# HCL Quantitative Aptitude Questions And Answers

HCL Quantitative Aptitude MCQs : This section focuses on "Quantitative Aptitude" for HCL placement drives. These Quantitative Aptitude MCQs are asked in previous HCL placements/recruitment exams and will help you to prepare for upcoming HCL drives.

1. The ratio of two numbers is 3:4 and their H.C.F is 4. Their L.C.M is:

A. 12

B. 16

C. 24

D. 48

View Answer

Ans : D

Explanation: Let the numbers be 3x and 4x. Then, their H.C.F= x. So, x=4

So, the numbers are 12 and 16.

L.C.M of 12 and 16 is 48

2. Two numbers when divided by a certain divisor leave remainders of 431 and 379 respectively. When the sum of these two numbers is divided by the same divisor, the remainder is 211. What is the divisor?

A. 499

B. 549

C. 599

D. 600

View Answer

Ans : C

Explanation: Let A be the divisor,

X and Y be the two numbers.

X = Ax + 431

Y = Ay + 379

X + Y = Az + 211 = A(x + y) + 810
or A (z - x - y) = 810 - 211 = 599

3. What decimal of 10 hours is a minute?

A. 0.025

B. 0.256

C. 0.0027

D. 0.00126

View Answer

Ans : C

Explanation: Decimal of 10 hours in a minute

= 10 / (60 x 60)

= 0.0027

4. A basket contains 3 blue, 5 black and 3 red balls. If 3 balls are drawn at random what is the probability that all are black?

A. (2/11)

B. (1/11)

C. (7/11)

D. (8/11)

View Answer

Ans : A

Explanation: Ways of selecting 3 black balls out of 5 - 5C3

Total ways of selecting 3 balls - 11C3The required probability =

5C3 / 11C3 = 10/165 = 2/33

5. If log y 1369y=3 then what is the value of y?

A. 35

B. 37

C. 33

D. 39

View Answer

Ans : B

Explanation: log y raised to the power of 3;

y^3 = 1369y

y^2 = 1369

y =37.

6. Find the lowest common multiple of 2/3, 3/5, 4/7, 9/13

A. 36

B. 30

C. 60

D. 54

View Answer

Ans : A

Explanation: Required L.C.M= (L.C.M of 2, 3, 4, 9)/(H.C.F of 3, 5, 7, 13)= 36/1=36

7. A and B together have Rs.1,210. If (4/15)th of A amount is equal to (2/5)th of B amount, then how much amount does B have?

A. 489

B. 484

C. 490

D. 450

View Answer

Ans : B

Explanation: Let B has x amount

A = 1210 - x

(4/15)(1210 - x) = (2/5)x

(2/3)(1210 - x) = x

2420 - 2x = 3x

5x = 2420 or x = 484

8. A can do a work in 10 days and B in 15 days. If they work on it together for 3 days, then the work that is left is :

A. 0.1

B. 0.2

C. 0.4

D. 0.5

View Answer

Ans : D

Explanation: Let the total work to be done is, say, 30 units.

A does the work in 10 days,

So A’s 1-day work = (30 / 10) = 3 units

B does the work in 15 days,

So B’s 1-day work = (30 / 15) = 2 units

Therefore, A’s and B’s together 1-day work = (3 + 2) = 5 units

In 3 days,

work done = 5 * 3 = 15 units

amount of work left = 30 – 15 = 15 units

Therefore the % of work left after 3 days = (15 / 30) * 100% = 50%

9. 15.002 × ? × 25.0210 = 7113.018

A. 9

B. 30

C. 19

D. 50

View Answer

Ans : C

Explanation: 15.002 × ? × 25.0210 = 7113.018

? = (7113.018/(15.002*25.0210))

? = 7100/15*25

? = 18.93

10. A man driving his bike at 24 Kmph reaches his office 5 minutes late. Had he driven 25 faster on an average he would have reached 4 minutes earlier than the scheduled time. How far is his office?

A. 24 Km

B. 72 Km

C. 18 Km

D. 30 Km

View Answer

Ans : C

Explanation: Let x km be the distance between his house and office.

While travelling at 24 kmph,he would take x/24hours.

While travelling at 25 % faster speed,

i.e. 24+[25%of24] = 24×(1/4)=30kmph,he would take x/30 hours.

Now as per the problem ,time difference=5 min late + 4min early =9 min

⇒[x/24] − [x/30] = 9 min

⇒ x= 18 km

Discussion