TCS 安置文件 |多选题 9

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. a+b除以12余数为8，a-b除以12余数为6。如果a>b，ab除以6余数是多少？
   一）3
   b) 1
   c) 5
   d) 4

   Answer: b) 1

   Solution:
   According to the question,
   a + b = 12k + 8
   => $$(a+b)^2 = 144k^2 + 64 + 192k$$
   a – b = 12l + 6
   => $$(a-b)^2 = 144l^2 + 36 - 144l$$
   Subtracting both the equations we get,
   ab = $$36(k^2 - l^2) + 48k - 36l + 7$$
   Now all the terms of ab is divisible by 6, except 7. So the remainder left is 1.

2. 有一组 26 个问题。每答错一题扣五分，答对一题加八分。假设所有问题都回答了，得分为0，正确回答的问题有多少？
   一）12
   b) 10
   c) 11
   d) 13

   Answer: b) 10

   Solution:
   This can be easily solved using hit and trial method.
   Let’s consider the first option. If 12 questions in all are answered correctly, then the total score = 12 * 8 = 96 marks.
   If 12 questions are answered correctly, then 14 questions were wrongly answered. So total deductions = 14 * 5 = 70 marks.
   So total score = 96 – 70 = 26 which is not correct.
   Let’s consider the second option. If 10 questions in all are answered correctly, then the total score = 10 * 8 = 80 marks.
   If 10 questions are answered correctly, then 16 questions were wrongly answered. So total deductions = 16 * 5 = 80 marks.
   So total score = 80 – 80 = 0
   Hence 10 is the correct option.

3. 一天，拉梅什从家迟到了 30 分钟，以比平时慢 25% 的速度开车，迟到了 50 分钟到达市场。 Ramesh 从家里到市场通常需要多长时间（以分钟为单位）？
   一）20
   b) 40
   c) 60
   d) 80

   Answer: a) 60

   Solution:
   Let the usual speed of Ramesh be ‘s’
   Let the distance between home and market be ‘d’
   So usual time took = d/s
   Time took on that particular day = d/(3s/4)
   So according to the question,
   d/s(4/3 – 1) = 20
   or, d/s = 60

4. 三个容器 A、B 和 C 分别装有比例为 1:5、3:5、5:7 的牛奶和水的混合物。如果容器的容量是 5:4:5 的比例，如果三个容器都混合在一起，请找出牛奶与水的比例。
   一）54:115
   b) 53:113
   c) 53:115
   d) 54:113

   Answer: c) 53:115

   Solution:
   Using the weighted average formula we can calculate the weight of milk,
   => [5*(1/6) + 4*(3/8) + 5*(5/12)]/(5+4+5) = 53/168
   So weight of water = 168 – 53 = 115
   So the ratio of milk to water = 53:115

5. 阿曼参加了一场橙色赛跑。比赛中，20个橙子以4米的间隔排列，第一个橙子距起点24米。要求阿曼一次一个地把橙子带回起点。他要跑多远才能带回所有的橙子？
   一）1440
   b) 2440
   c) 1240
   d) 2480

   Answer: d) 2480 

   Solution:
   Since every orange is placed at a difference of 4 meters and the first potato is placed at 24 meters from the starting position. Every orange is placed at 24m, 28m, 32m, 36m, ….20 terms.
   Now to bring ever orange one at a time, Aman needs to cover the double of the distance = 48, 56, 64, …20 terms.
   So putting the values in the sum of AP formula, a = 48, d= 8, n = 20.
   Total distance travelled = 20/2 [2 * 48 + (20-1)*8]
   = 2480 meters

6. 有两副牌，每副牌包含 20 张牌，上面写着 1 到 20 的数字。从每一副牌中随机抽取一张牌，得到数字 x 和 y log x + log y 为正整数的概率是多少。 （日志被带到基地 10。）
   a) 7/400
   b) 29/100
   c) 3/200
   d) 1/80

   Answer: a) 7/400

   Solution:
   We know that log x + log y = log xy
   for log xy to be positive, we have the following choices:
   (1, 10), (10, 1), (10, 10), (5, 20), (20, 5), (2, 5), (5, 2)
   So the probability = 7/400

7. 有一个可容纳10人的锥形帐篷。每个人需要 6 平方米的空间坐下和 30 立方米的空气呼吸。圆锥的高度是多少？
   a) 72 m
   b) 15 m
   c) 37.5 m
   d) 155 m

   Answer: b) 15 m

   Solution:
   All the persons are to sit on the ground forming the base of the cone.
   Total base covered = pi * $$r^2$$ = 6*10 = 60 sq-meter.
   The total volume of the tent will be equal to the total air to breathe by the 10 people = 30*10 = 300 cubic meter
   So, 1/3(pi * $$r^2$$ * h) = 300
   => h = 15 meters.

8. 找出可以整除125的143的最大幂！确切地。
   一）11
   b) 8
   c) 9
   d) 7

   Answer: c) 9

   Solution:
   We can write 143 = 11 × 13.
   So the highest power of 13 should be considered in 125!, which is 9 (13 * 9 = 117)
   The highest power of 11 in 125! is 12 (11 * 11 = 121 and remaining 1).
   That means, 125! = 11^12×13^9×…
   So only nine 13’s are available. So we can form only nine 143’s in 125!. So maximum power of 143 is 9.

9. 一辆车在下午 6:00 开始。从起点以18 m/s的速度到达目的地。它在那里等待了 40 分钟，然后以 28 m/s 的速度返回。计算到达目的地所需的时间。
   a) 晚上 9:44
   b) 晚上 8:32
   c) 晚上 7:30
   d) 晚上 9:30

   Answer: a) 9:44 pm

   Solution:
   Let the distance covered be D m
   The time to cover the starting distance = D/18 secs.
   The time taken for the reverse journey = D/28 secs.
   According to the quesiton,
   D/18 – D/28 = (40 × 60)
   On solving this we get,
   D = 2400 × 252/5 = 120960 m
   No the total time taken = (D/18) + (D/28) + 2400 = 13440 seconds
   = 3 hours and 44 minutes
   Therefore, the bus reaches back at 9:44 PM

10. 房子的价值每年都会贬值，折旧是年初价值的 3/4。如果踏板车的初始值为 Rs。 40, 000. 3 年末的价值是多少？
    一）卢比。 19000
    b) 卢比。 16875
    c) 卢比。 17525
    d) 卢比。 18000

    Answer:  b) 16875

    Solution:
    This is the question of succession depreciation.
    the starting amount = Rs. 40000
    This reduces by 3/4 th of its initial value every year = (40, 000) * (3/4)^3 = 16875