# Goldman Sachs Quantitative Aptitude Questions And Answers

Goldman Sachs Quantitative Aptitude MCQs : This section focuses on "Quantitative Aptitude" for Goldman Sachs Exam. These Quantitative Aptitude MCQs are asked in previous Goldman Sachs placements/recruitment exams and will help you to prepare for upcoming Goldman Sachs drives.

1. If 5 students utilize 18 pencils in 9 days, how long at the same rate will 66 pencils last for 15 students?

A. 10 days

B. 12 days

C. 11 days

D. None of these

View Answer

Ans : C

Explanation: Required number of days = 9 X 5/15 X 66/18 = 11 days

2. The tax on a commodity is diminished by 10 % and its consumption increased by 10 %. The effect on the revenue derived from it changes by K %. Find the value of K.

A. 1

B. 2

C. -1

D. -2

View Answer

Ans : C

Explanation: Directly using the formula, when a value is increased by R% and then decreased by R%, then net there is R2/100% decrease.

Putting R = 10, we get 1 % decrease.

3. A portion of a 30 m long tree is broken by a tornado and the top strikes the ground making an angle of 30° with the ground level. The height of the point where the tree is broken is equal to

A. 30/√3 m

B. 10 m

C. 15 m

D. 60 m

View Answer

Ans : B

Explanation: The height of the tree = 30m.

So, x + y = 30m

Also sin30∘ 1/2 = x/y x = 2y.

So 3x = 30,

x = 10 m.

4. Fill pipe A is 3 times faster than second Fill pipe B and takes 32 minutes less than Fill pipe B. When will the cistern be full if both pipes are opened together?

A. 25 minutes

B. 24 minutes

C. 30 minutes

D. 12 minutes

View Answer

Ans : D

Explanation: Let the time taken by A to fill the pipe is = A min.

So the time taken by B to fill the pipe B is = B min.

According to the given condition B = 3A;

and given that B – A = 32 min.

So solving we get A = 16 min, B = 48 min.

Let the total work be 48 units.

So time taken by them together is 48/4 = 12 min.

5. A year is selected at random. What is the probability that it contains 53 Mondays if every fourth year is a leap year?

A. 5/28

B. 3/22

C. (1/7)

D. 6/53

View Answer

Ans : A

Explanation: Let P(L) = probability of selecting a leap year. P(NL) = probability of selecting a non - leap year. P(M) = probability of getting 53 Mondays in a year.

P(M)=P(M|L)P(L)+P(M|NL)P(NL) (By law of probability)

= 2/7 * 1/4 + 1/7 * 3/4 => 5/28

6. A man walks 6 km at a speed of 1 1/2 kmph, runs 8 km at a speed of 2 kmph and goes by bus another 32 km. Speed of the bus is 8 kmph. Find the average speed of the man.

A. 4(5/6)

B. 3(5/6)

C. 5(7/6)

D. None of these

View Answer

Ans : B

Explanation: Man walked 6 km at 1.5 kmph, again he walked 8 km at speed of 2 kmph and 32 km at a speed of 8kmph time taken individually:

=> 6/1.5 = 4 m

=> 8/2 = 4 m

=> 32/8 = 4 m

Average speed of man= total distance/ total time

=> 46/12 = 3 (5/6)

7. A boat can travel at a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.

A. 2 hours

B. 3 hours

C. 4 hours

D. 5 hours

View Answer

Ans : C

Explanation: Speed downstream = (13 + 4) km/hr = 17 km/hr

Time is taken to travel 68 km downstream = (68/17)=4 hours

8. What least number must be added to 1056, so that the sum is completely divisible by 23 ?

A. 2

B. 3

C. 18

D. 21

View Answer

Ans : A

Explanation: Required number = (23 - 21)

= 2.

9. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?

A. 0.45

B. 45 (5/11) %

C. 54 (6/11)%

D. 0.55

View Answer

Ans : B

Explanation: Number of runs made by running = 110 - (3 x 4 + 8 x 6)

= 110 - (60) => 50.

Required percentage= ((50/110) * 100) = 45(5/11)%

10. Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case .

A. 4

B. 7

C. 9

D. 13

View Answer

Ans : A

Explanation: Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)

= H.C.F. of 48, 92 and 140 = 4.

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