** Numerical Ability **

Question 1

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

A) 1.8

B) 1.2

C) 1.5

D) 2.4

Answer

#### A) 1.8

Explanation

#### 20 km in 1 hr and 10 mins break for refueling.

Remaining time = (2hrs – 1hr 10 mins) = 50 mins = 5/6 hrs

He has to cover (50 – 20) = 30km in 5/6 hrs

Therefore, his speed will be = (30 / (5/6)) = 30 * 6/5 = 36 km/hr

Therefore increase in speed with reference to previous one = 36 / 20 = 1.8

Question 2

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

A) 3/35

B) 2/35

D) 2/175

B) 4/35

Answer

#### B) 2/35

Explanation

#### Mean Proportional between 2/7 & 32/343 = sqrt((2/7) * (32/343)) = sqrt(64/2401) = 8/49

Mean Proportional between 2 & 1/5000 = sqrt((2) * (1/5000)) = sqrt(1/2500) = 1/50

Mean Proportional between 8/49 & 1/50 = sqrt((8/49) * (1/50)) = sqrt(4/1225) = 2/35

Question 3

#### If the positive square root of ((90^0.5)+(80^0.5) is multiplied by (2^0.5 – 1), and the product is raised to the power of four, the result would be__

A) 1600

B) 11520000

C) 10

D) 100

Answer

#### C) 10

Explanation

#### ((90^0.5 + 80^0.5)^0.5 * (2^0.5-1)) ^ 4

= ((90^0.5 + 80^0.5) * (3 – 2*(2^0.5)) ) ^ 2

= (9*(10^0.5) – 6*(10^0.5) + 6*(20^0.5) – 8*(10^0.5)) ^ 2

= (10^0.5)^2 = 10

Question 4

#### 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) 16

C) 15 ½

D) 19 1/21

B) 15

Answer

#### A) 16

Explanation

#### 240*(x+10)/100 = 160*(180*x/100)/100

=> (24x+240)/10 = 288x/100

=> 240x+2400 = 288x

=> x = 50

15% of (x+20) = 15% of 70 = 10.5

25% of x = 25% of 50 = 12.5

%Diff = (12.5 – 10.5)*100/12.5 = 16

Question 5

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

A) 120 * pi

B) 60

C) 60 * pi

D) 120

Answer

#### B) 60

Explanation

#### Diameter = 30 cm

Radius = 15 cm

Area of the Pizza = pi*15*15 = 225*pi sq. cm

Circumference of the Pizza = 2*pi*15 = 30*pi cm

30*pi cm has the area = 225*pi

8 cm will have the area = 225*pi * 8 / (30*pi) = 60

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

Question 6

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

A) 15

B) 27

C) 9

D) 21

Answer

#### A) 15

Explanation

#### The rule of divisibilty by 24 is that the number should be divisible by 3 and 8.

As the last three digits are 352 and that is divisible by 8, the number is divisible by 8.

Now, the number is divisible by 3, so

1+n+n+3+5+2 = 2n+11, is divisible by 3.

So the possible values of n = 2, 5, 8

So sum is = (2+5+8) = 15

Question 7

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

A) 8

B) 10.2

C) 0

D) 9.2

Answer

#### A) 8

Explanation

#### Mean = (8+9+12+15+16+20+24+30+32+34)/10=20

Mean Deviations:

20-8 =12

20-9=11…

Mean Deviation=12+11+8+5+4+0+10+12+14/10=8

Question 8

#### 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) 3024

B) 3075

C) 3126

D) 2975

Answer

#### B) 3075

Explanation

#### Let the rate of interest is = r%

Using CI formula ->

51840 = 30000 * (1 + r/100)^3

=> 216/125 = (1+r/100)^3

=> 1 + r/100 = 6/5

=> r = 20

Again using CI formula for quarter -> 6 months = 1/2 years

Amount = 30000*(1+(20/4)/100)^(4*1/2) = 30000*(1+1/20)^2 = 33075

Interest = (33075 – 30000) = 3075

Question 9

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

A) 0.01

B) 10

C) 0.1

D) 1

Answer

#### C) 0.1

Explanation

#### 240% of 1.5 = 3.6

0.025% of 3.6 = 0.0009

(0.0009/0.9)*100 = 0.1

Question 10

#### 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) 3

B) 2.4

C) 1.2

D) 0.9

Answer

#### B) 2.4

Explanation

#### 2*pi*r*h = 5*(2*pi*r^2)

=> 6 = 5r

=> r = 1.2

=> d = 2r = 2.4cm

Question 11

#### 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% loss

B) 2.5% gain

C) 4% gain

D) 4% loss

Answer

#### D) 4% loss

Explanation

#### Let list price = 100x

12.5% discount on List Price.

Selling Price = 87.5x

Let Cost Price = y

y*112/100 = 87.5x

=> y = 78.125x

If given 25% discount Selling Price = 75x

%Loss = (78.125x-75x)*100 / 75x = 4%

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

Question 12

#### The Variation in temperature throughout the day in a desert town was studied on the basis of the records of the maximum and minimum temperatures which was 36 and 8 degree centigrade respectively. What was the standard deviation in degree centigrade?

A) 28

B) 22

C) 14

D) 12

Answer

#### A) 28

Explanation

#### Mean =(36+8)/2=22

Standard Deviation = sqrt([{(22–36)²+(22–8)²}/2]) = sqrt([2×14²/2]) = 14

Question 13

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

A) 12

B) 16

C) 14

D) 15

Answer

#### B) 16

Explanation

#### In One Week 4km of railway track was completed by 96 men.

Time left = (4-1) = 3 weeks

Work left = (18-4) = 14km

Let, x men completes the rest work in rest time.

