1000+ Latest TCS NQT Aptitude Questions and Answers 2021 – 2022

Find the Latest TCS NQT Aptitude Questions and Answers 2021. There are a total of 26 questions asked and the time is 40 mins given for you to solve i.e; less than 2 min for each question. TCS NQT Aptitude Test Paper is the 1st section of the test.

Here, you will find all the similar types of previous year questions based on the TCS NQT Syllabus and all the basic details for the aptitude questions asked in the TCS NQT exam.

Aptitude Test For TCS NQT
Number of Questions 26 Questions
Time Limit 40 Mins
Difficulty High
IMPORTANT NOTE:
  1. There will be no negative marking.
  2. TCS NQT is adaptive this year
  3. You will not get any extra rough paper in the exam as a calculator and Rough Paper will be available on your Desktop Screen. You are not allowed to move your eyes down while giving the examination.

TCS NQT Aptitude Syllabus

Topics are –

  • Number System

    • Number System, LCM & HCF
    • Divisibility
    • Numbers & Decimal Fractions
  • Mensuration

    • Geometry
    • Area, Shapes & Perimeter
  • Arithmetic Ability

    • Percentages
    • Allegations and Mixtures
    • Ratios, Proportion
    • Work and Time
    • Speed Time and Distance
    • Profit and Loss
    • Averages
    • Equations
    • Ages
    • Probability
    • Clocks & Calendar
    • Series and Progressions
  • Elementary Statistics

    • Mean, Median, Mode, Variance, and Standard Deviation
  • Data Interpretation

    • Pie Charts
    • Tabular DI
    • Graphical DI
  • Simplifications & Approximations

    • Simplifications and Calculations (Rational and Irrational Number)

TCS NQT Aptitude Questions and Answers 2021

1. John is faster than Peter. John and Peter each walk 24 km. Sum of the speeds of John and Peter is 7 km/h. Sum of time taken by them is 14 hours. Find John’s speed.

a.    5 km/h
b.    4 km/h
c.    7 km/h
d.    8 km/h

Show answer
b.    4 km/h
Explanation
We know that John’s speed is greater than Peter’s speed and the sum of their speed is 7.
So the combinations are = (6, 1), (5, 2), (4, 3)
Now checking from the options if John’s speed is equal to 4, then Peter’s speed is 3,
or, the time is taken by them = 24/4 + 24/3 = 14 hours.

2. There is a city where all 100% votes are registered. Among this 60% votes for Congress and 40% votes for BJP. Ram, gets 75% of congress votes and 8% of BJP votes. How many votes did Ram get?

a.      42.8 %
b.      56.6 %
c.      48.2 %
d.      64.4 %

Show answer
c.      48.2 %
Explanation
Let the total number if votes = 100. So Ram gets,
75% of 60 = 60 * 0.75 = 45 votes
8% of 40 = 40 * 0.08 = 3.2 votes
Thus total number of votes that Ram gets = 48.2 %

3. There are two bags containing white and black marbles. In the first bag there are 8 white marbles and 6 black marbles and in the second bag, there are 4 white marbles and 7 black marbles. One marble is drawn at random from any of these two bags. Find the probability of this marble being black.

a.      7/154
b.      22/77
c.      7/54
d.      41/77

Show answer
d.      41/77
Explanation
Probability of drawing a black ball from the first bag is = 6C1 / 14C1
Probability of drawing a black ball from the second bag is = 7C1 / 11C1
Total probability = 1/2 * (6C1/14C1) * (7C1/11C1) = 41/77

4. A team of 11 is needed to be formed who are to be selected from 5 men and 11 women, with the restriction of selecting not more than 3 men. In how many ways can the selection be done?

a.      2256
b.      1543
c.      1444
d.      1245

Show answer
a.      2256
Explanation
Selecting 0 men and 11 women = 5C0 * 11C11 = 1
Selecting 1 men and 10 women = 5C1 * 11C10 = 55
Selecting 2 men and 9 women = 5C2 * 11C9 = 10 * 55 = 550
Selecting 3 men and 8 women = 5C3 * 11C8 = 10 * 165 = 1650
So total number of ways = 1650 + 550 + 55 + 1 = 2256 way

5. The rejection rate for Audi production was 4 per cent, for Mercedes it was 8 per cent and for the 2 cars combined it was 7 per cent. What was the ratio of Audi production?

