IBM安置文件|定量分析集– 3

这是IBM定量适度的模型文件。本安置文件将介绍IBM安置中要求的能力，并且严格遵循IBM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IBM展示位置的机会。

1. 两位数的数字与通过互换其数字的位置获得的数字之间的差是45。该数字的两位数之间的差异是什么? 。
   1. 5
   2. 7
   3. 6
   4. 都不是

   回答：

   5
   

   解释：

   Let the ten’s digit be x and unit’s digit be y
   Then (10x + y) – (10y + x) = 45
   9(x – y) = 45
   x – y = 5

2. 两个数字的比例为5：7。如果其LCM为105，则它们的平方有什么区别?
   1. 216
   2. 210
   3. 72
   4. 840

   回答：

   216
   

   解释：

   Let ‘h’ be the HCF of the two numbers.
   => The numbers are 5h and 7h.
   We know that Product of Numbers = LCM x HCF
   => 5h x 7h = 105 x h
   => h = 3
   So, the numbers are 15 and 21.
   Therefore, difference of their squares = 21² – 15² = 441 – 225 = 216

3. 分别工作的三个人A，B和C可以分别在10、12和20天内完成工作。他们决定一起工作，但是两天后，A离开了工作，又过了一天，B也离开了工作。如果他们在整个工作中总共得到了两颗紫胶，那就找出最高份额和最低份额的区别。
   1. 70000
   2. 60000
   3. 10000
   4. 20000

   回答：

   70000
   

   解释：

   Let the total work be LCM(10, 12, 20) = 60 units
   => Efficiency of A = 60/10 = 6 units / day
   => Efficiency of B = 60/12 = 5 units / day
   => Efficiency of C = 60/20 = 3 units / day
   Since the number of working days are different for each person, the share of each will be calculated in the ratio of the units of work done.
   Now, A works for 2 days and B works for 3 days.
   => Work done by A = 2 x 6 = 12 units
   => Work done by B = 3 x 5 = 15 units
   => Work done by C = 60 – 12 – 15 = 33 units
   Therefore, ratio of work done = 12:15:33 = 4:5:11
   So, A’s share = (4/20) x 2, 00, 000 = Rs 40, 000
   B’s share = (5/20) x 2, 00, 000 = Rs 50, 000
   C’s share = (11/20) x 2, 00, 000 = Rs 1, 10, 000
   Therefore, difference of the highest and lowest share = Rs 1, 10, 000 – 40, 000 = Rs 70, 000

4. 两个数字的HCF为11，而LCM为385。如果两个数字的相差不超过50，那么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

5. 三个管道A，B和C连接到水箱。单独工作，它们分别需要10个小时，20个小时和30个小时。一段时间后，A关闭，再过2个小时，B也关闭。 C会再工作14个小时，以使水箱完全充满。找到关闭管道A的时间(以小时为单位)。
   1. 1个
   2. 1.5
   3. 2个
   4. 3

   回答：

   2
   

   解释：

   Let the capacity of the tank be LCM (10, 20, 30) = 60
   => Efficiency of pipe A = 60 / 10 = 6 units / hour
   => Efficiency of pipe B = 60 / 20 = 3 units / hour
   => Efficiency of pipe C = 60 / 30 = 2 units / hour
   Now, all three work for some time, say ‘t’ hours.
   So, B and C work for 2 more hours after ‘t’ hours and then, C works for another 14 hours.
   => Combined efficiency of pipe A, pipe B and pipe C = 11 units/hour
   => Combined efficiency of pipe B and pipe C = 5 units/hour
    
   So, we have 11 x t + 5 x 2 + 14 x 2 = 60
   => 11 t + 10 + 28 = 60
   => 11 t = 60 – 38
   => 11 t = 22
   => t = 2
    
   Therefore, A was closed after 2 hours.

6. 一名警察在100米的距离内看到一个小偷，并开始追赶他。小偷看到了他，也开始逃跑。如果小偷以8 km / hr的速度奔跑，而警察以10 km / hr的速度奔跑，请在警察抓住他之前找出小偷覆盖的距离。
   1. 250米
   2. 400米
   3. 450米
   4. 350米

   回答：

   400 meters
   

   解释：

   We can safely assume that the policeman is running in the same direction as the thief.
   Speed of policeman w.r.t thief = (10 – 8) = 2 km/hr.
   Time taken by policeman to cover the 100m distance between him and the thief = (100/1000) / 2 = 1/20 hr.
   Therefore, the distance covered by thief in 1/20 hrs = 8 × 1/20 = 2/5 km = 400 meters.

7. 一条船在静止的水中以13 km / h的速度行驶。如果溪流速度为4 km / h，向下游行驶68 km需要多少时间?
   1. 5小时
   2. 4小时
   3. 6小时
   4. 3小时

   回答：

   4 h
   

   解释：

   Speed of the boat downstream = 13 + 4 = 17 km/h.
   Therefore, time taken to go 68 km downstream = (68/17) = 4 h.

8. 糖的价格降低了10％。结果，月销售额增加了30％。找出每月收入增加的百分比。
   1. 17％
   2. 19％
   3. 18％
   4. 都不是

   回答：

   17 %
   

   解释：

   Let the price of sugar be Rs 100 and monthly sales be 100 units. Then,
   total revenue = 100 × 100 = Rs 10000.
   And, new revenue = 90 × 130 = Rs 11700.
   Increase in revenue = 11700 – 10000 = Rs 1700.
   Hence, percentage increase in revenue = (1700/10000) × 100% = 17%.

