TCS NQT Numerical Ability Questions

Introduction:

The TCS National Qualifier Test (TCS NQT) is an Ability Test that evaluates a candidate's skills and competencies.

There are three sections in TCS NQT Aptitude Exam:

1. The Numerical Ability
2. Reasoning Ability 
3. Verbal Ability section.

The TCS Aptitude section is also known as the Cognitive Ability Test.

Click here to know more about the TCS NQT Recruitment Process.

You can also practice a wide range of TCS Interview Questions from our problem list section.

TCS NQT Aptitude Questions

Part A contains the TCS NQT Aptitude Questions in the Numerical Ability segment (Cognitive Skills). There will be a total of 26 questions asked. They must be completed in  40 minutes.

Below are some categories of questions from the TCS NQT Aptitude questions of the Numerical Ability section that were asked in the previous TCS NQT Exam.

Type 1: Ratio and Proportion

Q1. Given 0.006, 1.2, and 6/25, what is the fourth proportional?

1. 48
2. 4.8
3. 36
4. 3.6

Answer: option a

Explanation: 

Here, we use one of the properties of proportion that is:  product of means = product of extremes, so, 

Let x be the fourth proportional:

 0.006: 1.2 :: 6/25 : x

=> x = (1.2 * 6/25)/0.006

=> x = 48

Type 2: Average

Q2. The index numbers of five commodities are 121, 123, 125, 126, and 128, respectively, with weights of 5, 11, 10, 8, and 6. What is the weighted average index number?

1. 123.8
2. 124.6
3. 124.2
4. 125.2

Answer: option b

Explanation: 

The weighted average index number = 121*5 + 123*11 + 125*10 + 126*8 + 128*65

= 124.6

Q3: To get the mean and median to be equal, what value should the highest quantity in data 65, 52, 14, 32, 26, 35, 18, 38 be replaced?

1. 51
2. 53
3. 64
4. 66

Answer: option b

Explanation: 

Sorting the given data in ascending order: 14, 18, 26, 32, 35, 38, 52, 65

Median = (32+35)/2 = 33.5

We replace 65 with ‘x’, also we are given that mean and median must be equal 

Mean = 33.5

=> (14 + 18, + 26 + 32 + 35 + 38 + 52 + x) / 8 = 33.5

Solving for ‘x’ we get x = 53.

Type 3: Time and Distance

Q4. Two ants are moving at average speeds of 2 and 3 mm per second in opposite directions. The former is 1 cm long, whereas the latter is 1.2 cm long. How long will it take for them to cross paths?

1. 0.4
2. 2.8
3. 4.4
4. 1.5

Answer: option c

Explanation: 

Here, we have used the concept of relative speeds to solve this problem. Relative distance for ant1 to cover w.r.t ant2 = 1cm + 1.2cm = 2.2cm

Also, the relative speed of ant1 w.r.t ant2 = 2 + 3 mm per second = 5 mm per second. 

Hence the time taken for the ants to cross paths is 22/5 = 4.4mm per second.

Q5. A man who has to walk 11 km finds that he has covered two-ninth of the remaining distance in 30 minutes. What is his speed in km/h?

1. 4
2. 4.8
3. 4.2
4. 4.5

Answer: option a

Explanation: 

Let the remaining distance = ‘d’ km

Distance covered by him in 30mins (0.5 hr) = 2d/9 km

Now, d + 2d/9 = 11 (total distance given in the question = 11km)

=> d = 9km

Therefore distance covered by him in 30 mins = 2*9/9 = 2 km

We know that speed = distance / time

So, speed = 2km/0.5hr = 4km per hour.

Type 4: Profit and Loss

Q6. A retailer paid Rs P for 25 identical toys and then sold some of them for Rs P. How many did he sell if he estimated his profit at 8% and used the selling price as the base instead of the cost price?

1. 20
2. 21
3. 23
4. 24

Answer: option c

Explanation: 

Let the cost price (CP) of 25 toys be 'P.'

Therefore, CP of each toy = P/25

Let the number of toys sold be 'x.'

CP of the toys sold = x*P/25

Selling price (SP) of the toys sold = P

In this case, profit percentage was calculated with selling price as the base.

So, Profit % = (SP-CP)/SP * 100

8/100 = (P - x*P/25)/ P

Solving for ‘x’ we get x = 23.

Type 5: Simple Interest

Q7: On January 1, 2016, Rs.12500 was invested at a rate of 4% simple interest p.a. On July 1, 2016, how much interest in rupees will have accrued at the end of the day?

1. 240
2. 250
3. 400
4. 500

Answer: option b

Explanation: 

P (principal amount) = Rs. 12500

T (time of for which money was invested) = 6 months = 0.5 yrs

R (rate of interest) = 4%

SI (simple interest earned) = P*T*R/100

= (12500*1/2*4)/100

= 250.

Type 6: Time and Work

Q8. Work is assigned to 6 men and 12 women, and they could complete it in 3 days. It was also observed that they could do seven times as much work as a man and a woman together. In how many days would 14 women have done the work?  

1. 10
2. 6
3. 12
4. 9

Answer: option d

Explanation: 

Let the efficiency of men and women doing the work be ‘m’ and ‘w’

Total work = 3 * (6m + 12w) as it takes 3 days to complete the work.

Also it is given that 6m + 12w = 7(m+w)

=> m/w = 5

Let m = 5 units, this implies that w = 1 unit

Total work = 3(6*5+12*1) = 126 units

Time taken by 14 women to complete the work is 126/4 = 6 days.

Type 7: HCF and LCM

Q9: What is the difference between m and n if the HCF of 180 and 432 is (180m + 432n), where m and n are integers?

1. 3
2. 7
3. 8
4. 9

Answer: option b 

Explanation: 

We find the HCF of 180 and 432 to be 36

180m + 432n = 36 - - - - - - - - - - - (1)

We can write 432 = 2(180) + 72

=> 72 = 432 - 2(180)

Also we can write 180  = 2(72) + 36

=> 180 = 2(432 - 2(180)) + 36

=> 180 = 2(432) - 4(180) + 36

=> 36 = 5(180) - 2(432) - - - - - - - - - - (2)

Comparing (1) and (2)

m = 5, n = -2

m-n = 5-(-2) = 7