a.      5/1
b.      7/1
c.      2/1
d.      3/1

Show answer
d.      3/1
Explanation
Using the simple weighted average formula we get,
(4x + 8y)/(x+y) = 7
or, 4x + 8y = 7x + 7y
or, a/b = 3/1

6. Anil works for 8 straight days and rest on the 9th day. If he starts his work on Monday, then on which day he gets his 12th rest day?

a.      Wednesday
b.      Thursday
c.      Monday
d.      Sunday

Show answer
a.      Wednesday
Explanation
Anil works for 8 days and rests on 9th day. In total 9 days are to be processed 12 times = 12 * 9 = 108.
If we calculate according to the week, we get 108 / 7 = remaining 3 days. So if Anil starts working on Monday, he will rest on the third day of the week which is Wednesday.

7. 10 programmers are able to type 10 lines in 10 minutes. How many programmers are required to type 60 lines in 60 minutes?

a.      30
b.      50
c.      10
d.      None of the above

Show answer
c.      10
Explanation
This is a simple question of logical reasoning. If 10 programmers can type 10 lines of code in 10 minutes then to type 60 lines of code, in 60 minutes, the same 10 coders will be required, since the lines of code and time are in proportion

8. When a + b is divided by 12 the remainder is 8, and when a – b is divided by 12 the remainder is 6. If a > b, what is the remainder when ab divided by 6?

a.      1
b.      3
c.      7
d.      5

Show answer
a.      1
Explanation
According to the question,
a + b = 12k + 8
=> $(a+b)^2 = 144k^2 + 64 + 192k$
a – b = 12l + 6
=> $(a-b)^2 = 144l^2 + 36 - 144l$
Subtracting both the equations we get,
ab = $36(k^2 - l^2) + 48k - 36l + 7$
Now all the terms of ab is divisible by 6, except 7. So the remainder left is 1.

9. One day, Ramesh started 30 minutes late from home and driving at 25% slower than the usual speed, reached the market 50 minutes late. How much time in minutes does Ramesh usually take to reach the market from home?

a.      40
b.      60
c.      50
d.      80

Show answer
b.      60
Explanation
Let the usual speed of Ramesh be ‘s’
Let the distance between home and market be ‘d’
So usual time took = d/s
Time took on that particular day = d/(3s/4)
So according to the question,
d/s(4/3 – 1) = 20
or, d/s = 60

10. Three containers A, B and C are having mixtures of milk and water in the ratio of 1:5, 3:5, 5:7 respectively. If the capacities of the containers are in the ratio 5:4:5, find the ratio of milk to water, if all the three containers are mixed together.

a.      53:113
b.      54:115
c.      54:113
d.      53:115

Show answer
d.      53:115
Explanation
Using the weighted average formula we can calculate the weight of milk,
=> [5*(1/6) + 4*(3/8) + 5*(5/12)]/(5+4+5) = 53/168
So weight of water = 168 – 53 = 115
So the ratio of milk to water = 53:115

11. Aman participates in an orange race. In the race, 20 oranges are placed in a line of intervals of 4 meters with the first orange 24 meters from the starting point. Aman is required to bring the oranges back to the starting place one at a time. How far would he run in bringing back all the oranges?

a.      2340
b.      1240
c.      2480
d.      1540

Show answer
c.      2480
Explanation
Since every orange is placed at a difference of 4 meters and the first potato is placed at 24 meters from the starting position. Every orange is placed at 24m, 28m, 32m, 36m, ….20 terms.
Now to bring every orange one at a time, Aman needs to cover the double of the distance = 48, 56, 64, …20 terms.
So putting the values in the sum of AP formula, a = 48, d= 8, n = 20.
Total distance travelled = 20/2 [2 * 48 + (20-1)*8] = 2480 meters

12. There are two decks of cards each deck containing 20 cards, with numbers from 1 to 20 written on them. A card is drawn at random from each deck, getting the numbers x and y What is the probability that log x + log y is a positive integer. (Logs are taken to the base 10.)

a.      3/200
b.      7/400
c.      29/100
d.      5/400

Show answer
b.      7/400
Explanation
We know that log x + log y = log xy
for log xy to be positive, we have the following choices:
(1, 10), (10, 1), (10, 10), (5, 20), (20, 5), (2, 5), (5, 2)
So the probability = 7/400

13. There is a conical tent which can accommodate 10 persons. Each person requires 6 sq. meter space to sit and 30 cubic meters of air to breathe. What will be the height of the cone?

a.      15 m
b.      72 m
c.      52 m
d.      21.5 m

Show answer
a.      15 m
Explanation
All the persons are to sit on the ground forming the base of the cone.
Total base covered = pi * $r^2$ = 6*10 = 60 sq-meter.
The total volume of the tent will be equal to the total air to breathe by the 10 people = 30*10 = 300 cubic meter
So, 1/3(pi * $r^2$ * h) = 300
=> h = 15 meters.

