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。建议解决以下每一个问题，以增加通过 TCS 面试的机会。
1. 有一组 30 个号码。前 10 个数字的平均值等于后 20 个数字的平均值。最后 20 个数字的总和是多少？
a) 前十个数字之和的两倍
b) 前 10 个数字的总和。
c) 最后十个数字之和的两倍
d) 无法确定。
Answer: a) Twice the sum of the first ten numbers解决方案：
Let the sum of the first 10 numbers is equal to ‘x’
Let the sum of the last 20 numbers is equal to ‘y’
According to the question:
x/10 = y/20
Therefore, y = 2x
2. 有一个叫Metron的小镇，这里的车辆前后轮大小不一。镇上遵循的计量单位是米。该车前轮周长133米，后轮周长190米。那么当前轮比后轮多转九圈时，小车行驶的距离是多少米？
一种。 1330
湾572
C。 399
d. 3990
Answer: d) 3990解决方案：
At first, we calculate the LCM of 133 and 190 which is 1330. So, the front wheels take 10 rounds to cover 1330 meters and the rear wheels take 7 rounds to cover the same. So to take 9 extra revolutions the vehicle would have travelled 1330 * 3 = 3990 meters.
3. 设一个数 x 除以 406 余数为 115。这个数除以 29 的数是多少？
Answer: 28解决方案：
According to the question, the number is equal to 406x + 115.
Since 406 is divisible completely by 29, therefore any multiple of 406 that is 406x when divided by 29 leaves remainder 0. Now 115, when divided by 29, leaves remainder 28.
4. 将形成一个字母数字序列。由两个字母和两个数字组成的序列应形成不重复。它可以通过多少种方式形成？
一种。 65000
湾64320
C。 58500
d. 67600
Answer: c) 58500解决方案：
The first can be filled in 26 ways.
The second place can be filled in 25 ways.
The third place can be filled in 10 ways.
The last digit can be filled in 9 ways
5. 根据特定的编码语言，A=0, B=1, C=2, …, Y=24, Z=25 那么ONE+ONE（仅以字母形式）可以编码吗？
a) 达比
b) 国际开发署
c) BDAI
d) ABDI
Answer: c) BDAI解决方案：
This is a 26 base question. Just like there is the Decimal system consisting of 10 digits from 0 to 9, the Base 26 system consist of 26 alphabets where A = 0, B = 1, Z = 25 and so on.
Let’s calculate, O N E + O N E
For E(4),
=> E + E
=> 4 + 4
=> 8
=> I
For N(13),
=> 13 + 13
=> 26
On converting 26 to Base 26 we get 1 0. Keeping 0(A) and taking 1 as carry
For O(14),
=> O + O + 1
=> 29
Dividing 29 by 26 we get 1(B) 3(D)
So answer is BDAI
6.按照problemsolvingproblemsolvingproblemsolving的顺序……2015年的字母表是什么？
a) p
b) 克
c) r
d) n
Answer: d) n解决方案：
‘problemsolving’ consist of 14 letters. On dividing 2015 by 14 we get 13. So the 13th letter is n and hence the answer.
7. 101102103104105106107…148149150除以9的余数是多少？
Answer: 2解决方案：
The divisibility rule for 9 is that the sum of all digits of a number should be divisible by 9. Let’s calculate the sum of the digits:
There are 50 1’s (unit place) = 50
There are 10 1’s (tens place) = 10
There are 10 2’s (tens place) = 20
There are 10 3’s (tens place) = 30
There are 10 4’s (tens place) = 40
There is one 5 (tens place) = 5
For each number 1 to 9, there are 5 sets of sum 45(1+2+…+9) = 225
=> So sum of all digits = 380
=> 380 / 9 = 2 (Answer)
8. 准备 30 升 78% 的浓酸溶液。 90%的浓酸需要与75%的浓酸溶液混合多少升才能得到结果？
一）10
b) 6
c) 3
d) 4
Answer: b) 6解决方案：
Let’s apply the weighted-average formula.
Let there be n1 litre of 90% acid solution and n2 litre of 75% solution
Therefore,
=> 78 = ((90 * n1) + (75 * n2))/(n1 + n2)
We get,
=> n1/n2 = 1 / 4
So 30 litres needed to be divided in the ratio of 1:4, which gives us 6 litre as the answer.
9.拉姆将在八年后32岁。四年后，Ram 父亲的年龄将是 Ram 年龄的两倍，而两年前，他母亲的年龄将是他年龄的两倍。拉姆的父亲和母亲现在的年龄是多少？
Answer: Father's age = 52, Mother's age = 46解决方案：
Ram will be 32 years old in next 8 years. So his present age is 32 – 8 = 24 years old
After 4 years Ram will be 28 years old. So his father will be 28 * 2 = 56 years old.
Therefore, fathers present age is 56 – 4 = 52 years old
Two years ago Ram was 22 years old. So his mothers age the was 22 * 2 = 44 years old
Therefore mothers present age is 44 + 2 = 46 years old.
10.一个班级男生人数与女生人数相等。如果 12 名女生入学，男生和女生的人数是女生的两倍，那么学生总数是多少？
Answer: 48解决方案：
Let ‘b’ be the number of boys and ‘g’ be the number of girls. According to the question:
=> b / (g – 12) = 2 / 1
Since b = g;
we get g = 24.
So the total number of students = 24 + 24 = 48