Open In App
Related Articles

HCL Placement Paper | Quantitative Aptitude Set – 1

Improve
Improve
Improve
Like Article
Like
Save Article
Save
Report issue
Report
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 smaller 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


Last Updated : 10 Mar, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads