# TCS Smart Hiring Numerical Ability Questions 2022

## What is TCS Smart Hiring Numerical Ability Questions round 2022?

TCS Smart Hiring Numerical Ability section is one of the most important section of TCS Smart Hiring Written Test.

This part will assess your numerical skills as well as your ability to analyze data and create algorithms quickly. This section cover the majority of the aptitude subjects. Here on this page we are going to give the complete detail along with the updated syllabus Numerical Ability section of TCS Smart For Freshers.

### TCS Smart Hiring Numerical Ability section For Freshers

Numerical Ability Test ForTCS Smart Hiring
Number of Questions26 Questions
Time Limit40 Mins
DifficultyHigh
Negative MarkingNo

To know more, click the button below!

Total Question

Total Time

Marking

Negative Marking

## Preparation Topics​ 1. A shopkeeper deals in milk and 45 litre mixture is to be distributed in Milk & Water in the ratio of

4 : 1. If 4 litre milk & 3 litre water will be added in the mixture then what will be the new ratio of water and milk? 5 : 6

5 : 6

19.77%

3 : 10

3 : 10

41.86%

4 : 5

4 : 5

22.09%

7 : 8

7 : 8

16.28%

In the mixture of 45 litre,

Milk = $\frac{45}{5}\times 4$ = 36ltr, Water = $\frac{45}{5}\times 1$ = 9ltr

New ratio,

= 9 + 3 : 36 + 4

= 12 : 40 = 3 : 10

Hence, option B is correct.

2. Narendra Modi has good analytical skills. He faces a problem. The problem is “The H.C.F and L.C.M of two numbers are 11 and 385 respectively. If one number lies between 75 and 125” , then according to the question that number is 77

77

44.78%

79

79

26.87%

84

84

20.9%

89

89

7.46%

Product of numbers = 11 x 385 = 4235

Let the numbers be 11a and 11b . Then , 11a x 11b = 4235  =>  ab = 35

Now, co-primes with product  35 are (1,35) and (5,7)

So, the numbers are ( 11 x 1, 11 x 35)  and (11 x 5, 11 x 7)

Since one number lies 75 and 125, the suitable pair is  (55,77)

Hence , required number = 77

3. During a class inspection, the inspector asked an eighth grader, the question was how many numbers in between 333 and 777 are divisible by 2, 3 and 7 both together ? 7

7

16.13%

8

8

25.81%

9

9

17.74%

10

10

40.32%

LCM of 2, 3, 7 is 42.

=> (777 – 333)/42 =10.57 => 10 Numbers.

4. The average of five consecutive odd numbers is 53. What is the difference between the highest and lowest numbers : 8

8

54.39%

7

7

24.56%

6

6

10.53%

5

5

10.53%

Let the numbers be x-4, x-2, x, x+2 and x+4

So, difference between highest and lowest number is (x+4)-(x-4)=8

5. 7 men can repair a road in 12 hours. How many men are required to repair the road in 3 hours ? 40

40

3.39%

36

36

11.86%

28

28

76.27%

32

32

8.47%

Explanation :

$\frac{m \times T}{W} = constant$

where, M= Men (no. of men)

T= Time taken

So, here we apply

$\frac{M_{1} \times T_{1}}{W_{1}} = \frac{M_{2} \times T_{2}}{W_{2}}$

Given that, $M_{1} = 7$ men, $T_{1} = 12$ hours ; $T_{2} = 3$ hours, we have to find $M_{2} =?$

Note that here, $W_{1} = W_{2} = 1$ road, ie. equal workload.

Clearly, substituting in the above equation we get, $M_{2} = 28$ men.

6. In a lottery, there are 15 prizes and 30 blanks. A lottery is drawn at random. What is the probability of getting a prize? 1/3

1/3

43.4%

1/4

1/4

11.32%

1/2

1/2

43.4%

1

1

1.89%

Total number of possible outcomes = 15+30 = 45

Total number of prizes = 15

p(E) = n(s) / n(E) = 15 / 45 = ⅓

7. The bus in which South Indian Star Rajinikanth used to work as conductor runs at the speed of 54 kmph Excluding stoppages, and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour? 11

11

21.43%

12

12

7.14%

10

10

33.33%

9

9

38.1%

Speed of the bus excluding stoppages = 54 kmph

Speed of the bus including stoppages = 45 kmph

Loss in speed when including stoppages = 54 - 45 = 9kmph

In 1 hour, the bus covers 9 km less due to stoppages.

