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

**Answer:** **Option C**

### 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

**Answer:** ** Option A**

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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

**Answer: Option B**

**(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

**Answer:** **Option D**

(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

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**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. 0.8

B. 0.08

C. 0.4

D. 0.04

^{(1/3)}

**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. 6

B. 12

C. 9

D. 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. 250

B. 400

C. 500

D. 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 m**^{2}. 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. 1.45

B. 2.8

C. 2.1

D. 2.45

^{2}. 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?

**Solution **:** Option D**

Area of Rectangle = 726m^{2}

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.**

Students/Marks | A | B | C | D |

X | 78 | 85 | 72 | 76 |

Y | 65 | 68 | 64 | 73 |

Z | 82 | 76 | 81 | 75 |

** 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. 20

B. 23

C. 21

D. 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. 14

B. 12

C. 20

D. 26

**Solution:** **Option D**

In given Question:-

Correct answer =5x and wrong answer=2y

Correct answer =4x and wrong answer= 1y

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

By adding x+y

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. 3.84

B. 3.6

C. 3.25

D. 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 50000

B. Rs 49000

C. Rs 47500

D. 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

**View All TCS NQT Papers of 2020: Click Here**