Sets - Any shortcuts to solve this problem?

[beater]

In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?
a) 13
b) 10
c) 9
d) 8
e) 7

B

[logitech]

This solution is dedicated to my formula-lover friend Vivek:

A+B+C= (AUBUC)+2(AnBnC)+ [exactly two classes]

25+25+34=68+(2x3)+ X

X=10

[hksam07]

sorry...but why do u have to multiply this part (AnBnC) (three students registered for all three classes) by 2?

[kanha81]

Should not the answer be 13?

total students = H + M + E - [exactly 2 classes] - 3
68 = 25 + 25 + 34 - X - 3
71 = 84 - X
X = 13

[A] what is OA?

[Stuart@KaplanGMAT]

Should not the answer be 13?

total students = H + M + E - [exactly 2 classes] - 3
68 = 25 + 25 + 34 - X - 3
71 = 84 - X
X = 13

[A] what is OA?

Your formula is almost correct, but not quite.

The people in all 3 groups have been counted 3 times, but we only want to count them once. So, instead of subtracting the number of triple group people, we need to subtract twice that number.

So, the answer is:

Total students = H + M + E - [exactly 2 classes] - 2(all 3 classes)

68 = 25 + 25 + 34 - X - 2(3)

68 = 78 - X

X = 10

[mjjking]

How do we know whether e have to count them twice or not?

[Stuart@KaplanGMAT]

How do we know whether e have to count them twice or not?

In the end, we want to count everyone once.

If someone appears in 2 groups, and we add those together, we've counted them twice; accordingly, we need to subtract them once to get a "true" count.

If someone appears in 3 groups, and we add those together, we've counted them three times; accordingly, we need to subtract them twice to get a "true" count.

[sureshbala]

When you add 25+25+34 we get 84. But given that total strength is 68. We got an extra of 16 because the people who take exactly two subjects (say X)are counted twice and the number of people who take all the three (say Y) are counted thrice.

Hence x+2Y = 16

Given Y = 3

So X =10

(Note: This result will be evident once you go through the basics of venn diagrams)