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TCS NQT Numerical Ability 2022

What's is TCS NQT Numerical Ability 2022?

TCS NQT Numerical Ability 2022 is one of the most important section of TCS NQT Written Test.

This part will assess your numerical skills as well as your ability to analyze data and create algorithms quickly. The TCS NQT Numerical Ability 2022 parts cover the majority of the aptitude subjects. Here on this page we ae going to give the complate detail along with the updated syllabus of TCS NQT Numerical Ability For Freshers.

TCS NQT Numerical Ability For Freshers

Aptitude Test For TCS NQT
Number of Questions 26 Questions
Time Limit 40 Mins
Difficulty High
Negative Marking No
To know more, click the button below!

26 Question

Total Question

40 minutes

Total Time

1 Marks

Marking

No

Negative Marking

Latest Numerical Ability Questions for TCS NQT 2022

1. Nabanita bought two copies of "Harry Potter" for her sister and resold them for 0.8 and 1.4 times their original cost. What was Nabanita's profit or loss as a percentage of her investment?

10% loss

10% loss

17.27%

5% gain

5% gain

17.27%

10% gain

10% gain

58.18%

5% loss

5% loss

7.27%

Let the C.P of each copy be Rs 100
ATQ one copy is sold at 0.8 x 100 = 80
One copy is sold at 1.4 x 100 = 140
Total money spent in buying the copies = Rs 200
Total money earned after selling the copies = Rs (80 + 140) = Rs 220
Total gain of Rs 20. Therefore gain% = 10%.

2. Madhumita and her friends ordered a pancake from the nearby Cafeteria. A pancake has a 30 cm diameter. What is the area (in square centimeters) of the upper surface of a pancake sector with an arc length of 8 cm?

120 pi

120 pi

20.73%

120

120

13.41%

60 pi

60 pi

28.05%

60

60

37.8%

Area of the sector when the arc is given = (L x r)/2
Where L = length of arc and r = radius of the circle
Here L = 8 cm and r = 15 cm
Thus, Area = \frac{8 \times 15}{2} = 60 sq. cm

3. “If the HCF of 180 and 432 is represented as (180a + 432b), where a and b are integers, then what is the difference between a and b?” Aliya taught HCF and LCM to one of her students and requested them to solve this problem on it.

7

7

47.62%

3

3

33.33%

8

8

9.52%

9

9

9.52%

Prime factors of 180 and 432 is :-
→ 180 = 2 * 2 * 3 * 3 * 5
→ 432 = 2 * 2 * 2 * 2 * 3 * 3 * 3
HCF = 2 * 2 * 3 * 3 = 36.
Now , given that, HCF is written as 180a+432b.
So,
→ 180a + 432b = 36 .
→ 36(5a + 12b) = 36
dividing both sides we get,
→ 5a + 12b = 1
Now, we have given that, m and n both are integers.
So,
putting values of m , we get :-
a = 0, b = (1/12) => b = not an integer.
a = 1 , b = (-1/3) => b = not an integer.
a = 2 , b = (-3/4) => b = not an integer.
a = 3 , b = (-7/6) => b = not an integer.
a = 4 , b = (-19/12) => b = not an integer.
a = 5 , b = (-2) => b = an integer.
Hence,
→ Difference between a and b = a - b = 5 - (-2) = 5 + 2 = 7

4. Abhishekh lends Rs 3000 to Amitabh and a specified sum to Ranbeer for a business purpose at a rate of 6% per annum Simple Interest. What is the amount lent to Ranbeer if Abhishekh receives Rs 1650 in interest from Amitabh and Ranbeer after 5 years?

Rs 3250

Rs 3250

17.54%

Rs 2500

Rs 2500

64.91%

Rs 3300

Rs 3300

10.53%

Rs 2750

Rs 2750

7.02%

For Amitabh and Ranbeer 6% for 5 year = 30% for each
30% for 3000 = 900
=> Total interest - 900 = Ranbeer
1650 - 900 = 30% of x (given to Ranbeer)
30% of x = 750
x = Rs 2500

5. In a Delhi-based society, ten percent of people did not vote in an election between two candidates to pick the society's President. A total of 20% of the residents were found to be ineligible. The victorious candidate received 60 percent of the valid votes and a 2,880 vote majority. What is the total number of people who voted in the election?

18,000

18,000

21.05%

25,000

25,000

18.42%

20,000

20,000

42.11%

24,000

24,000

18.42%

Let the total number of residents be x.
As 10% of the residents did not cast their votes
Then, Votes polled =90% of x.
20% of the votes polled were found invalid
Valid votes =80% of (90% of x).
60% of [80% of (90% of x)]−40% of [80% of (90% of x)] = 2880
=>8% of [90% of (90% of x)]=1620
Solving this equation,
x= 20000

6. Aman and Bibhash take on a venture worth Rs. 720. Aman can complete it in 12 days, while Bibhash can complete it in 16 days. What should Chandan's portion of the money gained be if they hire Chandan to assist them and complete the assignment in six days?

