# TCS NQT Numerical Ability Questions and Answers Set 1 : TCS NQT Placement Papers – 1

This is the official TCS NQT questions of TCS NQT Numerical Ability (Quantitative aptitude) of October 2020 Papers. We have collected the questions of TCS NQT through various sources. Here the TCS NQT Placement paper of Numerical ability. you can download the paper in PDF, Just go on share button in your browser and click on the print.

TCS NQT Numerical Ability Consist of 26 questions, which you have solve in 40 Minutes. The general syllabus of TCS NQT Numerical ability consist of

• Number System
• Probability
• Time Speed and Distance
• Equations
• Geometry, Mensuration and Progressions
• Ratios and Proportions
• Functions, Logarithms and Graphs
• Clocks, Direction Sense
• Profit and Loss, Averages, Mixtures and Alligations
• Permutation and Combination
• Puzzles
• Base Conversion

## TCS NQT Numerical Ability Questions Set -1

### Q1.What is the sum ( in RS) which when divided among X Y Z in the proportion 3:5:7 provides rupees 8000 more to Z then what it would have done to him when the proportion is 11 : 15 : 19?

A.180000
B.120000
C. 175000
D. 135000

X:Y:Z                 X:Y:Z
3: 5: 7                 11: 15:19
Z = 7a/15                Z = 19a/45
So as per Question (7a/15) – (19a/45) = 8000
=> 2a/45 = 8000
a = 18000.

### Q2. A man has to travel 50 km in 2 hours. He could cover 20 km in one hour, and then had to stop for 10 minutes for refueling. By What factor should he increase his speed with reference to that during the first hour so as to be able to complete the journey as per schedule?

A.1.5
B.1.8
C.1.2
D.2.4

Total time he has = 2 hrs
total distance he has to covered = 50 km
in first 1 hr he completed 20 km
so S1= 20 km/hr
so remaining distance = 50-20 = 30 Km
and remaining time = 1 hr
now he spent 10 minutes for refueling
so total remaining time he has = 60-10 = 50 minutes
he has to complete 30 km in this time
speed = 30/(10/60) km/hr
speed = 36 km/hr
for making a factor of initial speed as 20 km/hr
multiply 36 with 20/20
speed = 36x(20/20)
=(1.8x S1 )km\hr
= 1.8 km/hr  (As S1 is 20km/hr)

### Q3. X is four times as efficient as Y in respect of doing a particular work. Working together they complete the work in 16 days. In how many days y working alone will be able to half the work?

A.80
B.20
C.40
D.60

Efficiency of X = 4a
Efficiency of Y = a
Time taken by both to complete a work = 16 days
(1/4a)+1/a=1/16
(1+4)/4a=1/16
5/4a=1/16
a=20
Hence X will do the work in 20 days
And Y will do the work in 80 days
Thus Y will do half of work in 40 days

Note –
Efficiency and time taken are inversely proportional
So, if a = 20
then efficiency of X is 4 times of Y
So Y will do the task in 80 days and will complete half of  the task in 40 days

### Q4. If 5^X 3^Y = 225 x 405, find the value of  X2Y – 3X

A.27
B.81
C.125
D.25

Prime Factorization of 225 = 5 x 5 x 3 x 3
Prime Factorization of 405 = 3 x 3 x 3 x 3 x 5
5336 = 5^X 3^Y
So X = 3 and Y = 6
Now,X2Y – 3X =  32(6) – 3(3)
312 – 9 = 33 = 27

### Q5. The collection of numbers which comprise the data given below is arranged in ascending order.(3,7,9, N – 1,15,18,19,20)If the median of the data is 12.5, what is the value of N?

A.10.5
B.11.5
C.11
D.12

Median of the even data = (sum of middle terms) / 2

= [(N – 1) + 15] / 2 = 12.5

### Q6. How much percentage is (0.025% of 240% of 1.5) of 0.9?

A.10
B.0.01
C.0.1
D.1

0.025% of 240% of1.5 = 0.0009
0.01 % of 9 = 0.0009

### Q7. After purchasing two copies of the same book, X sold them respectively at 0.8 and 1.4 times their cost prices. What was the percentage gain earned or loss incurred by X?

