TCS Placement paper 2014 (9)

This placement paper is collected from on-campus recruitment (2013-2014). TCS already declared that it would organize its off-campus online aptitude test on 8th and 9th March 2014. All TCS off-campus 2014 aspirants go through all set of TCS placement paper posted in interviewtime.in , as these are the exact questions from TCS Seesame ( online test application) portal.

1. Two dice are thrown. What is the probability of getting a sum of 5 before a sum of 7 appears?

Ans. 0.4

Explanation:No. of events results in sum of 5 : (1,4),(2,3),(3,2),(4,1) = 4
No. of events results in sum of  7 : (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) =6
Total possibilities = 10
Probability of getting a sum of 5 before 7 appears is 4/10 = 0.4

2. Sachin can do a piece of work in 16 days, Amit can do the same work in 12 days while Ram can do it in 32 days. All of them started to work together but Sachin leaves after 4 days. Amit leaves the job 3 days before the completion of the work. How long the work last?
 
Ans. 9 days.

Explanation: Let, the work will be complete in x days.
Sachin worked for 4 days.
Amit worked for (x-3) days
Ran worked for x days.
Therefore, 4/16+(x-3)/12+x/32=1
By solving we get,
11x=96
x= 96/11 days (9 days approx..) Only Option 9 days is given 

3. If A speaks the truth 60% of the times, B speaks the truth 50% of the times. What is the probability that at least one will tell the truth, narrating the same incident?

Ans. 0.8

Explanation. Probability of A speaks truth 6/10.
Therefore, Probability of A speaks false =(1- 6/10)=4/10.
Probability of B speaks truth =5/10
Probability of B speaks false =(1-5/10)=5/10
Therefore, Probability that neither of them tell truth = 4/10*5/10=2/10
Therfore, Probability that at least one will tell the truth=(1- 2/10)=8/10=0.8



4. If three dies are thrown simultaneously probability of getting sum as 10?

Ans. 27/216

5. How many words with or without meaning, can be formed by using all the letters of the word, 'DELHI',using each letter exactly once?

Ans. 120

Explanation:No. of permutation possible taking all at a time.
Therefore, ⁵P₅=5!=120

 6. In how many ways can we distribute 10 pencils to 4 children so that each child gets atleast one pencil?

Ans. 84

Explanation: No, of way to distribute 10 pencils to 4 children.
By formula , we know
Number of ways of distributing n identical objects to r distinct objects so that each get atleast one ⁽ⁿ⁻¹⁾C_(r−1) .Therefore, we the solution is  ⁽¹⁰⁻¹⁾C₍₄₋₁₎=⁹C₃ = 84

7. How  many number plate with 1,2,3,4,5 digits can  be created, if repetition of numbers is not allowed?

Ans. 325

Explanation: No. of no. plate possible with 1 digit = 5
No. of no. plate possible with 2 digits = 5 * 4 =20
No. of no. plate possible with 3 digits = 5*4*3 = 60
No. of no. plate possible with 4 digits = 5*4*3*2=120
No. of no. plate possible with 5 digits = 5*4*3*2*1=120
Total no. of number plate possible =120+120+60+20+5=325

8. 70, 54, 45, 41,?

Ans: 40

Explanation: 70 - 54 = 16 = 4²
54 - 45 = 9 = 3²
45 - 41 = 4 = 2²

Therefore, now difference should be 1² . So, 41-1 = 40.

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