# TCS NQT Quantitative Aptitude Numerical Question Paper Set 4 | TCS Placement Papers Set 9

TCS NQT Quantitative Aptitude / Numerical Ability Question Paper of October 2020, and February 2021, Here the complete latest question papers of TCS Off campus recruitment drive for the engineering BE/BTech, MBA, as well BSC other graduates. You can download the TCS question paper in PDF format by click on the print button at the end of the page.

## TCS NQT Aptitude Questions and Answers Set 4

### Question 1: Two gears which work in tandem have 16th and 20th teeth‘ respectively. If the former returns 15 times in 8 seconds.How many turns will the latter take in one-third of a minute.

A. 30
B.40
C. 20
D. 10

Let us assume that, the later takes x turns in (1/3)rd of a minute.

Then,

Teeth Time in sec. Turns

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16 8 15

20 20(1/3 of 1 min.) x

we know that,

• Turn ∝ Time.
• Turn ∝ (1/Teeth).

Applying chain rule now, we get,

→ (15/x) = (20/16) * (8/20)

→ (15/x) = (1/2)

→ x = 15 * 2

→ x = 30 turns. (Ans.)

Hence the later takes 30 turns in 1/3rd of a minute.

### Question 2: 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.16
C.17
D.18

(x+10)% of 240 = x% of 180 + {60 % of 180x/100}

solving this to get

x = 50

15% of (x+20) = 25% of x – { y % of 25x/100 }

apply x value here, to get y value

15% of (50+20) = 25% of 50 – { y % of (25* 50)/100 }

10.5 = 12.5 – {y % of 12.5 }

{y % of 12.5 } = 2

y * 12.5/100 = 2

y = 200/12.5

y = 16

### Question 3: If the HCF of 180 and 432 is expressed as (180m + 432n) , where m and n are integers , then what is the difference between m and n?

A.7
B.3
C.8
D.9

Prime factors of 180 and 432 is :-

→ 180 = 2 * 2 * 3 * 3 * 5

→ 432 = 2 * 2 * 2 * 2 * 3 * 3 * 3

HCF = 2 * 2 * 3 * 3 = 36.

Now , given that, HCF is written as 180m+432n .

So,

→ 180m + 432n = 36 .

→ 36(5m + 12n) = 36

dividing both sides we get,

→ 5m + 12n = 1

Now, we have given that, m and n both are integers.

So,

putting values of m , we get :-

m = 0, n = (1/12) => n = not an integer.

m = 1 , n = (-1/3) => n = not an integer.

m = 2 , n = (-3/4) => n = not an integer.

m = 3 , n = (-7/6) => n = not an integer.

m = 4 , n = (-19/12) => n = not an integer.

m = 5 , n = (-2) => n = an integer.

Hence,

→ Difference between m and n = m – n = 5 – (-2) = 5 + 2 = 7

### Question 4 : The median of frequency distribution table is _

1. 62
2. 61
3. 61.5
4. 62.5

Sum of frequencies

12+26+30+8+13+11

=100

So,

n/2=50

Class-interval Frequency CF
50 – 54 _________12______12

55- 59_________26______38

60 – 64_________30______68

65 – 69________8______76

70 – 74 ________13_______89

75 – 79 _________11_______100

so just upper nearby class of n/2 is 60-64

So, 60-64 is median class

l= 60

cf= 38

f =30

h = 4

put all these values in the formula

Median=l+[(n/2-cf)/f]Xh

=60+[(50-38)/30]X4

=60+[(12/30)]X4

=60+1.6

61.6

Median = 61.6

### Question 5 :Raju lends Rs 3000 to Bharath and a certain sum to Charan at the same time of 6% per annum Simple Interest. If after 5 year, Raju altogether receives Rs 1650 as interest from Bharath and Charan, What is the sum lent to Charan?

A.Rs 3250
B.Rs 2500
C.Rs 3300
D.Rs 2750

Case 1) :- sum Lends to Bharath .

