in a class of 80 students, 65 studied English, 60 studied Hindi and 55 studied Sanskrit. Find the maximum number of students who studied all the three languages.

Answer 1

Normally, these types of problems are solved by using a venn diagram. In this case it would be a three circle venn diagram with the following seven possible regions:

1) English Only (E)

2) Hindi Only (H)

3) Sanskrit Only (S)

4) English and Hindi (EH)

5) Hindi and Sanskrit (HS)

6) English and Sanksrit (ES)

7) All Three (A)

In this problem, we have limited information as we want to determine the maximum amount of students that could possibly study all three languages. So we have to think logically and make some assumptions about how to maximize region #7, while still conforming to the info that we do have. Let's start with the most general equations. Here is what we know from the problem statement:

E + EH + ES + A = 65 (total # students studying English)

H + EH + HS + A = 60 (total # students studying Hindi)

S + ES + HS + A = 55 (total # students studying Sanskrit)

E + H + S + EH + HS + ES + A = 80 (total # students in the class)

We have seven unknown variables above but only four equations. So we cannot solve this as-is. Let's think about the upper bound of this problem. What if all 55 students studying Sanskrit also studied the other two languages? This would imply the following:

A = 55

S = ES = HS = 0 (from third eq. above, cannot have negative students)

E + EH + 55 = 65

H + EH + 55 = 60

E + H + EH + 55 = 80

If you solve the above system of three equations, you will get:

E = 20

H = 15

EH = -10

While this answer works out mathematically, it does not make sense logically. So we need to adjust our thinking. Instead, what if we assume that nobody studies only two languages? Either they study all three or just one. This eliminates some of the "overlap" in the venn diagram which should maximize our "all three" value.

EH = HS = ES = 0

Simplifying our original four equations:

E + A = 65

H + A = 60

S + A = 55

E + H + S + A = 80

Now, if we solve this new system of equations with four variables and four unknowns, we get:

E = 15

H = 10

S = 5

A = 50

We can kind of see by inspection, that 50 is the maximum value. Remember we tried an upper bound of A = 55 and it didn't work out logically, 50 is not much less.