# Percentages stumped me.

• Jun 3rd 2008, 04:01 PM
cody94
Percentages stumped me.
G'day from Australia. Cody here.
this one stumped me, how may I approach it please?
"a maths exam contained only 2 questions, prob 1 solved by 70%
prob 2 by 60% of students,every pupil solved at least 1 problem.
9 students solved both problems. How many pupils took the exam?
Thank you kindly for any assistance.
• Jun 3rd 2008, 04:41 PM
TheEmptySet
Quote:

Originally Posted by cody94
G'day from Australia. Cody here.
this one stumped me, how may I approach it please?
"a maths exam contained only 2 questions, prob 1 solved by 70%
prob 2 by 60% of students,every pupil solved at least 1 problem.
9 students solved both problems. How many pupils took the exam?
Thank you kindly for any assistance.

Let P be the number of pupils that took the exam.

We know that 0.7P pupils got question 1 correct and that 0.6P got question 2 correct. Since 9 students go both problems correct they are counted twice (once for question 1 and again for question 2). So we have overcounted by 9 students. So we get the equation

\$\displaystyle 0.7P+0.6P-9=P \iff 0.3P=9 \iff P=30\$

So the number of pupils that took the exam was 30.

I hope this helps. Good luck.
• Jun 3rd 2008, 04:49 PM
galactus
Hello Cody. This is Cody

Since 70% solved problem 1 and 60% solved problem 2 and 9 solved both, then this appears to be a P(A)+P(B)-P(A and B)

Let P(A)=.70 and P(B)=.60

Then A and B=9

So, 9 must be the portion that is the 30 %

9 is 30% of what number?. 9/.30=
• Jun 3rd 2008, 04:58 PM
cody94
Thank you very much, Galactus and MTset,
I can see both avenues now.
The "overcounting by 9" was the trick for me.
appreciate you time.
Cody.