HCL放置纸|定量能力集– 4

这是有关“定量智能”的HCL模型论文。这份安置文件将涵盖HCL安置中要求的才能，并且严格遵循HCL论文中提出的问题的模式。建议解决以下每个问题，以增加清除HCL位置的机会。

1. 找到一个正数，当它增加16时，它等于该数的倒数的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。两个数中的30是其HCF。是15。如果其中一个数字是30，那么另一个数字是什么?
   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. 以下哪项是最大的?
   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. 雇用了三个朋友A，B和C在一家面包店做糕点。他们单独工作，可以在一小时内分别制作60、30和40个糕点。他们决定一起工作，但是由于缺乏资源，他们不得不轮班工作30分钟。找到制作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。 A，B和C单独工作分别需要12、15和20分钟。另一根管D是废管，可以在30分钟内单独排空装满的水箱。如果同时打开所有水管，加满水箱的总时间(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. 两列火车的速度之比为7：8。如果第二列火车在4小时内覆盖400公里，请找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小时内横渡一定距离，然后在1½小时内返回。如果溪流以3 km / h的速度行驶，请找出静止水中摩托艇的速度。
   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. 巴拉克花费6650卢比购买一些商品，并获得6％的回扣。此后，他需缴纳10％的营业税。他的总支出是多少?
   1. 价钱：6870.10卢比
   2. Rs 6876.10
   3. Rs 6865.10
   4. Rs 6776.10

   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. 一个盒子里有10p，25p和50p硬币，比例为4：9：5，总和为206卢比。每种盒子有多少枚硬币?
   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的年龄比例分别为6：5。 7岁以后，Shyam的年龄将是32岁。拉姆(Ram)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. 两个连续奇数之和是什么，它们的平方差为56? 。
    1. 30
    2. 28岁
    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. 两个数字之比为3：5。如果它们的LCM为75。这些数字的总和是多少?
    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. 一个人雇用了一组20人从事建筑工作。这20个人每天工作8小时，可以在28天内完成工作。这项工作按时开始，但在18天后，发现仍有三分之二的工作仍在进行中。为了避免罚款并按时完成工作，雇主必须雇用更多的工人，并且还将工作时间增加到每天9个小时。如果所有男性的效率相同，则找到更多的男性雇佣人数。
    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以填充一个水箱。 A，B和C单独工作分别需要10、15和20个小时。 A在7 AM打开，B在8 AM打开，C在9 AM打开。考虑到管道C只能连续工作3个小时，并且需要1个小时的站立时间才能再次工作，因此在什么时候可以将罐完全充满。
    1. 下午12:00
    2. 下午12:30
    3. 下午1:00
    4. 下午1时30分

    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