认知展示纸|才能套装4

这是用于Cognizant技术解决方案面试的能力准备的模型介绍文件。这份安置文件将涵盖在CTS招聘活动中提出的才能问题，并且还将严格遵循在CTS面试中提出的问题模式。建议解决以下每个问题，以增加清除CTS面试的机会。

1. 一列以5公里/小时，125米长的速度行驶的火车在10秒内经过了一名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. 总和为Rs的总单利。在5年内以每年9％的利率达到4016.25。金额或本金是多少?
   a)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. 在两名候选人之间的选举中，一个人获得了有效选票总数的55％，而获得了无效选票的20％。在计算总票数的当天结束时，发现总数为7500。因此，获胜候选人所获得的有效票数是：
   a)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)周日
   c)周二
   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. 整数n除以4得出3的余数。当2n除以4时，余数将是多少?
   a)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 m的比赛中，Aman需要36秒才能完成比赛，而Bijay则需要45秒。阿曼在比赛中击败比耶的距离是多少?
   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. 从序列中识别奇数：835、734、642、751、853、981、532
   a)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. 在一个由6名男性和4名女性组成的小组中，将选择4名。可以通过多少种不同的方式来选择他们，以便在该小组中至少有一个男人在场?
   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. 一个盒子包含15个大理石，其中4个是白色，5个是红色和6个是蓝色。从袋子中随机抽出三个球。它们全部为红色的概率是：
   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天内完成工作，具体取决于他们的工作能力。如果X每隔三天得到Y和Z的协助，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.