2 + 1 = 28
28 / 1 + 2 = 30
30 / 2 + 1 = 16

9. A boat takes 25 hours for travelling downstream from point A to point B and coming back to point C midway between A and B. If the velocity of the stream is 5 km/hr and the speed of the boat in still water is 10 km/hr, what is the distance between A and B?

A. 146 km
B. 150 km
C. 100 km
D. 122 km

Answer – B. 150 km

Explanation:
Downstream speed = 10+5 = 15
Upstream speed = 10-5 = 5
Now total time is 25 hours
If the distance between A and B is d, then distance BC = d/2
Now distance/speed = time, so
d/15 + (d/2)/5= 25
Solve, d = 150 km
Therefore, the distance between A and B is 150km.

10. A boat takes 150 min less to travel 40 km downstream than to travel the same distance upstream. The speed of the stream is 4 km/hr. What is the downstream speed?

A. 12 km/hr
B. 10 km/hr
C. 8 km/hr
D. 16 km/hr

Answer – D. 16 km/hr

Explanation:
Let speed of boat in still water = x km/hr
So speed upstream = x-4, and speed downstream = x+4
Now given:
Time to travel 40 km downstream = time to travel 40 km upstream – 150/60
So 40/(x+4) = 40/(x-4) – 5/2
8/(x-4) – 8/(x+4) = 1/2
x+4 – (x-4)/(x2 – 16) = 1/16
solve, x = 12
so downstream speed = 12+4 = 16
Therefore, the downstream speed is 16 km/hr

11. In a stream running at 2 km/hr, a motorboat goes 6 km upstream and back again to the starting point in 2 hours. Find the speed of boat in still water.

A. 12 km/hr
B. 8 km/hr
C. 10 km/hr
D. 9 km/hr

Answer – B. 8 km/hr

Explanation:
Distance = time * [B^2 – R^2] / 2*B
6 = 2 * [B^2 – 4^2] / 2*B
B^2 – 6B – 16 = 0
(B-8)(B+2) = 0
So B = 8
Therefore, the speed of the boat in still water is 8 km/hr.

12. Out of the total investment, A invested 1/4th, B invested 1/3rd of the remaining and C the remaining. B earned Rs 10,000 after a year. Find the yearly profit of all.

A. Rs 45,000
B. Rs 40,000
C. Rs 58,500
D. Rs 49,600

Answer – B. Rs 40,000

Explanation:
Let total investment = Rs x
Then A’s = (1/4)*x
Remaining = 3/4th of x
So B’s investment = (1/3)*(3x/4) = x/4
And C’s = x – (x/4 + x/4) = x/2
so ratio of profits = x/4 : x/4 : x/2 = 1 : 1 : 2
so 1/4 * x = 10,000
x = 40,000
Therefore, the yearly profit of all is Rs 40,000

13. A dealer sold two ACs at Rs. 5940 each. On selling one AC he gained 10% and on selling the other he lost 10%. Find the dealer’s gain or loss percent?

A. 1% gain
B. 1% loss
C. 2% loss
D. 2% gain

Answer – B. 1% loss

Explanation:
5940 = (110/100)*cp1, cp1 = 5400
5940 = (90/100)*cp2, cp2 = 6600
So, CP = 5400 + 6600 = 12000
And selling price = 5940 + 5940 = 11880
% loss = (120/12000)*100 = 1

14. A shopkeeper buys 60 cycles and marks them at 20% above the cost price. He allows a discount of 10% on the market price for cash sale and 5% discount for credit sales. If three-fourths of the cycles are sold at cash and remaining for credit, the total profit is Rs. 11400. What is the cost price of a cycle?

A. 2000
B. 4000
C. 1000
D. 1500

Answer – A. 2000

Explanation:
Marked price = (120/100)*CP
cash sales = 45 and credit sales =15
(120/100)*cp*90/100*45 + (120/100)*cp*95/100*15 – 60*cp = 11400
Cost Price = 2000

15. A man buys some quantity of rice for Rs 4800. He sells one-third of it at a profit of 10%. At what percent gain should he sell the remaining two-thirds so as to make an overall profit of 20% on the whole transaction?

A. 15%
B. 20%
C. 25%
D. 10%

Answer – C. 25%

Explanation:
(1/3)*4800*110/100 + (2/3)*4800*(x/100) = (120/100)*4800
x = 125 so he should sell the remaining at 25% profit

16. P, Q, and R invest rupees 2000, 8000 and 10000 respectively in a business. At the end of the year, the balance sheet shows a loss of 40% of the initial investment. Find the share of loss of Q.

A. 2200
B. 2800
C. 3000
D. 3200

Answer – D. 3200

Explanation:
Total loss after one year = 20000*40/100 = 8000
share of Q = (4/10)*8000 = 3200

17. 12 percent of the voters in an election did not cast their votes. In this election, there are only two candidates. The winner by obtaining 45% of the total votes and defeated his rival by 2000 votes. The total number of votes in the election

A. 75000
B. 100000
C. 50000
D. 125000

Answer – B. 100000

Explanation:
12% percent didn’t cast their vote. 45% of total votes get by the winning candidates, so the remaining 43% will be scored by his rival. So,
(45/100 -43/100)*P = 2000
P = 100000

18. A number is first decreased by 25%. The decreased number is then increased by 20%. The resulting number is less than the original number by 40. Then the original number is?

A. 100
B. 200
C. 300
D. 400

Answer – D. 400

Explanation:
Let the number is a
a – (75/100)*a*(120/100) = 40
Then, we will get a = 400

19. A sum of money lent at compound interest for 2 years at 20% per annum would fetch Rs.723 more if the interest was payable half yearly than if it was payable annually. The sum is ____

A. Rs. 15000
B. Rs. 30000
C. Rs. 45000
D. Rs. 20000

Answer – B. Rs. 30000

Explanation:
sum – Rs.x
C.I. compounded half yearly = (4641/10000)x
C.I. compounded annually = (11/25)x
(4641/10000)x – (11/25)x = 723
x = 30000

20. Leela takes a loan of Rs. 8400 at 10% p.a. compounded annually which is to be repaid in two equal annual installments. One at the end of one year and the other at the end of the second year. The value of each installment is?

A. 4200
B. 4140
C. 4840
D. 5640

Answer – C. 4840

Explanation:
8400 = x*(210/121) => 4840

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