** Numerical Ability **

Question 1

#### For a grouped frequency distribution having eight classes, the upper class boundaries of the lowest and the highest classes are 10 and 66 respectively. What is the lower class boundary of the highest class?

A. 57

B. 60

C. 58

D. 59

Answer

#### C. 58

Explanation

#### upper class boundary of lowest class = 10

upper class boundary of highest class = 66

Total classes = 8

66 – 10 = 56

class interval = 56 / 8 + 1 = 8

lower boundary of the highest class = 66 – 8 = 58

Question 2

#### If (√7+ √5) / (√7 − √5)= a − b√35, then the value a – 2b is

A. 4

B. 5

C. 8

D. 7

Answer

#### C. 8

Explanation

#### => (√7+ √5)^2 / 2 = a – b√35

=> (12 + 2√35) / 2 = a – b√35

=> 6 + √35 = a – b√35

a = 6, b = -1 => a-2b = 6+2=8

Question 3

#### The average marks (out of 100) scored by the students of four towns A, B, C and D of a district, in English and Hindi at a secondary level examination held in 2019 are presented through a Bar Graph shown below.

How much percentage (correct up to two decimal places) are the average marks in Hindi above that of the highest marks among the towns in English?

A. 18.33

B. 14.33

C. 16.67

D. 20.67

Answer

#### C. 16.67

Explanation

#### Avg. marks in Hindi = (60+80+70+70)/4 = 70

Highest marks in English = 60

%above = (70-60)*100/60 = 16.67

Question 4

#### Raju buys 3 goats and 2 sheeps for Rs.11600. When he sells the goats at 20% profit and the sheep at 10% loss, he earns a total profit of Rs.1000. The cost of one sheep is _____.

A. Rs.2600

B. Rs.4600

C. Rs.2400

D. Rs.2200

Answer

#### D. Rs.2200

Explanation

#### Let, CP of one goat is Rs. x and CP of one sheep is Rs. y

3x + 2y = Rs.11600 …(1)

Now, given that, he sells goat for 20% profit and sheep at 10% loss.

SP of 3 goats = (3x * 120/100) = Rs.3.6x

SP of 2 sheep = (2y * 90)/100) = Rs.1.8y

Total Profit is = Rs.1000

Total SP = 11600 + 1000 = Rs.12600

3.6x + 1.8x = 12600 …(2)

Solving Eqn.(1) and Eqn.(2) we get,

=> 10(3.6x + 1.8x) – 12(3x + 2y) = 10*12600 – 12*11600

=> 36x – 36x + 18y – 24y = 126000 – 139200

=> (-6y) = (-13200)

=> y = Rs.2200

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Question 5

#### The sum of two numbers is 2604 and their HCF is 124. Which is the smaller between them if their difference is the least possible?

A. 1116

B. 1240

C. 620

D. 496

Answer

#### B. 1240

Explanation

#### x + y = 2604

HCF is 124

For difference to be minimum y=x+124 (Assuming y to be larger number)

Then we get

=> x+x+124=2604

=> 2x=2480

=> x=1240 (minimum number is x)

**Also Checkout: TCS NQT Sample Question Paper**

Question 6

#### What sum (in Rs) given on loan for two years with scheme of return on the basis of compound interest at a yearly rate of 10% will correspond to repayment through equal monthly installments of Rs.9075?

A. 165000

B. 180000

C. 189000

D. 198000

Answer

#### B. 180000

Explanation

#### Total repayment = Rs. 9075 * 24 = Rs. 2,17,800

Let, loan amount = Rs. P

ATQ, P*(1+10/100)^2 = 217800

=> P*(121/100) = 217800

=> P = 180000

Question 7

#### The capacities of three containers X, Y and Z are 1, 2 and 4 liters respectively. Initially, X is empty, while Y and Z are full of water and milk respectively. X is filled from Y, Y is replenished from Z and X is emptied into Z. If this process is repeated once more, then what will be the ratio

of milk in Y to water in Z?

A. 3:5

B. 1:1

C. 4:3

D. 4:5

Answer

#### B. 1:1

Explanation

#### X(1) Y(2) Z(4)

Initially M=0 W= 0 M=0 W = 2 M = 4 W = 0

X is filled from Y M=0 W= 1 M=0 W = 1 M = 4 W = 0

Y is filled from Z M=0 W= 1 M=1 W = 1 M = 3 W = 0

X is emptied in Z M=0 W= 0 M=1 W = 1 M = 3 W = 1

Milk and water in Y are in 1:1 ratio hence 1 litre will have 0.5 Water & 0.5 milk

X is filled from Y M=0.5 W= 0.5 M=0.5 W = 0.5 M = 3 W = 1

Milk and water in Z are in 3:1 ratio hence 1 litre will have 0.25 Water & 0.75 milk

Y is filled from Z M=0.5 W= 0.5 M=1.25 W = 0.75 M = 2.25 W = 0.75

X is emptied in Z M=0 W= 0 M=1.25 W = 0.75 M = 2.75 W = 1.25

Milk in Y = 1.25

Water in Z = 1.25

1.25 : 1.25 = 1 : 1

ratio of milk in Y to water in Z = 1 : 1

Question 8

#### The average score in Mathematics of a class increases by 10% if the total marks secured by a number of students who form 20% of the class strength and whose average score is 48 is not included in the calculation. What is the average score?

