**HCL Aptitude Questions & Answers For Freshers:**Candidates looking for HCL Sample Aptitude Questions with Solutions will be pleased to hear the news. HCL Aptitude Questions and Answers with Explanation can be found in this article. Any placement exam includes an aptitude section.

As a result, students can practice HCL Aptitude Questions and Solutions to get a sense of the level of complexity of questions that might be given in the online exam. You should be able to breeze through the exam with the help of these questions. Please keep in mind that these are HCL practice questions, not the real ones. No one can provide you with exact questions from the HCL online test. Each exam's questions may differ.

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## HCL Aptitude Questions & Answers For Freshers Complete Overview

- Number of Questions: 15 Questions
- Time Limit: 15 mins

Topics: Profit and Loss, Time and Work. Averages. Percentages, Time and Distance, Time and Distance, Quadratic Equations. Races and Games. Areas. Compound Interest. Simple Interest. Simplification.

The questions could also be about Partnership, Boats & Streams, Volume & Surface Area, Number Problems, Odd Man Out, Probability, Permutation, and other related themes. By practicing HCL Aptitude Questions and Answers with Explanation, applicants can cover all of the topics included in the Aptitude section.

The questions could also be about Partnership, Boats & Streams, Volume & Surface Area, Number Problems, Odd Man Out, Probability, Permutation, and other related themes. By practicing HCL Aptitude Questions and Answers with Explanation, applicants can cover all of the topics included in the Aptitude section.

## HCL Aptitude Questions & Answers For Freshers By Sections

**1. The marked price of an article is 20% above the cost price. When the selling price of an article is increased by 30% the profit doubles. If the market price of an article is 480, the original selling price is.**

A. 531.15

B. 537.14

C. 571.4

D. 582.12

Answer – C. 571.4

Explanation:

Given MP = 120/100*CP. So, CP = 400.

SP -400 = P (Profit)

(130/100)*SP – 400 = 2P

Solving both equation we get, SP = 4000/7 = 571.4

**2. The average expenditure of Sharma for January to June is Rs. 4200 and he spent Rs. 1200 in January and Rs.1500 in July. The average expenditure for the months of February to July is:**

A. 4250

B. 4500

C. 3500

D. 2750

E. 3250

Answer – A. 4250

Explanation:

Total Expenditure(Jan – June) = 4200 * 6 = 25200

Total Expenditure(Feb – June) = 25200 – 1200 = 24000

Total Expenditure(Feb – July) = 24000 + 1500 = 25500/6 = 4250.

Therefore, Average expenditure for months of February to July is 4250

**3. The average presence of students of a class in a College on Monday, Tuesday and Wednesday are 32 and on Wednesday, Thursday, Friday and Saturday are 30. if the average number of students on all the six days is 26 then the number of students who attended the class on Wednesday is?**

A. 50

B. 80

C. 40

D. 70

E. 60

Answer – E. 60

Explanation:

32 * 3 + 30 * 4 – 26 * 6 = 96 + 120 – 156 = 60

**4. The average weight of all the 11 players of CSK is 50 kg. If the average of the first six lightest weight players of CSK is 49 kg and that of the six heaviest players of CSK is 52 kg. The average weight of the player which lies in the sixth position in the list of players when all the 11 players of CSK are arranged in the order of increasing or decreasing weights.**

A. 54 kg

B. 53 kg

C. 56 kg

D. 52 kg

E. 50 kg

Answer – C. 56 kg

Explanation:

Average of First six players = 49 * 6 = 294

Average of Last six players = 52 * 6 = 312; Average of all players = 50 * 11 = 550

Average weight of sixth player = 294 + 312 – 550 = 56.

