TCS安置纸| MCQ 3

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. 如果f(x)= ax ^ 4 – bx ^ 2 + x + 5且给定f(-3)= 2，则f(3)=? (a ^ b =升为b的幂)
   a)3
   b)8
   c)1
   d)-2

   Answer: b) 8

   解决方案：

   We can directly solve:
   => f(-3) = a(-3)^4 – b(-3)^2 + (-3) + 5 = 2
   => 81a – 9b + 2 = 2
   => 81a – 9b = 0
   Now solving f(3),
   => f(3) = 81a – 9b + 8
   => f(3) = 0 + 8 = 8(Answer)

2. 有一家巧克力工厂，将巧克力分配给一个班级。它会向50名学生提供为期30天的巧克力课程，同时请记住，所有学生都可获得相同数量的巧克力。在最初的10天里，只有20名学生在场。多少学生被容纳到小组中，以便所有巧克力都被消耗掉?
   a)70
   b)55
   c)60
   d)45

   Answer: d) 45

   解决方案：

   Let each student get 1 chocolate each, so the total number of chocolates = 50 * 30 = 1500 chocolates.
   For first 10 days 20 students were present, so total chocolates consumed = 20 * 10 = 200 chocolates.
   Chocolates left = 1300. These are to be distributed for the next 20 days. Therefore in each day 1300 / 20 chocolates were to be consumed which = 65 chocolates per day.
   So the required answer = 65 – 20 = 45 chocolates.

3. 给定log(0.318)= 0.3364和log(0.317)= 0.3332，找到log(0.319)?
   a)0.3396
   b)0.3394
   c)0.3393
   d)0..390

   Answer: a) 0.3396

   解决方案：

   => log(0.319) = log(0.318) + (log(0.318) – log(0.317))
   = 0.3364 + (0.3364 – 0.3332)
   = 0.3364 + 0.0032
   = 0.3396 (Answer)

4. 有一组20名学生，其中18名是男孩，2名是女孩。他们应以圆形方式坐下，以使两个女孩始终被男孩隔开。可以通过几种方式安排学生?
   a)12
   b)18！x2
   c)17×2！
   d)17！

   Answer: b) 18!x2

   解决方案：

   There are in all 20 places out of which if one girl sits in one position then the other girl may sit either to her left or right skipping one place, which is to be filled by a boy. So total number of ways the boys can sit = 18! ways and girls may alternate there sits so the total answer would be = 18! * 2 ways.

5. 拉姆(Ram)参加考试。在论文A中，他在70分中得18分。在论文B中，他在30分中得14分。那么他在哪篇论文中表现更好?
   a)纸A
   b)纸B

   Answer: b) Paper B

   解决方案：

   We just need to calculate the percentage he scored in each paper.
   In paper A: (18/70) * 100 = 25.7%
   In paper B: (14/30) * 100 = 46.6% (Answer)

6. 10小时后，凌晨2点从18N 10E的地方起飞，降落在36N 70W的地方。目的地的当地时间是几点?
   a)6:00上午
   b)上午6:40
   c)上午7:40
   d)7:00上午
   e)上午8:00

   Answer: b) 6:40 a.m

   解决方案：

   Let’ calculate the difference in the number of latitudes = 70 + 10 = 80 degrees towards east.
   We know 1 degree = 4 min, so 80 degrees = 80 * 4 = 320 mins
   320 mins = 5 hr 20 minutes
   Now the plane landed 10 hours later so time of landing = 12 hrs according to the starting place
   So time at destination = 12 hrs – 5 hrs 20 min = 6 hr 40 mins(Answer)

7. 沿着铁轨以9 km / hr的速度进行的田径运动。轨道长240m，在以相同的方向以45km / hr的速度行驶的120m长的火车之前。火车将花费多少时间才能完全越过运动员?
   a)3.6秒
   b)18秒
   c)72秒
   d)36秒

   Answer: d) 36 sec 

   解决方案：

   Let’s try to find the relative speed = 45 – 9 = 36km/hr
   = 36 * 5/18 = 10m/s
   Now the total distance needed to be covered by the train to completely cross the athlete = 240 + 120 = 360m
   So time = dist/speed = 360/10 = 36 seconds

8. A需要3天才能完成工作，而B需要2天。他们俩都完成了工作并赚取了卢比。 150. A的份额是多少?
   a)卢比。 70
   b)卢比。 30
   c)卢比。 60
   d)卢比。 75

   Answer: c) 60

   解决方案：

   A completes 1/3rd of work in one-day and B completes 1/2 of work in one day. So the ratio of there work is:
   A:B = 2:3
   So A’s share = (2/5)*150 = 60 rupees(Answer)

9. Ram和Shyam的工资之比为2：3。如果两个人的工资都分别增加4000卢比，则新比例变为40:57。 Shyam目前的工资是多少?
   a)卢比。 17，000
   b)卢比。 20，000
   c)卢比。 25、500
   d)这些都不是

   Answer: d) None of these

   解决方案：

   Let Rams and Shyams salary be ‘x’. The ratio of there salary according to the question is 2x:3y
   According to the question,
   (2x+4000):(3x+4000) = 40:57
   On solving we get 3x = 34000
   Therefore, the present salary of Shyam is Rs. 34000

10. SI的总金额将以10％的年利率每年2％递增吗?
    a)6％
    b)5 2/3％
    c)4％
    d)6 2/3％

    Answer: c) 4%

    解决方案：

    Let the sum of money be Rs ‘x’. So SI = 2x/5
    So, rate = (SI*100)/(P*Time)
    => (2x*100)/(5*x*10)
    => 4 % (Answer)