A.10% loss
B.5% gain
C.10% gain
D.5% loss

Let the C.P of each copy be Rs 100
ATQ one copy is sold at 0.8 x 100 = 80
One copy is sold at 1.4 x 100 = 140
Total money spent in buying the copies = Rs 200
Total money earned after selling the copies = Rs (80 + 140) = Rs 220
Total gain of Rs 20. Therefore gain% = 10%.

### Q8.A file of cadets consisting of ten rows and five columns measures 420 m in length along the direction of their marching. How much time (in hours and minutes) would it take to march for a stretch of 3 km, if the stride of each cadet is 80 cm and he takes 57 strides per minute?

A.1 hr 10 min
B.1 hr 20 min
C.1 hr 24 min
D.1 hr 15 min

So total distance need to be covered by last cadet

=> D = 420

Speed = (80/100) X 5.7 X(1/60) per sec

= 76/100 m/sec
=> T=D/S=(3420 X 100)/76

=4500Sec
= 75 min = 1 hr 15 min

### Q9. Two vessels X and Y of capacities one and two litres respectively are completely filled with mixtures of two chemicals A and B. The ratio by volume of the chemicals A and B in X and Y are 3:2 and 4:5 respectively. The contents of A and B are mixed and the combination is kept in a vessel C of capacity of four litres. How many litres of Chemical A should be added to the combination so as to make the ratio of A to B equal to 1:1?

A.1/270
B.1/67
C.1/68
D.1/135

In X
A:B
(600/1000):(400/1000)

In Y
A:B

(8/9):(10/9)

X                         Y

(6/10; 4/10) Lt +(8/9; 10/9)Lt

A=(6/10+8/9); B=(4/10+10/9)

A:B=(54+80/90):(36+100/90)

134:136

67:68

(67/68):1

=(67/68)+(1/68):1
= 1/68
= We will get 1:1 => 1/68

### Q10. The diameter of a pizza is 30 cm. What is the area(in cm^2) of the upper surface of a sector of the pizza whose arc length is 8 cm?

A.120
B.120π
C.60π
D.60

Area of sector when arc is given = (L x r)/2
Where L = length of arc and r = radius of the circle
Here L = 8 cm and r = 15 cm

Therefore, Area = (8 X 15)/2
= 60cm2

### Q11. In a competitive exam, 5 marks are awarded for every correct answer and for every wrong answer, 2 marks are deducted. Sathwik scores 32 marks in the examination. If 4 marks had been awarded for each correct answer and 1 mark had been deducted for each incorrect answer, Sathwik would have scored 34 marks. If Sathwik attempted all the questions, how many questions were there in the test ?

A.14
B.12
C.20
D.26

In given Question:-
So,
5x -2y = 32
4x-2y =34——*2
Now
5x-2y =32
8x-4y=68
By solving
3x=36
x=12
Now to find ‘y’
5*12 -2y =32
60 -2y= 32
y=14
12+24 =26

Questions in the test =26

### Q12. An item was sold at a profit of 12% after giving a discount of 12.5% on the list price. What would be the gain or loss percentage if a discount of 25% is given on the list price ?

A.2.5% gain
B.2.5% loss
C.4% loss
D.4% gain

Cp ——–> 100%
Sp ——–> 112% (12% profit)
But Sp is 12.5% of M.P => 12.5% discount
=> 87.5% of M.P = 112
=> list price = 128
25% discount on M.P = 25% of 128 = 32
=> 128 – 32 = 96 = S.P
C.P = 100 => SP = 96 => 4% loss

### Q13. The mean of a set of data is 5. What will be the mean if ten is subtracted from each data ?

A.-5
B.5
C.10
D.-15

let the number of data be n
so sum of data = mean * number of data
= 5 n
now 10 is subtracted  to each data
so now sum becomes = 5n-10n =-5n
mean = sum of data /no. of data
=-5n/n = -5
so new mean becomes -5

### Q14. If (x+10)% of 240 is 60% more than x% of 180, then 15% of (x+20) is what percent less than 25% of x ?

A.15
B.19 1/21
C.16
D.15 1/2

(x + 10)% of 240 = x% of 180 + 60% of 180
if you solve this expression x value = 50
15% of x+20 is 10.5
25% of x is 12.5
so how much less = 2/12.5 * 100 = 16

