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。
- 呼叫中心代理有一个按姓名字母顺序排列的 305 个电话号码列表,但 Anuj 没有任何姓名。他需要迅速联系丹麦曼克,向他传达信息。如果每个电话需要 2 分钟才能完成,并且每个电话都被接听,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.有一个办公室,由38人组成。其中10人喜欢打高尔夫球,15人喜欢踢足球,20人既不打高尔夫球也不踢足球。有多少人既喜欢高尔夫又喜欢足球?
一)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 + B – AUB = 10 + 15 -18 = 7
3. 如果掷骰子 2 次,得到一对总和等于 3 或 4 的数字的概率是多少?
一)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. 店主一串蛋糕要12卢比。 Anuj 与店主讨价还价,多买了两个,这使得它们比最初的要价低了 1 卢比。 Anuj 收到了多少个 12 卢比的蛋糕?
一)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、Ram一个人可以在2天内完成1/4的工作。仅 Shyam 就可以在 4 天内完成 2/3 的工作。那么哪些部分的工作必须由阿尼尔在 2 天内完成,才能让他们在 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不断将数字加倍,然后从中减去 1。如果他选择 3 作为初始数字并重复操作 30 次,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 =
5 * 2 – 1 = 9 =
9 * 2 – 1 = 17 =
Proceeding in the similar fashion, on 30 times we get
7. Ram 一个人可以在 7 天内粉刷一堵墙,而他的朋友 Roy 一个人可以在 9 天内粉刷同一面墙。他们可以在多少天内一起工作来粉刷墙壁? (四舍五入你的答案)
一)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、两堵长6米和11米的竖墙相距12米。找到两面墙的顶部距离?
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 =
= 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(这可以针对 m 和 n 的各种值实现)?
一)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. 一个白色立方体(有六个面)要在两个不同的面上涂成蓝色。有多少种不同的方式可以实现这一点(如果通过适当旋转立方体,一幅画可以被带到另一幅画上,则两幅画被认为是相同的)?
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.