Hence, time in which the bus stops per hour

= Time taken to cover 9 km

$= \frac{distance}{speed} = \frac{9}{54} hour = \frac{1}{6} hour = \frac{60}{6} min = 10 min$

8. A sum of Rs. 427 is to be divided among A, B and C such that 3 times A’s share, 4 times B’s share and 7 times C’s share are all equal. The share of C is : 102

102

21.21%

98

98

12.12%

76

76

27.27%

84

84

39.39%

3A = 4B = 7C = k,Then A = k/3, B = k/4 and C= k/7.

A : B : C = k/3 : k/4 : k/7 = 28:21 :12.

Cs share = Rs. [427 x (12/61)] = Rs. 84

9. Lewis Hamilton was testing his brand new,“Mercedes-AMG Cigarette Racing Boat”. His boat did a remarkable run of 6 km upstream and back again to the starting point in 33 minutes, when the stream was running at 2 kmph. Find the speed of his new motorboat in still water. 8 kmph

8 kmph

24.14%

12 kmph

12 kmph

27.59%

18 kmph

18 kmph

27.59%

22 kmph

22 kmph

20.69%

Let the speed of the motorboat in still water be x kmph. Then,

Speed downstream = (x + 2) kmph; Speed upstream = (x – 2) kmph.

$\therefore \frac{6}{x+2} + \frac{6}{x-2} = \frac{33}{60} \Leftrightarrow 11x^2 – 240x – 44 = 0$

$\Rightarrow 11x^2 – 242x + 2x – 44 = 0 \Leftrightarrow (x – 22)(11x + 2) = 0 \Leftrightarrow x = 22$

Hence, the speed of a motorboat in still water = 22 kmph.

Hence, option D is correct.

10. Darren Criss's present age is two-fifth of the age of his mother. After 8 years, he will be one-half of the age of his mother. How old is the mother at present ? 28

28

13.04%

40

40

56.52%

32

32

13.04%

36

36

17.39%

Let the mother's present age be x years.

Then, the person's present age = $\frac{2}{5} x$ years.

$\frac{2x}{5}+8=\frac{1}{2}(x+8)$

$2(2x + 40) = 5(x + 8)$

=> x = 40.

11. What was the day of the week on the 28th day of the 5th month of 2006? Monday

Monday

21.74%

Thursday

Thursday

13.04%

Sunday

Sunday

52.17%

Friday

Friday

13.04%

28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) = 6 odd days

(31[jan] + 28 [Feb]+ 31[Mar] + 30[April] + 28[May] ) = 148 days = (21 weeks + 1 day) = 1 odd day.

Total number of odd days = (0 + 0 + 6 + 1) = 7 = 0 odd days.

Given day is Sunday

12. Amrendra Bahubali was good in studies in his childhood. He wants to do a problem which is “A number is decreased by 20% and then increased by 20%.  The number so obtained is 20 less than the original number.” What was the original number ? 1200

1200

38.46%

1000

1000

15.38%

500

500

38.46%

2000

2000

7.69%

Let the original number be x.

Final number obtained = 120% of (80% of x) =(120/100 * 80/100 * x) = (96/100)x.

x-(96/100)x=20

=> x =500

13. Ravi bought some articles, and sold them at a certain price , having some certain profit. If he would have sold them for twice the S.P, the profit would have been tripled.

Find the profit% 50%

50%

21.74%

100%

100%

34.78%

75%

75%

26.09%

25%

25%

17.39%

Let C.P. be Rs. x and S.P. be Rs. y.

Then, 3(y - x) = (2y - x)    y = 2x.

Profit = Rs. (y - x) = Rs. (2x - x) = Rs. x.

Profit % = $\frac{x}{x} * 100% = 100%$

14. 100 people at a party shake hands once with everyone else in the room.How many handshakes took place? 4950

4950

37.5%

5500

5500

16.67%

5000

5000

33.33%

6950

6950

12.5%

There are 100 people in the room ,so our n value is 100

No. of handshakes   = $^{100}C_{2}$

= 4950

15. m and n are two whole numbers and p is their LCM. If m= 13, p= 39 and the HCF of these two numbers is 1 then find the value of n ? 13

13

18.18%

3

3

54.55%

507

507

9.09%

23

23

18.18%

LCM or least common multiple, is the least number which is a multiple of both the original numbers. Hence, the LCM would factor out the HCF of the numbers.

