Problem Solving

In a class of 100
80 have atleast one pen
65 have atleast one pencil
70 have atleast one slate

If a is max no of classmates having all three
and b min no of classmates having all three,

a-b is:

A)35
B)20
C)15
D)50
E)60


Please explain your method .

Thanks .

Here we are looking at extreme cases, so we assume best case / worst case sort of scenarios to arrive at these extremes.

You have to assume here that a student can only have 1 of each, eg. he cannot have 2 pens, or 3 slates. He can however have 1 pen, 1 slate. or 1 pencil, 1 pen, 1 slate, etc. This is not explicitly mentioned, but without this assumption you cannot solve the problem.

Out of the 3 values, 65, the number with atleast one pencil, is the lowest. This indicates that no more than 65 could have all three, since (100-65) = 35 is the number who do not have a pencil and thus will not be counted towards those with all 3.

Why did we choose 65? This is because if you pick say 70, then we forget that only 65 have pencils, and thus the other 5 could not possibly have one.
These 65 out of 80 that have a pen, COULD be the same ones that have a pencil, and also have a slate, and 65 out of 70 with a slate COULD have a pen and a pencil as well. Thus the lease value of the 3 sets is the maximum possible value of the intersection between all 3 sets.

So a = 65

Now let's look at the other extreme. We can infer from the data that:
20 do not have a pen ; i.e. they have a pencil and/or a slate
35 do not have a pencil ; i.e. they have a pen and/or a slate
30 do not have a slate ; i.e. they have a pen and/or a pencil

The scenario we will consider now, is that all these students i.e. all (20+35+30) = 85 of them, account for all the students who have 1 out of the 3 objects or 2 out of the 3 objects. This is actually a special case, where we are assuming that we are not overcounting and that all 3 of these sets of students are unique and have no overlap.

Thus (100 - 85) = 15 of the students have all 3 objects. b = 15

a - b = 50

Sorry for the long explanation

Rahul,

I'm confused as to why you subtracted 50 from 65. Why do we have to assume that the 100 students HAVE TO own one of the 3 things. Can't it be that its none? Can you elaborate. Thanks.

Ex: 100 people in a class, 60 with eraser, 60 with pencil.
a) Possibility one is 20 with both and 40 with only eraser and 40 with only pencil
b) Extreme: 60 with both eraser and pencil and 40 with nothing.

Are you assuming all 100 have to own something because its not solvable without this assumption? (Or is it solvable?)

Rahul@gurome wrote:Since 65 is the smallest number, so a maximum of 65 class mates can have all the 3 items, a = 65
80 have at least one pen implies 20 do not have pen
70 have at least one slate implies 30 do not have slate
There is a possibility that out of 65 class mates with pencil, 20 + 30 = 50 do not have pen + slate.
So, a minimum of 65 - 50 = 15 class mates should have all the 3 items.
So, b = 15
Hence, a - b = 65 - 15 = 50

The correct answer is (D).