Rs. 270

Rs. 270

11.11%

Rs. 360

Rs. 360

22.22%

Rs. 80

Rs. 80

11.11%

Rs. 90

Rs. 90

55.56%

Aman’s work = 1/12 th of total work
Bibhash’s work = 1/16th of total work
Combined work of Aman and Bibhash in 1 day = 1/12 + 1/16 = 7/48
Given that they together finished their work in 6 days = ½ + 1/16 - 1/6 = -1/48
Thus, Chandan can finish the work in 48 days.
Chandan’s share = \frac{1}{48} \times 6 \times 720 = Rs 90

7. In four days, X can complete 25% of a road, while Y can complete one-third of the same work. How long will it take Z to finish the road in the same amount of time that X and Y would if they worked together?

25/6

25/6

7.41%

24/7

24/7

29.63%

17/5

17/5

29.63%

None of these

None of these

33.33%

X can do 25% of a work in = 4 days.
So,
→ X can do 1% of a work in = (4/25) days.
Then,
→ X can do 100%(total work) of a work in = (4/25) * 100 = 16 days.
Similarly, given that,
→ Y can do (1/3) of a work in = 4 days.
Then,
→ Y can do whole work in = 4 * (3/1) = 12 days.
now, given that,
Z can complete the work in the same time as X andY would take working together.
So,
→ Time taken by Z to complete the whole work = 1 / {(1/16 + (1/12)} days.
Now, we have to find how many days all three of them will complete the same work.
So,
→ per day work of X + per day work of Y + per day work of Z = Per day work of all three.
Putting all values we get:-
→ (1/16) + (1/12) + {(1/16) + (1/12)}
→ (1/16 + 1/16) + (1/12 + 1/12)
→ (2/16) + (2/12)
→ (1/8) + (1/6)
→ (3 + 4)/24
→ (7/24)
Hence,
→ Time taken by all three to complete total work = (24/7) days.
∴ If all the three work together, they will take (24/7) days to complete the same work.

8. The average weight of 18 pupils of South Point High School in section 6C is 45 kg. A new pupil joins the group, weighing 64 kg. What is the new group's average weight (in kilograms)?

47

47

17.24%

46

46

68.97%

48

48

6.9%

45

45

6.9%

Let total number of weight of 18 students = x
Average weight = x/18=45
=> x=810
Therefore total weight of 18 students is 810kg
Now,
Adding 64 we get 874 kg total weight
So, the average weight of 19 students is 874/19=46kg

9. Two beetles, 1.1 centimetre and 1.3 cm in length, crawl in opposing directions at 3 and 5 mm per second, respectively. How long will it take them to pass each other?

2.8

2.8

13.79%

3

3

62.07%

1.5

1.5

17.24%

4

4

6.9%

Relative speed of ants = 3 + 5 mmps = 8mmps
Effective distance = 1.1 + 1.3 = 2.4 cm = 22 mm
Time = distance/speed = 24/8 = 3

10. Two coworkers in a multinational corporation have a combined age of 74. After a year, the age difference between them will be 9:10. What was the age ratio of their ages nine years ago?

7:6

7:6

7.41%

3:2

3:2

14.81%

4:3

4:3

18.52%

13:15

13:15

59.26%

Let’s take X & Y as their ages.
Sum of their ages
x + y = 74
After 1 year, their ratio is 9/10
X +1 /Y+1 =9/10
10x + 10 =9y + 9
10x = 9y + 9 -10
10x = 9y-1
X = 9y -1 /10
Now putting value of x in equation 1
(9y -1 /10) +y -74
9y -1 +10y = 740
19y =741
y= 741 /19 =39
And x = 74 -y
= 74 - 39 =35
Ratio of ages nine years ago = 35-9 /39-9
=26/30
=13:15

11. When M works alone, it takes 75 days longer than N to accomplish a project, but when they work together, it only takes 20 days. If a particular sum has been set aside as payment for this work, how should it be divided between M and N?

