## 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:

- The Numerical Ability
- Reasoning Ability
- 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?**

- 48
- 4.8
- 36
- 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?**

- 123.8
- 124.6
- 124.2
- 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?**

- 51
- 53
- 64
- 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?**

- 0.4
- 2.8
- 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?**

- 4
- 4.8
- 4.2
- 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?**

- 20
- 21
- 23
- 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?**

- 240
- 250
- 400
- 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? **** **

- 10
- 6
- 12
- 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?**

- 3
- 7
- 8
- 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