HCL 安置纸 |定量能力集 – 5

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. 两个数的乘积为108，平方和为225。数之差为：
   .
   1. 5
   2. 4
   3. 3
   4. 都不是

   回答：

   57
   

   解释：

   Let the numbers be x and y.
   Then xy = 108 and x² + y² = 225
   (x –y)² = x² + y² – 2xy
   (x –y)² = 225 – 216
   (x –y)² = 9
   Therefore (x –y) = 3

2. 以下哪项的除数最多？
   1. 99
   2. 101
   3. 176
   4. 182

   回答：

   176
   

   解释：

   99 = 1 * 3 * 3 * 11
   101 = 1 * 101
   176 = 1 * 2 * 2 * 2 * 2 * 11
   182 = 1 * 2 * 7 * 13
   Clearly, 176 has the most number of divisors.

3. 两数之比为3:2。如果数字的 LCM 是 60，那么较小的数字是？
   1. 20
   2. 30
   3. 40
   4. 50

   回答：

   20
   

   解释：

   say, 1st number =3x
   2nd number =2x
   LCM of numbers = 6x
   given LCM = 60
   => x6 = 60
   =>x = 10

4. 6 男 10 女受雇修建一条 360 公里长的公路。他们每天工作 6 小时，在 15 天内完成了 150 公里的道路。 15 天后，又有两名男性被雇佣，四名女性被解雇。此外，工作时间也增加到每天 7 小时。如果 2 男 3 女的日常工作能力相等，求完成工作所需的总天数。
   1. 19
   2. 35
   3. 34
   4. 50

   回答：

   34
   

   解释：

   We are given that the daily working power of 2 men and 3 women are equal.
   => 2 Em = 3 Ew
   => Em / Ew = 3/2, where ‘Em’ is the efficiency of 1 man and ‘Ew’ is the efficiency of 1 woman.
   Therefore, the ratio of efficiency of man and woman = 3 : 2.
   If ‘k’ is the constant of proportionality, Em = 3k and Ew = 2k.
   Here, we need to apply the formula
   => (M_i E_i) D₁ H₁ / W₁ = (M_j E_j) D₂ H₂ / W₂, where
   => (M_i E_i) = (6 x 3k) + (10 x 2k)
   => (M_j E_j) = (8 x 3k) + (6 x 2k)
   D₁ = 15 days
   D₂ = Number of days after increasing men and reducing women
   H₁ = 6 hours
   H₂ = 7 hours
   W₁ = 150 km
   W₂ = 210 km

   So, we have
   38k x 15 x 6 / 150 = 36k x D₂ x 7 / 210
   => 38k x 6 = 12k x D₂
   => D₂ = 19 days
   Therefore, total days required to complete the work = 15 + 19 = 34 days

5. 两管A和B分别可以在10小时和30小时内装满一个罐。由于油箱泄漏，加满油箱需要 2.5 小时以上。仅泄漏就需要多长时间才能清空水箱？
   1. 20小时
   2. 25小时
   3. 30小时
   4. 35小时

   回答：

   30 hours
   

   解释：

   Let the capacity of the tank be LCM (10, 30) = 30 units
   => Efficiency of pipe A = 30 / 10 = 3 units / hour
   => Efficiency of pipe B = 30 / 30 = 1 units / hour
   => Combined efficiency of both pipes = 4 units / hour
   Now, total time taken by A and B working together to fill the tank if there was no leak = 30 / 4 = 7.5 hours
   => Actual time taken = 7.5 + 2.5 = 10 hours

   The tank filled by A and B in these 2.5 hours is the extra work done to compensate the wastage by the leak in 10 hours.
   => 2.5 hours work of A and B together = 10 hours work of the leak
   => 2.5 x 4 = 10 x E, where ‘E’ is the efficiency of the leak.
   => E = 1 unit / hour

   Therefore, time taken by the leak alone to empty the full tank = 30 / 1 = 30 hours

6. Max 以 9 公里/小时的平均速度完成他的旅程。他以 6 公里/小时的速度行驶了前 9 公里，并需要 1·5 小时才能完成剩余的距离。找出他完成剩余距离的速度。
   1. 11 公里/小时
   2. 12 公里/小时
   3. 13 公里/小时
   4. 14 公里/小时

   回答：

   12 km/h
   

   解释：

   Let the required speed be x km/h.
   Total time taken to finish his journey = (9/6 + 1·5) = 3 hours.
   Total distance = 9 + 1·5x km.
   Given, average speed = 9 km/h.
   Therefore, (9 + 1·5x)/3 = 9
   => 9 + 1·5x = 27
   => 1·5x = 18
   => x = 12 km/h.