14. A 7-digit number is to be formed with all different digits. If the digits at the extreme right and extreme left are fixed to 5 and 6 respectively, find how many such numbers can be formed?

a.      120
b.      620
c.      30240
d.      6720

Show answer
d.      6720
Explanation
If the digits at extreme left and right are fixed as 5 and 6, then the number of digits left = 8
So the in-between 5 places can be filled in 8 * 7 * 6 * 5 * 4 ways
= 6720 ways

15. If VXUPLVH is written as SURMISE, what is SHDVD written as?

a.      PBASA
b.      PEASA
c.      PEBSB
d.      None of the above

Show answer
b.      PEASA
Explanation
It is a question of coding-decoding where,
V is written as S (V – 3 = S)
X is written as U (X – 3 = U)
and so on.
Similarly, SHDVD will be written as PEASA

16. In a football match, 16 teams participate and are divided into 4 groups. Every team from each group will play with each other once. The top 2 winning teams will move to the next round and so on the top two teams will play the final match. So how many minimum matches will be played in that tournament?

a.      15
b.      40
c.      13
d.      43

Show answer
d.      43
Explanation
Total matches to be played = 4C2 = 6 matches.
So total number of matches played in the first round = 6 * 4 = 24 matches
Now top two teams from each group progress to the next round. These 8 teams are to be divided into 2 groups.
Total matches played in the second round = 6 × 2 = 12 matches
Now 4 teams progress to the next round. Total matches played in the third round = 6 * 1 = 6matches
From this round, 2 teams progress to the next round. And final will be played between them.
Total matches = 24 + 12 + 6 + 1 = 43

17. Find the greatest power of 143 which can divide 125! exactly.

a.      9
b.      12
c.      6
d.      8

Show answer
a.      9
Explanation
We can write 143 = 11 × 13.
So the highest power of 13 should be considered in 125!, which is 9 (13 * 9 = 117)
The highest power of 11 in 125! is 12 (11 * 11 = 121 and remaining 1).
That means, 125! = 11^12×13^9×…
So only nine 13’s are available. So we can form only nine 143’s in 125!. So maximum power of 143 is 9.

18. There are 12 letters and exactly 12 envelopes. There is one letter to be inserted randomly into each envelope. What is the probability that exactly 1 letter is inserted in an improper envelope?

a.      0
b.      1
c.      10!
d.      None of these

Show answer
a.      0
Explanation
This is a question of very common sense in which,
12 letters are to be inserted in 12 envelopes, 1 in each, so if one letter is inserted into a wrong envelope there has to be another letter which is to be inserted into another wrong envelope. So the probability of this is 0.

19. Salim bought a certain number of oranges at a rate of 27 oranges for rupees 2 times M, where M is an integer. He divided these oranges into two equal halves, one part of which he sold at the rate of 13 oranges for Rs M and the other at the rate of 14 oranges for Rs M. He spent and received an integral no of rupees, but bought the least number of oranges. How many did he buy?

a.      102680
b.      980
c.      1890
d.      9828

Show answer
d.      9828
Explanation
Let Salim buy 2x number of oranges.
So he buys 27 oranges at a price of 2M.
He buys 1 orange at a price of 2M/27
or, x oranges cost him Rs. 2Mx/27
Now he sells x oranges at the rate of 13 oranges for Rs. M
So he sells 1 orange at Rs. M/13
and x oranges at Rs Mx/13
The same goes for 14 oranges which are Mx/14,
According to the question, 2Mx/27, Mx/13, Mx/14 are integers
So, x oranges must be divisible by 27, 13 and 14
The lcm of 27, 13 and 14 = 4914 or 2x = 9828

20. The marked price of a shirt was 40% less than the suggested retail price. Ram purchased the coat for half of the market price at the 15th-anniversary sale. What percent less than the suggested retail price did Ram pay?

a.      70%
b.      30
c.      80%
d.      50%

Show answer
a.      70%
Explanation
Let the retail price of the shirt be Rs. 100
So according to the question, the market price will be = 100*0.6 = 60
Purchased price of Ram = 60/2 = 30
which is 70% less than the retail price.

