Question

✰ QUESTION :

➽ In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, those whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

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Answer 1

(Minor edit in the calculation. The correct answer is 101.)

Required Knowledge

No required knowledge. (There is a way to solve without knowing the sets.)

Before We Solve

If students opt for any of the following, the number for the students is divisible by 2, 5, or 7. This means if they are not divisible, they opt for none of the three subjects.

Solution

Method 1.

Step 1.

So let's find how much students opt for none of the three subjects.

Let's find students who opt for

• Physics, multiples of 2
• Chemistry, multiples of 5
• Math, multiples of 7

If we divide 120 by 2, the number of students is 60.

$\implies 120\div 2=\mathrm{Quotient\ 60,\ Remainder\ 0}$

If we divide 120 by 5, the number of students is 24.

$\implies 120\div 5=\mathrm{Quotient\ 24,\ Remainder\ 0}$

If we divide 120 by 7, the number of students is 17.

$\implies 120\div 7=\mathrm{Quotient\ 17,\ Remainder\ 1}$

Step 2.

But, we cannot add them all because we didn't consider the number of students who opt for two subjects.

Let's find students who opt for,

• Physics and Chemistry, multiples of 10
• Chemistry and Math, multiples of 35
• Math and Physics, multiples of 14

If we divide 120 by 10, the number of students is 12.

$\implies 120\div 14=\mathrm{Quotient\ 12,\ Remainder\ 0}$

If we divide 120 by 35, the number of students is 3.

$\implies 120\div 35=\mathrm{Quotient\ 3,\ Remainder\ 15}$

If we divide 120 by 14, the number of students is 8.

$\implies 120\div 14=\mathrm{Quotient\ 8,\ Remainder\ 8}$

Step 3.

There are only two steps further. Let's find the students who opt for three subjects.

• All subjects, multiples of 70

If we divide 120 by 70, the number of students is 1.

$\implies 120\div 70=\mathrm{Quotient\ 1,\ Remainder\ 50}$

Step 4.

Here, we will use the Venn diagram, which involves the relation between groups. The picture is attached.

A set can be likened to a boundary.

_________________________

Set U is the boundary of 120 students.

Set A is the boundary of students who opt for Physics.

Set B is the boundary of students who opt for Chemistry.

Set C is the boundary of students who opt for Math.

_________________________

We are almost done. Let's put the numbers in the diagram, and see how many students opt for none.

(Tip: Start putting the numbers from the intersection with the most sets. Then, it should be excluded because the same students should not appear more than once.)

Students who opt for at least one subject

$=41+10+7+11+2+7+1$

$=79$

Students who opt for none

$=180-79=\boxed{101}$

Method 2.

Formula:-

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)$

The results are given above. Let's substitute the numbers into this formula.

$\implies n(A\cup B\cup C)=60+24+17-12-3-8+1$

$\implies n(A\cup B\cup C)=79$

We need students who opt for none of the above.

$\implies n(U)-n(A\cup B\cup C)=180-79=\boxed{101}$

We get the same result. This verified our method is correct.

_________________________

This is the required answer.

Answer 2

Given :-

In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, those whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math.

To Find :-

How many opt for none of the three subjects

Solution :-

We know that

n(A∪B∪C) = nA + nB + nC - n(A ∪ B) - n(B∪C) - n(C ∪ A) + n(A ∪ B ∪ C)

Let the set A be Physics

Set B be Chemistry

Set C be  Math

According to the question

No. of students who opt for physics = n/2 = 120/2 = 60

No. of students who opt for chemistry = n/5 = 120/5 = 24

No. of students who opt for Maths = n/7 = 120/7 = 17 (approx)

Now

A ∪ B = 120/10 = 12

B ∪ C = 120/14 = 8(approx)

C ∪ A = 120/35 = 3(approx)

Now

A ∪ B ∪ C = 60 + 24 + 17 - (12 + 8 + 3) + 1

A ∪ B ∪ C = 101 - (23) + 1

A ∪ B ∪ C = 102 - 23

A ∪ B ∪ C = 79

No. of student who opt for none = n - 79 = 120 - 79 = 41