TCS placement paper 2013 batch

TCS important questions(new pattern ) faced on 13 sep 2012

1.Raj  writes a number.He sees that athe number of two digits exceeds four times the sum of its digits by 3. If the number is increased by 18,the result is the same as the number formed by reversing the digits. Find the number

Ans:35

2.a,b,c are non negative integers such that 28a+28b+31c=365, then a+b+c is

Ans:12

3.If x^y denotes x raised to the power y, find last two digits of (1941^3843)+(1961^4181)

Ans:82

Find the last two digits of 41²⁷⁸⁹
In no time at all you can calculate the answer to be 61 (4 ×— 9 = 36). Therefore, 6 will be the tens digit and one will be the units digit)
Find the last two digits of 71⁵⁶⁷⁴⁷
Last two digits will be 91 (7 ×— 7 gives 9 and 1 as units digit)
Now try to get the answer to this question within 10 s:
Find the last two digits of 51⁴⁵⁶  ×— 61⁵⁶⁷
The last two digits of 51⁴⁵⁶ will be 01 and the last two digits of 61⁵⁶⁷ will be 21. Therefore, the last two digits of 51⁴⁵⁶  ×— 61⁵⁶⁷ will be the last two digits of 01 ×—21 = 21
Last two digits of numbers ending in 3, 7 or 9

3.P(x)=(x^2012+x^2011+x^2010+…..+x+1)^2-x^2012

Q(x)=x^2011+x^2010+…+x+1

The remainder when P(x) is divided by Q(x) is ?

Ans:0

4.A circle has 29 pts arranged in a clock wise manner numbered from 0 to 28 as shown in figure.

A bug moves clockwise around the circle,according to the following rule (i.e) if it is in the position i then it moves i+1+r position for next one second… so that in 2012^th sec what position the bug will be

Ans:15

Then sums from coding and decoding that alphabets are arranged consecutively starts from A=0…..Z=26 like that…

5.At the end of 1994 rohit was half an old as his grand mother.The sum of years in which they were born is 3844. How old rohit was at the end of 1999.

6.when number are written in base b,we have 12*25=333, the value of b is

7.how many polynomial functions f of degree>=1 satisfy f(x^2)=[f(x)]^2=f(f(x))?

Ans:1