**5. If m and n are two whole numbers and if m^n= 49. Find n^m, given that n ≠ 1**

A. 94

B. 561

C. 128

D. 118

E. None of these

Answer – C. 128

Explanation:

49 = 7^2 = m^n

n^m = 2^7 = 128

**6. The greatest possible length which can be used to measure exactly the lengths 1m 92cm,3m 84cm ,23m 4cm**

A. 32

B. 36

C. 34

D. 23

E. None of these

Answer – A. 32

Explanation:

192 = 4^2×2^2×3

384 = 4^2×2^2×6

2304 = 4^2×2×6^2

HCF = 4^2×2 = 16×2 = 32

**7. HCF of 4/3, 8/6, 36/63 and 20/42**

A. 4/126

B. 4/8

C. 4/36

D. 4/42

E. None of these

Answer – A. 4/126

Explanation:

HCF of numerator(4,8,36,20) = 4

LCM of denominator(3,6,63,42) = 126

**8. Find the LCM of 3/8, 9/32, 33/48, 18/72**

A. 3/8

B. 8/33

C. 198/8

D. 8/3

E. None of these

Answer – C. 198/8

Explanation:

LCM of numerator(3,9,33,18) = 198

HCF of denominator(8,32,48,72) = 8

Therefore LCM = 198/8.

**9. A number of students in 4th and 5th class is in the ratio 6: 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both classes are boys?**

A. 62.5%

B. 52.6%

C. 55.8%

D. 53.5%

E. 54.8%

Answer – E. 54.8%

Explanation:

Total students in both = 6x+11x = 17x

Boys in class 4 = (60/100)*6x = 360x/100

Boys in class 5 = (52/100)*11x = 572x/100

So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x

% of boys = [9.32x/17x] * 100 = 54.8%.

**10. Consider two alloys A and B. 50 kg of alloy A is mixed with 70 kg of alloy B. A contains brass and copper in the ratio 3 : 2, and B contains them in the ratio 4 : 3 respectively. What is the ratio of copper to brass in the mixture?**

A. 7: 5

B. 5: 11

C. 4: 9

D. 5: 7

E. 8: 5

Answer – D. 5: 7

Explanation:

Brass in A = 3/5 * 50 = 30 kg, Brass in B = 4/7 * 70 = 40 kg

Total brass = 30+40 = 70 kg

So copper in mixture is (50+70) – 70 = 50 kg

So copper to brass = 50: 70

**11. The ratio of A and B is in the ratio 5: 8. After 6 years, the ratio of ages of A and B will be in the ratio 17: 26. Find the present age of B.**

A. 65

B. 77

C. 60

D. 72

E. None of these

Answer – D. 72

Explanation:

A/B = 5/8 , A+6/B+6 = 17/26

Solve both, B = 72

Therefore, the present age of B is 72.

**12. A bag contains 25p, 50p and 1Re coins in the ratio of 2: 4: 5 respectively. If the total money in the bag is Rs 75, find the number of 50p coins in the bag.**

A. 40

B. 45

C. 50

D. 25

E. None of these

Answer – A. 40

Explanation:

2x, 4x, 5x

(25/100)*2x + (50/100)*4x + 1*5x = 75

x = 10, so 50 p coins = 4x = 40

**13. What is the difference between the selling price of an article costing 1000 rupees when a discount of 20% is given in the article and when two successive discounts of 10% are given in the article?**

A. 10

B. 20

C. 30

D. 40

Answer – A. 10

Explanation:

(80/100)* 1000 = 800

1000*(90/100)*(90/100) = 810.

Therefore, when two successive discounts of 10% are given in the article is 10.

**14. Priya bought 10 tables at the rate of 600 each. She spends 1600 rupees on transportation and 400 on the packaging. At what price should Priya sell a table to make a profit of 20%.**

A. 860

B. 920

C. 960

D. 1020

Answer – C. 960

Explanation:

Total cost = 600*10 + 1600 + 400 = 8000 (For 10 tables)

CP of one table = 8000/10 = 800.

SP = 800*120/100 = 960

**15. If an article is sold for 270 at a loss of 10% then, to make a profit of 15%, at what price article should be sold.**

A. 315

B. 325

C. 335

D. 345

Answer – D. 345

Explanation:

270 = (90/100)*CP. So Cp = 300.

So, SP = 300*(115/100) = 345

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