# TCS Questions on Equations with Answers – 1

1.For real number x, int(x) denotes integer part of x.int(x) is the largest integer less than or equal to x.int(1,2)=1,int(-2,4)=-3.  Find the value of int(1/2)+int(1/2+ 100)+int(1/2+2/100)+….+int(1/2+99/100)
a.100
b.120
c.130
d.150
Answer:d
Explaination:
int (1/2) = 0
int (1/2 + 100 ) = 100
into (1/2 + 2/100) = 0
……
int ( 1/2 + 50/100 ) = 1
int (1/2 + 51 /100) = 1
…….
int (1/2 + 99/100) = 1
So 100 +  1 + 1 + …..50 times = 150

2.There are 100 wine glasses.  I offered my servant to 3 paise for every broken glass to be delivered safely and forfeit 9 paisa for every glass broken at the end of day. He recieved Rs.2.40 .how many glass did he break.
a.20
b.73
c.5
d.8
Answer:c
Explaination:If a glass has been broken, he has to loose 3 paisa + 9 paise = 12 paise
Assume K  glasses got broken
100 x 3 – 12 x K = 240 ⇒K=5

3.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
Answer: C
Explaination: Put n = 0
Then f(f(0))+f(0) = 2(0) + 3  f(1) + 1 = 3  f(1) = 2
Put n = 1
f(f(1)) + f(1) = 2(1) + 3 f(2) + 2 = 5 f(2) = 3
Put n = 2
f(f(2)) + f(2) = 2(2) + 3  f(3) + 3 = 7  f(3) = 4
……
f(2012) = 2013

4.The average mark obtained by 22 candidates in an examination is 45. The average of the first ten is 55 while the last eleven is 40 .The marks obtained by the 11th candidate is ?
a.1
b.2
c.3
d.0
Answer : d
Explaination:It is clear that 22 x 45 = 10 x 55 + K + 11 x 40   K = 0

5:If f(1)=4 and f(x+y)=f(x)+f(y)+7xy+4,then f(2)+f(5)=?
a.150
b.135
c.115
d.125
Answer : d
Explaination: Let x =1 and y = 1
f(1 + 1) = f(1) + f(1) + 7 x 1 x 1 + 4  f(2) = 19
Let x =2 and y = 2
f(2 + 2) = 19 + 19 + 7 x 2 x 2 + 4  f(4) = 70
Let x = 1 and y = 4
f( 1 + 4) = 4 + 70 + 28 + 4 = 106
f(2) + f(5) = 125

In the equation A + B + C + D + E = FG where FG is the two digit number whose value is 10F + G and letters A, B , C , D , E, F and G each represent different digits. If FG is as large as possible. What is the value of G?
a) 4
b) 2
c) 1
d) 3
Answer: b
6.In the equation A + B + C + D + E = FG where FG is the two digit number whose value is 10F + G and letters A, B , C , D , E, F and G each represent different digits. If FG is as large as possible. What is the value of G?
a) 4
b) 2
c) 1
d) 3
Answer: B
Explaination:FG is as large as possible and all the 7 numbers should be different.
By trial and Error method,
9 + 8 + 7 + 6 + 5 = 35…5 is getting repeated twice.
9 + 8 + 7 + 6 + 4 = 34…4 is getting repeated
9 + 8 + 7 + 5 + 4 = 33…3 repeats
9 + 8 + 6 + 5 + 4 = 32
None of the numbers repeat in the above case and 32 is the maximum number FG can have. The value of G is 2

7.How many two digit numbers are there which when subtracted from the number formed by reversing it’s digits as well as when added to the number formed by reversing its digits, result in a perfect square.
a.1

b.2
c.3
d.5Answer:a
Explaination:Let the number xy = 10x + y
Given that, 10x+y – (10y – x) = 9(x-y) is a perfect square
So x-y can be 1, 4, 9.  ——– (1)
So given that 10x+y +(10y +x) = 11(x+y) is a perfect square.
So x+y be 11. Possible options are (9,2), (8,3),(7,4),(6,5) ———(2)
From the above two conditions only (6,5) satisfies the condition
Only 1 number 56 satisfies.

8.Four times the first of three consecutive odd integers is 6 more than twice the third. The third integer is?

a.12
b.11
c.0
d.13
Answer:b
Explanation:Let three consecutive odd integeres be x,x+2,x+4 4x=6+2(x+4) x=7 x+4=11

9.If 43 times of two digit numbers is 34 times of two digit no .and sum of ( number and reverse of the number) is 14 then what is the number?
a.68
b.66
c.67
d.69
Answer:a

Explanation:
Go with options Answer is 68 also 43*68=34*86 i.e answer will be 68