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Quantitative Aptitude Question on Maxima and Minima

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

A

48

B

53

C

47

D

52

Answer

47

Explanation

Solution

Step 1 : Define the variables
Let (n),(s),(d),and (t) represent the number of students who like none, exactly one, exactly two, and all three drinks respectively.

Step 2: Write the equations
We have two equations based on the information given:
Equation 1:$$(n+s+d+t=100)
Equation 2:(s+2d+3t=205)2:(s+2d+3t=205)

Step 3 : Express (d) in terms of (n) and (t)
Subtract Equation 1 from Equation 2 to eliminate (s):
[s+2d+3t−(n+s+d+t)=205−100][s+2d+3t-(n+s+d+t)=205-100]
Simplify:
[d+2t−n=105][d+2t-n=105]
This equation relates the number of students who like exactly two drinks (d) with the number of students who like none (n) and all three (t)

Step 4 : Find maximum and minimum values of (t)
We know that (0 leq n,s,d,t leq 100) and (n+s+d+t=100).(n+s+d+t=100).
To find the maximum and minimum possible values of (t), we consider extreme cases:
a) Maximum: Let (t=52)(t=52) (all students like all three drinks).
Solve Equation 3 for (d):
[d+2(52)−n=105[d + 2(52)-n=105 implies d−n=1]d-n=1]
Since (d) and (n) should be non-negative, the only possible solution is (d=1)(d=1) and (n=0).(n=0).
b) Minimum: Let (t=5)(t=5) (only a few students like all three drinks).
Solve Equation 3 for (d):
[d+2(5)−n=105[d+2(5)-n=105 implies d−n=95]d-n=95]
Again, the only possible solution is (d=95)(d=95) and(n=0). (n=0).

Step 5 : Calculate the difference
The difference between the maximum and minimum possible values of (t) is:
[47=52−5][47=52-5]
Therefore, the difference between the maximum and minimum number of students who like all three drinks is 47.