### Q15. What is the diameter (in cm) of a solid right circular cylinder whose height is 6 cm and the area of the curved surface is five times the combined area of the two flat surfaces

A.2.4
B.0.9
C.1.2
D.3

Here h = 6 cm
C.S.A = 2πrh
2πrh = 5(2πr^2)
2 x π x r x 6 = 10 x π x r^2
12 x π x r = 10 x π x r^2
12 = 10r
1.2 = r
Then diameter is 2.4

### Q16. A sum invested on simple interest  grows to Rs 22500/- and Rs 25500/- is seven and  nine years respectively. What is the rate percentage of the interest ?

A.7.5
B.9.6
C.12.5
D.13.5

Total Interest of two years= 25500-22500=3000
Interest of one year= 3000/2 = 1500
Total Interest = 9 * 1500 = 13500
Principle = 25500 – 13500 = 12000
R= I * 100/P *t
= 13500 *100/1200*9 = 12.5

A.0.5
B.5
C.0.2
D.2

1/5= 0.2

### Q18. The variation in temperatures throughout the day in a desert town was studied on the basis of the record of minimum and maximum temperatures which were 8 and 36 degrees centigrade respectively. What was the standard deviation in degree centigrade?

A.12
B.22
C.14
D.28

min —> 8 degrees

Max —–> 36 degrees

(8+36)/2 = 22

= 14

### Q19. A sum of Rs. 30000 invested in a scheme where the interest gets compounded annually and grows to Rs. 51840 in three years. How much interest in Rs. would have got accrued in six months in the same scheme had the interest been compounded quarterly?

A.3075
B.2975
C.3024
D.3126

51840 = 30000 x (1+(r/100))^3

51840/30000= (1+(r/100))^3

=> 1+(r/100) = 3√1.728

=> 1+(r/100) = 1.2

=> r/100 = 0.2

=>r = 20%

Rate of interest CI per quarterly = r/4 = 5%

=> for 6 months => 5% of 30000 = 1500 (1st quarter)

5% of 30000 + 5% of 1500 = 1500 + 75 (2nd quarter)

=> total of 6 months = 3075.

### Q20. 96 men were engaged for a project of constructing a railway track of the length of 18 km in four weeks. After one week it was observed that the work of 4 km was completed. How many additional men should be engaged for timely completion of the project?

A.16
B.14
C.15
D.12

4 km was completed by 96 men
Remaining work is 14km in three weeks so per week 14/3
For 4km——— 96men
For 14/3 km ———– ?
= 14/3 x 96 x ¼ = 112 so additionally 16 men needed.

### Q21. If n is an integer such that 1nn352 is a six-digit number exactly divided by 24, what will be the sum of the possible value of n?

A.21
B.27
C.9
D.15

Possible values of n are 2, 5 and 8 so sum of all possible values is 2 + 5 + 8 = 15

### Q22. What is the mean deviation of the data 8,9,12,15,16,20,24,30,32,34?

A.10.2
B.8
C.0
D.9.6

Step 1
Find the mean
8+9+12+15+16+20+24+30+32+34/10
=200/10
=20

Find the distance of each value from that mean

Step 3:- Find the mean of those distances

12+11+8+5+4+0+4+10+12+14/10
80/10=8

### Q23. The marks scored by a student out of 100 in English and Mathematics at four consecutive monthly tests held in 2019 are presented through a Bar Graph shown below. In each case the marks are a multiple of five.

In which month is the difference of marks scored in the two subjects the highest ?

1. Jan
2. Mar
3. Apr
4. Feb Answer : Feb (it is clear from the graph.)

• A
• D
• B
• C

### Q25. What is the mean proportional (MP) between the MPs of(2/7 & 32/343) and ( 2 & 1/5000)

A. 2/175
B.3/35
C.4/35
D. 2/35

a:x :: x:b
a/x :: x/b
ab=x^2
x=√ab
(2/7) X(32/343) = x^2
x = 8/49……..(i)
2 X (1/5000) = x^2
x= 1/50……(ii)
Mean Proportion (i) and (ii)
(8/49) X (1/50) = x^2
x= 2/35