9. A，B和C的当前年龄比例为4：5：9。九年前，他们的年龄总和为45岁。找到他们现在的年龄
   1. 15、20、35
   2. 20、24、36
   3. 20、25、45
   4. 16、20、36

   回答：

   16, 20, 36
   

   解释：

   Let the current ages of A, B and C be ax years, 5x years and 9x respectively.
   Then (4x-9) + (5x-9) + (9x-9) =45
   => 18x – 27 = 45
   => 18x = 72
   => x = 4
   Present ages of A, B and C are 4x = 16, 5x = 20, 9x = 36 respectively.

10. Vinod和Ashok的当前年龄分别为3：4。 5年后，他们的年龄比例分别变为7：9。 Ashok的当前年龄是多少?
    1. 40年
    2. 28年
    3. 32岁
    4. 36年

    回答：

    40 years
    

    解释：

    Let the present age of Vinod and Ashok be 3x years and 4x years respectively.
    Then (3x+5) / (4x+5) = 7 / 9

    => 9(3x + 5) = 7(4x + 5)
    => 27x + 45 = 28x + 35
    => x = 10
    => Ashok’s present age = 4x = 40 years

11. 一个两位数的数字使得两位数的乘积为12。当从该数字中减去9时，这些数字将颠倒过来。数字是： 。
    1. 34
    2. 62
    3. 43
    4. 26

    回答：

    43
    

    解释：

    Let the ten’s and unit’s digit be x and y.

    Then 10x + – 9 = 10 x + x

    10×2 + 12 -9x = 120 + x2
    9×2 – 9x – 108 = 0
    x2 –x – 12 = 0
    x2 –4x + 3x – 12 = 0
    (x – 4) (x + 3) = 0
    Therefore x = 4
    Hence the required no. is 43

12. 在每种情况下，除以17、23、35、59剩下相同余数的最大数是哪一个?
    1. 2个
    2. 3
    3. 6
    4. 12

    回答：

    6
    

    解释：

    Required Number = HCF (23-17, 35-23, 59-35, 59-17)
    = HCF (6, 12, 24, 42)
    = 6

13. 两个数字的比例为5：7。如果其LCM为105，则它们的平方有什么区别?
    1. 261
    2. 210
    3. 72
    4. 840

    回答：

    216
    

    解释：

    Let ‘h’ be the HCF of the two numbers.
    => The numbers are 5h and 7h.
    We know that Product of Numbers = LCM x HCF
    => 5h x 7h = 105 x h
    => h = 3
    So, the numbers are 15 and 21.
    Therefore, difference of their squares = 21² – 15² = 441 – 225 = 216

14. 一个人和一个人单独完成工作的时间比两个人在一起分别多了18天和8天。如果两者都可以一起工作，请查找所需的天数。
    1. 12
    2. 8
    3. 16
    4. 36

    回答：

    12
    

    解释：

    Let the time required to complete the work by A and B together = n days
    => Time required by A alone = n + 18 days
    => Time required by B alone = n + 8 days
    Therefore, n² = 18 x 8 = 144
    => n = 12
    Hence, A and B require 12 days to complete the work if they work together.

15. 与单独工作相比，两条管道A和B分别需要9个小时和6.25个小时才能填满一个游泳池。如果两个人一起工作，则找到填充池所花费的总时间。
    1. 6
    2. 6.5
    3. 7
    4. 7.5

    回答：

    7.5
    

    解释：

    Let the time taken if both were working together be ‘n’ hours.
    => Time taken by A = n + 9
    => Time taken by B = n + 6.25
     
    In such kind of problems, we apply the formula :
    n² = a x b, where ‘a’ and ‘b’ are the extra time taken if both work individually than if both work together.
    Therefore, n² = 9 x 6.25
    => n = 3 x 2.5 = 7.5
     
    Thus, working together, pipes A and B require 7.5 hours.

16. 保罗必须行驶24公里。步行1小时40分钟后，他发现自己已经走了剩下的5/7距离。找出保罗的速度(以米/秒为单位)。
    1. 5/3 m /秒
    2. 7/5 m /秒
    3. 2/3 m /秒
    4. 8/5 m /秒

    回答：

    5/3 m/s
    

    解释：

    Let the required speed be x km/hr.
    Distance covered by Paul in 1 hr 40 min = x × 100/60 = 5x/3 km.
    Remaining distance = (24 – 5x/3) km.
    Therefore, 5x/3 = 5/7 × (24 – 5x/3)
    => 7/5 × 5x/3 = 24 – 5x/3
    => 7x/3 = (72 – 5x)/3
    => 7x = 72 – 5x
    => 12x = 72 => x = 6.
    Paul’s speed in meters per second = 6 × 5/18 = 5/3 m/s.

17. 彼得的快艇在静止的水中以9公里/小时的速度行驶。他划船到距离105公里的地方，然后回到起点。如果溪流速度为1.5 km / h，请找出彼得花费的时间。
    1. 24小时
    2. 21小时
    3. 23小时
    4. 22小时

    回答：

    24 h
    

    解释：

    Upstream speed = 9 – 1.5 = 7.5 km/h.
    Downstream speed = 9 + 1.5 = 10.5 km/h.
    Therefore, time taken = 105/7.5 + 105/10.5 = 14 + 10 = 24 h.

18. 杰克消耗了他75％的工资。后来他的薪水增加了20％，支出增加了10％。找到他的储蓄增加的百分比。
    1. 51％
    2. 60％
    3. 50％
    4. 55％

    回答：

    50 %
    

    解释：

    Let Jack’s original salary be Rs 100. Then,
    his expenditure = Rs 75,
    his savings = Rs 25.
    Now, his new salary = Rs 120. So,
    new expenditure = (110/100) × 75 = Rs 165/2,
    new savings = 120 – 165/2 = Rs 75/2.
    Increase in savings = 75/2 – 25 = Rs 25/2.
    Therefore, percentage increase in savings = (25/2)/25 × 100% = 50%.