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。
- 有一组30个数字。前10个数字的平均值等于后20个数字的平均值。最近20个数字的总和是多少?
a)前十个数字之和的两倍
b)前10个数字之和。
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 - 有一个叫做Metron的小镇，车辆前后轮的大小不同。城镇中紧随其后的度量单位是米。汽车前轮的周长为133米，后轮的周长为190米。那么，当前轮比后轮多旋转九圈时，小车行进的距离以米为单位是多少?
一种。 1330
b。 572
C。 399
d。 3990Answer: 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 metres 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 metres.
- 让数字“ x”除以406会得到余数115。当数字除以29时，数字将是多少?
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. - 将形成一个字母数字序列。由两个字母后跟两个数字组成的序列应无重复。它可以以几种方式形成?
一种。 65000
b。 64320
C。 58500
d。 67600Answer: 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。
- 根据一种特定的代码语言，A = 0，B = 1，C = 2，…，Y = 24，Z = 25可以将ONE + ONE(仅字母形式)编码吗?
a)DABI
b)CIDA
c)BDAI
d)ABDIAnswer: 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 - 在解决问题的顺序中解决问题的顺序…第2015个字母是什么?
a)p
b)g
c)
d)nAnswer: 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.
- 当数字101102103104105106107…148149150除以9时，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) - 准备30升78％的浓酸溶液。需要将多少升90％的浓酸与75％的浓酸溶液混合才能得到结果?
a)10
b)6
c)3
d)4Answer: 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. - 从现在起的八年内，Ram将32岁。在4年内，Ram的父亲的年龄将是Ram的年龄的两倍，而在两年前，他母亲的年龄将是其年龄的两倍。 Ram的父母现在的年龄是几岁?
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. - 在一堂课中，男孩的数量等于女孩的数量。如果12个女孩入学时，男孩的数量仍然是女孩的两倍，那么学生的总人数是多少?
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