This is an HCL model paper for Quantitative Aptitude. This placement paper will cover aptitude that is asked in HCL placements and also strictly follows the pattern of questions asked in HCL papers. It is recommended to solve each one of the following questions to increase your chances of clearing the HCL placement.

**Find the greatest number that will divide 355, 54 and 103 so as to leave the same remainder in each case**.- 4
- 7
- 9
- 13

**Answer:****7****Explanation:**Required number = H.C.F. of |a -b|, |b – c| and |c – a|

= H.C.F. of |355 – 54|, |54 – 103| and |103 – 355|

= 301, 49, 252

= 7**Six bells commence tolling together and toll at intervals of 3, 6, 9, 12, 15 and 18 seconds respectively. In 60 minutes, how many times do they toll together ?**- 10
- 20
- 21
- 25

**Answer:****21****Explanation:**L.C.M. of 3, 6, 9, 12, 15 and 18 is 180.

So, the bells will toll together after every 180 seconds(3 minutes).

In 60 minutes, they will toll together (60/3)+1 = 21 times.**The smallest 5 digit number exactly divisible by 11 is:**- 11121
- 11011
- 10010
- 11000

**Answer:****10010****Explanation:**The smallest 5-digit number 10000.

10000 when divided by 11, leaves a remainder of 1

Hence add (11 – 1) = 10 to 10000

Therefore, 10010 is the smallest 5 digit number exactly divisible by 11- 474
- 534
- 500
- 368

**Answer:****474****Explanation:**Therefore the given expression = (121 + 353) = 474

**What decimal of 10 hours is a minute?**- 0.025
- 0.256
- 0.0027
- 0.00126

**Answer:****0.0027****Explanation:**Decimal of 10 hours in a minute

= 10 / (60 x 60)

= 0.0027**‘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 :**- 10%
- 20%
- 40%
- 50%

**Answer:****50%****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 unitsB does the work in 15 days,

So B’s 1-day work = (30 / 15) = 2 unitsTherefore, 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 unitsTherefore the % of work left after 3 days = (15 / 30) * 100% = 50%

**A pump can fill a tank with water in 1 hour. Because of a leak, it took 1.5 hours to fill the tank. The leak can drain all the water of the tank in:**- 2 hours
- 2.5 hours
- 3 hours
- 3.5 hours

**Answer:****3 hours****Explanation:**Pump fills the tank in 1 hour

Time taken by Pump to fill due to leak = 1.5 hour

Therefore, in 1 hour, the amount of tank that the Pump can fill at this rate = 1 / (1.5) = 2/3Amount of water drained by the leak in 1 hour = (1 – (2/3)) = 1/3

Therefore, the tank will be completely drained by the leak in (1 / (1/3)) = 3 hours

**2 pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. Both pipes are opened. The tank will be filled in just 15 minutes, if the B is turned off after:**- 5 min
- 6.5 min
- 7 min
- 7.5 min

**Answer:****7.5 min****Explanation:**Let the total work to be done is, say, 60 units.

A fills the tank in 20 minutes,

So A’s 1-minute work = (60 / 20) = 3 unitsB fills the tank in 30 minutes,

So B’s 1-minute work = (60 / 30) = 2 unitsTherefore, A’s and B’s together 1-minute work = (3 + 2) = 5 units

Let the time when A and B both are opened be x minutes

and Since the total time taken to fill the tank is 15 minutesTherefore, an expression can be formed as

5x + 3(15 – x) = 15

=> x = 7.5Therefore, the B is turned off after 7.5 minutes

**In an IPL match, the current run rate of CSK is 4.5 in 6 overs. What should be the required run rate of CSK inorder to achieve the target of 153 against KKR?**- 7
- 8
- 8.5
- 9

**Answer:****9****Explanation:**Current run rate = 4.5 in 6 overs

Runs already made = 4.5 * 6 = 27Target = 153

Runs still required = 153 – 27 = 126

Overs left = 14Therefore required run rate = 126 / 14 = 9

**The average of 10 numbers is 0. Of them, how many can be samller than zero, at most?**- 0
- 1
- 9
- 10

**Answer:****9****Explanation:**Let the 9 numbers be smaller than zero and let their sum be ‘s’

Now, in order to get the average 0, the 10th number can be ‘-s’

Therefore, average = (s + (-s))/10 = 0/10 = 0

**Which is not the prime number?**.- 43
- 57
- 73
- 101

**Answer:****57****Explanation:**A positive natural number is called prime number if nothing divides it except the number itself and 1.

57 is not a prime number as it is divisible by 3 and 19 also, apart from 1 and 57.**If the average of four consecutive odd numbers is 12, find the smallest of these numbers?**- 5
- 7
- 9
- 11

**Answer:****9****Explanation:**Let the numbers be x, x+2, x+4 and x+6

Then (x + x + 2 + x + 4 + x + 6)/4 = 12

∴ 4x + 12 = 48

∴ x = 9**Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are:**- 6, 27
- 8, 36
- 38, 171
- 20, 90

**Answer:****38, 171****Explanation:**Let the numbers be 2X and 9X

Then their H.C.F. is X, so X = 19

∴ Numbers are (2×19 and 9×19) i.e. 38 and 171**HCF of two numbers is 11 and their LCM is 385. If the numbers do not differ by more than 50, what is the sum of the two numbers ?**- 132
- 35
- 12
- 36

**Answer:****132****Explanation:**Product of numbers = LCM x HCF

=> 4235 = 11 x 385Let the numbers be of the form 11m and 11n,

such that ‘m’ and ‘n’ are co-primes.

=> 11m x 11n = 4235

=> m x n = 35

=> (m, n) can be either of (1, 35), (35, 1), (5, 7), (7, 5).

=> The numbers can be (11, 385), (385, 11), (55, 77), (77, 55).But it is given that the numbers cannot differ by more than 50.

Hence, the numbers are 55 and 77.

Therefore, sum of the two numbers = 55 + 77 = 132**A person employed a group of 20 men for a construction job. These 20 men working 8 hours a day can complete the job in 28 days. The work started on time but after 18 days, it was observed that two-thirds of the work was still pending. To avoid penalty and complete the work on time, the employer had to employ more men and also increase the working hours to 9 hours a day. Find the additional number of men employed if the efficiency of all men is the same.**- 40
- 44
- 64
- 80

**Answer:****44****Explanation:**Let the total work be 3 units and additional men employed after 18 days be ‘x’.

=> Work done in first 18 days by 20 men working 8 hours a day = (1/3) x 3 = 1 unit

=> Work done in last 10 days by (20 + x) men working 9 hours a day = (2/3) x 3 = 2 unitHere, we need to apply the formula

M1 D1 H1 E1 / W1 = M2 D2 H2 E2 / W2,

where

M1 = 20 men

D1 = 18 days

H1 = 8 hours/day

W1 = 1 unit

E1 = E2 = Efficiency of each man

M2 = (20 + x) men

D2 = 10 days

H2 = 9 hours/day

W2 = 2 unitSo, we have

20 x 18 x 8 / 1 = (20 + x) x 10 x 9 / 2

=> x + 20 = 64

=> x = 44Therefore, number of additional men employed = 44