1:4

1:4

46.15%

1:5

1:5

23.08%

2:5

2:5

19.23%

2:3

2:3

11.54%

Let us assume that N alone can complete the project alone in x days.
Then, M will complete the work alone in (x + 75) days.
Given that, both take 12 days to complete the project .
So,
→ 1/x + 1/(x + 75) = 1/20
→ (x + 75 + x)/x(x + 75) = 1/20

→ (2x + 75) * 20 = x² + 75x
→ 40x + 1500 = x² + 75x
→ x² + 75x - 40x - 1500 = 0
→ x² + 35x - 1500 = 0
→ x² + 60x - 25x - 1500 = 0
→ x(x + 60) - 25(x + 60) = 0
→ (x + 60)(x - 25) = 0
Putting both equal to zero , we get,
→ x = (-60) or 25.
since days cant be negative.
Therefore,
→ Time taken by Y alone = 25 days.
→ Time taken by X alone = 25 + 75 = 100 days.
Now, we know that,
Efficiency is indirectly proportional to time..
Wages are distributed among workers in ratio of their efficiency of work per day.
So,
→ Time taken by M : Time taken by N = 100 : 25 = 4 : 1 .
Then,
→ Efficiency of M : Efficiency of N = 1 : 4.
Hence,
→ Wages of M : Wages of N = 1 : 4

12. Divyanshi combined the M and N alloys. She combined 80 kilogrammes of mixture M with 50 kilogrammes of mixture N. What is the amount of lead in the new alloy if mixture M has a 3:7 lead-to-copper ratio and alloy N has a 7:3 lead-to-copper ratio?

24 kg

24 kg

23.81%

35 kg

35 kg

14.29%

56 kg

56 kg

14.29%

59 kg

59 kg

47.62%

Alloy M :-

Lead : copper = 3 : 7
Lead = 80 * (3/10) = 24 kg .
Copper = 80 * (7/10) = 56 kg.
Alloy N :-

Lead : copper = 7 : 3
Lead = 50 * (7/10) = 35 kg .
Copper = 50 * (3/10) = 15 kg.
Therefore,

→ Lead in new alloy = Lead in alloy M + Lead in alloy N = 24 + 35 = 59 kg.

13. A seller on Flipkart's Big Billion Sale marks up a dress by 40% over its cost price. What will be the loss or profit percent on selling it if he gives a 25% discount on the advertised price?

Loss , 10%

Loss , 10%

15.38%

Loss, 2%

Loss, 2%

3.85%

Profit, 1.5%

Profit, 1.5%

26.92%

Profit, 5%

Profit, 5%

53.85%

Let the cost price be 100
Then marked price = 100 + ( 100* 40/100)
= 140
Discount percent = 25%
Price after discount = 105
Profit percentage = 5/100 * 100
= 5% Profit

14. With the selling price of a book as the benchmark, a bookseller computed his profit at 12%. What would his actual profit percentage have been if the selling price had been 20% higher?

21.23

21.23

17.39%

20.5

20.5

21.74%

22.5

22.5

43.48%

24

24

17.39%

Assume the cost price was 100
Then after 12% profit the selling price would have been =100+12 = 112
Now,
Profit at 20% would be= 112 x 20 /100 =22.5

15. Ranita put a particular amount of money into a LIC, and the interest compounded quarterly grew to Rs 140000 and Rs 157304 in 15 and 21 months, respectively. What is the annual percentage rate of interest?

24

24

21.05%

6

6

26.32%

15

15

36.84%

12

12

15.79%

Let r is the rate of interest per annum and p be the principal
157304=P(1+r/400)^{7} (A)
140000=P(1+r/400)^{5} (B)
Dividing (A) by(B)
\frac{157304}{14000}=(1+r/400)^{2}
1.1236=(1+r/400)^{2}
(1+r/400)=✓1.1236
1+r/400=1.06
r/400=1.06–1=0.06
r=24% per annum

16. Madhulika put a specific amount of money into ICICI Mutual Funds. What will the compound interest on 12000 at the same rate after two years if there is a 45 percent increase in the amount in three years at simple interest?

15600

15600

11.76%

1800

1800

23.53%

3600

3600

23.53%

3870

3870

41.18%

Let us assume that, Principal is Rs. x and Let rate of interest is R% annually .
Than from given data we have :-
Principal = Rs.x
Rate = R% per annum.
Time = 3 years.
SI = 45% of x .
we know that ,
SI = (P * R * T) / 100
Putting all values we get,
→ 45% of x = (x * R * 3)/100
→ (45 * x)/100 = (x * R * 3)/100
→ 45 = 3R
dividing both sides by 3 ,
→ R = 15% .

Now, we have to find compound interest of Rs.12,000 at the same rate as 15% after 2 years.
we know that,
CI = P[{1 + (R/100)}^{time} - 1]
Putting values now we get,

→ CI = 12000[{ 1 + (15/100)}² - 1]
→ CI = 12000[{1 + (3/20)}² - 1]
→ CI = 12000[(23/20)² - 1]
→ CI = 12000[(529/400) - 1]
→ CI = 12000[(529 - 400)/400]
→ CI = 12000 * (129/400)
→ CI = 30 * 129
→ CI = Rs.3,870

17. Aishwarya rode the bus to her friend's house and then cycled back, discovering that it took 20 minutes longer than taking the bus, despite the fact that the distance traveled in both cases was the same, i.e. 5 km. If the average speed of a bus and a cycle is 3:1, what is the average speed of a cycle?

