HCL Placement Paper | Quantitative Aptitude Set – 1

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.

  1. Find the greatest number that will divide 355, 54 and 103 so as to leave the same remainder in each case.
    1. 4
    2. 7
    3. 9
    4. 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

  2. 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 ?
    1. 10
    2. 20
    3. 21
    4. 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.

  3. The smallest 5 digit number exactly divisible by 11 is:
    1. 11121
    2. 11011
    3. 10010
    4. 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



    1. 474
    2. 534
    3. 500
    4. 368

    Answer:

    
    474
    

    Explanation:

    As

    Therefore the given expression = (121 + 353) = 474

  4. What decimal of 10 hours is a minute?
    1. 0.025
    2. 0.256
    3. 0.0027
    4. 0.00126

    Answer:

    
    0.0027
    

    Explanation:

    Decimal of 10 hours in a minute
    = 10 / (60 x 60)
    = 0.0027

  5. ‘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 :
    1. 10%
    2. 20%
    3. 40%
    4. 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 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%

  6. 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:
    1. 2 hours
    2. 2.5 hours
    3. 3 hours
    4. 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) = 2/3

    Amount 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

  7. 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:
    1. 5 min
    2. 6.5 min
    3. 7 min
    4. 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 units

    B fills the tank in 30 minutes,
    So B’s 1-minute work = (60 / 30) = 2 units

    Therefore, 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 minutes

    Therefore, an expression can be formed as
    5x + 3(15 – x) = 15
    => x = 7.5

    Therefore, the B is turned off after 7.5 minutes



  8. 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?
    1. 7
    2. 8
    3. 8.5
    4. 9

    Answer:

    
    9
    

    Explanation:

    Current run rate = 4.5 in 6 overs
    Runs already made = 4.5 * 6 = 27

    Target = 153
    Runs still required = 153 – 27 = 126
    Overs left = 14

    Therefore required run rate = 126 / 14 = 9

  9. The average of 10 numbers is 0. Of them, how many can be samller than zero, at most?
    1. 0
    2. 1
    3. 9
    4. 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

  10. Which is not the prime number?.
    1. 43
    2. 57
    3. 73
    4. 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.

  11. If the average of four consecutive odd numbers is 12, find the smallest of these numbers?
    1. 5
    2. 7
    3. 9
    4. 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

  12. Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are:
    1. 6, 27
    2. 8, 36
    3. 38, 171
    4. 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

  13. 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 ?
    1. 132
    2. 35
    3. 12
    4. 36

    Answer:



    
    132
    

    Explanation:

    Product of numbers = LCM x HCF
    => 4235 = 11 x 385

    Let 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

  14. 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.
    1. 40
    2. 44
    3. 64
    4. 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 unit

    Here, 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 unit

    So, we have
    20 x 18 x 8 / 1 = (20 + x) x 10 x 9 / 2
    => x + 20 = 64
    => x = 44

    Therefore, number of additional men employed = 44



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