A. 90

B. 60

C. 75

D. 80

Answer

#### D. 80

Explanation

#### Let average score = x, Total students = 100

Total marks = 100x

0% of Total students = 20 students

Their average score = 48 marks

Total marks not included = 48 * 20 = 960 marks

Marks left now = (100x – 960) marks

Students left = 100 – 20 = 80 students

New average score = {(100x – 960)/80} marks

Given that, average score is increased by 10%

(100x – 960) / 80 = (110x/100)

=> 5(100x – 960) = 4 * 110x

=> 500x – 5 * 960 = 440x

=> 500x – 440x = 5 * 960

=> 60x = 5 * 960

=> x = 80

Question 9

#### If 25% of a number is equal to the three-fifths of another number, what will be the ratio of the first number to the second number?

A. 5:12

B. 12:5

C. 12:15

D. 15:12

Answer

#### B. 12:5

Explanation

#### First Number = x, Second Number = y

ATQ, x/4 = 3y/5

=> x/y = 12/5 = 12:5

Question 10

#### A hollow spherical ball of thickness 1 cm and external radius 5 cm is melted and then from the solid so obtained, without any loss of material, 61 identical spherical balls are obtained. What is the diameter (in cm) of each ball?

A. 1.5

B. 3

C. 1

D. 2

Answer

#### D. 2

Explanation

#### Internal radius = 5-1 = 4cm

Volume of the melted material = (4/3)*pi*(5^3-4^3) = (4/3)*pi*61 cc

61 indentical ball are obtained from this.

Volume of each ball = (4/3)*pi cc

So, radius of each ball = 1cm

Diameter of each ball = 2cm

Question 11

#### Four distributions AddaJobs, Peepinsta, PhraseSeven, Offlineprepare9i are given below:

AddaJobs: 4, 2, 1, 3, 3, 5, 6, 9, 8, 7

Peepinsta: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13

PhraseSeven: 5, 6, 4, 5, 7, 5, 8, 1, 5, 2

Offlineprepare9i: 2, 6, 6, 11, 10, 9, 10, 8, 7

What is the mean of the two modes in the only bimodal distribution among the above?

A. 5

B. 3

C. 6

D. 8

Answer

#### D. 8

Explanation

#### Offlineprepare9i is the only bimodal distribution here.

It has 6 and 10 two times, which makes them the mode.

The mean of the mode = (6+10)/2 = 8

**Also Checkout: TCS NQT Sample Question Paper**

Question 12

#### What is the value of 1.59 X 1.59 + 8.46 X 0.53 + 9 X 0.47 X 0.47 ?

A. 9.025

B. 6.25

C. 9

D. 4

Answer

#### C. 9

Explanation

#### 1.59 * 1.59 + 8.46 * 0.53 + 9 * 0.47 * 0.47

= 9 * 0.53 * 0.53 + 9 * 0.94 * 0.53 + 9 * 0.47 * 0.47

= 9 * (0.53^2 + 2 * 0.47 * 0.53 + 0.47^2)

= 9 * (0.53+0.47)^2

= 9 * 1 = 9

Question 13

#### What is the simplified value of: (7^2 + 24^2) / [20^2 – (1/2)(5^2 – 1) + 10{(0.3)^2 + (0.1)^2} – 2 x 7]^2?

A. 1/5

B. 1/225

C. 1/125

D. 1/25

Answer

#### B. 1/225

Explanation

#### (7^2 + 24^2) / [20^2 – (1/2)(5^2 – 1) + 10{(0.3)^2 + (0.1)^2} – 2 x 7]^2

= (49 + 576) / [400 – (1/2)24 + 10{0.09 + 0.01} – 14]^2

= (625) / [400 – 12 + 1 – 14]^2

= (625) / (375 * 375) = 1/225

Question 14

#### Four men and two women can do a piece of work together in one day. If a woman is twice as efficient as a man, in how many days can a woman working alone do the work?

A. 4

B. 6

C. 8

D. 2

Answer

#### A. 4

Explanation

#### Let, Efficiency of a man is x unit/day.

Efficiency of a woman = 2x unit/day.

4 men completes = 4x unit of work / day.

2 women completes = 4x unit of work / day.

1 one day total unit of work completes by 4 men and 2 women is = 4x + 4x = 8x units. = Total work

Time taken by a woman working alone to do the complete work = 8x/2x = 4 days.

Question 15

#### The median of the following data is ______.

A. 24.266

B. 24.266

C. 24.123

D. 24.166

Answer

#### D. 24.166

Explanation

#### C.I. (m) (f) (d) fd c.f.

0-10 5 10 -2 -20 10

10-20 15 15 -1 -15 25

20-30 25 12 0 0 37

30-40 35 15 1 15 52

40-50 45 8 2 16 60

Question 16

#### A train starting from station X was to arrive at station Y at 6:06 PM. It could travel at 62.5% of its usual speed and reach Y at 7 PM. At what time did it start from X?

