Finding the data regarding the four racket games. 
In a vijantkhand sports stadium, athletes choose from $4$ different 
  racket games (apart from athletes which is compulsory for all)
  These are tennis, table tennis, squash and badminton. It is known 
  that $20\%$ of thee athletes practising there are not choosing any
  of the racket sports. The four games given here are played by 
  $460,\ 360,\ 360 $ and $440$ students respectively. The number
  of athletes playing exactly $2$ racket games for any combination
  of two racket games is $40$. There are $60$ athletes who play all $4$
  games but in strange coincidence, it was noticed that the number of people
  playing exactly $3$ games was also equal to $20$ for each combination
  of $3$ games.

What is the number of athletes in the stadium  ?
$a.)\ 1140 \ \ \ b.) 1040 \ \ \ c.)\ 1200 \ \ \ \color{green}{d.)\ 1300} $
What is the number of athletes in the stadium who play either 
only squash or only tennis ?
$a.)\ 120 \ \ \ b.) 220 \ \ \ \color{green}{c.)\ 340} \ \ \ d.)\ 440 $
How many athletes in the stadium perform only athletics ?
$a.)\ 160 \ \ \ b.)\ 1040 \ \ \ \color{green}{c.)\ 260} \ \ \ d.)\ 220 $
If all the atheletes were compulsory asked to add one game to their 
existing list (except those who were already playing in all four games )
then what will be the number of athletes who would be playing all $4$ 
games after this change ?
$a.)\ 80 \ \ \ b.)\ 100 \ \ \ c.)\ 120 \ \ \ \color{green}{d.)\ 140} $
As there are $4$ sets I applied the formula
$(T\cup Tt \cup S\cup B)
= T+Tt +S+B-(T\cap Tt)-(S\cap Tt)-(B\cap Tt)-(T\cap S)-(T\cap B)-(S\cap B) 
+(T\cap Tt \cap S)+(T\cap Tt \cap B)+(T\cap B \cap S)+(B\cap Tt \cap S)-(T\cap Tt \cap S\cap B) \\
\dfrac{80x}{100} = 460+360 +360+440-40\times 6+20\times 4-60 \\
x=1750 $
which is not in the options I look for a short and simple way.
I have studied maths upto $12$th grade.
 A: If everyone increases their number of games by one (except those already playing four games), then the number of people playing four games is equal to the number playing three games before added to the number playing four games to start with.
You just have to add the four lots of 20 who were playing three games (=80) to the 60 who were already playing four games to get a total of 140.
A: Your idea for the first sub-question is correct... with a small catch: The numbers given in the problem statement are not the sizes of the intersections. For example, the set $T \cap S$ corresponds to athletes who certainly chose both tennis and squash but could have also chosen badminton and/or table tennis additionally, whereas the problem statement tells us that the number of athletes playing tennis and squash and nothing else is $40$.
This actually makes the problem simpler -- for example, to calculate the number of athletes who chose tennis only, we can simply take all the $460$ athletes who chose tennis and subtract those who chose one additional sport ($3\times 40$, since there are three possible additional sports), two additional sports ($3\times 20$, since there are three choices of two extra sports too) and all three additional sports ($60$); yielding $460-3\times 40-3\times 20-60=460-240=220$.
