Joanna bought only $0.15 stamps and$0.29 stamps. How many : GMAT Data Sufficiency (DS)
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# Joanna bought only $0.15 stamps and$0.29 stamps. How many

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Joined: 09 Jun 2010
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Joanna bought only $0.15 stamps and$0.29 stamps. How many [#permalink]

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28 Jul 2010, 05:49
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Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy ? 1 she bought$4.40 worth of stamps
2 She bought an equal number of $0.15 stamps and 0.29 stamps . I was thinking we need the total stamps as an addtional information so I picked E ! pleas help me understand thanks Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93377 [1] , given: 10557 Re: D.S [#permalink] ### Show Tags 28 Jul 2010, 05:59 1 This post received KUDOS Expert's post xmagedo wrote: Joanna bought only$0.15 stamps and $0.29 stamps. How many$0.15 stamps did she buy ?
1 she bought $4.40 worth of stamps 2 She bought an equal number of$0.15 stamps and 0.29 stamps .

I was thinking we need the total stamps as an addtional information so I picked E !

pleas help me understand
thanks

Joanna bought only $0.15 stamps and$0.29 stamps. How many $0.15 stamps did she buy? Let $$x$$ be the # of$0.15 stamps and $$y$$ the # of $0.29 stamps. Note that $$x$$ and $$y$$ must be an integers. Q: $$x=?$$ (1) She bought$4.40 worth of stamps --> $$15x+29y=440$$. Only one integer combination of $$x$$ and $$y$$ is possible to satisfy $$15x+29y=440$$: $$x=10$$ and $$y=10$$. Sufficient.

(2) She bought an equal number of $0.15 stamps and$0.29 stamps --> $$x=y$$. Not sufficient.

So when we have equation of a type $$ax+by=c$$ and we know that $$x$$ and $$y$$ are non-negative integers, there can be multiple solutions possible for $$x$$ and $$y$$ (eg $$5x+6y=60$$) OR just one combination (eg $$15x+29y=440$$). Hence in some cases $$ax+by=c$$ is NOT sufficient and in some cases it is sufficient.

For more on this type of questions check: gmat-prep2-92785.html?hilit=stamps

Hope it helps.
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29 Jul 2010, 01:44
Statement 1: equation 15x+29y=440
since both 15x and 440 are multiples of 5 then y=5, 10, 15, 20, ...
since 290y<440 then y<20 then y=5, 10, 15
since 15x mod 3 = 0, 29 mod 3 = 2, 440 mod 3 = 2, then y mod 3 = 1, or y=10
sufficient

Statement 2: clearly not sufficient

Notation: (a mod b) gives remainder when divide a by b
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Re: D.S   [#permalink] 29 Jul 2010, 01:44
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