benejo wrote:

Jack and Lily are playing quiz. They start with 50 chocolates each and Jack asks questions. For each correct answer, Jack gives Lily 4 chocolates and for each incorrect answer, Lily gives three chocolates to Jack. If at the end of the quiz, Lily gave only one-third answers correct and is left with 36 chocolates, how many questions did Jack ask her?

A. 7

B. 14

C. 21

D. 36

E. 43

Source:

ExpertsGlobalLily started with 50 chocolates. She ends up with 36, i.e., a net loss of chocolates, which we can also call "points"

(36-50) = -14 = the SUM of her points from the questions

A weighted average accounts for the net loss. There are twice as many negative values as there are positive values.

Let \(x\) = the number of questions

\(\frac{1}{3}\)x were correct

Each correct answer = +4 chocolates to Lily

\(\frac{2}{3}\)x were incorrect

Each incorrect answer = -3 chocolates from Lily

Call -3 and +4 "point value"

She had a net loss of chocolates: (36 - 50) = -14

(weight of incorrect answers)* (point value) +

(weight of correct answers) * (point value)

must sum to -14

\((\frac{2}{3}x * -3) + \frac{1}{3}x * 4 = -14\)

\(-\frac{6}{3}x + \frac{4}{3}x = -14\)

\(-6x + 4x = -42\)

\(-2x = -42\)

\(x = 21\)Answer C

Thinking in terms of deficits is odd. I'd check this answer.

21 questions: 7 correct, 14 incorrect

(7) * (4) = +28 chocolates

(14) * (-3) = -42 chocolates

(50 + 28 - 42) = 36 chocolates. That works