H =25

M =25

E =35


TOTAL = H + M + E -2(H AND M AND E) - ( H AND M ) - ( H AND E)

- (M AND E)

68 = 25 + 25 + 34 - 2(3) - ( H AND M ) - ( H AND E)

- (M AND E)


SOLVING FOR 2 SUBJECT INTERSECTIONS

ANS = 10

We count all three subjects twice as it isan intersection b/w three sets.

[/b]

Forget about normal set approaches. There are more ways of doing this, which will infact help you save time in quants. Here is the approach.

Let number of students registered for 1 class only be I, 2 classes only be II and all 3 classes only be III than

I+II+III= 68---(1)

Now if you add 25 from History, 25 from Math and 34 from English, you actually add I, you add two twice (say if 2 people are in History and Math than you account 2 while adding history and while adding math, hence twice) and you add III thrice (same reason as above)

Hence I+2II+3III = 84 ----(2)

(2)-(1)
II+2III=16

we know that III =3 hence

II+6=16
II=10

hence B

Note: Eqn 1 and Eqn 2 formed in this question is a standard one and will apply to any kind of set problems in the above format. When you see this question you can straight away put these two eqns and finding value after that should be 20 secs job.

Let me know if you have any queries.

Thanks,
Quant-Master