TCS安置纸| MCQ 5

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. 在除以460时，必须将以下哪个数字加到5678上以给出提示35?
   a)980
   b)618
   c)955
   d)797

   Answer: d) 797

   Solution:
   Let the number added to 5678 be x to give a remainder 35 and quotient k when divided by 460.
   So, 5678 + x = 460k + 35
   or, 5643 + x = 460k
   So 5643 + x must be divisible by 460
   Ananysing from the options, we get on adding 797 to 5643, the number 6440 is divisible by 460.

2. 拉哈曼(Rahaman)去了一家文具店，以100卢比的价格买了18支铅笔。他为每支灰色铅笔多付了1卢比，而不是每支黑色铅笔。一支灰色铅笔的价格是多少?他买了几支灰色铅笔?
   a)5、10卢比
   b)6、10卢比
   c)5、8卢比
   d)Rs.6、8

   Answer: b) Rs. 6, 10

   Solution:
   The best way is to analyse from the mentions.
   Let’s take option b in which 10 pencils are bought at Rs.6 each. So total cost of grey pencils = 6 * 10 = Rs.60. So Rahaman is left with 40 rupees. He buys 8 black pencils at Rs 5 each which is 1 rupee less than what he had spent in buying the grey ones. Thus satisfying the conditions.

3. 每个人四个人一次掷四个骰子。找出至少两个人会掷相同号码的概率吗?
   一种。 13/18
   b。 5/18
   C。以上都不是
   d。 1295/1296

   Answer: a) 13/18 ways

   Solution:
   Total possible outcomes = $$6^4$$ = 1296
   Number of ways in which no two people get same number = 6*5*4*3 = 360 ways
   The probability of no two people getting the same number = 360 / 1296 ways = 5/18 ways
   So the probability of at least two people getting the same number = 1 – 5/18 = 13/18 ways

4. 拉姆说：“如果您给我一半的钱，我将得到75卢比。” Shyam说：“如果您给我三分之一的钱，我将获得75卢比。Shyam拥有多少钱4
   c)48
   d)60

   Answer: d) 60

   Solution: Let Ram and Shyam be denoted by ‘R’ and ‘S’ respectively
   According to the question,
   Eqn 1. R + S/2 = 75
   Eqn 2. R/3 + S = 75
   Therefore, solving both the equations we get, R = 45 and S = 60.

5. 拉姆去市场买苹果。如果他可以讨价还价并将每只苹果的价格降低2卢比，他就可以用自己的钱购买30个苹果，而不是20个苹果。他有多少钱?
   一种。 100卢比
   b。 Rs.50
   C。 120卢比
   d。 Rs.150

   Answer: c) 120

   Solution: Let the price per orange be Rs. x.
   So total money Ram has in buying at original price = 20x.
   On reducing the price by 2 rupees each the total money must be (x-2)*30
   According to the question,
   20x = (x-2)*30
   On solving this we get x = 6 or the total money = Rs. 120

6. 使用数字1、2、3和5可以形成小于500的多少个正整数，每个数字仅使用一次。
   a)68
   b)34
   c)66
   d)52

   Answer: b) 34

   
   Solution:
   Single digit numbers can be formed in 4 ways.
   2 digit number can be formed in 4 * 3 = 12 ways
   3 digit number less than 500 can be formed in 3 * 3 * 2 = 18 ways.
   Total number of ways = 18 + 12 + 4 = 34 ways

7. 一个男孩进入一家商店，以y卢比买了x本书。当他要离开时，簿记员说：“如果再买10本书，您可以花2卢比买到所有的书，一打也能省下80美分”。那么x和y是什么?
   a)(5，1)
   b)(10，1)
   c)(15，1)
   d)无法确定。

   Answer: a) (5, 1)

   Solution:
   x number of books cost him y rupees.
   So, 1 book will cost him y/x rupees.
   12 books will cost him rupees 12 y/x.
   The shopkeeper says,
   x + 10 books cost him 12 rupees
   1 book will cost him 12/(x+10) rupees
   12 books will cost him 24/(x+10) rupees
   We know that 80 cents = 4/5 of a dollar,
   So, 12y/x – 24/(10+x) = 4/5
   Analysing the given choices, we get (5, 1) satisfies the equation.

8. 等边三角形的周长等于正六边形。找出他们的面积比例?
   一种。 3：2
   b。 1：6
   C。 2：3
   d。 6：1

   Answer: c) 2:3

   Solution:Let the side of the equilateral triangle be a unit and that of the regular hexagon be b unit.
   So perimeter of the triangle = 3a and perimeter of the hexagon is 6b unit.
   or, 3a = 6b
   or a/b = 2/1
   The area of the equilateral tr
   iangle = $$\sqrt(3)a^2/4$$
   The area of the regular hexagon = $$3 * \sqrt(3)a^2/2$$
   or, $$\sqrt(3)a^2/4$$ : $$3 * \sqr t(3)a^2/2$$
   Solving this and subsituting a/b we get the answer as 2 : 3

9. 在给定的序列中：70、54、45、41……。下一个数字是多少?
   a)40
   b)36
   c)35
   d)38

   Answer: a) 40

   Solution:The series goes like:
   70 – 54 = 16 (4^2)
   54 – 45 = 9 (3^2)
   45 – 41 = 4 (2^2)
   41 – 40 = 1 (1^1)

10. 每隔七年出版一系列故事书。当第七本书出版时，出版年代的总和是13524。第一本书出版于哪一年?
    a)1910年
    b)1911年
    c)2002年
    d)1932年

    Answer: b) 1911

    Solution:We get the series of publications as n, n+7, n+14, n+21, n+28, n+35, n+42.
    Sum of publications = 13524 = 7/2[2n + (7-1)*7] (Using the sum of AP formula)
    We get, n = 1911 (answer)