TCS安置纸| MCQ 10

这是用于能力倾向准备的TCS模型放置论文。这份安置文件将涵盖在TCS招聘活动中提出的才能问题，并且还将严格遵循在TCS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TCS面试的机会。

1. 从2到98(包括两者)的所有偶数都将相乘。产品的单位位数是多少?
   a2
   b)0
   c)6
   d)4

   Answer: b) 0

   Solution:
   Let us look at the sequence of the multiplications,
   2 * 4 * 6 * 8 *10* 12 * 14 * 16 * 18 *…* 98
   If we look closely we will find that the units place of every number forms a sequence of 2, 4, 6 ,8 and 0 multiplying to a number whose units place is always 0 and in all we get 0. So the unit digit of final number = 0.

2. 10位程序员能够在10分钟内键入10行。 60分钟内需要多少位程序员才能键入60行?
   a)10
   b)16
   c)60
   d)以上都不是

   Answer: a) 10

   Solution:
   This is a simple question of logical reasoning. If 10 programmers can type 10 lines of code in 10 minutes then to type 60 lines of code, in 60 minutes, the same 10 coders will be required, since the lines of code and time are in proportion

3. Anil连续工作8天，在第9天休息。如果他在星期一开始工作，那么哪一天他有第12个休息日?
   a)星期四
   b)周二
   c)星期三
   d)星期五

   Answer: c) Wednesday

   Solution:
   Anil works for 8 days and rests on 9th day. In total 9 days are to be processed 12 times = 12 * 9 = 108.
   If we calculate according to the week, we get 108 / 7 = remaining 3 days. So if Anil starts working on Monday, he will rest on third day of the week which is Wednesday.

4. 过度捕捞是一个严重的环境问题。科学家们能够确定，如果拖网渔网的网眼尺寸为“ x”厘米(正方形网眼)，则进入网中的鱼被网捕获的百分比以二次方程式表示， 100 – 0.04x ^ 2- 0.24x。例如，如果网格大小为零，则将捕获100％进入网中的鱼。一名涉嫌使用非法大小的网的拖网渔船用方形网眼拖网，将网扔到Lakshadweep和海岸警卫队附近的海底，逮捕了船员。随后查看了所捕捞的鱼的大小，并估计，拖网渔船使用的网至少会捕捞进入网的鱼的97.8％。拖网渔船使用的网的x的最大值是多少?
   a)7
   b)4.5
   c)6
   d)5

   Answer: d) 5

   Solution:
   According to the question,
   for few values of x, the total fish caught is 97.8%. So
   => 100 – 0.04x^2- 0.24x = 97.8
   => 0.04x^2 + 0.24x = 2.2
   => 4x^2+ 24x = 220
   => x^2+ 6x – 55 = 0
   Solving, we get x = 5 and -11
   So, the value of x = 5 has to be positive and hence the answer.

5. 奥迪产品的拒绝率为4％，梅赛德斯的拒绝率为8％，两辆车的拒绝率为7％。奥迪的产量比例是多少?
   a)4/1
   b)2/1
   c)3/1
   d)7/1

   Answer: c) 3/1

   Solution:
   Using the simple weighted average formula we get,
   (4x + 8y)/(x+y) = 7
   or, 4x + 8y = 7x + 7y
   or, a/b = 3/1

6. 需要组成一个11人小组，从5名男性和11名女性中选拔，但最多只能选出3名男性。可以通过几种方式进行选择?
   a)1121
   b)1565年
   c)1243
   d)2256

   Answer: d) 2256

   Solution:
   Selecting 0 men and 11 women = 5C0 * 11C11 = 1
   Selecting 1 men and 10 women = 5C1 * 11C10 = 55
   Selecting 2 men and 9 women = 5C2 * 11C9 = 10 * 55 = 550
   Selecting 3 men and 8 women = 5C3 * 11C8 = 10 * 165 = 1650
   So total number of ways = 1650 + 550 + 55 + 1 = 2256 way

7. 有两个装有白色和黑色大理石的袋子。在第一个袋子中，有8个白色大理石和6个黑色大理石，在第二个袋子中，有4个白色大理石和7个黑色大理石。从这两个袋子中的任何一个中随机抽取一块大理石。找出这种大理石变黑的可能性。
   a)7/54
   b)7/154
   c)41/77
   d)22/77

   Answer: c) 41/77

   Solution:
   Probability of drawing a black ball from the first bag is = 6C1 / 14C1
   Probability of drawing a black ball from the second bag is = 7C1 / 11C1
   Total probability = 1/2 * (6C1/14C1) * (7C1/11C1) = 41/77

8. 有一个城市，所有100％的选票都已注册。在这60％的国会赞成票和40％的BJP赞成票中。 Ram，获得75％的国会投票和8％的BJP投票。拉姆获得了几张选票?
   a)48.2％
   b)56.6％
   c)42.8％
   d)64.4％

   Answer: a) 48.2 %

   Solution:
   Let the total number if votes = 100. So Ram gets,
   75% of 60 = 60 * 0.75 = 45 votes
   8% of 40 = 40 * 0.08 = 3.2 votes
   Thus total number of votes that Ram gets = 48.2 %

9. 约翰比彼得快。约翰和彼得各走24公里。约翰和彼得的速度总和为7 km / h。他们花费的时间总计为14个小时。找到约翰的速度。
   a)4公里/小时
   b)5公里/小时
   c)3公里/小时
   d)7公里/小时

   Answer: a) 4 km/h

   Solution:
   We know that John’s speed is greater than Peter’s speed and the sum of there speed is 7.
   So the combinations are = (6, 1), (5, 2), (4, 3)
   Now checking from the options if John’s speed is equal to 4, then Peter’s speed is 3,
   or, the time taken by them = 24/4 + 24/3 = 14 hours.

10. 如果f(x)= 2x + 2，f(f(3))的值是多少?
    a)8
    b)64
    c)16
    d)18

    Answer: d) 18

    Solution;
    f(f(3)) = 2(f(3)) + 2
    => 2(2(3) + 2) + 2
    => 16 + 2 = 18