21. HCF of 2472, 1284, and a 3rd number, is 12. If their LCM is 8*9*5*103*107, then what is the number?

a.      2^2*3^2*5^1
b.      2^2*3^2*7^1
c.      2^2*3^2*8103
d.      None of the above.

Show answer
a.      2^2*3^2*5^1
Explanation
2472 = $2^3*3*103$
1284 = $2^2*3*107$
HCF = $2^2*3$
LCM = $2^3*3^2*5*103*107$
HCF of the number is the highest number which divides all the numbers. So N should be a multiple of 22×3
LCM is the largest number that is divided by the given numbers. As LCM contains 32×5 these two are from N.
So N = [Tex]$2^2×3^2×5^1$[Tex]

22. What will be the 55th word in the arrangement of the letters of the word PERFECT?

a.      CEPFRET
b.      CEPRFET
c.      CEPERFT
d.      CEPFERT

Show answer
d.      CEPFERT
Explanation
Let’s arrange the word PERFECT in dictionary order = CEEFPRT
Here,
CEE(4!)=24
CEF(4!)=24
CEPF(3!)=6
So the 55th word is CEPFERT.

23. There is a hall consisting of 23 people. They are shaking hands together. So how many hands shakes possible if they are in a pair of cyclic sequence?

a.       22
b.       33
c.       251
d.      253

Show answer
d.      253
Explanation
Since there are 23 people, number of handshakes possible = 23C2 = 253 handshakes.

24. Assume that f(1)=0 and f(m+n)=f(m)+f(n)+4(9mn-1). For all natural numbers (Integers>0)m and n. What is the value of f(17)?

a.      4831
b.      4832
c.      5508
d.      5436

Show answer
b.      4832
Explanation
We need to use f(1) to calculate the value of f(17)
f(17) can be written as f(1+16)
f(16) can be written as f(8+8)
f(8) can be written as f(4+4)
f(4) can be written as f(2+2)
f(2) can be written as f(1+1)
f(1) = 0, so f(2) = f(1+1) = f(1)+f(1)+4(9*1*1-1) = 32.
or, f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204.
or, f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980
or, f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260
or, f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832

25. In the given series 11, 23, 47, 83, 131, … What is the next number?

a.      169
b.      191
c.      175
d.      121

Show answer
b.      191
Explanation
The given series follows the order of multiple of 12
23 – 11 = 12
47 – 23 = 24
83 – 47 = 36
131 – 83 = 48
x – 131 = 60
or x = 191

26. Rahaman went to a stationery shop and bought 18 pencils for Rs.100. He paid 1 rupee more for each grey pencil than for each black pencil. What is the price of a grey pencil and how many grey pencils did he buy?

a.      Rs.5, 8
b.      Rs.6, 10
c.      Rs.6, 8
d.      Rs.5, 10

Show answer
b.      Rs.6, 10
Explanation
The best way is to analyse from the mentions.
Let’s take option b in which 10 pencils are bought at Rs.6 each. So total cost of grey pencils = 6 * 10 = Rs.60. So Rahaman is left with 40 rupees. He buys 8 black pencils at Rs 5 each which is 1 rupee less than what he had spent in buying the grey ones. Thus satisfying the conditions.

27. There is a rectangular ground 17 × 8 m surrounded by a 1.5 m width path. The depth of the path is 12 cm. Sand is filled and find the quantity of sand required.

a.      6.05
b.      7.05
c.      10.08
d.      5.5

Show answer
c.      10.08
Explanation
Area of the inner rectangle = 17 * 8 = 136 meter-square
Area of the outer rectangle = (17 + 2*1.5) * (8 * 2*1.5) = 220 meter-square
So area of the remaining path = 220 – 136 = 84 meter-square
So sand required to fill the path = 84 * (12/100) = 10.08 meter-square

28. A boy entered a shop and bought x number of books for y rupees. When he was about to leave the bookkeeper said, “if you buy 10 more books, you can have all the books for 2 rupees and you will also save 80 cents a dozen”. So what are x and y?

a.      (15, 1)
b.      (10, 1)
c.      (5, 1)
d.      Cannot be determined.

