HCL 安置纸 |定量能力集 – 4

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. 找出一个正数，当增加 16 时，它等于该数倒数的 80 倍。
   1. 20
   2. -4
   3. -10
   4. 4

   回答：

   4
   

   解释：

   Let the number be x.

   Then x + 16 = 80 * (1/x)

   x² + 16x – 80 = 0
   x² + 20x – 4x – 80 =0
   (x + 20) (x -4)
   Therefore x = 4

2. LCM。两个数字是 30 和他们的 HCF。是 15。如果其中一个数字是 30，那么另一个数字是多少？
   1. 30
   2. 25
   3. 15
   4. 20

   回答：

   15
   

   解释：

   Say another number = x

   Product of two numbers = product of HCF and LCM
   => x.30 = 15*30
   => x=15

3. 以下哪个是最大的？
   1. 7/8
   2. 15/16
   3. 23/24
   4. 31/32

   回答：

   31/32
   

   解释：

   LCM (8, 16, 24, 32) = 96
   7/8 = 84/96
   15/16 = 90/96
   23/24 = 92/96
   31/32 = 93/96
   Hence, 31/32 is the largest of all.

4. 三个朋友 A、B 和 C 受雇在一家面包店制作糕点。单独工作，他们可以在一个小时内分别制作 60、30 和 40 个糕点。他们决定一起工作，但由于缺乏资源，他们不得不轮班工作 30 分钟。找出制作 185 个糕点所需的时间。
   1. 4个小时
   2. 3 小时 45 分钟
   3. 4 小时 15 分钟
   4. 5个小时

   回答：

   4 hours 15 minutes
   

   解释：

   It is given that A, B and C make 60, 30 and 40 pastries respectively in an hour.
   => In 30 minutes, they will make 30, 15 and 20 pastries respectively.
   So, in one cycle of 1 hour 30 minutes where each works for 30 minutes, pastries made = 30 + 15 + 20 = 65
   Now, in 2 cycles (3 hours), 130 pastries would be made.
   In the next 30 minutes, A would make 30 pastries.
   So, total time elapsed = 3 hours 30 minutes and pastries made = 130 + 30 = 160
   In the next 30 minutes, B would make 15 pastries.
   So, total time elapsed = 4 hours and pastries made = 160 + 15 = 175
   In the next 15 minutes, C would make 10 pastries.
   So, total time elapsed = 4 hours 15 minutes and pastries made = 175 + 10 = 185
   Therefore, total time taken = 4 hours 15 minutes

5. 打开三个管道 A、B 和 C 以填充水箱。单独工作，A、B 和 C 分别需要 12、15 和 20 分钟。另一根管子 D 是一条废液管，单独工作 30 分钟就可以清空装满的水箱。如果所有管道同时打开，则填充水箱所需的总时间（以分钟为单位）是多少？
   1. 5
   2. 6
   3. 7
   4. 8

   回答：

   6
   

   解释：

   Let the capacity of the cistern be LCM(12, 15, 20, 30) = 60 units.
   => Efficiency of pipe A = 60 / 12 = 5 units / minute
   => Efficiency of pipe B = 60 / 15 = 4 units / minute
   => Efficiency of pipe C = 60 / 20 = 3 units / minute
   => Efficiency of pipe D = 60 / 30 = 2 units / minute
   => Combined efficiency of pipe A, pipe B, pipe C and pipe D = 10 units / minute
    
   Therefore, time required to fill the cistern if all the pipes are opened simultaneously = 60 / 10 = 6 minutes

6. 两列火车的速度比为7:8。如果第二列火车在 4 小时内行驶了 400 公里，求出第一列火车的速度。
   1. 69.4 公里/小时
   2. 78.6 公里/小时
   3. 87.5 公里/小时
   4. 40.5 公里/小时

   回答：

   87.5 km/h
   

   解释：

   Let the speed of the two trains be 7x and 8x.
   Then, 8x = 400 / 4
   => 8x = 100 => x = 12.5 km/h.
   Hence, speed of the first train = 7x = 7 × 12.5 = 87.5 km/h.

7. 摩托艇在 1 小时内穿过一定距离，并在 1.5 小时内返回。如果溪流以 3 公里/小时的速度运行，请找出摩托艇在静水中的速度。
   1. 10 公里/小时
   2. 15 公里/小时
   3. 12 公里/小时
   4. 都不是

   回答：

   15 km/h
   

   解释：

   Let the speed of motorboat in still water be x km/h. Then,
   Downstream speed = (x + 3) km/h.
   Upstream speed = (x – 3) km/h.
   Then, (x + 3) × 1 = (x – 3) × 3/2
   => 2x + 6 = 3x – 9
   => x = 15.
   So, the speed of motorboat in still water is 15 km/h.