Work Done is inversely proportional to time.

Applying chain rule now,

=> M1 * D1 * W2 = M2 * D2 * W1

=> 96 * 1 * 14 = x * 3 * 4

=> 12x = 96 * 14

=> x = 8 * 14

=> x = 112

Therefore, Additional Men = 112 – 96 = 16

Question 14

#### In a competitive exam, 5 marks are awarded for every correct answer and for every wrong answer, 2 marks are deducted. Sathwik and 32 marks in this examination. If 4 marks has 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) 20

B) 14

C) 12

D) 26

Answer

#### D) 26

Explanation

#### Let Sathwik answered x right questions and y wrong questions.

Then total number of questions = x+y

The marks Sathwik scores in the examination = 5x-2y

5x-2y = 32 …i

Now if 4 marks are awarded for every correct answer and for every wrong answer, 1 marks are deducted.

The marks Sathwik scores in the examination = 4x-y

4x-y = 34 …ii

Now Solving Equation (i) & Equation (ii) ->

x=12, y=14

Total questions = (12+14) = 26

Question 15

#### 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 do half the

work?

A) 40

B) 80

C) 60

D) 20

Answer

#### A) 40

Explanation

#### Explanation:

Let, Efficiency of Y = a

Efficiency of X = 4a

Time taken to complete the work = 16 days

=> 1/4a + 1/a = 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

Question 16

#### 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) 11

B) 12

C) 11.5

D) 10.5

Answer

#### A) 11

Explanation

#### The Median of a even series is = (N-1 + 15)/2 = 12.5

N = 10 + 1 = 11

Question 17

#### A file of cadets consisting of ten rows and five columns measures 420m 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 the each cadet is 80cm and he takes 57 strides per minute?

A) 1 hr 15 min

B) 1 hr 24 min

C) 1 hr 10 min

D) 1 hr 20 min

Answer

#### A) 1 hr 15 min

Explanation

#### Stride of Each Cadet = 80cm = 0.8m

Each Cadet takes 57 strides in 1 minute

So in 60 seconds he cover = (57 * 0.8)m

In 1 second he cover = (57 * 0.8) / 60 = 0.76m

Speed to cover the distance = 0.76m/s.

Total distance to be covered = 420m + 3000m = 3420 m.

Total time taken to march for a stretch of 3km = (3420/0.76) = 4500s = (5/4)hrs = 1 hour + (1/4 * 60) = 1hr 15mins

Question 18

#### What is the sum (in Rs) which, when divided among X, Y, Z in the proportion 3 : 5 : 7 provided Rs. 8000 more to Z than what it would have done to him when the proportion is 11 : 15 : 19?

A) 120000

B) 180000

C) 135000

D) 175000

Answer

#### B) 180000

Explanation

#### Let, The Sum = x

For Initial Distribution, The dividend for z=7x/(7+3+5)=7x/15

For New Distribution, The dividend for z=19x/(19+11+15)=19x/45

=> 7x/15–19x/45=8000

=> 2x = 8000*45

=> x = 180000

Question 19

#### What is the value of (0.0000128)1/7?

A) 0.2

B) 5

C) 2

D) 0.5

Answer

#### A) 0.2

Explanation

#### (0.0000128)^(1/7) = (0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2 * 0.2) ^ (1/7) = 0.2

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

Question 20

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

A) 7.5

B) 9.6

C) 13.5

D) 12.5

Answer

#### D) 12.5

Explanation

#### Total interest in two years = 25500-22500 = 3000

Total Interest in 9 years = 9*3000/2 = 13500

Principal, P = 22500 – 13500 = 12000

By, formula of Simple interest ->

SI = P*R*T/100

=> 13500 = 12000*R*9/100

=> R = 12.5

Question 21

#### If 5^X3^Y = 225 x 405, find the value of X^2Y-3X

A) 81

B) 27

C) 125

D) 25

Answer

#### B) 27

Explanation

#### 225 = 5^2 * 3^2

405 = 5^1 * 3^4

Their product = 5^3 x 3^6

So, X=3 and Y=6

3^(12-9) = 3^3 = 27

Question 22

#### The mean of set data is 5. What will be the mean if 10 is subtracted from each data.

A) 5

B) 10

C) -5

D) -15

Answer

#### C) -5

Explanation

#### Let, Total data points = n

Mean of Data set = 5

Total of Data set = 5n

If 10 is subtracted from each data point, 10n will be subtracted as a whole.

New Total = 5n-10n = -5n

New Mean = -5n/n = -5

Question 24

#### 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) 5% gain

B) 10% gain

C) 5% loss

D) 10% loss

Answer

#### B) 10% gain

Explanation

#### Let the CP of each copy = 100

ATQ, One copy is sold at = 0.8 x 100 = 80

Other copy is sold at = 1.4 x 100 = 140

Total money spent = 200

Total money earned = (80+140) = 220

Total gain = 20

Therefore gain% = (20/220) x 100 = 10%

Question 25

#### 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 is a multiple of five. In which month is the difference of marks scored in the two subjects are highest?

A) Apr

B) Mar

C) Feb

D) Jan

Answer

#### C) Feb

Explanation

#### Difference in Feb is the highest which is 20 marks.

Question 26

#### 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) 1/135

B) 1/67

C) 1/68

D) 1/270

Answer

#### A) 1/135

Explanation

#### In Vessel X:

Chemical A = 3/5

Chemical B = 2/5

In Vessel Y:

Chemical A = 8/9

Chemical B = 10/9

Now In vessel C,

Chemical A : Chemical B = (3/5+8/9):(2/5+10/9) = 67:68

We have to add Chemical A = 68/(67+68) – 67/(67+68) = 1/135

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