IBM 安置文件 |定量分析套装 – 3

这是 Quantitative Aptitude 的 IBM 模型论文。这份实习论文将涵盖 IBM 实习中要求的能力，并严格遵循 IBM 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 IBM 安置的机会。

1. 一个两位数和把它的数字位置互换得到的数的差是45。这个数的两位数有什么区别？ .
   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。如果他们的 LCM 是 105，那么他们的平方之间有什么区别？
   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. A、B、C 三个人单独工作，分别可以在 10、12 和 20 天内完成一项工作。他们决定一起工作，但两天后，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两个 lacs，请找出最高和最低份额的差异。
   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 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

5. 三个管道 A、B 和 C 连接到一个储罐。单独工作，他们分别需要10小时、20小时和30小时。一段时间后，A 关闭，再过 2 小时，B 也关闭。 C 再工作 14 小时，使水箱完全装满。找出管道 A 关闭后的时间（以小时为单位）。
   1. 1
   2. 1.5
   3. 2
   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 米外看到一个小偷，开始追他。小偷看到他，也开始逃跑。如果小偷以 8 公里/小时的速度奔跑，而警察以 10 公里/小时的速度奔跑，则在警察抓住小偷之前找出小偷走过的距离。
   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. 一艘船在静水中以 13 公里/小时的速度行驶。如果溪流的速度为 4 公里/小时，向下游行驶 68 公里需要多长时间？
   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. 糖价下降10%。因此，每月销售额增加了 30%。找出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、C现在的年龄比例是4:5:9。九年前，他们的年龄总和是45岁。找出他们现在的年龄（以年为单位）
   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现在的年龄比例分别为3:4。 5年后，他们的年龄比分别变为7:9。 Ashok 现在的年龄是多少？
    1. 40年
    2. 28岁
    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. 一个两位数的数字的乘积是 12。当从数字中减去 9 时，数字是相反的。号码是： 。
    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 17、23、35、59 以留下相同余数的最大数是哪个？
    1. 2
    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。如果他们的 LCM 是 105，那么他们的平方之间有什么区别？
    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. A 单独和 B 单独完成一项工作分别比两者一起工作多 18 天和 8 天。如果两者一起工作，请找出所需的天数。
    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. 单独工作时，两个管道 A 和 B 分别需要比它们一起工作时多 9 小时和 6.25 小时来填充水池。如果两者一起工作，请找出填充池所需的总时间。
    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，他发现自己已经走完了剩余距离的 5/7。找出 Paul 的速度（以米/秒为单位）。
    1. 5/3 米/秒
    2. 7/5 米/秒
    3. 2/3 米/秒
    4. 8/5 米/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. 彼得的快艇在静水中以 9 公里/小时的速度行驶。他划到距离105公里的地方，然后回到起点。如果溪流的速度为 1.5 公里/小时，请找出彼得所用的时间。
    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. 杰克消耗了他工资的 75%。后来他的工资增加了 20%，他的支出增加了 10%。找出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%.