7. 一列火车在 10 秒内穿过一根杆子。如果火车的长度是 100 米，那么火车的速度是多少公里？
   1. 34
   2. 36
   3. 30
   4. 32

   回答：

   36
   

   解释：

   V = 100/10 = 10 m/s = 10*3600/1000 = 36Km/hr

8. 杰克和罗伯特出现在一次考试中。罗伯特比杰克少了9分。杰克的分数是他们分数总和的 56%。计算他们的个人分数。
   1. 22 和 33
   2. 41 和 35
   3. 40 和 35
   4. 42 和 33

   回答：

   42 and 33
   

   解释：

   Let Robert’s score be x. Then, Jack’s score = x+9.
   Now, x+9 = 56% of [(x+9) + x]
   => x+9 = 14/25 × (2x + 9)
   => 25 × (x+9) = 14 × (2x+9)
   => 25x + 225 = 28x + 126
   => 3x = 99 => x = 33.
   Therefore, Robert scored 33 marks and Jack scored 42 marks.

9. 在图书馆里，计算机、物理和数学书籍的比例是5:7:8。如果藏书量分别增加了 40%、50% 和 75%，求出新的比例：
   1. 3:9:5
   2. 7:5:3
   3. 2:3:4
   4. 2:5:4

   回答：

   2:3:4
   

   解释：

   40% increase will lead to a factor of 140 and similiarly 150 and 175

   so the new ratio is

   (5*140):(7*150):(8*175)

   on solving we get 2:3:4

10. 一个人现在的年龄是他母亲年龄的三分之一。 12年后，他的年龄将是他母亲年龄的一半。他妈妈现在几岁？
    1. 30
    2. 34
    3. 38
    4. 36

    回答：

    36
    

    解释：

    Let the present ages of son and his mother are x years and 3x years.
    Then (3x + 12) = 2( x + 12)
    => 3x + 12 = 2x + 24
    => x = 12
    => Present age of mother = 3x = 36 years

11. 21个结果的平均值是20。^第1个10个的平均值是24个，最后10个的平均值是14个。第11个的结果是： 。
    1. 42
    2. 44
    3. 46
    4. 40

    回答：

    40
    

    解释：

    11’th result = sum of 21 results – sum of 20 results
    = 21 x 20 – (24 x 10 + 14 x 10)
    = 420 – (240 + 140)
    = 420- 380 = 40

12. 完全数 n 是等于其除数之和的数。以下哪个是完全数？
    1. 6
    2. 9
    3. 15
    4. 21

    回答：

    9
    

    解释：

    6 is divisible by 1, 2 and 3.
    And, 6 = 1 + 2 + 3.

13. 三个数字的比例为 2 : 3 : 4 并且它们的 LCM 是 240。它们的 HCF 是：
    1. 40
    2. 20
    3. 30
    4. 10

    回答：

    20
    

    解释：

    Let the numbers be 2x, 3x and 4x
    LCM = 12x
    12x=240
    => x=20
    H.C.F of 40, 60 and 80=20

14. 一座体育场将在 1500 天内建成。承包商雇用了 200 名男性、300 名女性和 750 台机器人机器。 600 天后，75% 的工作仍有待完成。由于担心延误，承包商拆除了所有女性和 500 台机器人机器。此外，他雇用了更多的人，其效率与早期雇用的人相同。这导致工作加速，体育场提前 50 天建成。如果一天内有 6 名男性、10 名女性和 15 台机器人具有相同的工作产出，则求出额外雇用的男性人数。
    1. 1100
    2. 1340
    3. 1300
    4. 1140

    回答：

    1140
    

    解释：

    Let the total work be 4 units.
    => Work done in first 600 days = 25% of 4 = 1 unit
    => Work done in next 850 days = 75% of 4 = 3 unit
    Also, we are given that the daily work output of 6 men, 10 women and 15 robotic machines are same.
    => 6 Em = 10 Ew = 15 Er
    => Em : Ew : Er = 5 : 3 : 2, where ‘Em’ is the efficiency of 1 man, ‘Ew’ is the efficiency of 1 woman and ‘Er’ is the efficiency of 1 robotic machine.
    Therefore, ratio of efficiency of man, woman and robotic machine = 5:3:2.
    If ‘k’ is the constant of proportionality, Em = 5k, Ew = 3k and Er = 2k.
    Here, we need to apply the formula
    => (M_i E_i) D₁ H₁ / W₁ =(M_j E_j) D₂ H₂ / W₂, where
    => (M_i E_i) = (200 x 5k) + (300 x 3k) + (750 x 2k)
    => (M_j E_j) = (200 x 5k) + (m x 5k) + (250 x 2k), where ‘m’ is the additional men employed
    D₁ = 600 days
    D₂ = 850 days
    H₁ = H₂ = Daily working hours
    W₁ = 1 unit
    W₂ = 3 units
    So, we have
    3400k x 600 / 1 = (1500 + 5m)k x 850 / 3
    => 3400k x 1800 = (1500 + 5m)k x 850
    => 1500 + 5m = 7200
    => 5m = 5700
    => m = 1140
    Therefore, additional men employed = 1140

15. 两个管道 A 和 B 与第三个管道 C 交替工作以填充游泳池。单独工作，A、B 和 C 分别需要 10、20 和 15 小时。找出填充池所需的总时间。
    1. 7 小时 14 分钟
    2. 6 小时 54 分钟
    3. 5 小时 14 分钟
    4. 8 小时 54 分钟

    回答：

    66 hours 54 minutes
    

    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