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?
Answers
Total number of students: 120
even: 60 -> Odd= 60
Divisible by 5: 120/5=24. Half of them are odd=12.
Divisible by 7: 120/7=17 ( we only care about the integer value). The number of odd numbers is either 8 or 9. To check, multiply 17*7. Since the product is odd, we know that the number of odd numbers is 9.
Divisible by both 7 and 5: 35,70,105. We only want the number of odd numbers=2.
Odd numbers=60
Odd numbers that are divisible by 5=12
Odd numbers that are divisible by 7=9
Odd numbers that are divisible by 7 and 5=2
Number of people who did not take any of the 3 classes=60-12-9+2=41
(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}⟹120÷2=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}⟹120÷5=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}⟹120÷7=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}⟹120÷14=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}⟹120÷35=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}⟹120÷14=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}⟹120÷70=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.
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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.
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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=41+10+7+11+2+7+1
=79=79
Students who opt for none
=180-79=\boxed{101}=180−79=
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)n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩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⟹n(A∪B∪C)=60+24+17−12−3−8+1
\implies n(A\cup B\cup C)=79⟹n(A∪B∪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}⟹n(U)−n(A∪B∪C)=180−79=
101
We get the same result. This verified our method is correct.
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