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Placement Questions & Answers :: TCS NQT

1. If 19 and 1140 are the respective HCF and LCM of two numbers, which are greater than 19 then what will be the possible number of such pair?

Answer: 3

Explanation:

Product of HCF and LCM = product of the numbers
Then, product of the numbers = 19 x 1140
Let 19a and 19b be the numbers.
19a x 19b = 19 x 1140
ab = 19 x 1140 / 19 x 19 = 60
If ab = 60 then (a,b) = (1,60), (2,30), (3,20), (4,15), (5,12) and (6,10).
Since a and b are co-primes then (a,b) = (1,60), (4,15) and (5,12)
Hence the number of such pairs = 3a

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IBM TCS NQT 

2. find the number between 100 to 400 which is divisible by either 2,3,5,7..

Answer: 210

Explanation:

LCM of 2,3,5,7=210

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Accenture TCS NQT 

3. In a circular racetrack of length 100 m, three persons A, B and C start together. A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively. While C runs in the opposite at 15 m/s. When will all the three meet for the first time after the start?

Answer: 100

Explanation:

Since the track is a circular track A and B will meet every 50 seconds. i.e. 100 / (10-8).
Since it is a multiple of 50 they will be meeting at the starting point every 50 Seconds.
If you multiply 15 x 50 you will get 750 and after the second 50 it will be 1500.
All of them will meet at the starting point after 100

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TCS TCS NQT 

4. At a stadium a certain rows are to be reserved to accommodate 12 cricket players 24 football players and 32 soccer players. How many minimum rows need to be reserved if same number of players of only one category are to be seated in each rows

Answer: 4

Explanation:

Here is no explanation for this answer

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Syntel Inc. TCS NQT 

5. What is the largest number that will divide 90207, 232585 and 127986 without leaving a remainder?

Answer: 257

Explanation:

Given numbers are 90207,127986 and 232585.

To Get the greatest number which divides given numbers without leaving a remainder.

i.e find the HCF of given numbers.
90207=3*3*3*13*257
127986=2*3*83*257
232585=5*181*257
=> 257 is the largest number.

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TCS TCS NQT 

6. Water is continuously poured from a reservoir to a locality at the steady rate of 10,000 liters per hour. When delivery exceeds demand the excess water is stored in a tank. If the demand for 8 consecutive three-hour periods is 10000, 10000, 45000, 25000, 40000, 15000, 60000 and 35000 liters respectively, what will be the minimum capacity required of the water tank (in 1000 liters) to meet the demand?

Answer: 40

Explanation:

Here is no explanation for this answer

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iGate TCS NQT 

7. In the town of Unevenville, it is a tradition to have the size of the front wheels of every cart different from that of the rear wheels. They also have special units to measure cart wheels which is called uneve. The circumference of the front wheel of a cart is 133 uneves and that of the back wheel is 190 uneves. What is the distance traveled by the cart in uneves, when the front wheel has done nine more revolutions than the rear wheel?

Answer: 3990

Explanation:

LCM of 133 and 190 is 1330.


So, to cover this distance, front wheel takes 10 rounds, and back wheel takes 7 rounds.


So, for 3 rounds extra, 1330 uneves distance has to be travelled.


To take 9 rounds extra, 1330 * 3 = 3990 uneves has to be traveled.

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TCS TCS NQT 

8. There is a toy train that can make 10 musical sounds. It makes 2 musical sounds after being defective. What is the probability that same musical sound would be produced 5 times consecutively?

Answer: 1/32

Explanation:

1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32

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TCS TCS NQT 

9. After the typist writes 12 letters and addresses 12 envelopes, she inserts the letters randomly into the envelopes (1 letter per envelope). What is the probability that exactly 1 letter is inserted in an improper envelope?

Answer: 0

Explanation:

Here is no explanation for this answer

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TCS TCS NQT 

10. A 5-digit number is formed by the digits 2,4,5,6,8 (each digit used exactly once) . What is the probability that the number formed is divisible by 4 ?

Answer: 2/5

Explanation:

A no. is divisible by 4 if last two digit is divisible by 4
so numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
last two places are fixed by 24,28,48,52,56,64,68,84
so remaining 3 places can be filled in 3! ways for each
total=8*3!=48=n(E)
n(S) = 5!
so
= n(E)/n(S)
=48/120=2/5

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Capgemini TCS NQT