HCL 安置纸 |定量能力集 – 1

这是 Quantitative Aptitude 的 HCL 模型论文。这份实习论文将涵盖 HCL 实习中要求的能力，并严格遵循 HCL 论文中提出的问题模式。建议解决以下每个问题，以增加清除 HCL 安置的机会。

1. 找出能整除 355、54 和 103 的最大数，以便在每种情况下留下相同的余数。
   1. 4
   2. 7
   3. 9
   4. 13

   回答：

   7
   

   解释：

   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. 六个钟开始一起收费，并分别以 3、6、9、12、15 和 18 秒的间隔收费。在 60 分钟内，他们一起收费多少次？
   1. 10
   2. 20
   3. 21
   4. 25

   回答：

   21
   

   解释：

   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. 能被 11 整除的最小 5 位数字是：
   1. 11121
   2. 11011
   3. 10010
   4. 11000

   回答：

   10010
   

   解释：

   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

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

   回答：

   474
   

   解释：

   As
   

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

5. 10 小时的小数点是几分？
   1. 0.025
   2. 0.256
   3. 0.0027
   4. 0.00126

   回答：

   0.0027
   

   解释：

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

6. “A”可以在 10 天内完成一项工作，“B”可以在 15 天内完成。如果他们一起工作3天，那么剩下的工作是：
   1. 10%
   2. 20%
   3. 40%
   4. 50%

   回答：

   50%
   

   解释：

   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%

7. 一个泵可以在 1 小时内将一个水箱注满水。由于泄漏，加满油箱需要 1.5 小时。泄漏可以排出水箱中的所有水：
   1. 2小时
   2. 2.5小时
   3. 3小时
   4. 3.5小时

   回答：

   3 hours
   

   解释：

   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

8. 2 根管子 A 和 B 可以分别在 20 分钟和 30 分钟内装满一个罐。两个管道都打开了。如果 B 在以下情况下关闭，则水箱将在 15 分钟内加满：
   1. 5分钟
   2. 6.5 分钟
   3. 7 分钟
   4. 7.5 分钟

   回答：

   7.5 min
   

   解释：

   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

9. 在 IPL 比赛中，CSK 的当前运行率为 4.5 分 6 分。为了达到 153 对 KKR 的目标，CSK 所需的运行率应该是多少？
   1. 7
   2. 8
   3. 8.5
   4. 9

   回答：

   9
   

   解释：

   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

10. 10 个数字的平均值为 0。其中，最多可以有多少个比零更小？
    1. 0
    2. 1
    3. 9
    4. 10

    回答：

    9
    

    解释：

    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

11. 哪个不是质数？ .
    1. 43
    2. 57
    3. 73
    4. 101

    回答：

    57
    

    解释：

    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.

12. 如果连续四个奇数的平均值是 12，请找出这些数字中最小的一个？
    1. 5
    2. 7
    3. 9
    4. 11

    回答：

    9
    

    解释：

    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

13. 两个数字的比例为 2:9。如果他们的 HCF 为 19，则数字为：
    1. 6、27
    2. 8、36
    3. 38, 171
    4. 20, 90

    回答：

    38, 171
    

    解释：

    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

14. 两个数字的 HCF 是 11，它们的 LCM 是 385。如果数字相差不超过 50，那么这两个数字的总和是多少？
    1. 132
    2. 35
    3. 12
    4. 36

    回答：

    132
    

    解释：

    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

15. 一个人雇用了 20 名男子从事建筑工作。这 20 名每天工作 8 小时的人可以在 28 天内完成这项工作。工作按时开始，但在 18 天后，据观察，三分之二的工作仍未完成。为了避免受到惩罚并按时完成工作，雇主不得不雇用更多的男人，并将工作时间增加到每天 9 小时。如果所有男性的效率都相同，求出额外雇佣的男性人数。
    1. 40
    2. 44
    3. 64
    4. 80

    回答：

    44
    

    解释：

    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