TCS 安置文件 |多选题 10

这是一份用于能力准备的 TCS 模型安置论文。这份安置文件将涵盖 TCS 招聘活动中提出的能力问题，并严格遵循 TCS 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 TCS 面试的机会。

从 2 到 98（包括两者）的所有偶数都将相乘。产品的个位数是多少？
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.

编程需要懂一点英语
10 个程序员可以在 10 分钟内输入 10 行。需要多少程序员在 60 分钟内输入 60 行？
一）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

编程需要懂一点英语
Anil 连续工作 8 天，第 9 天休息。如果他周一开始工作，那么他的第 12 天休息日是哪一天？
a) 星期四
b) 星期二
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.

编程需要懂一点英语
过度捕捞是一个严重的环境问题。科学家们能够确定，如果拖网渔船的网目尺寸为“x”厘米（方形网眼），则进入网中的鱼被网捕获的百分比以二次方程的形式表示， 100 – 0.04x^2- 0.24x。例如，如果网孔尺寸为零，则进入网中的鱼将 100% 被捕获。一艘带有方形网眼的拖网渔船涉嫌使用非法尺寸的网，将网扔到 Lakshadweep 和海岸警卫队附近的海底，逮捕了船员。后来查看了捕获的鱼的大小，并估计对于拖网渔船使用的网，至少有 97.8% 的鱼会被捕获。拖网渔船使用的网的 x 的最大值是多少？
一）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.

编程需要懂一点英语
奥迪生产的废品率为 4%，梅赛德斯为 8%，两辆车加起来为 7%。奥迪生产的比例是多少？
一）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

编程需要懂一点英语
需要组成一个11人的队伍，从5男11女中选拔，限制选男不超过3人。有多少种选择方式？
一）1121
乙）1565
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

编程需要懂一点英语
有两个袋子，里面装着白色和黑色的弹珠。第一个袋子里有 8 个白色弹珠和 6 个黑色弹珠，第二个袋子里有 4 个白色弹珠和 7 个黑色弹珠。从这两个袋子中的任何一个中随机抽取一个大理石。求这个大理石是黑色的概率。
一）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

编程需要懂一点英语
有一个城市，所有 100% 的选票都已登记。在这其中 60% 的选票支持国会，40% 的选票支持 BJP。 Ram，获得 75% 的国会选票和 8% 的 BJP 选票。拉姆得了多少票？
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 %

编程需要懂一点英语
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是 7 公里/小时。他们花费的时间总和是 14 小时。求约翰的速度。
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.

编程需要懂一点英语
如果 f(x) = 2x + 2 f(f(3)) 的值是多少？
一）8
乙) 64
c) 16
d) 18
```
Answer: d) 18
```
Solution;
f(f(3)) = 2(f(3)) + 2
=> 2(2(3) + 2) + 2
=> 16 + 2 = 18

编程需要懂一点英语