# TCS NQT Foundation Aptitude

## TCS NQT Foundation Aptitude Questions with Solutions

The TCS NQT Advanced Aptitude 2023 component’s sorts of questions, degree of difficulty, and significance are all covered in this article.

You have 75 minutes(shared) to complete the 20 questions in the TCS NQT Foundation Aptitude test.

**Important Note: **

- There will be no negative marking.
- Full time Graduates from B.E / B.TECH / M.E / M.TECH of 2023 batch

**TCS NQT Foundation Aptitude Test**

Aptitude Test for |
TCS NQT Foundation |
---|---|

Number of Questions | 20 questions |

Time Limit | 75 Mins (shared) |

Mode Conducted | Online/Written |

**TCS NQT Foundation Aptitude Syllabus**

The TCS ITP Aptitude Exam has a medium to high level of difficulty. To practice the sample questions, visit our sample quiz section.

**TCS NQT FOUNDATION APTITUDE QUESTIONS AND ANSWERS:-**

Topics are –

- Number System, LCM & HCF
- Divisibility
- Numbers & Decimal Fractions
- Geometry
- Area, Shapes & Perimeter
- Percentages
- Allegations and Mixtures
- Ratios, Proportion
- Work and Time
- Speed Time and Distance
- Profit and Loss
- Averages
- Equations
- Ages
- Series and Progressions
- Mean, Median, Mode, Variance, and Standard Deviation
- Pie Charts
- Tabular DI
- Graphical DI
- Simplifications and Calculations (Rational and Irrational Number)
- Probability
- Clocks & Calendar

**Important: **We recommend that you practice the example questions from our TCS NQT Foundation Aptitude Dashboard in order to achieve a good score. Because the TCS NQT Foundation Aptitude will be difficult.

**Practice Questions: **

**Question** **1. ** The mean of a set of data is 5. What will be the mean if ten is subtracted from each data?

a)5

b)10

c)-5

d)-10

**Answer:** -5

**Explanation:** let the number of data be n

so sum of data =mean\times number of data mean ×number of data

= 5n

now 10 is subtracted to each data

so now sum becomes = 5n-10n =-5n

mean = sum of data /no. of data

= n−5n

= -5

so new mean becomes -5

**Question** 2. In a certain code language if EXTREME = 97 and SEVEN = 70 then CLOCK = ?

a)25

b)84

c)49

d)90

**Answer:** 49

**Explanation:**

In this the sum of the place values is added to the number of letters in the word and given as its value.

Sum of place values of EXTREME =5 + 24 + 20 + 18 + 5 + 13 + 5 = 90 and 90 + 7 = 97

Sum of place values of SEVEN = 19 + 5 + 22 + 5 + 14 = 65 and 65 + 5 = 70

Similarly, Sum of place values of CLOCK = 3 + 12 + 15 + 3 + 11 = 44 and 44 + 5 = 49

**Question 3.** With what value should the highest quantity in the data : 65, 52, 14, 26, 18, 35, 32, 38 be replaced so that the mean and medium become equal?

a)40

b)50

c)53

d)67

**Answer:** 53

**Explanation:** Let’s sort the given data – 14, 18, 26, 32, 35, 38, 52, 65

mean = 35

median=32+35/2

=33.5

Difference = 35 – 33.5 = 1.5

Value = 1.5 *8 = 12

65-12 = 53

**Question 4.** A pipe can fill a cistern in 16 hours. After 1/4 the cistern is filled, three more similar pipes are opened. What is the total time taken to fill the cistern completely?

a)4 hours

b)9 hours

c)7 hours

d)None of the above

**Answer:** 7 hours

**Explanation:** One pipe can fill the cistern in 16 hours

Amount of work done by a pipe in 1 hour = 1/16

Amount of work done by a pipe in 4 hour = 1/4

Therefore ¼ of the work is done by a pipe alone in : 4 hours ………(1)

Work left after 1/4th of the cistern is filled = ¾

Work done by 4 pipes together in 1 hour = 4/16 = 1/4

Work done by 4 pipes together in 3 hour = ¾

Therefore ¾ of the work is done by 4 pipes in : 3 hours …………(2)

Total time taken to fill the cistern completely = 7 hours

**Question5.** If the mean of the data 5,7,10,14,x,25,32,19 is x, then what is the value of x?

a)14

b)16

c)18

d)20

**Answer:** 16

**Explanation:** Mean= number of observations/sum of observations.

x= 5+7+10+14+x+25+32+19/8

112+x=8x

7x=112

x=16

**Question 6.** FPR, GPT, HPV, IPX What is the next word ?

a) EPY

b) JQZ

c) JPY

d) JPZ

**Answer:** JPZ

**Explanation:** The first letters are in alphabetical order F,G,H,I so the next letter will be J

The second letter is constant P

In a third letter, there is a gap of one in the alphabetical series R, T, V, X, So the next letter will be Z

So, the next word becomes: JPZ

**Question 7**. How many liters of a 90% solution of concentrated acid needs to be mixed with a 75% solution of concentrated acid to get a 30 liter solution of 78% concentrated acid?

a) 6

b) 7

c) 8

d) 9

**Answer:** 6

**Explanation: **The concentration is given which is wrt 100, hence we can take x lt of90% and (30-x)of 75%

x*90 + (30-x)*75 = 30*78

x = 6 ltr

**Question 8.** Rajeev buys goods worth Rs 7000. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @10%. Find the amount he will pay for the goods.

a) 7238

b) 7300

c) 7200

d) 7569

**Answer:** 7238

**Explanation: **Rebate = 6% of Rs 7000 = Rs \frac{6}{100} \times 7000 = Rs 420

Sales tax = 10% of Rs (7000 – 420) = Rs \frac{10}{100} \times 6580 = Rs 658

Final amount = Rs (6580 + 658) = Rs.7238

**Question 9.** A team won 80% of the games it played. It played 5 more games of which it won 3 and lost 2. Its loss percentage changed to 25%. How many games did it play overall?

a) 14

b) 16

c) 20

d) 25

**Answer:** 20

**Explanation:** if game played=x

then lost game=20% of x =x/5

now they played 5 more games in which they lost 2

so, (x+5)*25%=x/5+2

x=15

so total game he played=15+5 = 20

**Question 10.** There are five boxes in a cargo hold. The weight of the first box is 200 kg and the weight of the second box is 20% higher than the weight of the third box, whose weight is 25% higher than the first box’s weight. The fourth box at 350 kg is 30% lighter than the fifth box. Find the difference in the average weight of the four heaviest boxes and the four lightest boxes.

a) 45

b) 55

c) 65

d) 75

Answer: 65

Explanation:

weight of 1st box=200 kg

weight of 3rd box = 200 + 25% of 200 = 250 kg

weight of 2nd box = 250 + 20% of 250 = 300 kg

weight of 4th box = 350 kg

weight of 5th box = 350 + 30% of 350 =500 kg

avg weight of four heaviest boxes= 1400/4=350 kg

and avg weight of four lightest box=1100/4=275 kg

diff=350 – 275= 75 kg