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Re: A stationery store sells pens and pencils. If the price of each item i [#permalink]
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Bunuel wrote:
A stationery store sells pens and pencils. If the price of each item is expressed as a non-zero whole number of cents, how much do 5 pencils cost?

(1) 5 pencils and 3 pens cost 30 cents.
(2) 4 pencils and 4 pens cost 32 cents.


This is a classic C trap question. Let x = the cost of a pencil and y = the cost of a pen.

S1: We can write this as 5x+3y = 30. At first glance it may look unsolvable because there are two variables; however, there is only one solution since the prompt said we can only use "non-zero whole numbers", i.e. integers. This equation only works when x=3 and y=5; therefore 5x = 15. SUFFICIENT.

S2: We can write this as 4x + 4y = 32 and then simplify it to x + y = 8. There are many possible values of x. INSUFFICIENT.

ANSWER: A
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Re: A stationery store sells pens and pencils. If the price of each item i [#permalink]
Expert Reply
Bunuel wrote:
A stationery store sells pens and pencils. If the price of each item is expressed as a non-zero whole number of cents, how much do 5 pencils cost?

(1) 5 pencils and 3 pens cost 30 cents.
(2) 4 pencils and 4 pens cost 32 cents.


Solution:
Pre Analysis:
  • Let us assume the cost of each pen and pencil be x and y cents
  • Where x and y are positive integers (non-zero whole numbers)
  • We are asked the value of 5y or the value of y

Statement 1: 5 pencils and 3 pens cost 30 cents
  • According to this question, \(3x+5y=30\)
  • We might think that this is one equation with two variables and thus not solvable. But the concept of special equation comes here.
    \(⇒y=\frac{30-3x}{5}\)
    \(⇒y=6-\frac{3x}{5}\)
  • The only possible value of x here is \(x=5\). Thus, \(y=6-\frac{3x}{5}=6-\frac{3\times 5}{5}=6-3=3\)
  • Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: 4 pencils and 4 pens cost 32 cents
  • According to this statement, \(4x+4y=32\)
    \(⇒y=\frac{32-4x}{4}⇒y=8-x\)
  • Here, multiple values of \(x=1,2,3,4,5,6\) and \(7\) are possible and corresponding different values of y
  • Thus, statement 2 alone is not sufficient

Hence the right answer is Option A

Originally posted by SaquibHGMATWhiz on 10 Nov 2022, 10:34.
Last edited by SaquibHGMATWhiz on 16 Aug 2023, 00:17, edited 1 time in total.
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Re: A stationery store sells pens and pencils. If the price of each item i [#permalink]
Expert Reply
A stationery store sells pens and pencils. If the price of each item in cents has a positive integer value, how much do 5 pencils cost?

(1) 5 pencils and 3 pens cost 30 cents.

This implies \(5x+3y=30\). This equation has only one positive integer solution: \(x=3\) and \(y=5\), so the price of 5 pencils is \(5*3=15\) cents. Sufficient.

(2) 4 pencils and 4 pens cost 32 cents.

This implies \(4x+4y=32\), which gives \(x+y=8\). This equation has more than one positive integer solution, for example: \(x=1\) and \(y=7\) or \(x=3\) and \(y=5\). Not sufficient.


Answer: A
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Re: A stationery store sells pens and pencils. If the price of each item i [#permalink]
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