TCS - 2015 - 2

TCS – 2015 - 2

1.       Rajiv can do a piece of work in 10 days , Venky in 12 days and Ravi in 15 days.  They all start the work together, but Rajiv leaves after 2 days and Venky leaves 3 days before the work is completed.  In how many days approximately the work is completed?
(a) 5
(b) 6
(c) 9
(d) 7
Ans: (d)

Rajiv in one day does 1/10 work.

Venky in one day does 1/12 work

Ravi in one day does  1/15 work.   All the three in one day does – 1/10 + 1/12 + 1/15 = ¼ work.

They work for 2 days and the work completed is ¼ * 2 = ½.   Remaining work is ½

Venky and Ravi in one day do – 1/12 + 1/15 = 3/20 work.

Hence to complete the remaining ½ work they would have taken -> ½ * 20/3 = 10/3 or 3 1/3 days. Venky after doing work for 1/3 day leaves.

Thus in 1/3 days the work done by both is 3/20 * 1/3 = 1/20 work.

Remaining work is ½ - 1/20 -> 9/20

To complete this Ravi will take ->  (9/20) / (1/15) -> 9/20 * 15 -> 6 ¾ days -> roughly 7 days.    

2.       On a 26 question test, five points were deducted for each wrong answer and eight points were added for each correct answer.  If all the questions were answered, how many were correct, if the score was zero?
(a) 10
(b) 12
(c) 11
(d) 13
Ans: (a)

Let the number of correctly answered questions be ‘x’. Then we have an equation:

8x – 5(26 – x) = 0.

Solving, we get the value of ‘x’ as 10.

3.        X = 101102103104105106107......146147148149150  (From numbers 101-150).  Find out the remainder when this number is divided by 9.

Ans: 2

                The rule of divisibility for 9 is the sum of the digits is to be divisible by 9.

To arrive at the sum of the digits we calculate the sum in the unit, tenth and the hundredth places separately.

(a)    Sum of the digits in the hundredth place is – 1 * 50 = 50

(b)   Sum of the digits in the tenths place is:

1*10 + 2*10 + 3*10 + 4*10 + 5*1 = 105. (Note the initial 0*9 will give only ‘0’ value)

(c)    Sum of the digits in the Unit’s place is:   (1 + 2 + 3 + 4 + …………+ 9) * 5 = 225.

Totalling (a), (b) and (c) we get …380. This when divided by 9 will leave a reminder 2.

4.       A number is 101102103104...150. As in the earlier question what is the reminder if

divided by 3?  

Ans: 2

The rule of divisibility for 3 is the same as that of 9 the reminder will remain the same as 2

5.       7^1+7^2+7^3+.......+7^205.

In the above series how many numbers have 3 in the unit place? 

Ans: 51

The unit digits for 7^1, 7^2, 7^3 and 7^4 are  7, 9, 3 and 1. These digits repeat again and again. In other words, the unit digits in 7 to the power repeat after every fourth power and each can be called one cycle.

Thus in 7^205 we have 205/4 -> 51 completed cycles.

Thus 51 powers of 7 will have unit digit as 3.

Hence the answer is 51.

6.       In paper A, one student got 18 out of 70 and in paper B he got 14 out of 30. In which paper he did fare well?

Ans: Paper B

A simple question to answer.

Percentage of marks scored in Paper A is 18/70*100 = 25.7%

Percentage of marks scored in Paper B is 14/30*100 = 46.6%

In paper B he fared well.

7.       Find the total no of divisors of 1728 (including 1 and 1728)

Ans: 28.

The number of factors of a given number N is given by a^p*b^q*c^r ….

Where a, b, c,…. Are prime numbers and (p+1)*(q+1)+(r+1)*……. are factors.

Applying the above we have 1728 = 2^6*3^3

Hence the factors are (6+1) * (3+1) = 28

8.       The sum of two numbers is 45. Sum of their quotient and reciprocal is 2.05. Find the product of the numbers.

Ans: 500

Let x and y be the numbers.

We are given that x + y = 45 (i) and (x/y) + (y/x) = 2.05 (ii)

From (ii) we have (x² + y²) / xy = 2.05. expanding

[(x+y)² – 2xy] / xy = 2.05 -> (x+y)²  = 2.05xy + 2xy = 4.05xy

(x+y) = 45. We now have  ->   45*45/ 4.05 = xy

Thus xy = 500.

9.       A number when divided by 406 leaves remainder 115. What will be the reminder if the number is divided by 29?

Ans: 28

Let the number be N and the quotient X.

Then N = 406X + 115.  Since 406 is exactly divisible by 29, to arrive at the answer

We have to divide the reminder 115 alone by 29.

Thus 115/29 reminder is 28.

10.   The number of multiples of 10 which are less than 1000, which can be written as a sum of four consecutive integers is
(a) 50
(b)100
(c) 150
(d) 216

Ans: (a)

We can write 10 as (1 + 2 + 3 + 4) four consecutive numbers.

Now adding 5 to each of these digits we get:

6 + 7 + 8 + 9 = 30 (again consecutive four digits.)

Again adding 5 to each of the above digits we get:

11 + 12 + 13 + 14 = 50

From the above we observe that 10, 30, 50, 70 ……that are multiple of 10 have consecutive four digits.

Thus we have 50 numbers that can be written in four consecutive digits between 1 and 1000.

Please note that 20, 40, 60 …… all cannot be written through four consecutive numbers.