Show answer
c.      (5, 1)
Explanation
x number of books cost him y rupees.
So, 1 book will cost him y/x rupees.
12 books will cost him rupees 12 y/x.
The shopkeeper says,
x + 10 books cost him 12 rupees
1 book will cost him 12/(x+10) rupees
12 books will cost him 24/(x+10) rupees
We know that 80 cents = 4/5 of a dollar,
So, 12y/x – 24/(10+x) = 4/5
Analyzing the given choices, we get (5, 1) satisfies the equation.

29. Ram goes to the market to buy apples. If he can bargain and reduce the price per apple by Rs.2, he can buy 30 apples instead of 20 apples with the money he has. How much money does he have?

a.      Rs.50
b.      Rs.120
c.      Rs.100
d.      Rs.110

Show answer
b.      Rs.120
Explanation
Let the price per orange be Rs. x.
So total money Ram has in buying at original price = 20x.
On reducing the price by 2 rupees each the total money must be (x-2)*30
According to the question,
20x = (x-2)*30
On solving this we get x = 6 or the total money = Rs. 120

30. Two vertical walls of the length of 6 meters and 11 meters are at a distance of 12 meters apart. Find the top distance of both walls?

a.      12 meters
b.      13 meters
c.      11 meters
d.      14 meters

Show answer
b.      13 meters
Explanation

Let’s consider this figure,

We need to find the distance of AB,
We know AC = 12 m and BC = 11-6 = 5 m
So applying pythagoras theorem we get,
AB =

$\sqrt(12^2 + 5^2)$

= 13 metres

TCS-NQT-premium-preparation-material

31. A series of storybooks were published at an interval of seven years. When the seventh book was published the total sum of the publication year was 13524. In which year was the first book published?

a.      1911
b.      1932
c.      2002
d.      2012

Show answer
a.      1911
Explanation
We get the series of publications as n, n+7, n+14, n+21, n+28, n+35, n+42.
Sum of publications = 13524 = 7/2[2n + (7-1)*7] (Using the sum of AP formula)
We get, n = 1911 (answer)

32. For f(m, n) =45*m + 36*n, where m and n are integers (either positive or negative). What is the minimum positive value for f(m, n) for all values of m, n (this may be achieved for various values of m and n)?

a.      16
b.      11
c.      10
d.      9

Show answer
d.      9 
Explanation
To get the minimum value of f(m, n), put m = 1 and n = -1, we get
f(, n) = 9

33. Mr. Mehta chooses a number and keeps on doubling the number followed by subtracting one from it. If he chooses 3 as the initial number and he repeats the operation 30 times then what is the final result?

a.      (2^30) – 2
b.      (2^31) – 1
c.      (2^30) – 1
d.      None of these

Show answer
d.      None of these
Explanation

According to the question,
3 * 2 – 1 = 5 =

2^2+1$

5 * 2 – 1 = 9 =

2^3+1$

9 * 2 – 1 = 17 =

2^4+1$

Proceeding in the similar fashion, on 30 times we get

2^{31}+1$

34. Given, log(0.318) = 0.3364 and log(0.317) = 0.3332, find log(0.319)?

a.      0.3393
b.      0.3396
c.      0.3693
d.      0.0393

Show answer
b.      0.3396
Explanation
=> log(0.319) = log(0.318) + (log(0.318) – log(0.317))
= 0.3364 + (0.3364 – 0.3332)
= 0.3364 + 0.0032
= 0.3396 (Answer)

35. A call center agent has a list of 305 phone numbers of people in alphabetical order of names, but Anuj does not have any of the names. He needs to quickly contact Danish Mank to convey a message to him. If each call takes 2 minutes to complete, and every call is answered, what is the minimum amount of time during which he can guarantee to deliver the message to Danish?

a.      18 minutes
b.      20 minutes
c.      205 minutes
d.      250 minutes

Show answer
a.      18 minutes 
Explanation
We need to search for a particular name in a phone book. So we need to apply a method in which we can easily search a number in a minimum count. So we divide the list into two equal halves, i.e., 305/2 = 152.5 or let’s take 152. Now we can decide whether to check for Danish in the upper or lower half of 152. This is decided by the starting letter of the name in a page. Proceeding in the similar manner we get,
152/2 = 76
76/2 = 38
38/2 = 19
19/2 = 9
9/2 = 4
4/2 = 2
2/2 = 0
So we get 0 at the 9th time, hence this is the minimum number of the count to find Danish. So total time taken = 9 * 2 = 18 minutes.

