In a class of $100$ students, $73$ like coffee, $80$ like tea and $52$ like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is

Looking at this problem, we need to find how many students could like all three drinks, considering the given constraints.

We have 100 students total, and we know:

• 73 like coffee
• 80 like tea
• 52 like lemonade

We need to find the difference between the maximum and minimum possible number of students who like all three drinks.

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Let's define variables based on how many drinks each student likes:

• $a$ = students who like none of the drinks
• $b$ = students who like exactly one drink
• $c$ = students who like exactly two drinks
• $d$ = students who like all three drinks

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All students must be accounted for:

$a + b + c + d = 100$

When we count 73 + 80 + 52 = 205, we're counting:

• Each student who likes exactly 1 drink: counted once
• Each student who likes exactly 2 drinks: counted twice
• Each student who likes all 3 drinks: counted three times

So: $b + 2c + 3d = 205$

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$(b + 2c + 3d) - (a + b + c + d) = 205 - 100$

$c + 2d - a = 105$

This gives us: $a = c + 2d - 105$

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Since only 52 students like lemonade, at most 52 students can like all three drinks.

Let's check if $d = 52$ is possible:

If $d = 52$, then $a = c + 2(52) - 105 = c - 1$

Since $a \geq 0$, we need $c \geq 1$

From equation 1: $a + b + c + 52 = 100$, so $a + b + c = 48$

If $c = 1$ and $a = 0$, then $b = 47$

Therefore, maximum $d = 52$

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We need the smallest possible value of $d$ where all variables remain non-negative.

From $a = c + 2d - 105$:

For $a \geq 0$: $c + 2d \geq 105$

For $c \geq 0$: we need $c$ to be non-negative

From equation 1: $b = 100 - a - c - d = 100 - (c + 2d - 105) - c - d = 205 - 2c - 3d$

For $b \geq 0$: $205 - 2c - 3d \geq 0$, so $2c + 3d \leq 205$

The minimum occurs when we minimize $d$ while satisfying $c + 2d \geq 105$.

If $d = 5$: $c + 10 \geq 105$, so $c \geq 95$

Let's check: $c = 95$, $d = 5$, $a = 95 + 10 - 105 = 0$

Then $b = 100 - 0 - 95 - 5 = 0$

Therefore, minimum $d = 5$

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The difference between maximum and minimum possible number of students who like all three drinks is:

$52 - 5 = 47$