TCS Latest Placement paper 2014 Set-1

This set of questions are asked at on-Campus placement drive for the session 2013-14. There are few questions that are repeated from the previous year's paper (2012-13) as told by many students. We are able to get some questions from them.

1. Ahmed, Babu,  Chitra,  David  and  Eesha  each  choose  a  large  different  number.  Ahmed says, “ My number is not the largest and not the smallest”. Babu says, “My number is not the largest and not the smallest”. Chitra says, “My number is the largest”. David says, “My number is the smallest”. Eesha says, “ My number is not the smallest”.

Exactly  one  of  the  five  children  is  lying.  The  others  are  telling  the  truth.  Who  has  the largest number?

Ans. Eesha

Explanation: Ahmed and Babu cannot lie because each of them say two facts (not the largest,Not the smallest) and there is no chance for both the facts to be wrong. David says “My number is smallest”. If David lies, one of the remaining four should lie. But  exactly one  person  lies  in  this  problem.  So  David  says  the  truth.If  David’s statement is true, Eesha’s statement is also true. The one who lies is Chitra and Eesha has the largest number.

2. A cow and a horse are bought for Rs.200000. The cow  is sold at a profit of 20% and the horse is sold at a loss of 10%. The overall gain is Rs.4000. The cost price of the cow is

Ans. 80,000

Explanation: Let the cost price of cow and horse is C and H Respectively

C + H = 200000  -  (1)

1.2C + .9H = 204000  -  (2)

Solving equation (1) & (2)

C = 80000.

3. If X^Y denotes X raised to the power Y, Find the last two digits of ( 1941 ^ 3843 ) + ( 1961^4181).

Ans.  82

Explanation: 1941^2 ends in 81. 1941^3 ends in 21, 1941^4 ends in 61, 1941^5 ends in 01 and 1941^6 ends in 41 and this cycle keeps repeating. Similarly the cycle for 1961 powers is 61, 21, 81, 41, 01 and the cycle repeats. After adding up the final two digits of these numbers for their respective powers, we find that the sum is 82.

4. George can do some work in 8 hours, Paul can do the same work in 10  hours while Hari can  do the  same  work  in  12  hours. All the  three  of them  start  working  at  9  a.m  while George stops work  at 11 a.m and remaining two complete the work. Approximately at what time will the work be finished?

Ans. 1 pm

Explanation: Total number of work to be done= 120 Units (LCM of 8,10,12)

George’s one hour work = 120/8 = 14 Units

Paul’s one hour work = 120/10 = 12 Units

Hari’s one hour work = 120/12 = 10 Units

Units of work finished at 11 AM = (14+12+10)*2 = 74

Remaining work to be done = 120-74 = 46 units

One hour Paul + Hari work = 22 units

Approximately they will take two hours to finish the work

So the work will get finished at 1 PM

5. If M is 30% of Q, Q is 20% of P and N is 50% of P then M/N =

Ans. 3/25

Explanation: M is 30% of Q

Q is 20% of P

Nis 50 % of P

Then M/N = ?

Let P= 100

N = 50

Q = 20

M = 6

M/N = 6/ 50 =3/25.

6. In a office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on the top of the pile in the secretary’s inbox. When there is time, the secretary takes the top letter off the pile and type’s it. If there are five letter inall , and the boss delivers in the order of 1 2 3 4 5, which of the following could NOT be the order in which secretary types them.

Ans. 4 5 2 3 1

Explanation: Going by the options and checking logically which order is  possible.

7. There  are  5  sweets  –  Jumun,  Kulfi,  Peda,  Laddu  and  Jilabi  that   I  wish  to  eat  on  5 consecutive days –  Monday through Friday, one sweet a day, based on the following self imposed constraints:

1)  Laddu is not eaten on Monday

2)  If Jamun is eaten on Monday, then Laddu must be eaten on Friday

3)  If Laddu is eaten on Tuesday, Kulfi should be eaten on Monday

4)  Peda is eaten the day following the day of eating Jilabi

 Based on the above, peda can be eaten on any day except?

Ans. Monday

Explanation: Peda  can  be  had  only  after  having  Jilabi.  So  Peda  can  never  be  had  on  the starting day, which is Monday.

8. At 12.00 hours Jake starts to walk from his house at 6 kms an hour. At 13.30 hours, Paul follows him from Jake’s  house on his bicycle at 8 kms per hour. When will Jake be 3 kms behind Paul?

Ans. 19:30 hours

Explanation: Jake starts at 12.00 and covers 6 km/h.

Paul starts at 1.30 and covers 8 km/h.

Relative speed between Jake & paul is 2 kmph, where Paul stating Jake is 9 km ahead  of Paul.  From  13.30  hours  paul  takes  4.30  hrs  to  meet  Jake.  Again  he  needs  1.30  hrs  to  lead  Jake  by  3  km  Relative  speed.  Totally he  takes  6  hrs.  so 13.30+6 = 19.30 hrs.

9. Jake can dig a well in 16 days. Paul can dig the same well  in 24 days. Jake, Paul and Hari together dig the well in 8 days. Hari alone can dig the well in ?

Ans. 48 days.

Explanation: Given that speed of Jake is greater than Paul.

Distance = 24 km

Sum of their speed is 7 km/h = J+P

So possible speed ratio between J & P is

Go by Option

6:1 Not in option

5:2 = (24/5)+(24/2) ≠ 14 Hours

4:3 = (24/4)+(24/3) = 14 Hours

So Jake’s speed is 4 km/h.

10. If  a lemon and an apple together cost Rs. 12.00, a tomato and a lemon  cost Rs. 4.00 and an apple cost Rs.8.00 more than a tomato or a lemon then which of the following can be Page  2/4 the price of a lemon?

Ans. 2

Explanation: Let cost of a Lemon is L

Let cost of a Apple is A

Let cost of a Tomato is T

L+A = 12 – (1)

T+L = 4  -  (2)

A = 8+L  -  (3)

A = 8+T  -  (4)

Sub (3) in (1)

L+8+L = 12

L = 2, A = 10, T = 2.

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