LCM x HCF =  m x n

i.e. n = LCM x HCF / m

∴ The value of n = 3

16. Buddh International CIrcuit (BIC) has decided to increase the capacity of its public stands which are in the shape of square.If they decide that each side of a square stand is increased by 25%, find the percentage change in its area. 56.25%

56.25%

47.37%

23.45%

23.45%

21.05%

15.54%

15.54%

15.79%

67.12%

67.12%

15.79%

To solve this question, we can apply the net% effect formula

Net % effect =

Here, x = y = 25%

Applying the net% effect, we get

= 50 + 6.25 = 56.25%

Hence, the total percentage change in area will be 56.25%

17. Ashwin put a sum of Rs.49000 into two separate LIC schemes. The first scheme pays interest at a pace of 5% per year, while the second scheme pays interest at a value of 12% per year. Find the amount deposited in scheme 2 if Ashwin's cumulative interest received after one year is Rs.4900. 33210

33210

20%

35000

35000

30%

42001

42001

15%

None of the above

None of the above

35%

Ans: Let the total investment in 2nd scheme be Rs.x, then

(49000 - x) \times \frac{5}{100} + x \times \frac{12}{100} = 4900

x= 35000

18. A man takes 5 hours 40 minutes in walking to a certain place and riding back. He would have taken 3 hours less by riding both ways. What would be the time he would take to walk both ways? 4 hours 35 minutes

4 hours 35 minutes

17.39%

8 hours 35 minutes

8 hours 35 minutes

30.43%

10 hours

10 hours

13.04%

8 hours 40 minutes

8 hours 40 minutes

39.13%

1 Bike + 1 Walk = 5 hours 40 minutes

2 Bike = 5 hours 40 minutes – 3 hours = 2 hours 40 minutes

Hence, 1 way Bike journey takes time = 2 hours 40 minutes ÷ 2 = 1 hour 20 minutes

So, 1 way walk should take (5 hours 40 minutes) – (1 hour 20 minutes) = 4 hours 20 minutes

2 way walk would take time = 4 hours 20 minutes × 2 = 8 hours 40 minutes

19. Sum of the twice the age of Saif Ali Khan and his Father age is 79. Sum of the twice the age of Father and Saif Ali Khan's age is 104. The average age of Saif Ali Khan, his Father and his Mother is 32. Then what is the age of his Mother? 32

32

19.05%

33

33

9.52%

34

34

14.29%

35

35

57.14%

2S+F = 79

2F+S = 104

S =18, F = 43

18+43+M/3 = 32

M =35

20. Virat Kohli's salary is 50% of  Pujara's salary which is 40% of Raina's salary. What percentage of Raina's salary is Kohli's salary ? 10

10

14.29%

20

20

57.14%

15

15

23.81%

18

18

4.76%

Virat Kohli's Salary  = 50% of Pujara = 50% of (40% of Raina

= 50/100[(40/100)*100] % of Raina

= 20% of Raina ×

## Detailed Analysis of Numerical Ability Section For Freshers​

Go through the table given below. You will get all the information about each and every topic that comes under this section.

TopicsNo. of questions in testDifficultyImportance
Ratio and Proportion0 – 2HighHigh
Clocks and Calendars0 – 1MediumLow
Profit and Loss0 – 2MediumHigh
Number System0 – 2LowMedium
LCM and HCF0 – 1LowLow
Percentage0 – 2MediumLow
Mensuration0 – 3HighHigh
SI and CI0 – 2HighHigh
Work and Time0 – 2HighHigh
Speed, Time and Distance0 – 2HighHigh
Data Interpretation0 – 2HighHigh
Number Series0 – 2MediumMedium
Equations0 – 1MediumMedium
Elimentary Statistics0 – 4MediumHigh
Averages0 – 1MediumLow
Mixtures and Allegations0 – 1HighLow
Simplifications0 – 4MediumHigh
Surds and Indices0 – 1MediumLow

#### Is there any negative marking for this section?

No, there will be no negative marking for TCS Smart Numerical Ability Questions 2022 Section.  You can attempt all the questions but you need to practice well for this round.

#### What will be the cut of Numerical ?

The cut off for Numerical Ability section is high. The cutoff will be between 75 and 80 percent, or 20 and 22 questions.