TCS - Recent Questions - 8

TCS – Recent Questions - 8

1. Cara, a blue whale participated in a weight loss program at the biggest office. At the end of every month, the decrease in weight from original weight was measured and noted as 1, 2, 6, 21, 86, 445, 2676. Though Cara made a steadfast effort, the weighing machine showed an erroneous weight once. What was that weight? 
a) 2676                  b) 2                        c) 445                    d) 86

Ans: (d)

The series goes in the following manner:

1x 1+1=2
2 x 2+2=6
6 x 3+3=21
21 x 4+4=88 and not 86
88 x 5+5 = 445
445*6+6 = 2676  (The mistake in the weighing was 86 that should actually be 88)

2. A school has 120, 192 and 144 students enrolled for its science, arts and commerce courses. All students have to be seated in rooms for an exam in such way that each room has students of the same course and all rooms have equal number of students. What is the least number of rooms needed?

Ans: 19.

Since each room shall have students of the same course we have to find the HCF of

120, 192, and 144. This comes to 24

Now to seat 120 students we need 120/24 = 5 Rooms. Similarly

For seating 192 students we need   192/24 = 8 Rooms

For seating 144 students we need   144/24 = 6 Rooms.

Totally we need  5 + 8 + 6 = 19 rooms.

3. What is the 32nd word of "WAITING" in a dictionary?
Ans: AGNTIWI

Let us first arrange the letters in the word WAITING alphabetically.

This will be:    A  G   I   I   N  T   W

Starting with letter A the other six letters can be arranged in 6!/2! Ways. This will give an answer 360 beyond our requirement.

Starting with the next two letters AG, the other five letters can be arranged in 5!/2! Equals 60 again

beyond our requirements.

Next starting with AGI the remaining four letter can be arranged in 4! Ways -> 24 ways……… (i)

If we now take AGN then the other four letters can be arranged in 4!/2! -> 12 ways ………….. (ii)

Adding (i) and (ii) we get totally 36 ways. Since our requirement is to find 32^nd word

We start with AGNI then the other three letters can be arranged in 3! Ways -> 6 ways………..,(iii)

Adding (i) and (iii) we now get 24 + 6 -> 30 words

The 31^st word will be AGNTIIW …………..31^st word

The next word will be AGNTIWI which will be the 32^nd word and our answer

  

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

Ans: 84 ways
Number of ways of distributing n identical objects to r distinct persons so that each get at least one object is given by = (n−1)C(r−1)    Substituting the value of n as 10 and that of r as 4, we have (10−1)C(4−1)=9C3 ->  84ways.

5. 60% of male in a town and 70% of female in a town are eligible to vote. Out of which 70% of male and 60% of female who are eligible to vote voted for candidate A. what is the value of votes in % did A get?

Ans: 65% (approx.)

Let us assume the male and female population as 100 each.

Eligible to vote ->  Male 60% = 60.  Female 70% = 70.

Total eligible to vote ->  60 + 70 = 130.

70% of 60 voted for candidate A -> 42 males

60% of 70 voted for candidate A -> 42 females.

Thus totally 42 + 42 = 84 eligible voters voted for candidate A

Hence the percentage of votes secured by A ->  84/130 * 100 = 64.61% say 65%

 
6. George and Mark can paint 720 boxes in 20 days while Mark and Harry paint in 24 days and Harry and George in 15 days. George works for 4 days, Mark for 8 days and Harry for 8 days. The total number of boxes painted by them is how many?

Ans: 348 boxes

G + M in one day can paint -> 720/20 = 36 boxes.

M + H in one day can paint -> 720/24 = 30 boxes

H + G in one day can paint -> 720/15 = 48 boxes

Totally 2(G + M + H) in one day can paint 114 boxes -> (G + M + H) in one day -> 114/2 = 57 boxes

G in one day -> (G+M+H) – (M+H) -> 57 – 30 = 27 boxes. In 4 days G will paint -> 27 x 4 = 108 boxes.

M in one day ->(G+M+H)  -  (H+G) -> 57 – 48 =   9 boxes. In 8 days M will paint ->  9 x 8 =   72 boxes.

H in one day -> (G+M+H) – (G+M) -> 57 – 36 = 21 boxes. In 8 days H will paint -> 21 x 8 =168 boxes.

Totally (108 + 72 + 168) = 348 boxes are painted.

 
7. 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) 21                      c) 36                       d) None of these

Ans: (d)

Since the child can make either 1 step or 2 steps we can use the Fibonacci series method to arrive at the address.

The series will go like this:  1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ………..

None of the choice answers falls in this series and hence the answer is (d)

8. A boy buys 18 sharpeners, (Brown/white) for Rs.100.  For every white sharpener, he pays one rupee more than the brown sharpener. What is the cost of white sharpener and how many  white sharpener did he buy?
a) 5, 13
b) 5, 10
c) 6, 10
d) None of these
Ans: (c)

Let us assume that B is the number of brown sharpeners and W the number of white sharpeners. Let ‘x’ be the price of one Brown sharpener. Then the price of one White sharpener is (x+1)

We now have an equation -> W(x+1) + Bx = 100 ………….. (1)

We also know ->   W + B = 18 ->  B = (18 – W)

Now substituting the value B in (1) above we have

W(x+1) + x(18 – W) ->   Wx + W + 18x – Wx  = 100 ->  W + 18x 100 …………… (2)

Now we shall arrive at the answer from the choices:

Choice (a)  If W is 13 then we have 13 + 18x = 100 -> x = 87/18 -> 4.83  This is not the answer.

Choice (b)  if W is 10 then we have 10 + 18x = 100 -> x = 90/18 -> 5

If x = 5 then cost of white sharpener is 5+1 = 6.

Thus we have an answer the cost of white sharpener as 6 and the number of sharpeners as 10.

Choice (c) is the answer.

9. Alphabets letters from A to Z are consecutively assigned with number  1 for A and 26 for Z.  By 27th letter we mean A, 28th B etc.  
In a strange country a military general sends this secret message according to the following codification scheme.  In codifying a sentence, the 1st time a letter occurs it is replaced by the nth letter from it.   2nd time if the letter occurs it is replaced by n^2 letter from it. 3rd time if occurs  it is replaced by n^3 letter from it.   In other words if ‘m’ is the number of the letter then in the code this letter will be referred as the letter bearing no m + n. For example if the number of the letter is 4 ie D then in the code it is referred as F ie 4 + 2 = 6 where 2 is the value of ‘n’. In the message from the general the value of n=6.  What is the code word for ABBATIAL
a) GHNNZOOR
b) GHKJZOHR
c) GHHGZOGR
d) GHLKZOIR
Ans: (d)
A should be coded as 1+6 = G (it occurred for first time)
B should be coded as 2+6 = H (it occurred for first time)
B Should be coded as 2 + 36 = 38 - 26 = 12 = L (it occurred for second time)

Now from the choice only (d) fulfils this sequence GHL. Hence the answer is (d)

10. Of a set of 30 numbers, average of 1st 10 numbers is equal to average of last 20 numbers.  The sum of last 20 numbers is?
a) 2 x sum of last 10 numbers
b) 2 x sum of 1st 10 numbers
c) sum of 1st 10 numbers
d) Cannot be determined
Ans: (b)

Let the average of first 10 numbers be ‘a’. Then the sum of the numbers is = 10a
Average of the last 20 numbers also be ‘a’. Then their sum = 20a

We have now 10a = 20a

Hence option (b) is the answer.