TCS NQT Placement Paper

TCS NQT Numerical Ability Questions And Answers – TCS Placement Paper I

TCS NQT Previous Year Question Paper and Answers: TCS NQT is a National Qualifier Test to hire freshers. Here the last year TCS Numerical ability Question paper and answers. TCS Placement paper of previous 10 years.

TCS NQT Numerical Ability Test Pattern & Syllabus

TCS NQT numerical ability section consist of 26 question to be solved in 50 minutes, The TCS NQT questions mainly will be asked from the following topics.

  • Number System
  • Arithmetic
  • Elementary Statistics
  • Data Interpretation

The Main Sub-Topics of TCS NQT Numerical ability is

  • Averages
  • Mixtures & Allegations
  • Percentages
  • Permutations & Combinations
  • Profit and loss
  • Algebra
  • Ratio and proportions
  • Probability
  • Time, speed, distance
  • Time and Work
  • Functions
  • Geometry
  • Numbers

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TCS NQT Numerical Aptitude Question Paper

1.A student’s average ( arithmetic mean) test score on 4 tests is 78. What must be the students
score on a 5th test for the students average score on the 5th test to be 80?

(1) 80
(2) 84
(3) 86
(4) 88

Answer: (4) 88
4 tests sum=78*4=312
5 tests sum=80*5=400
then 5th test score=400-312=88

2. Mr. A took the sum of natural numbers. By mistake he added a number twice and find the sum to be 1000. Find the repeated number.
(1) 20
(2) 15
(3) 10
(4) 18

Answer: 3) 10
Sum of first 44 natural no is 990
Hence 10 is calculated twice.

3. A team of 12 is needed to be formed who are to be selected from 7 men and 12 women, with the restriction of selecting not more than 3 men. In how many ways can the selection be done?
(1) 9171
(2) 8243
(3) 7835
(4) 9967

Answer: With the restrictions we can form teams with 3 mens ,2 mens ,1 men ,0men
Ways =( 7C3*12C9 ) + (7C2 *12C10). + 
(7C1*12C11) +(7C0*12C12)
=7700 + 1386 + 84 +1
= 9171

4. Jake can paint a house in 20 days. Paul can paint the same house in 15 days. Jake, Paul and Harry together paint in 5 days. Harry alone can paint the house in

(1) 24 days
(2) 18 days
(3) 20 days
(4) 12 days

Answer: Lcm of 20,15,5 is 60 
Hence efficient of jake =60/20=3 units/day
Paul=60/15=4 units/day
Jake+paul+harry=60/5=12units per day 
We get harry’s efficiency =12-(4+3)=5
Therefore harry can complete in 60/5=12 days

5. Out of a group of ducks, 7/2 times the square root of the total number are outside the pond. The remaining 15 are inside the pond. Find the total number of ducks.

(1) 16
(2) 25
(3) 36
(4) 49

Answer: 7/2 * ✓36= 21 are outside the pond
And remaining 15 are inside the pond.
So the no. Is 36.

6. When an Rs. 10 per kg variety of an element is mixed with an Rs. 15 per kg variety of the same element, then what can be the price of the mixture?

(1) Rs. 15.1/kg
(2) Rs. 11.2/kg
(3) Rs. 9/kg
(4) Data insufficient

Answer: 2.Rs. 11.2/kg It is the law of averages and the price of the mixture would lie between the minimum and the maximum values which in this case are Rs. 10/kg and Rs 15/kg. Option 2, Rs 11.2/kg is the value that is between and hence is the answer.

7. Sebastian purchased an infinity stone and saved Rs. 200 when a discount of 25% was given. What is the Cost Price of the infinity stone?

(1) Rs. 1000
(2) Rs. 900
(3) Rs. 800
(4) Cannot be determined

Answer: Let marked price be=x
So (100-25)%x=x-200
75x = 100x -20000
X=800

8. The Simple Interest on a sum is Rs. 300 and the Compound Interest is Rs. 360 in the second year. What would be the Compound Interest for one year, compounded annually, on a different sum of Rs 6000 if the rate percent p.a. is the same as that of on the first sum?