→ P = Rs.3000

→ Rate = 6% per annum.

→ Time = 5 years.

→ SI = (P * R * T) / 100

→ SI = (3000 * 6 * 5) / 100

→ SI = 30 * 30

→ SI = Rs.900

So,Raju recieved Rs.900 as interest from Bharath.

Case 2) :- Let sum lends to Charan is Rs.x.

→ P = Rs. x

→ Rate = 6% per annum.

→ Time = 5 years.

→ SI = (P * R * T) / 100

→ SI = (x * 6 * 5) / 100

→ SI = 30x /100

→ SI = Rs.(3x/10)

So, Raju recieved Rs.(3x/10) as interest from Charan.

Now, given that, Raju altogether receives Rs.1650 as interest from Bharath and Charan.

Therefore,

→ 900 + (3x/10) = 1650

→ (3x/10) = 1650 – 900

→ (3x/10) = 750

→ 3x = 750 * 10

→ 3x = 7500

dividing both sides by 3,

→ x = Rs.2500

Hence sum lends to Charan= Rs 2500

### Question 6 : What is the sum (in Rs) which when divided among A, B, C, D in the proportion 2:3:5:8 provides Rs 8420 less than what it provides to him when the proportion is ½ : ⅓ : ⅕ : ⅛ ?

A.25020
B.37530
C.12510
D.17540

Let us assume that, total sum is Rs.x.

Than,

→ share of D when divide in proportion 2:3:5:8 is = 8/(2 +3 + 5 + 8) of x = (8/18) of x = Rs.(4x/ 9)

now, Ratio in second time :-

→ A:B:C:D=1/2:1/3: 1/5: 1/8

Taking LCM of denominators ,

→ A:B:C:D=(60: 40: 24:15) / 120 = 60: 40: 24:15.

So,

→ Share of D when divided in proportion 60 : 40 : 24 : 15 = 15/(60 + 40 + 24 + 15) of x = (15x/139)

now, given that, he gets Rs.8420 less in second share.

Therefore,

→ (4x/9) – (15x/139) = 8420

→ (139*4x – 135x) /1251 = 8420

→ 556x – 135x = 8420 * 1251

→ 421x = 8420 * 1251

dividing both sides by 421,

→ x = 20* 1251

→ x = Rs.25,020

Hence the required sum is Rs.25,020

### Question 7: What is the simplified value of:

[(0.397)/(1.2X1.2 +0.66+1.1X1.1)]=?

A.1
B.0.1
C.0.01
D.0.001

[(0.397)/(3.31)]

0.11

### Question 8: A, B, C are three cups having capacities of 120, 180 and 200 cc respectively which are completely filled with tea of three different varieties. They are all mixed in a separate vessel of capacity more than that A, B, C taken together, and then the mixture is poured successively into A,B, C. Then how much (in cc) of A and B’s tea will be there in C?

a.56(A), 64(B)
b.52(A), 68(B)
c.44(A), 76(B)
d.48(A), 72(B)

given that, all cup are completely filled with tea of three different varieties.

So,

→ Tea in cup A = 120cc

→ Tea in cup B = 180cc

→ Tea in cup C = 200cc

now, all three cups mixed in a separate vessel .

Than,

→ Ratio of varieties of tea in vessel = A : B : C = 120 : 180 : 200 = 6 : 9 : 10 .

Now, again the mixture is poured successively into A, B, C.

Than, we can conclude that, Now, mixture has Ratio of varieties of tea in 6 : 9 : 10.

Therefore,

→ In C , quantity of A = 200 * (6/25) = 48cc.

→ In C, quantity of B = 200 * (9/25) = 72cc.

Hence, quantity of A and B in C are 48 cc and 72 Respectively.

### Question 9: In a factory, the coefficients of variation of the wages of urban and rural workers are 52% and 60% respectively and their standard deviations are 2600 and 2640 respectively. If 60% of the workers are urban. What is the overall average wage(in Rs) of the workers?

