认知安置纸|能力集 4

这是 Cognizant Technology Solutions面试的能力准备模板。这份安置文件将涵盖 CTS 招聘活动中提出的能力问题，并严格遵循 CTS 面试中提出的问题模式。建议解决以下每个问题，以增加通过 CTS 面试的机会。

1. 一列速度为 5 公里/小时、长度为 125 米的火车在 10 秒内通过一个与火车行驶方向相同的人。火车的速度是：
   a) 50 公里/小时
   b) 54 公里/小时
   c) 55 公里/小时
   d) 60 公里/小时

   Answer: a) 50 km/hr

   Solution:
   The relative speed of the train to man = (125 / 10) m/s
   = 25 / 2 m/s
   = (25/2 * 18/5) km/hr
   = 45 km/hr
   Let the relative speed of the train be x km/hr.
   Therefore, x-45 = 5 or x = 50 km/hr

2. 一笔总金额为卢比。 4016.25 以 9 % 的速度在 5 年内。金额或本金金额是多少？
   一）卢比。 4462.50
   b) 卢比。 8032.50
   c) 卢比。 8900
   d) 卢比。 8925

   Answer: d) Rs. 8925

   Solution:
   We know, SI = PTR/100
   or, P = (SI * 100) / TR
   or, P = (4016.25 * 100) / 9*5
   or, P = 8925 (answer)

3. 在两位候选人之间的选举中，一位获得了总有效票的 55%，获得了 20% 的无效票。最后统计总票数的时候，总票数是7500张，那么获胜的候选人得到的总有效票数是：
   一）2800
   b) 3300
   c) 3100
   d) 2700

   Answer: d) 2700

   Solution:
   Since 20% of the votes were invalid, 80% of the votes were valid = 80% of 7500 = 6000 votes were valid
   Since one candidate got 55% of the total valid votes, then the second candidate must have 45% of the votes = 0.45 * 6000 = 2700 votes

4. 2008 年 1 月 1 日是星期二。 2009 年 1 月 1 日是哪一天？
   a) 星期四
   b) 星期日
   c) 星期二
   d) 星期三

   Answer: a) Thursday

   Solution:
   In such type of questions, one needs to identify the type of year, i.e., whether the year is a normal year or is it a leap year.
   So the year 2008 was a leap year. So, it has to have 2 odd days. The year following 2008 is 2009 so the first day of the year would be two days ahead of what it was in 2008. So 1st Jan 2009 would be a Thursday.

5. 一个整数 n，除以 4 的余数为 3。 2n 除以 4 的余数是多少？
   一）0
   b) 1
   c) 2
   d) 4

   Answer: c) 2

   Solution:
   According to the question,
   n = 4q + 3
   therefore, 2n = 8q + 6
   or, 2n = 4(2q + 1 ) + 2
   Thus, we get when 2n is divided by 4, the remainder is 2.

6. 在 100 米的比赛中，Aman 用时 36 秒完成比赛，Bijay 用时 45 秒。阿曼在比赛中击败比杰的距离是多少？
   a) 20 米
   b) 25 米
   c) 22.5 米
   d) 9 米

   Answer: b) 20 meters

   Solution:
   The difference in the time of the race completion = 45 – 36 = 9 sec.
   So the distance covered by Bijay in 9 sec = 100/45 * 9 = 20 meters.
   Therefore Aman beats Bijay by 20 meters

7. 从系列中识别奇数：835、734、642、751、853、981、532
   一）532
   b) 853
   c) 981
   d) 751

   Answer: d) 751

   Solution:
   Looking at the series closely we see that in each number, the difference between the first and last digit of each number is the middle number, except 751

8. 在 6 男 4 女一组中，将选出 4 名。可以通过多少种不同的方式来选择他们，使得小组中至少应该有一个人？
   a) 209 种方式
   b) 194 种方式
   c) 205 种方式
   d) 120 种方式

   Answer: a) 209 ways

   Solution:
   A group of 4 has to be selected with at least one man So this can be done in
   (1 man and 3 women), (2 men and 2 women), (3 men and 1 women) and 4 men.
   The number of ways in which this can be done is
   (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
   On solving this we get 209 ways in which these combinations can be obtained.

9. 一个盒子里有 15 个弹珠，其中 4 个是白色的，5 个是红色的，6 个是蓝色的。从袋子中随机抽取三个球。它们都是红色的概率是：
   a) 1/22
   b) 2/89
   c) 2/77
   d) 2/91

   Answer: d) 2/91

   Solution:
   The number of ways in which all the three balls would be red = 5C3 / 15C3
   = 10/455 = 2/91

10. X、Y 和 Z 可以分别在 20、30 和 60 天内完成一项工作，具体取决于他们的工作能力。如果 X 每隔三天由 Y 和 Z 协助，那么 X 将如何完成工作？
    a) 12 天
    b) 15 天
    c) 16 天
    d) 18 天

    Answer: b) 15 days

    Solution:
    We need t first count the amount of work done in 2 days by X
    X can do a piece of work in 20 days
    So, in 2 days he can do = 1/20 * 2 = 1/10

    Amount of work done by X, Y and Z in 1 day = 1/20 + 1/30 + 1/60 = 1/10
    So, amount of work done in 3 days = 1/10 + 1/10 = 1/5
    So the work will be completed in 3 * 5 = 15 days.