(1) Rs. 2400
(2) Rs. 1200
(3) Rs. 3000
(4) None of the above

Answer: 1200(r=20%)

9. 41ⁿ – 14ⁿ is divisible by ___ when n belongs to N.
1) 27
(2) 9
(3) 3
(4)All of the above

Answer: all of the above …because if power n will even or odd number will be 41-14 always.. i.e 27 and 27 is also divisible by 9 and 3

10. A fast train takes 3 hours less than a slow train for a journey of 900 km, If the speed of the slow train is 15 km/hr less than that of the fast train, the speeds of the slow train is

(1) 50 km/hr
(2) 75 km/hr
(3) 60 km/hr
(4)45 km/hr

Answer: Let the speeds are =x,x-15
900/x-15 – 900/x= 3
300x-300x+4500=x^2-15x
x^2-15x -4500=0
(x-75)(x+60)=0=>x=75kmph
Slower train speed=75-15=60kmph

11. 3/5th of the balls in a bag are red, the rest are green. If 3/4th of the red balls and 5/7th of the green balls are defective, find the total number of balls in the bag given that the number of non-defective balls is 111.

(1) 516
(2) 432
(3) 420
(4) 504

Answer: Let 5x are the balls
Red Green =3x,2x
Defective=3x*3/4,2x*5/7=9x/4,10x/7
Total defective =103x/28
Non-deffective =5x-103x/28=(140-103)x/28
37x/28=111=>Total balls 5x=5*111*28/37
=420

12. Eesha’s father was 34 years of age when she was born. her younger brother, Shashank, now that he is 13, is very proud of the fact that he is as tell as her, even though he is three years younger than her. Eesha’s mother, who is shorter than Eesha, was only 29 when Shashank was born .what is the sum of the ages of Eesha’s parents now?

(1) 92
(2) 76
(3) 66
(4) 89

Answer: Eeshas father=34(when eesha born)
eeshas present age=16
eeshas fathers present age=50(34+16)
shashanks present age=13
shashanks mother age present=29+13=42
sum of their ages=50+42=92

13. If A, B and C are three positive integers such that A is greater than B and B is greater than C, then which of the following is definitely true?

(1) A% of B is greater than B% of C.
(2) B% of A is greater than C% of B
(3) C% of A is greater than B% of C
(4) All of the above

Answer: ALL Conditions are true
Let assume A=20,B=10,C=5;
now, A% OF B=2 > B% of C=0.5. Hence A is TRUE. ie)(20/100)*10=2 && (10/100)*5=0.5.
now B % of C=2 > C% of B=0.5. Hence B is TRUE.
now C% of A=1 > B% of C=0.5. Hence C is TRUE.

Hence, all the 3 conditions are true;

14. The prime factorization of integer N is A x A x B x C, where A, B and C are all distinct prime integers. How many factors does N have?

(1) 9
(2) 12
(3) 15
(4) 27

Answer: 12 because N=A*A*B*C;N=A^2*B*C=(2+1)(1+1)(1+1);=12

15. Two finals are scheduled – The Wimbledon match and the World Cup Cricket at the same time. Anu wants to watch the Wimbledon finals and her brother Vinu wants to watch WCC final. They decide to roll a tetrahedral die twice. The tetrahedral is numbered 1,2,3,4 on its four sides and all numbers are equally likely to appear. Anu rolls first and then Vinu rolls. If the number on the first roll is strictly greater than the number on the second roll. Anu wins and gets to watch Wimbledon. What is the probability that Anu will get to watch Wimbledon?

(1) 7/16
(2) 9/16
(3) 3/8
(4) 1/2

Answer:

The probability that Anu will get to watch the Wimbledon match is 3/8

Step-by-step explanation:

Probability of any number appearing = 1/4

Anu wins in the following condition

1) If 1 occus on the first roll then there is no chance of her winning because in the next  roll the number appeaing will be either 1 or greater than 1

2) If 2 appears on 1st roll then Anu will win if 1 appears on the second roll

3) If 3 appears on the 1st roll then Anu will win if 1 or 2 appears on the second roll

4) If 4 appears on the first roll then Anu will win if 1, 2, 3 appears on the second roll

Therefore, the probability of Anu winning

16.1/4 of the tank contains fuel. When 11 liters of the fuel is poured into the tank, the indicator rests at the 1/2 mark. Find the capacity of the tank in liters.