8. Barack 花费 6650 卢比购买一些商品并获得 6% 的回扣。在此之后，他支付 10% 的销售税。他的总开支是多少？
   1. 6870.10 卢比
   2. 6876.10 卢比
   3. 6865.10 卢比
   4. 6776.10 卢比

   回答：

   Rs 6876.10
   

   解释：

   Rebate received by Barack = 6% of Rs 6650 = 6/100 × 6650 = 3/5 × 665 = Rs 399.
   Sales Tax paid by Barack = 10% of Rs (6650-399) = 10% of Rs 6251 = Rs 625.10.
   Therefore, Barack’s total expenditure = Rs (6251 + 625.10) = Rs 6876.10.

9. 在一个盒子里，有 10p、25p 和 50p 的硬币，比例为 4:9:5，总金额为 206 卢比。这个盒子有多少种硬币？
   1. 200、360、160
   2. 135, 250, 150
   3. 90, 60, 110
   4. 无法确定

   回答：

   200, 360, 160
   

   解释：

   Let the number of 10p, 25p, 50p coins be 4x, 9x, 5x respectively. Then,
   4x/10 + 9x/4 + 5x/2 = 206 (Since, 10p = Rs 0.1, 25p = Rs 0.25, 50p = Rs 0.5)
   => 8x + 45x + 50x = 4120 (Multiplying both sides by 20 which is the LCM of 10, 4, 2)
   => 103x = 4120
   => x = 40.
   Therefore,
   No. of 10p coins = 4 x 40 = 160 (= Rs 16)
   No. of 25p coins = 9 x 40 = 360 (= Rs 90)
   No. of 50p coins = 5 x 40 = 200 (= Rs 100)

10. 目前，Ram 和 Shyam 的年龄比分别为 6:5。 7 年后，Shyam 的年龄将是 32 岁。拉姆现在的年龄是多少？
    1. 32
    2. 40
    3. 30
    4. 36

    回答：

    30
    

    解释：

    Let the present age of Ram and Shyam be 6x years and 5x years respectively.

    Then 5x + 7 = 32
    => 5x = 25
    => x = 5
    => Present age of Ram = 6x = 30 years

11. 两个连续奇数的和，它们的平方差是 56 是多少？ .
    1. 30
    2. 28
    3. 34
    4. 32

    回答：

    28
    

    解释：

    Let the no. be x and (x +2).
    Then (x +2)2 – x2 = 56
    4x + 4 = 56
    x + 1 = 14
    x = 13
    Sum of numbers = x + (x +2) = 28

12. 将 252 表示为素数的乘积。
    1. 2*2*3*3*7
    2. 3*3*3*3*7
    3. 2*2*2*3*7
    4. 2*3*3*3*7

    回答：

    2 * 2 * 3 * 3 * 7
    

13. 两个数字的比例为 3 : 5。如果它们的 LCM 是 75。数字之和是多少？
    1. 25
    2. 45
    3. 40
    4. 50

    回答：

    40
    

    解释：

    1st number = 3x
    2nd number =5x
    LCM of 3x and 5x is 15x
    => 15x = 75
    => x = 5
    sum = 15+25 =40

14. 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, additional men employed = 44

15. 打开三个管道 A、B 和 C 以填充一个罐。单独工作，A、B 和 C 分别需要 10、15 和 20 小时。 A 早上 7 点开门，B 早上 8 点开门，C 早上 9 点开门。考虑到C管一次只能工作3小时，需要静置1小时才能再次工作，什么时候水箱会完全装满。
    1. 12 : 00 下午
    2. 中午 12 时 30 分
    3. 1 : 00 下午
    4. 1 : 30 下午

    回答：

    12 : 30 PM
    

    解释：

    Let the capacity of the tank be LCM (10, 15, 20) = 60
    => Efficiency of pipe A = 60 / 10 = 6 units / hour
    => Efficiency of pipe B = 60 / 15 = 4 units / hour
    => Efficiency of pipe C = 60 / 20 = 3 units / hour
    => Combined efficiency of all three pipes = 13 units / hour
     
    Till 9 AM, A works for 2 hours and B work for 1 hour.
    => Tank filled in 2 hours by A = 12 units
    => Tank filled in 1 hour by B = 4 units
    => Tank filled till 9 AM = 16 units
    => Tank still empty = 60 – 16 = 44 units
     
    Now, all three pipes work for 3 hours with the efficiency of 13 units / hour.
    => Tank filled in 3 more hours = 39 units
    => Tank filled till 12 PM = 16 + 39 units = 55 units
    => Tank empty = 60 – 55 = 5 units
     
    Now, C is closed for 1 hour and these remaining 5 units would be filled by A and B working together with the efficiency 10 units/hour.
    => Time taken to fill these remaining 5 units = 5 / 10 = 0.5 hours
     
    Therefore, time at which the tank will be completely filled = 12 PM + 0.5 hours = 12 : 30 PM