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
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