TCS安置纸| MCQ 4

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. 呼叫中心座席按姓名的字母顺序列出了305个电话号码，但Anuj没有任何姓名。他需要快速联系丹麦曼克，以向他传达信息。如果每个电话需要2分钟才能完成，并且每个电话都已接听，那么他可以保证将消息传递给丹麦语的最短时间是多少?
   a)206分钟
   b)610分钟
   c)18分钟
   d)34分钟

Answer: c) 18 minutes

Solution: 
We need to search for a particular name in a phone book. So we need to apply a method in which we can easily search a number in a minimum count. So we divide the list into two equal halves, i.e., 305/2 = 152.5 or let’s take 152. Now we can decide whether to check for Danish in the upper or lower half of 152. This is decided by the starting letter of the name in a page. Proceeding in the similar manner we get, 
152/2 = 76 
76/2 = 38 
38/2 = 19 
19/2 = 9 
9/2 = 4 
4/2 = 2 
2/2 = 0 
So we get 0 at the 9th time, hence this is the minimum number of the count to find Danish. So total time taken = 9 * 2 = 18 minutes.

2.有一个由38人组成的办公室。其中有10个人喜欢打高尔夫球，有15个人喜欢踢足球，有20个人既不高尔夫也不踢足球。有多少人喜欢高尔夫和足球?
a)10
b)7
c)15
d)18

Answer: b) 7

Solution: 
Let the number of people liking golf = ‘A’ 
Let the number of people liking football = ‘B’ 
Let the number of people liking either golf or football = A U B = 38 – 20 = 18 
People liking both golf and football = 

$A \cap B$$

= A + B – AUB = 10 + 15 -18 = 7

3.如果掷骰子2次，得到一对数字之和等于3或4的概率是多少?
a)6/36
b)5/36
c)1/9
d)1/12

Answer: b) 5/36

Solution: 
Total probability = 36 
We can get a sum of 3 or 4 in this many ways: 
=> (2, 1), (1, 2), (1, 3), (3, 1), (2, 2) = 5 
So probability = 5 / 36

4.店主为一堆蛋糕收取12卢比的费用。阿努(Anuj)向店主讨价还价，又多买了两个，这使他们只花了1卢比，比要价少了十几卢比。阿努(Anuj)用12卢比收到了多少个蛋糕?
a)10
b)14
c)18
d)16

Answer: d) 16

Solution: 
Let the number of cakes = ‘x’ or ‘x/12’ dozen 
So, x/12 cost Anuj 12 rupees, or 1 dozen cost him = 144/x rupees 
Now, he gets two extra = 144/(x+2) in 1 rupees less, 
=> 144/x – 144/(x+2) = 1 
=> On putting 16, the equation is satisfied, hence the answer.

5.仅Ram就能在2天内完成工作的1/4。仅Shyam就能在4天内完成工作的2/3。那么，Anil必须在2天内完成哪部分工作，以便他们在3天内共同完成工作?
a)1/8
b)1/20
c)1/16
d)1/12
答案： d)1/12

Solution: 
Ram alone can complete the work in 2*4 = 8 days. 
Shyam alone can complete the work in 4*(3/2) = 6 days. 
Taking the lcm of 8, 6, 3 = 24 
Capacity of Ram = 24/8 = 3 
Capacity of Shyam = 24/6 = 4 
Capacity of Anil = 8 – (4+3) = 1 
Now in 2 days Anil can do 2 unit of work = 2/24 = 1/12 part of the work

6. Mehta先生选择一个数字，并继续将该数字加倍，然后再减去一个。如果他选择3作为初始数字并且重复30次操作，那么最终结果是什么?
a)(2 ^ 30)– 1
b)(2 ^ 30)– 2
c)(2 ^ 31)– 1
d)这些都不是

Answer: d) None of these

Solution: 
According to the question, 
3 * 2 – 1 = 5 = 

$2^2+1$$

5 * 2 – 1 = 9 = 

$2^3+1$$

9 * 2 – 1 = 17 = 

$2^4+1$$

Proceeding in the similar fashion, on 30 times we get 

$2^{31}+1$$

7.单独的Ram可以在7天内刷墙，而他的朋友Roy则可以在9天内刷墙。他们可以在几天内一起粉刷墙壁? (舍弃答案)
a)3
b)5
c)4
d)7

Answer: c) 4

Solution: 
This can be solved by applying a simple formula = ab/(a+b) 
or, (9*7)/(9+7) 
or, 63/16 = 3.9375 = 4 (answer)

8.两个长度分别为6米和11米的垂直墙相距12米。找出两堵墙的最高距离?
a)15米
b)13米
c)12米
d)10米

Answer: b) 13 meters

Solution: 
Let’s consider this figure, 
 

We need to find the distance of AB, 
We know AC = 12 m and BC = 11-6 = 5 m 
So applying pythagoras theorem we get, 
AB = 

$$\sqrt(12^2 + 5^2)$$

= 13 metres

9.对于f(m，n)= 45 * m + 36 * n，其中m和n是整数(正数或负数)。对于所有m，n值，f(m，n)的最小正值是多少(这可以通过m和n的各种值来实现)?
a)18
b)12
c)9
d)16

Answer: c) 9

Solution: 
To get the minimum value of f(m, n), put m = 1 and n = -1, we get 
f(, n) = 9

10.将一个白色立方体(具有六个面)在两个不同的面上涂成蓝色。有几种不同的实现方式(如果在立方体适当旋转的情况下可以将一幅画带到另一幅画，则认为两幅画是相同的)?
a)30种方式
b)18种方式
c)4种方式
d)2种方式

Answer: d) 2

解决方案：

This can be achiededv in the following different ways;: 
First, painting on opposite faces can be achieved in 1 way. 
Second, painting on adjacent faces can be achieved in 1 way. 
Therefore in 2 ways.