Aptitude Questions for TCS NQT 2021

36. If a dice is rolled 2 times, what is the probability of getting a pair of numbers with a sum equal to 3 or 4?

a.      6/36
b.      5/36
c.      1/9
d.      1/12

Show answer
b.      5/36
Explanation
Total probability = 36
We can get a sum of 3 or 4 in this many ways:
=> (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5
So probability = 5 / 36

37. There is a set of 30 numbers. The average of the first 10 numbers is equal to the average of the last 20 numbers. What is the sum of the last 20 numbers?

a.      Sum of first 10 numbers.
b.      Twice the sum of the first ten numbers
c.      Twice the sum of the last ten numbers
d.      Cannot be determined.

Show answer
b.      Twice the sum of the first ten numbers
Explanation
Let the sum of the first 10 numbers is equal to ‘x’
Let the sum of the last 20 numbers is equal to ‘y’
According to the question:
x/10 = y/20
Therefore, y = 2x

38. If (4x/3 + 2P) = 12 for what value of P, x = 6?

a.      2
b.      1
c.      6
d.      3

Show answer
a.      2
Explanation
When x = 6, (4 * 6)/3 + 2P = 12
⇒ 8 + 2P = 12
⇒ 2P = 12 – 8 = 4
⇒ P = 2

39. Let a number ‘x’ when divided by 406 leave a remainder 115. What will be the number when the number is divided by 29?

a.      28
b.      30
c.      22
d.      40

Show answer
a.      28
Explanation
According to the question, the number is equal to 406x + 115.
Since 406 is divisible completely by 29, therefore any multiple of 406 that is 406x when divided by 29 leaves the remainder 0. Now 115, when divided by 29, leaves the remainder 28.

TCS NQT Digital Aptitude Questions

40. In how many ways can 3 integers be selected from the set {1, 2, 3, …….., 37} such that the sum of the three integers is an odd number?

a.      3886
b.      3876
c.      7638
d.      1938

Show answer
b.      3876
Explanation
There are 18 even and 19 odd numbers in the given set. For sum to be odd either all 3 numbers should be odd or 2 of them even and one odd. This is possible in 19C3 + (18C2 × 19C1) = 3876 ways

41. A sequence of an alpha-numeric is to be formed. The sequence consisting of two alphabets followed by two numbers is to be formed with no repetitions. In how many ways can it be formed?

a.      67600
b.      64320
c.      65000
d.      58500

Show answer
d.      58500
Explanation
The first can be filled in 26 ways.
The second place can be filled in 25 ways.
The third-place can be filled in 10 ways.
The last digit can be filled in 9 ways

42. Percent profit earned when an article is sold for Rs. 546/- is double the percent profit earned when the same article is sold for Rs. 483/-. If the marked price of the article is 40% above the cost price, what is the marked price of the article?

a.      Rs. 588/-
b.      Rs. 688/-
c.      Rs. 618/-
d.      Rs. 514/-

Show answer
a.      Rs. 588/-
Explanation
Let profit be P and C.P.= x
Now, x+2P=546
x+P=483
subtracting both,
P=63
x=483-63=420
M.P.= 1.4*420=588

43. At what rate per cent per annum will the SI on a sum of money be 2/5 of the amount in 10 years?

a.      4%
b.      5 2/3 %
c.      6 2/3 %
d.      6%

Show answer
a.      4%
Explanation
Let the sum of money be Rs ‘x’. So SI = 2x/5
So, rate = (SI*100)/(P*Time)
=> (2x*100)/(5*x*10)
=> 4 % (Answer)

44. The sum of three from the four numbers A, B, C, D are 4024, 4087, 4524 and 4573.  What is the largest of the numbers A, B, C, D?

a.      1712
b.      1650
c.      1164
d.      1211

Show answer
a.      1712
Explanation
a+b+c=4024
b+c+d= 4087
a+c+d=4524
a+b+d=4573
Combining all we get 3(a+b+c+d) = 17208
⇒ a + b + c +d  = 3736
Now we find individual values. a = 1649, b = 1212, c = 1163, d = 1712. So maximum value is 1712.

