TCS NQT Numerical Ability Questions & Answers Set 2 : TCS NQT Placement Papers 5

Previous Year October 2020 TCS NQT Numerical Ability Questions & Answers Set, Here the latest TCS quantitative aptitude questions with solutions. To download this questions in pdf click on the print button on share list.

TCS NQT Numerical Ability Questions Set -2:

Question 1: The Range and the standard deviation of a data are R & S respectively. With the shift of the origin of the data, change(s) occur in the value(s) of:

A. Both R & S
B. R only
C. Neither R nor S
D. S only

Question 2: A hollow cylinder of radius 5cm takes six turns about its axis on a horizontal plane,and thereby covers a distance which istimes its length.What is the area of the curved surface (in) of the cylinder?

A.100 π
B.120π
C. 80π
D. 50π

Distance covered ,
2*π*r*6
=6*π*L,
L=2r
Curved area=2*π*r*L
=2*π*r*2r
=4*π*5*5
=100π

Question 3 : What is the simplified value of (10/15*6)[22/2/5 of [(64-2*8) +2]-⅖?

A. 6
B. 4
C. 5
D. 2

(10/15*6)[22/(2/5) of {(64-2*8)+2}]-(2/5)

-> (4)       [ 22/ (2/5) of 50 ]      -2/5

-> (4)       [22/20]      -2/5

->4*       11/10     -2/5

->4

Question 4: Which one among the following has the least value?

A. √75 – √74
B. √74 – √73
C. √77 – √76
D. √76 – √75

(77+76)(77-76)=(153)(1)=153

(75+74)(75-74)=(149)(1)=149

(76+75)(76-75)=(151)(1)=151

(74+73)(74-73)=(147)(1)=147

Hence 147 is least among all

√76 – √75

Question 5: A man who has to walk 11 km, finds that in 30 minutes he has travelled two-ninth of the remaining distance. What is his speed in km/hr?

A.4.2
B.4
C.4.5
D.4.8

Solution : Option D

distance =11km= 11*2/9 = 22/9
Time =30 min = 30/60=½ hr
x= distance / time
x= 22 /9  / ½
x= 44/9
x= 4.8

Question 6: What is the fourth proportional of 0.006, 1.2 and 6/25?

A. 4.8
B. 3.6
C. 36
D. 48

Solution : Option D

Question 7: A sum was lent to X for three years by an organization, who fixed a yearly rate of 10% compound interest for repayment along with the condition of recovery in equal annual installments of Rs. 31944. What percentage (correct to 2 decimal places) above the borrowed amount had X to pay to the organization?

A. 16.52
B. 18.43
C. 20.63
D. 21.25

Solution :   Option D

= 31944[0.75 + 0.82 + 0.90]
= 31944[ 2.47 ]
P = 78901.68
Interest = 78901.68 – (3 x 31944)
= 16930
(16930/78901)*100=21.4

Question 8: The present ages of three brothers are in the proportion 12:14 :17. The difference between the ages of elder and the eldest is 6 years. What will be the proportion of their ages after 4 years?

A. 40:46:55
B. 14:16:19
C. 42:41:15
D. 14:16:18

Solution Option B

Let the common value of their ages be x.
Then proportion becomes 12x : 14x : 17x
According to problem,
Elder brother’s age is 14x
Eldest brother’s age is 17x
17x – 14x = 6
=> 3x = 6
=> x = 2
Thus the present age of,
Youngest brother is 24 years
Elder brother is 28 years
Eldest brother is 34 years
After 4 years the age of,
Youngest brother is 28 years
Elder brother is 32 years
Eldest brother is 38 years
Thus the proportion becomes,
28 : 32 : 38
which can be simplified and written as,
14 : 16 : 19

Question 9: The index numbers of five commodities are 121,123,125,126,128 and the weight assigned to these are respectively 5,11,10,8,6.Then what is the weighted average index number?

A. 123.8
B. 125.2
C. 124.6
D. 124.2

Solution :  Option D

P = commodity
V = weight
PV = 121×5+ 123×11 + 125 × 10+ 126×8+ 128 × 6
= 4984
P = 40
weighted average = PV / V
index number
= 4984 / 40
= 124.6

Question 10: With what value should the highest quantity in the data 65,52,14,26,18,35,32,38 be replaced so that the mean and median become equal?

A. 66
B. 53
C. 64
D. 51

Solution : Option B

The highest quantity 65 will be replaced by 53 to make the mean and median equal.

Question 11: What is the real value of ( 0.000314 + 0.000198)(1/3)A. 0.8B. 0.08C. 0.4D. 0.04

Solution Option B

=0.000314 + 0.000198 = 0.000512
=(0.000512)1/3
=(0.08)(3 * 1/3)
=0.08

Question 12: A work is assigned to 6 men and 12 women, they could complete it in 3 days. It was also observed that together they can do 7 times as much work a man and a woman can do. In how many days would 14 women have done the work?A. 6B. 12C. 9D. 10

Solution :  Option C

6m + 12w = 3 days
6m + 12w = 7 (m + w)
6m + 12w = 7m + 7w
6m + 12w = 6m + 1m +7w
6m + 5w + 7w =  6m + 1m + 7w
1 man = 5 woman
30w + 12w = 3 days
42w = 3 day
14 women =
= 9 days

Question 13: The mean and the standard deviation of a data which consists of a set of ten positive numbers are 8 and 2 respectively. If the sum of squares of 9 among the 10 members is 599,what is the 10th number

A. 9
B. 8
C. 7
D. 11

Solution :  Option A

Let us assume that the 10th number is x .
we know that,
standard deviation =
N = Number of terms.
Given that,
→ standard deviation = 2 .
→ Mean = 8 .
→ N = 10 .
→ sum of square of 9 numbers = 599.
→ 10th number = x .
So,
→ sum of square of all 10 numbers =
Putting all values now , we get :-
→ 2 = √[{(599 + x²)/10} – (8)²]
Squaring both sides , we get,
→ 4 = {(599 + x²)/10} – (8)²
→ 4 = {(599 + x²)/10} – 64
→ 4 + 64 = (599 + x²)/10
→ 68 * 10 = 599 + x²
→ x² = 680 – 599
→ x² = 81
Square root both sides we get,
→ x = 9

Question 14: With what value should the highest quantity in the data 65, 52, 14, 26, 18, 35, 32, 38 be replaced so that the mean and medium become equal?

