rizzy wrote:

Bunuel wrote:

Official Solution:

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

Denote \(x\) as the required number of correct answers. \(x\) must satisfy the equation \(\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}\) or \(350M + 350 = 320M + 10x\) or \(x = 3M + 35\).

Answer: B

Can you split out that formula and explain why each term is in there? I don't understand how you got to \(\frac{0.64*M*50 + x}{50M + 50} = \frac{7}{10}\)

Consider the required equation

0.64 * M * 50 + x = 7/10 * 50 ( M + 1) 0.64 * M * 50 = 50 M Questions were answered off which 64% were accurate

x = the least number of questions that the next student have to get right to bring the total of correct answers to 70%

7/10 * 50 ( M + 1) = total correct answers must be 70% . Here in (M+1) =

+1 is for the next studentHope it helps