TCS - Recent Questions - 4

TCS – Recent Questions – 4

1.       In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. Secretary takes the top letter and types it.  Boss delivers in the order 1, 2, 3, 4, 5. Which cannot be the order in which the secretary types?
a) 2, 4, 3, 5, 1
b) 4, 5, 2, 3, 1
c) 3, 2, 4, 1, 5
d) 1, 2, 3, 4, 5
Ans: (b)
Assuming the typist types letter No 4, the boss comes and place letter No 5 on top. There are only five letters and so, beneath the letter No 5 should be letter No 3. Hence the sequence under option (b) is not correct.

2.       Sum of two numbers is 50 & sum of their reciprocals 1/12. Find the two numbers.

Ans: The numbers are 20 & 30

X + y = 50 -> x = 50 – y

1/x + 1/y = 1/12.   Substituting the value of ‘x’ in this we have -> 1/50-y + 1/y = 1/12

Solving we get -> y² – 50y + 600 = 0

This can be written as y² – 30y – 20y + 600 = 0. This can further be written as

Y(y-30)  - 20(y-30) = 0 Thus we get two value for ‘y’ as 30 and 20.

Substituting either of the value we get value for ‘x’ as 20 or 30.

Hence, the two numbers are 20 and 30

3.       Two cars start from the same point at the same time towards the same destination which is 420 km away.  The first and second car travel at respective speeds of 60 kmph and 90 kmph.  After travelling for some time the speeds of the two cars get interchanged.  Finally the second car reaches the destination one hour earlier than the first.  Find the time after which the speeds get interchanged?

Ans: 4 hours.

Let the two cars be A and B and the total time taken by them to reach the destination be ‘X’ and ‘Y’.  Let the speeds of the car be interchanged after  ‘t’ hours. Now we have the following:

Car A:  60t + 90(X-t) = 420 -> 90X – 30t = 420. ------------- (I)

Car B:  90t + 60(Y-t) = 420 -> 60Y + 30t = 420. --------------(II)

Adding (I) and (II) we get -> 90X + 60Y = 840. We are informed the second car reaches the destination one hour earlier than the first car. Car B reaches One hour earlier than Car A.

So, we have (X – Y) = 1 or Y = X – 1. Substituting this value of Y we get

90X + 60(X-1) = 840 solving we get the value of ‘X’ as 6. Apply this value in equation (I) and we get the value of ‘t’ as 4 hours.

4.       Kate wanted to buy 2kgs of apples. The vendor kept the 2kg weight on the right side and weighed 4 apples for that.  She doubted on the correctness of the balance and placed 2 kg weight on the left side and she could weight 14 apples for 2 kg. If the balance was correct how many apples she would have got?
Ans: 9 apples

From the information it is apparent the left pan is heavier than the right pan. Let the difference in weight between the two pans be ‘w’

When the vendor weighed -> 2kg = 4 A (apples) But actually 4 A + w. ------- (1)

When Kate weighed ->  2 + w = 14 A -----------------------(2)

Equating the two and solving we get number of Apples as 18 distributed in both the pans. Thus she would have got 9 apples for 2kg.

5.       In a staircase, there are 10 steps. A child is attempting to climb the staircase. Each time she can either make 1 step or 2 steps.  In how many different ways can she climb the staircase?
a) 10
b) 22
c) 36
d) None of these
Ans: (d)
The question is best answered using the Fibonacci series, with the starting terms as 1, 2. The series then will go as 3, 5, 8, 13, 21, 34, 55, etc.

6.       Letters of alphabets are consecutively assigned with numbers with 1 assigned to A and 26 to Z.  By 27th letter we mean A 28th B. In general 26m+n, m and n negative integers is the same as the letters numbered n. 
A strange country military general sends this secret message keeping p=6 with the following codification scheme. In codifying a sentence, the 1st time a letter occurs it is replaced by the ‘p’ th letter from it. 2nd time if it occurs it is replaced by P^2 letter from it. 3rd time it occurs  it is replaced by p^3 letter from it.   What is the code word for ABBATIAL
a) GHNNZOOR
b) GHKJZOHR
c) GHHGZOGR
d) GHLKZOIR
Ans: (d)
First letter A should be coded as 1 + 6 = 7 -> G   (Occurring for the first time)

            Second letter B to be coded as 2 + 6 = 8 -> H      (Occurring for the first time)

           Third letter B should be coded as 2 + 36 = 38 – 26 = 12 -> L  (Second time occurrence)

           Fourth letter A to be coded as 1 + 36 = 37 – 26 = 11 -> K (Second time occurrence)

           Likewise for the other letters. Answer is  GHLKZOIR            

7.       If f(1)=4 and f(x+y)=f(x)+f(y)+7xy+4, then f(2)+f(5)=?

Ans : 125

                Let us keep x = 1 and y = 1

                f(1+1) = f(1) + f(1) + 7 x 1 x 1 + 4 => 4 + 4 + 7 + 4 -> f(2) -> 19.

               

                Let x =2 and y = 2

                F(2+2) = f(2) + f(2) + 7 x 2 x 2 + 4 => f(4) -> 19 +19 + 7 x 2 x 2 + 4 -> 70

                Let x = 1 and y = 4

                f(1 + 4) = f(1) + f(4) + 7 x 1 x 4 +4.  Substituting the value f(1) = 4 and f(4) = 70

We have  ->  4 + 70 + 28 + 4 = 106 ----- f(5)

Hence the value of f(2) + f(5)  = 19 + 106 ->  125.

8.       If f(f(n))+f(n)=2n+3 and f(0)=1, what is the value of f(2012)?
a) 2011
b) 2012
c) 2013
d) 4095
Ans: (c)

Let n = 0   then f{f(0)} + f(0) = 2(0) + 3 -> f(1) + 1 = 3 ->  f(1) = 3-1 = 2

Let n = 1   then f{f(1)} + f(1) = 2(1) + 3 -> f(2) + 2 = 2(1) + 3 = 5 -> f(2) 5 – 2 = 3

On the same basis

f(2012) is equal to 2013.

9.       The sum of three digits of a number is 17.  The sum of the square of the digits is 109.  If we subtract 495 from the number, the digits of the number are reversed.  Find the number.
Ans: 863

Let the digits of the number be  A, B, and C.   then,

A + B + C = 17 ----------- (1)

A² + B² + C² = 109 -------(2)       since the number is three digits

100A + 10B + C – 495 = 100C + 10B + A -----(3)

From (3) we get (A – C) = 5. Now applying trial and error method we can have the following combinations for the three digits A, B and C namely

(6, 10, 1),  (7, 8, 2), (8, 6, 3), (9, 4, 4)  Of these (8, 6, 3) satisfies the condition.

Hence the answer is 8, 6, 3

10.   5000 voted in an election between two candidates. 14% of the votes were invalid. The winner won by a margin approximately closer to 15%. Find the number of votes secured by the winner.
a)2650                   b)2564                  c)2473                   d)2360

Ans: (c)

Total votes polled 5000. Invalid votes 14% equalling to 700. Hence valid votes -> 4300

The margin of victory approx. 15%  ->  15% of 4300 -> 645.

Excluding the margin the remaining votes were equally shared by the two candidates. Thus the share of each is 4300 – 645 = 3655/2 -> 1828 (approx.)

Hence the votes secured by the winning candidate is his share plus the margin of victory votes ---  1828 + 645 = 2473.