TCS placement paper questions with complete solutions

The placement papers for TCS have seen a change in the type of questions asked after a long period of about 6 years. The new set of questions pose a challenge that is a touch higher than that presented in previous TCS placement papers. One has to prepare himself extensively in the 2012 TCS pattern to obtain a realistic chance of getting through to the second round of placement with a more than reasonable score. Read through the following set of questions to get a gist of what TCS has in store for campus recruit hopefuls.

1.Consider three positive integers x, y and z such that the three integers satisfy the following equation:
28x+30y+31z=365
What would be the sum of these three positive integers? i.e. What is the value of x+y+z?
(a) 13
(b) Less than or equal to 11
(c) 12
(d) Greater than 13

Solution: Unfortunately, we have only one equation to solve three unknown variables x, y and z. This data is not sufficient and the variables may have infinite values. But we can narrow down our options because we are given two conditions that the variables cannot have negative values and that they cannot take a decimal part in the value. Observe the coefficients of the three integer variables x, y and z carefully. The coefficient of y is 30, which means that regardless of the value of y, the unit's place of the result of 30y will always be zero. For eg: if y=5 then 30y=150. But on observing the RHS of the equation, one notices that the unit's place of 365 is 5. We can thus conclude that the unit's place of the RHS is contributed by appropriate selections of values for x and z, i.e the unit's place of the values of 28x and 31z should be such that they add up to give the result as 5 which is the unit's place of the RHS.
We can proceed on a trial and error basis. First assume x=1; i.e. 28x=28. To obtain a 5 as the resultant unit's place, we need to obtain the value of 31z such that it ends with 7. The most obvious choice of z in this case would be 7, i.e z=7 to give 31z a value of 217. 28x+31z would then give a value of 245. This would mean that 30y would be expected to provide a value of 120 to take the sum of the equation to 365, which would occur when y=4. Then according to the question x+y+z=1+4+7=12.

Next assuming x=2; i.e 28x=56. To obtain a 5 as the resultant unit's place, we need to obtain the value of 31z such that it ends with a 9. The most obvious choice of z in this case would be 9, i.e. z=9 to give 31z a value of 279. 28x+31z would then give a value of 335. This would mean that 30y would be expected to provide a value of 30 to take the sum of the equation to 365, which would occur when y=1. Then according to the question x+y+z=2+1+9=12. Any other values assumed for x will not provide a satisfactory combination of y and z values within the conditions put forward in the question. Hence we see in both cases the solution to the question would 12.
Answer: (c) 12

Shortcut: Looking closely at the equation, we can notice that the coefficients are the number of days in the months and the RHS of the equation is the number of days in a year. February is the month with 28 days. There are 4 months in the year with 30 days and 7 months consisting of 31 days. Hence x=1, y=4, and z=7. This would give a solution x+y+z=12 which is option (c).

2. Mahesh had a 100 rupee note with him which he used to pay for a certain commodity in a shop. The shopkeeper returned him the remaining balance. Mahesh noticed that the remaining balance exceeded four times the sum of its digits by 3. He also noticed that if the money returned by the shopkeeper were increased by 18, then it would result in a value that is the same as the number formed by reversing its digits. What is the price of the commodity?
(a) 65
(b) 58
(c) 51
(d) 43

Solution:
Assume 'x' to be the price of the commodity and 'y' to be the amount of balance returned by the shopkeeper. From the question we can obtain two equations for the balance returned 'y'. But first we must assume the balance 'y' to consist of two digits 'ab' (Please note that the remaining balance is not the product of ab but the variable 'a' is the tens digit of the balance and 'b' is its unit's digit) i.e.y=10a+b
(a) Balance exceeds four times the sum of its digits by 3.
10a+b=4(a+b)+3
(b) If balance is increased by eighteen, then the result is equal to value obtained by reversing the digits of the balance.
10a+b+18=10b+a

On solving the two equations, we get a set of simultaneous equations
2a-b= 1
a-b=-2
Substituting b=a+2 in the first equation we obtain the value of a=3, and further analysis will result in the value of 'b' to be b=5. Hence y=10a+b=30+5=35. Consequentially the value of x=100-y=100-35=65.
Answer: (a)65

Shortcut: From observing the question we notice that the easier of the two equations to verify would be the second one. We can observe the options and come to the conclusion that the remaining balance to be worked with will be the following which is obtained from the options.
(a)35 (b)42 (c)49 (d)57
When we reverse the digits of each option and observe if they are equal to the addition of 18 to the option, we notice that this is possible in only two of the four options i.e.
(a)35+18=53 and (d)57+18=75
Next we only have to verify the first equation for these two values alone i.e.
35= 4(3+5)+3
57= 4(5+7)+3
The RHS=LHS only in the first equation among the two and hence the balance obtained by Mahesh is 35 and subsequently the price of the commodity is Rs65.

3. There are three numbers such that their average is 10 more than the least number of the three and 15 less than the greatest number of the three. The median of the three numbers is given to be 5. Then what would be the sum of the three numbers in question?
(a) 5
(b) 20
(c) 30
(d) 25

Solution:
Assume the average to be a variable 'avg'. Then the least number would be avg-10 and the greatest number would be avg+15. It is given that the value of the middle number is 5. Hence according to the definition of average of numbers, we have:
(avg-10 + 5 + avg+15)/3 = avg
Solving this equation we obtain avg=10. Also, the sum of the three numbers is equal to 3*avg i.e 3*10=30.
Answer: (c) 30

4. In a certain code language X V T R P is coded as 24-22-20-18-16. How would L J H F E be coded as in the same code language?
(a) 12-10-8-6-5
(b) 12-10-8-6-4
(c) 10-12-4-6-10
(d) 22-20-18-16-15

Solution:
The code language can be observed to be as one which assigns the digits 1-26 to alphabets from a-z. In this corresponding manner, the value of L=12, J=10, H=8, F=6, E=5.
Answer: (a) 12-10-8-6-5

5. 5 friends seated around a table in a restaurant make the following statements to the waiter.
Alex: Exactly 1 of us is lying.
Bala: Exactly 2 of us are lying.
Calvin: Exactly 3 of us are lying.
Dalton: Exactly 4 of us are lying.
Ebenezer: Exactly 5 of us are lying.
Who can the waiter consider to be telling the truth?
(a) Alex
(b) Calvin
(c) Bala
(d) Ebenezer
(e) Dalton

Solution: If we consider Alex's statement which says that 'exactly one of them is lying' to be true, then a contradiction arises when we try to find out who that particular person is, that is lying. On observing the statements of the other four, all of them are contrasting from each others' statement hence there is more than one person lying, and Alex can be considered to be lying when he makes his statement. The same idea can be applied to the next two persons Bala and Calvin. The statement made by Ebenezer is self contradictory because the five people that he is mentioning to be lying includes him and thus we can't consider his statement to be true. Only Dalton's statement can be taken as true as the exact four people that he mentions to be lying are the other four people at the table.
Answer: (e)Dalton