A. 51
B. 53
C. 66
D. 64

Solution:  Option B

Let’s sort the given data – 14, 18, 26, 32, 35, 38, 52, 65
mean = 35
media=(32+35)/2
=33.5
Difference = 35 – 33.5 = 1.5
Value = 1.5 x 8 = 12
=65-12 = 53

Question 15: Two ants of length 1 cm and 1.2 cm. Crawl in opposite directions with an average speed of 2 and 3 mm per second respectively. How many seconds will they take to cross each other?

A. 2.8
B. 4.4
C. 0.4
D. 1.5

Solution :Option B

Relative speed of ants = 2 + 3 mmps = 5mmps
Effective distance = 1 + 1.2 = 2.2 cm = 22 mm
Time = distance/speed = 22/5 = 4.4

Question 16: A sum of Rs 12500 is invested on 1st January 2016 at 4% simple interest per annum. How much interest (in Rs) gets accrued on the end of the day on 1st July, 2016?A. 250B. 400C. 500D. 240

Solution: Option A

SI = Prt/100
P = 12500
R = 4%
t = 6 months = 0.5 year
SI = (12500 * 4 * 0.5) / 100
SI = 125 * 2 = 250

Question 17: Six square plots are connected end to end to obtain a rectangular plot of area 726 m2. If we take pi = 22/7 by what factor is the perimeter of this plot more than that of circumference of a circle of radius 10 m?A. 1.45B. 2.8C. 2.1D. 2.45

Solution :  Option D

Area of Rectangle = 726m2
Rectangle is comprised of 6 square plots
A * A = 726/6 = 121
A = 11
Now length of the rectangle will be 6 * 11 = 66 m
Perimeter of rectangle = 2(11 + 66) = 154
Perimeter of circle of radius 10 m = 2 pi r = 2 * 22/7 * 10 = 440/7 = 62.85
By factor the perimeter of Rectangular plot is more than circular plot = 154/62.85 = 2.45

Question 18: The table below presents the percentage marks obtained by the three students X, Y, and Z in the four components of assessment A, B, C, D, of a paper whose respective weightages are 4, 3, 2, 1.

What is the ratio of the average percentage of X in A, B to to that of Y,Z combined in CD
A. 28: 25
B. 326: 293
C. 342: 283
D. 81:73
Solution :Option B
Average Marks obtained by X in A and B = (78 + 85)/2
=163/2
Average marks obtained by Y in C and D =137/2
Average marks obtained by Z in C and D =156/2
Average of X and Y combined
{[(137/2)+(156/2)]/2}
=293/4-
> Ratio of X and Y, Z combined
->X : (Y and Z)
->163/2: 293/4
->326: 293

Question  19: A retailer purchased 25 identical toys for a price Rs P and sold some of these for price Rs P.  If he calculated his profit as 8% , with selling price as base instead of cost price then how many toys did he sell.A. 20B. 23C. 21D. 24

Solution : Option B

25 identical toys are Rs.P.
one toy = Rs.P/25
profit % = (sp-cp)/cp
In question said to change the base from cp to sp
8/100 = (p – xp/25) / p
solve for x,
then x=23.

Question 20: 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. 14B. 12C. 20D. 26

Solution: Option D

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

Question 21 : The List price of an item was kept 40% above what the shopkeeper had paid to the manufacturer. On selling the item a profit of 8.64% was earned allowing  two successive discounts, the first one which was 20% . What was the percentage rate of the second discount.A. 3.84B. 3.6C. 3.25D. 2.4

Solution : Option D

let the CP be 100
then the MP will be 140
if he earns profit of 8.64 percent then SP = 108.64
so total discount will be 140-108.64= 31.36
discount percentage = 31.36*100/ 140 = 22.4
1st discount = 20 %
2nd discount =22.4- 20 = 2.4 %

Question 22 : The index number of five commodities are 121, 123 ,125 , 126, 128 and the weights  assigned to these are respectively 5, 11 ,10,8,6 . Then , what is the assigned weighted average index number?

A.125.2
B.124.2
C.123.8
D.124.6

Solution :Option D

P = commodity
V = weight
PV = 121×5+ 123×11 + 125 × 10+ 126×8+ 128 × 6
= 4984
P = 40
weighted average = PV / V
index number= 4984 / 40
= 124.6

Question 23: A sales representative commision is 6% on all sales upto 15000 and 5% on all sales exceeding this. He remits Rs. 47350 to his company after deducting his commission. What were the total sales.A. Rs 50000B. Rs 49000C. Rs 47500D. Rs 50500

Solution :

47350 = x – 6% of 15000 – 5% of (x-15000)
47350 = x – 900 – 5% of x – 5% of 15000
47350 + 900 – 750 = x – 5% of x
47500 = 95% of x
=> x = 47500 x 100/95
x = 50000
Rs. 50000