8 km/h

8 km/h

14.29%

10 km/h

10 km/h

42.86%

6 km/h

6 km/h

19.05%

15 km/h

15 km/h

23.81%

Distance = Speed x Time
Speed of cycle = A km//hr
Speed of Bus = 3A km//hr
Distance = 5 km
Time taken by Cycle= 5/A hr
Time taken by Bus = 5/3A hr
5/A - 5/3A = 20/60
=> 5/A - 5/3A = 1/3
=> 15 - 5 = A
=> 10 = A

The average speed of the cycle is 10 km/hr

18. Aashay and his friends are going on a trip. His location is 50 kilometers away. He intends to finish this in two hours. He could travel 20 kilometers in an hour but had to stop for 10 minutes to refuel. By what factor should he raise his speed in comparison to that of the first hour in order to complete the journey on time?

1.5

1.5

5.56%

1.8

1.8

66.67%

1.2

1.2

11.11%

2.4

2.4

16.67%

Total time Aashay has = 2 hrs
Total distance he has to covered = 50 km
in first 1 hr he completed 20 km
So S1= 20 km/hr
So remaining distance = 50-20 = 30 Km
And remaining time = 1 hr
Now he spent 10 minutes for refueling
So total remaining time he has = 60-10 = 50 minutes
He has to complete 30 km in this time
Speed = 30/(10/60) km/hr
Speed = 36 km/hr
For making a factor of initial speed as 20 km/hr
Multiply 36 with 20/20
Speed = 36x(20/20)
=(1.8x S1 )km/hr
= 1.8 km/hr

19. Ramesh wants to combine two different kinds of fruit juice. In what proportion should he combine two varieties of fruit juice, each priced at Rs. 240 and Rs. 280, so that the total cost is Rs 254 per liter?

3:2

3:2

4.76%

7:6

7:6

14.29%

13:7

13:7

66.67%

4:3

4:3

14.29%

Let X & Y denote the amounts (in liter) of two varieties of fruit juice priced at Rs. 240 and Rs. 280 per kg respectively should be mixed so that the cost of the combination is Rs. 254 per kg.
Hence from above data we get following relation,
240*X + 280*Y = (X + Y)*254
or (254 - 240)*X = (280 - 254)*Y
or 14*X = 26*Y
or X/Y = 26/14 = 13/7
or X : Y = 13 : 7
Therefore it is evident from above that the required ratio of two varieties of fruit juice is priced at Rs. 240 and Rs. 280 per kg respectively should be mixed so that
the cost of the combination is Rs. 254 per litre = 13: 7

20. X makes a 15% profit on a set of sketch pens he sells to K. K then sells it to M, generating a 20 percent profit. If M made a payment of Rs.690, What was the original cost of the sketch pen set?

700

700

4.55%

690

690

13.64%

575

575

18.18%

500

500

63.64%

Let the original cost of the sketch pen set for X be “x”

X sold the item at 15% profit.
The cost for K is 1.15x, which is sold at a profit of 20%
The cost for M is 1.2(1.15x)
M paid 690
So, 690 = 1.2 \times 1.15x
Solving the equation ----
x= 500

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Preparation Topics​

Detailed Analysis of TCS NQT Numerical Ability For Freshers​

Go through the table given below. You will get all the information about each and every topic that comes under TCS NQT Numerical Ability round.
 Topics No. of questions in test Difficulty Importance
Ratio and Proportion 0 – 2 High High
Clocks and Calendars 0 – 1 Medium Low
Profit and Loss 0 – 2 Medium High
Number System 0 – 2 Low Medium
LCM and HCF 0 – 1 Low Low
Percentage 0 – 2 Medium Low
Mensuration 0 – 3 High High
SI and CI 0 – 2 High High
Work and Time 0 – 2 High High
Speed, Time and Distance 0 – 2 High High
Data Interpretation 0 – 2 High High
Number Series 0 – 2 Medium Medium
Equations 0 – 1 Medium Medium
Elimentary Statistics 0 – 4 Medium High
Averages 0 – 1 Medium Low
Mixtures and Allegations 0 – 1 High Low
Simplifications 0 – 4 Medium High
Surds and Indices 0 – 1 Medium Low

TCS NQT Numerical Ability For Freshers​

FAQ

Is there any negative marking for this section?

No, there will be no negative marking for TCS NQT Numerical Ability.  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 TCS NQT Numerical Ability is high. The cutoff will be between 75 and 80 percent, or 20 and 22 questions.
Our top video course will help you prepare for this. 

What would the TCS NQT NUMERICAL ABILITY DIFFICULTY LEVEL BE IN 2022? ​

The difficulty level for TCS NQT Numerical Ability ranges between Moderate to high. You need to prepare well for this round.

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