TCS 安置文件 |多选题 6

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。

1. 从安格斯发现的神秘蛋孵化的克鲁索生长速度很快，安格斯不得不把它从家里搬到湖边。给定克鲁索出生前几周的体重为 5、15、30、135、405、1215、3645。找出奇数体重。
   一）3645
   乙) 135
   c) 15
   d) 30

   Answer: d) 30

   Solution:
   Looking at the series closely we find that the 3rd number is oddly placed.
   The series is in the form:
   5 * 3 = 15
   15 * 3 = 45
   45 * 3 = 135
   135 * 3 = 405 and so on

2. 假设 f(1)=0 且 f(m+n)=f(m)+f(n)+4(9mn-1)。对于所有自然数 (Integers>0)m 和 n。 f(17) 的值是多少？
   一）5436
   b) 4831
   c) 5508
   d) 4832

   Answer: d) 4832

   Solution:
   We need to use f(1) to calculate the value of f(17)
   f(17) can be written as f(1+16)
   f(16) can be written as f(8+8)
   f(8) can be written as f(4+4)
   f(4) can be written as f(2+2)
   f(2) can be written as f(1+1)
   f(1) = 0, so f(2) = f(1+1) = f(1)+f(1)+4(9*1*1-1) = 32.
   or, f(4) = f(2+2) = f(2)+f(2)+4(9×2×2 – 1) = 32+32+4×35 = 204.
   or, f(8) = f(4+4) = f(4)+f(4)+4(9×4×4 – 1) = 204+204+4×143 = 980
   or, f(16) = f(8+8) = f(8)+f(8)+4(9×8×8 – 1) = 980+980+4×575 = 4260
   or, f(17) = f(1+16) = f(16)+f(1)+4(9×16×1 –1) = 4260+0+ 4×143 = 4832

3. 3000 卢比的总和分配给 P、Q 和 R。P 得到 Q 和 R 加起来的 2/3，R 得到 P 和 Q 加起来的 1/3，R 的份额是多少？
   一）750
   b) 850
   c) 800
   d) 700

   Answer: a) 750

   Solution:
   According to the question,
   case 1: P = 2(Q + R)/3
   or, (Q+R)/P = 3/2
   case 2: Also, R = (P+Q)/3
   or, (P+Q)/R = 3/1
   Simply using componendo-dividendo, we get,
   for case 1, (P+Q+R)/P = 3+2/2 = 5/2 = 20/8
   for case 2, (P+Q+R)/R = 3+1/1 = 4/1 = 20/5
   On solving we get, P = 8, Q = 7, R = 5
   or R’s share = 5/(8+7+5) * 3000 = 750

4. 在给定的系列 11, 23, 47, 83, 131, … 下一个数字是什么？
   145
   乙) 178
   c) 191
   d) 176

   Answer: c) 191

   解决方案：

   The given series follows the order of multiple of 12
   23 – 11 = 12
   47 – 23 = 24
   83 – 47 = 36
   131 – 83 = 48
   x – 131 = 60
   or x = 191

5. 如果一个数除以 357，余数是 5，如果这个数除以 17，余数是多少？
   一）9
   b) 3
   c) 7
   d) 5

   Answer: d) 5 

   Solution:
   Let the number be N when divided by 357 leaves remainder 5 and quotient q.
   So, N = 357k + 5 = 17 * 21 * k + 5
   So, 357 is exactly divisible by 17 so remainder is 5

6. 一根高 36m 的杆子在一条断了一定高度的路的边上。它掉下来的方式是杆子的顶部接触到道路的另一边。如果道路的宽度是12m，那么杆子断裂的高度是多少？
   一）12
   b) 16
   c) 24
   d) 18

   Answer: b) 16

   Solution:
   Let the point at which the pole broke be ‘x’ from the ground, so the length of the broken piece be (36-x).
   So applying Pythagoras theorem we get, $$(36-x)^2 = x^2 + 12^2$$
   => $$1296 + x^2 - 72x = x^2 + 144$$
   => 72x = 1296 – 144
   => x = 16

7. 有一个由23人组成的大厅。他们正在握手。那么，如果它们处于一对循环序列中，可能会握手多少次呢？
   一）23
   b) 22
   c) 253
   d) 250

   Answer: c) 253

   Solution:
   Since there are 23 people, number of handshakes possible = 23C2 = 253 handshakes.

8. 在地下室，有一些自行车和汽车。周二地下室有 182 个轮子。有多少辆自行车？
   一）20
   b) 19
   c) 18
   d) 16

   Answer: b) 19

   Solution:
   This is a very ambiguous question and must be calculated using the options.
   If there are 20 bicycles, there must be 20*2 = 40 wheels
   Remaining wheels = 182-40 = 142 wheels = 142/4 is not an integer so there cannot be 20 bicycles.
   Similarly checking for 19 bicycles = 19*2 = 38 wheels
   Remaining wheels = 182 – 38 = 144 = 144/4 = 36 cars hence this is the answer.

9. 有一个 17 × 8 m 的矩形地面，周围有 1.5 m 宽的路径。路径的深度为 12 厘米。沙子被填充并找到所需的沙子数量。
   a) 5.5
   b) 10.08
   c) 6.05
   d) 7.05

   Answer: b) 10.08

   Solution:
   Area of the inner rectangle = 17 * 8 = 136 meter-square
   Area of the outer rectangle = (17 + 2*1.5) * (8 * 2*1.5) = 220 meter-square
   So area of the remaining path = 220 – 136 = 84 meter-square
   So sand required to fill the path = 84 * (12/100) = 10.08 meter-square

10. 数字 272738 和 232342 除以两位数 n 时，余数分别为 13 和 17。求 n 的数字之和？
    一）5
    b) 4
    c) 7
    d) 8

    Answer: c) 7

    Solution:
    So according to the question, (272738 – 13) and (232342 – 17) are exactly divisible by n.
    So if we find the HCF of these two numbers, we get n,
    The HCF of 272725 and 232325 is 25
    So the sum of the digits = 7.