(1) 44
(2) 36
(3) 6
(4) 8

Answer: Let the capacity of the tank be x liters. Given, 1/4 of x + 11 = 1/2 of x

By solving we get the x value as 44 liters

17. You have been given a physical balance and 7 weights of 52, 50, 48, 44, 45, 46, and 78 Kgs.. Keeping weights on one pan and object on the other, what is the maximum you can weigh less than 183 Kgs.?

(1) 178
(2) 180
(3) 182
(4) 181

Answer: The highest 3 weights are 78+52+50= 180.

18. When 147 is divided by N the remainder is 4, When 255 is divided by N the remainder is 8
, When 622 is divided by N the remainder is 11, then Find N?

(1) 17
(2) 11
(3) 13
(4) 19

Answer: dividend=divisor x quotient + remainder
substract the remainder from dividend,
i.e.
147-4=143;255-8=247;622-11=611
HCF of 143, 247,611=13
Answer =13.

19. Out of a group of swans, 7/2 times the square root of the total number are playing on the shore of the pond. The remaining 2 are inside the pond. Find the total number of swans.

(1) 16
(2) 25
(3) 04
(4) 09

Answer: Let the number of swans = x2

x2= 7x/2 + 2 –> x2= (7x + 4)/2

2x2= 7x + 4, —>2x2– 7x – 4 = 0

The roots of x are 4, -1/2.Here -1/2 is not possible, so the x value will be 4.

The total number of swans is x2i.e 16.

20. There are 20 persons among whom two are sisters. Find the number of ways in which we can arrange them around a circle so that there is exactly one person between two sisters? Please note that the exact position on the circle does not matter (no seat numbers are marked on the circle), and only the relative positions of the people matter.

(1) 2! * 19!
(2) 2 * 18!
(3) 18!
(4) None of these

Answer: We can arrange 18 persons around a circle in (18 – 1)! = 17! ways. Now, there are exactly 18 places where we can arrange the two sisters. Also, the two sisters can be arranged in 2! Ways. Thus, the number of ways of arranging the persons subject to the given condition is (17!)(18)(2!) = 2*(18!)

21. Find the length of the longest pole that can be placed in an indoor stadium 24m long, 18m wide and 16m high.

(1) 36m
(2) 34m
(3) 30m
(4) 25m

Answer: 34.
sqrt(1156)=34
becoz max length is cuboid diagonal d=sqrt(L^2+B^2+H^2)
where L lenght
B breadth & H height

22. There are 10 points on a straight line AB and 8 on another straight line AC none of them being point A. how many triangles can be formed with these points as vertices?

(1) 680
(2) 720
(3) 816
(4) 640

To form a triangle we need 3 points

select 2 points from the 10 points of line AB & 1 from the 8 on AC
= (10C2)*(8C1)

select 2 points from the 8 points of line AC & 1 from the 10 on AB=
(8C2)*(10C1)

total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640

23. In a certain city, 60 percent of the registered voters are Party A supporters and the rest are Party B supporters. In an assembly election, if 75% of the registered Party A supporters and 20% of the registered Party B supporters are expected to vote for Candidate A, what percent of the registered voters are expected to vote for Candidate A?

(1) 20
(2) 40
(3) 53
(4) 57

Answer: let there is 100 reg voter
then 60 voters of congress and 40 bjp voters.
then from 60 voters 75% give vote to a so it is 45
then from 40 voters 20% give vote to a so it is 8
so sum is 45+8=53 people from 100 people so answer is 53%

24. It was the semester exam day, Vidhya caught the college bus. She enjoyed travelling by bus. Moving at 6 mph, the bus took Vidhya to college at the right time. She finished her exam and had a chit chat with her friends and suddenly she realized that it was 6 pm and she had missed the college bus. She decided to walk back home at 4 mph. What is her average speed for the day?

(1) 4 mph
(2) 5 mph
(3) 2.4 mph
(4) 4.8 mph

Answer: Use formula for average speed: 2xy/(x+y). Putting x=6 and y=4, we get average speed=4.8 mph.

25. A boat travels 300m upstream in 15 mins. its speed downstream is 8/5 time its upstream . which of the following option is correct?

(1) speed of current=12/10 of speed of boat
(2) speed of current=3/13 of speed of boat
(3) speed of current=9/13 of speed of boat
(4) None of the above

Answer: This question won’t even need the distance and time.
If speed of boat =x and speed of current=y then
(x+y)=8/5(x-y)
=>y+8y/5=8x/5-x
=>13y/5=3x/5
=>y=13x/3

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