Bunuel wrote:
After \(M\) students took a test, there was a total of 64% of correct answers. If the test contains 50 questions, what is the least number of questions that the next student have to get right to bring the total of correct answers to 70% ?
A. 3M + 20
B. 3M + 35
C. 4M + 15
D. 4M + 20
E. 4M + 45
If M tests were taken, and each test has 50 questions, then the total number of
questions answered =
50MWe're told that there was a total of 64% of
correct answers.
So, the total number of correctly-answered questions = 64% of
50M =
32MIf a NEW student takes the test, then there will be M+1 tests taken.
Since each test has 50 questions, the NEW total number of
questions answered =
50M + 50
Let C = the number of questions the NEW students answer correctly
So, the NEW number of correctly-answered questions =
32M + C
We want to bring the total of correct answers to 70%
That is, we want: (
32M + C)/(
50M + 50) = 70/100
Simplify right side: (
32M + C)/(
50M + 50) = 7/10
Multiply both sides by 10 to get: 10(
32M + C)/(
50M + 50) = 7
Multiply both sides by (
50M + 50) to get: 10(
32M + C) = 7(
50M + 50)
Expand: 320M + 10C =350M + 350
Subtract 320M from both sides: 10C =30M + 350
Divide both sides by 10 to get: C =3M + 35
Answer: B
Cheers,
Brent