A math exam contained two problems. Problem A was solved by 70% of the students. Problem B was solved by 60%. Every student solved atleast one of these problems. Nine students solved both the problems. How many students sat in the exam?
2006-09-13 00:13:45 · 6 answers · asked by Arnav 3 in Education & Reference ➔ Trivia
Think about two sets A and B.
A= students solved problem A
B=Students solved problem B
They intersect and make intersection cell C.
C= Students solved both problems.
If we say all students= x
A= 7x /10
B= 6x/10
C= 9
X= A+B - C
X = 7x/10+6x/10-9
x= 30
2006-09-13 02:40:08 · answer #1 · answered by Leprechaun 6 · 0⤊ 0⤋
Tough one. Each student did one, AND ONLY ONE, of the following:
A right, B wrong - say X%
A wrong, B right - say Y%
A right, B right - say Z%
(X+Z)% got A right; (Y+Z)% got B right
So X+Z = 70, Y+Z = 60, and X+Y+Z = 100.
Add the first two together: X+Y+2Z = 130.
Subtract the third: Z = 30. Bingo. 30% of the students got both questions right. If nine students were 30%, then 3 students were 10%, so 30 students sat the exam.
2006-09-13 00:24:50 · answer #2 · answered by bh8153 7 · 0⤊ 0⤋
Let the total number of students be x.
Therefore, the number of students who solver problem 1= 70x/100=7x/10 students
The number of students who got probelm b right= 60x/100=6x/10
the number of students who got both right= 9
therefore, the number of students who only solved problem 1= 7x/10 - 9
the number of student who only solved problem 2= 6x/10 - 9
Therefore, 7x/10 -9 +9+6x/10-9=x
13x/10 -9=x
13x/10 -x= 9
3x/10=9
x= 9X10/3
x= 30
30 students took the exam.
2006-09-13 00:42:51 · answer #3 · answered by mj 2 · 0⤊ 0⤋
You lost me at the word MATH. They need a new teach though if only nine passed the exam LMAO
2006-09-13 00:23:16 · answer #4 · answered by ? 4 · 0⤊ 1⤋
15 students took the exam.
(Why do you ask?)
2006-09-13 00:32:24 · answer #5 · answered by George 1 · 0⤊ 0⤋
sorry Hun i think I'm the dumbest person when it comes to math lol
2006-09-13 00:15:49 · answer #6 · answered by sweetlily 3 · 0⤊ 1⤋