A .4350
B .4200
C .4450
D .4600

Mean of urban  = a

Mean of rural  =  b

Coefficient of Variation = (Standard deviation /Mean) * 100 %

urban

=> 52 = ( 2600/a ) * 100

=> a = 5000

rural

66 = ( 2640 / b) * 100

=> b = 4000

60% of the workers are urban

40 %  of the workers are rural

=> overall average( in rs) of the workers​  =  5000 * 0.6  + 4000*0.4

= 3000 + 1600

= 4600

overall average( in rs) of the workers​ = 4600

### Question 10: If 2 numbers are respectively 25% and 40% less than the third number, what is the ratio of these 2 numbers?

A 3:4
B 4:5
C 5:4
D 3:5

Let the 3rd number be 100.

Then the first number is 25% of 3rd number = (25 X 100)/100

=25

So the first number=(100-25)=75

Similarly,

The second number is 40% of 3rd number =(40 X100)/100

=40

Hence the second number=(100-40)= 60

Hence ratio of both the numbers = 75/60 = 5/4

So, the ratio will be 5:4

### Question 11: X working alone takes 75 days more than Y to do a work, and working together they complete it in 20 days. If a certain sum has been earmarked as wages for this work, in what ratio should it get distributed among X and Y?

A: 1:4
B: 1:5
C: 2:5
D: 2:3

Let us assume that, Y alone can complete the work alone in xdays.

Than, X will complete the work alone in (x + 75) days.

given that, both takes 12 days to complete the work .

So,

1/x + 1/(x + 75) = 1/20

→ (x + 75 + x)/x(x + 75) = 1/20

→ (2x + 75) * 20 = x² + 75x

→ 40x + 1500 = x² + 75x

→ x² + 75x – 40x – 1500 = 0

→ x² + 35x – 1500 = 0

→ x² + 60x – 25x – 1500 = 0

→ x(x + 60) – 25(x + 60) = 0

→ (x + 60)(x – 25) = 0

putting both equal to zero , we get,

→ x = (-60) or 25.

since days cant be in negative.

Therefore,

→ Time taken by Y alone = 25 days.

→ Time taken by X alone = 25 + 75 = 100days.

Now, we know that,

• Efficiency is indirectly proportional to time..
• wages is distributed among works in ratio of their efficiency of work per day.

So,

Time taken by X : Time taken by Y = 100 : 25 = 4 : 1 .

Than,

Efficiency of X : Efficiency of Y = 1:4.

Hence,

wages of X : wages of Y = 1 : 4

### Question 12: 72 men are engaged for a task with a timeline given beforehand It is found that is  38 of the given time 36% of the task is accomplished. How many additional men are to be engaged so that the timeline can be met?

A: 6
B: 7
C: 8
D: 9

Let say Total Work  = 100W   and to Be completed in  100D Days

Let say a Man do work  A in 1 Day

38% of the given time  = 38D says

Work Done  = 72A * 38 D

36 % of the task is completed  => Work Done = 36W

72A * 38 D  = 36W

Work Left  =  100W – 36W = 64W

Days Left = 100D – 38D  = 62D

Let say Men now = M

So Work done  = MA62D

= 62M ( W/76)

62M ( W/76) = 64W

=> M = 64 x 76 / 62

=> M = 78.45

As time line need to met so we need to round up men

=> M = 79

Additional Men Required = 79 – 72 = 7

### Question 13: The monthly expenses in a boarding house are partly fixed and partly a multiple of the number of boarders. It is Rs 78660 and rs 94884 when the numbers of borders are 62 and 75 respectively. What is the monthly expense (in Rs), when the number of borders is 80?

A: 100926
B: 100804
C: 101124
D: 101534

monthly expenses = fixed + varying

let fixed amount be ‘f’ and varying be ‘x’

f + 62x = 78660

f + 75x = 94884

after solving we get f=1284 and x=1248

the monthly expense when no. of borders are 80 is

1284 + 1248(80) = 101124

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