A. 4:44 PM

B. 4:56 PM

C. 4:36 PM

D. 4:24 PM

Answer

#### C. 4:36 PM

Explanation

#### Let, the actual speed of the train be 100x km/hr and the actual time taken be y hours.

Distance covered = (100xy)km

now, given that, if train travels at 62.5% of its usual

New speed = 62.5% of 100x = 62.5x km/h

increased time = 7PM – 6:06 = 54 minutes = (54/60) = (9/10) hours

New total time = (y + 9/10) hours

Distance covered = (62.5x) * {y + (9/10)} km

=> 100xy = 62.5x * {y + (9/10)}

=> 100y = 62.5(10y + 9)/10

=> 1000y = 625y + 562.5

=> 1000y – 625y = 562.5

=> 375y = 562.5

=> y = 1.5 hours.

Train start from X at = 6:06 – 1:30 = 5:66 – 1:30 = 4:36 PM.

Question 17

#### In how many ways can X give Rs.500 to Y using only Rs.100 and Rs.20 notes, with the condition that she has only 16 notes of Rs.20, and being asked to use the notes of both the denominations?

A. 6

B. 3

C. 4

D. 2

Answer

#### B. 3

Explanation

#### 5 Rs.20 notes + 4 Rs. 100 notes = 500

10 Rs.20 notes + 3 Rs. 100 notes = 500

15 Rs.20 notes + 2 Rs. 100 notes = 500

therefore, 3 ways.

Question 18

#### A sales representative’s commission is 6% on all sales up to Rs. 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. 49000

B. Rs. 47500

C. Rs. 50500

D. Rs. 50000

Answer

#### D. Rs. 50000

Explanation

#### Let, Total Sales = x

Total Commission = y

Then according to the given conditions,

0.06*15000 + 0.05 (x – 15000) = y …1

47350 + y = x …2

On simplifying 1, we have

900 + 0.05x – 750 = y

Therefore, y – 0.05x = 150 ………E3

On simplifying E2, we have

x – y = 47350 ………E4

To solve this, adding E3 & E4, we get

0.95x = 47500

Therefore, x = 50000

Hence, the total sales is Rs. 50000

Question 19

#### A particular distribution is represented by two data points. If the range and the standard deviation of the distribution are R & S respectively, what is the relation between them?

A. S = √R

B. S = 2R

C. S = R

D. S = R/2

Answer

#### D. S = R/2

Explanation

#### Let, the points are a,b.

Range, R = a-b

SD, S = √[(1/2){(a-(a+b)/2)^2 + (b-(a+b)/2)^2}]

= √[(1/2){(a/2-b/2)^2 + (b/2-a/2)^2}]

= √[(1/2){(a/2-b/2)^2 + (a/2-b/2)^2}]

= √[{(a-b)/2}^2] = (a-b)/2

Therefore, S = R/2

**Also Checkout: TCS NQT Sample Question Paper**

Question 20

#### What will be the percentage increase in the area of a square, If its side is increased by 20%?

A. 44%

B. 40%

C. 36%

D. 20%

Answer

#### A. 44%

Explanation

#### Let, initial length of side be x

Increased length = x+20%(x) = 1.2x

Initial area =x^2

Increased area = (1.2x)^2 = 1.44x^2

Percent increase in area= (1.44x^2-x^2)*100/x^2 =0.44×100=44%

Question 21

#### The ratio is incomes of P and Q is 7:5 and the ratio of their expenditures is 4:3. IF at the end of the year, P and Q save Rs.3000 and Rs. 2000 respectively, what is Q’s income?

A. Rs.5000

B. Rs.4500

C. Rs.4000

D. Rs.7000

Answer

#### A. Rs.5000

Explanation

#### Ratios of income = 7:5 => 7x and 5x

Ratios of exp = 4:3 => 4y and 3y

Savings of P = 3000

Savings of Q = 2000

So, 7x – 4y = 3000

5x – 3y = 2000

On solving we get x = 1000 and y = 1000

Q’s income = 5 * 1000 = 5000

Question 22

#### Raju lends Rs. 3000 to Bharath and a certain sum to Charan at the same time at 6% per annum simple interest. If after 5 years, Raju altogether receives Rs. 1650 as the interest from Bharath and Charan, what is the sum lent to Charan?

A. Rs. 2500

B. Rs. 2750

C. Rs. 3250

D. Rs. 3300

Answer

#### A. Rs. 2500

Explanation

#### Case 1> Sum Lends to Bharath.

P = Rs.3000

R = 6% per annum.

T = 5 years.

SI = (3000 * 6 * 5) / 100 = Rs.900

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

Case 2> Let sum lends to Charan is Rs.x .

P = Rs. x

R = 6% per annum.

T = 5 years.

SI = (x * 6 * 5) / 100 = Rs.(3x/10)

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

Therefore,

900 + (3x/10) = 1650

=> (3x/10) = 1650 – 900

=> 3x = 7500

=> x = Rs.2500

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