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The following is a practice question from a GRE preparation book.

A case of $100$ candy bars is sold at a discount of $\$0.36$ per bar.

Quantity A: the discount per case
Quantity B: 36%

(a) A is greater
(b) B is greater
(C) The two quantities are equal
(d) The relationship cannot be determined from the information given


My approach: Quantity B is 36%, which is simply 0.36

Quantity A is 0.36 * 100 = 36

Therefore A > B.

However the answer I'm given is D. Where have I made a mistake?

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  • $\begingroup$ You don't know what a bar costs, so you don't know the percent discount. $\endgroup$
    – dxiv
    Commented Jan 7, 2018 at 7:24
  • $\begingroup$ Perhaps I don't understand the meaning of 'discount per case' clearly. Should not just mean the discount per 100 bars? Why is the % discount relevant? $\endgroup$
    – dnclem
    Commented Jan 7, 2018 at 7:37
  • $\begingroup$ Your answer would have been correct if Quantity B had said $\,\$0.36\,$, instead. But it does not, it says $\,36\%\,$, which is presumbaly to be interpreted as the percent discount off a box of candies. $\endgroup$
    – dxiv
    Commented Jan 7, 2018 at 7:40
  • $\begingroup$ Was subtle to me. Thanks. $\endgroup$
    – dnclem
    Commented Jan 7, 2018 at 7:50

1 Answer 1

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The discount per case means the price after discount. However, the discount per case is the discounted amount divided by the total amount. Since only the discounted amount is known ($36), you cannot find the discount per case. Per case indicates that you know how much the case costs in the first place.

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  • $\begingroup$ I am failing to understand how discount per case means the price after discount. How does 'discount per case' not mean 'discount per 100 bars', i.e. the total discount for 100 bars? $\endgroup$
    – dnclem
    Commented Jan 7, 2018 at 7:36
  • $\begingroup$ I think they want you to give it in terms of a percentage discount seeing that Quantity B is a percentage $\endgroup$
    – Harry Alli
    Commented Jan 7, 2018 at 7:40
  • $\begingroup$ That makes sense. Thanks. $\endgroup$
    – dnclem
    Commented Jan 7, 2018 at 7:51

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