TCS NQT Business Aptitude Questions

45. According to a particular code language, A=0, B=1, C=2, …, Y=24, Z=25 then can ONE+ONE (in the form of alphabets only) be coded?

a.      DABI
b.      BDAI
c.      CIDA
d.      ABDI

Show answer
b.      BDAI
Explanation
This is a 26 base question. Just like there is the Decimal system consisting of 10 digits from 0 to 9, the Base 26 system consist of 26 alphabets where A = 0, B = 1, Z = 25 and so on.
Let’s calculate, O N E + O N E
For E(4),
=> E + E
=> 4 + 4
=> 8
=> I
For N(13),
=> 13 + 13
=> 26
On converting 26 to Base 26 we get 1 0. Keeping 0(A) and taking 1 as carry
For O(14),
=> O + O + 1
=> 29
Dividing 29 by 26 we get 1(B) 3(D)
So answer is BDAI

46. A flight takes off at 2 a.m. from a place at 18N 10E and landed at 36N 70W, 10 hours later. What is the local time of the destination?

a.      7:40 a.m.
b.      6:40 a.m.
c.      6:00 a.m.
d.      7:00 a.m.

Show answer
b.      6:40 a.m.
Explanation
Let’ calculate the difference in the number of latitudes = 70 + 10 = 80 degrees towards east.
We know 1 degree = 4 min, so 80 degrees = 80 * 4 = 320 mins
320 mins = 5 hr 20 minutes
Now the plane landed 10 hours later so the time of landing = 12 hrs according to the starting place
So time at destination = 12 hrs – 5 hrs 20 min = 6 hr 40 mins(Answer)

47. A takes 3 days to complete work while B takes 2 days. Both of them finish work and earn Rs. 150. What is A’s share of the money?

a.      Rs. 70
b.      Rs. 60
c.      Rs. 50
d.      Rs. 20

Show answer
b.      Rs. 60
Explanation
A completes 1/3rd of work in one-day and B completes 1/2 of work in one day. So the ratio of there work is:
A:B = 2:3
So A’s share = (2/5)*150 = 60 rupees(Answer)

48. If f(x) = ax^4 – bx^2 + x + 5 and given f(-3) = 2, then f(3) = ? (a^b = a raised to power b)

a.       1
b.       4
c.       8
d.      -3

Show answer
c.       8
Explanation
We can directly solve:
=> f(-3) = a(-3)^4 – b(-3)^2 + (-3) + 5 = 2
=> 81a – 9b + 2 = 2
=> 81a – 9b = 0
Now solving f(3),
=> f(3) = 81a – 9b + 8
=> f(3) = 0 + 8 = 8(Answer)

49. A certain function f satisfies the equation f(x)+2*f(6-x)=x for all real numbers x.  The value of f(1) is

a.      1
b.      2
c.      3
d.    Cannot be determined

Show answer
c.      3
Explanation
Put x =1 ⇒ f(1)+2*f(6-1) = 1 ⇒ f(1) + 2*f(5) = 1
Put x = 5 ⇒ f(5)+2*f(6-5) = 5 ⇒ f(5) + 2*f(1) = 5
Put f(5) = 5 – 2*f(1) in the first equation
⇒ f(1) + 2*(5 – 2*f(1)) = 1
⇒ f(1) + 10 – 4f(1) = 1
⇒ f(1) = 3

50. What is the remainder when the number 101102103104105106107…148149150 is divided by 9?

a.      3
b.      2
c.      10
d.      11

Show answer
b.      2
Explanation
The divisibility rule for 9 is that the sum of all digits of a number should be divisible by 9. Let’s calculate the sum of the digits:
There are 50 1’s (unit place) = 50
There are 10 1’s (tens place) = 10
There are 10 2’s (tens place) = 20
There are 10 3’s (tens place) = 30
There are 10 4’s (tens place) = 40
There is one 5 (tens place) = 5
For each number 1 to 9, there are 5 sets of sum 45(1+2+…+9) = 225
=> So sum of all digits = 380
=> 380 / 9 = 2 (Answer)

51. There are a set of 20 students out of which 18 are boys and 2 are girls. They are to be seated in a circular manner so that the two girls are always separated by a boy. In how many ways can the students be arranged?

a.      18!x2
b.      17!
c.      17×2!
d.      13

Show answer
a.      18!x2
Explanation
There are in all 20 places out of which if one girl sits in one position then the other girl may sit either to her left or right skipping one place, which is to be filled by a boy. So the total number of ways the boys can sit = 18! ways and girls may alternate their sits so the total answer would be = 18! * 2 ways.

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