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08 Oct 2012, 02:53
Expert's post
16
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Difficulty:

55% (hard)

Question Stats:

60% (01:20) correct 40% (01:18) wrong based on 1332 sessions

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Max has $125 consisting of bills each worth either$5 or $20. How many bills worth$5 does Max have?

(1) Max has fewer than 5 bills worth $5 each. (2) Max has more than 5 bills worth$20 each.

Practice Questions
Question: 59
Page: 280
Difficulty: 600
[Reveal] Spoiler: OA

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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] ### Show Tags 08 Oct 2012, 02:53 Expert's post 7 This post was BOOKMARKED SOLUTION Max has$125 consisting of bills each worth either $5 or$20. How many bills worth $5 does Max have? Let $$x$$ be the # of 5$ bills and $$y$$ the # of 20$bills --> $$5x+20y=125$$ --> $$x=?$$ (1) Max has fewer than 5 bills worth$5 each. $$x<5$$ --> $$5x+20y=125$$ --> $$y=\frac{125-5x}{20}=\frac{25-x}{4}$$ as $$x<5$$ and $$y$$ must be an integer then only possible value for $$x$$ is 1. Sufficient.

(2) Max has more than 5 bills worth $20 each. $$y>5$$ --> $$5x+20y=125$$ --> $$x=\frac{125-20y}{5}=25-4y$$ as $$y>5$$ and $$x$$ must not negative then only possible value for $$y$$ is 6, hence $$x=1$$. Sufficient. Answer: D. _________________ Kudos [?]: 128459 [0], given: 12173 Director Joined: 24 Aug 2009 Posts: 500 Kudos [?]: 839 [2], given: 276 Schools: Harvard, Columbia, Stern, Booth, LSB, Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]

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08 Oct 2012, 03:16
2
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From stem 5x + 20y = 125
Question is x=?
Note :- x & y can take only integer values because we can not tear either $5 or$20 notes
1) x<5--> The only possible integer value is x=1 -->Sufficient
2) y>5--> The only possible integer value is y=6 & x = 1 -->Sufficient
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08 Oct 2012, 03:23
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Let x be $5 bill & y be$20 bill, 5x+20y =125, Find x?
ST1: Sufficient: x<5, Therefore x can be 0,1,2,3,4. Here only x=1 satisfy the given equation and for all other value y will not be an integer value.
ST2: Sufficient: y>5, y=6,7,8... Now for y =7 the total value exceeds 125. Therefore Y must be 6. And so x will be 1.

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11 Oct 2012, 14:20
1
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Expert's post
SOLUTION

Max has $125 consisting of bills each worth either$5 or $20. How many bills worth$5 does Max have?

Let $$x$$ be the # of 5$bills and $$y$$ the # of 20$ bills --> $$5x+20y=125$$ --> $$x=?$$

(1) Max has fewer than 5 bills worth $5 each. $$x<5$$ --> $$5x+20y=125$$ --> $$y=\frac{125-5x}{20}=\frac{25-x}{4}$$ as $$x<5$$ and $$y$$ must be an integer then only possible value for $$x$$ is 1. Sufficient. (2) Max has more than 5 bills worth$20 each. $$y>5$$ --> $$5x+20y=125$$ --> $$x=\frac{125-20y}{5}=25-4y$$ as $$y>5$$ and $$x$$ must not negative then only possible value for $$y$$ is 6, hence $$x=1$$. Sufficient.

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10 Jan 2014, 08:45
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Bunuel wrote:
Max has $125 consisting of bills each worth either$5 or $20. How many bills worth$5 does Max have?

(1) Max has fewer than 5 bills worth $5 each. (2) Max has more than 5 bills worth$20 each.

Practice Questions
Question: 59
Page: 280
Difficulty: 600

Basically, the stem gives us: 5*x + 20*y = 125, and asks us what x is.

1) tells us that x < 5, so we try to maximize for 20 to control what possible values x can take. For y = 5 we have x = 5 and for y = 6 we have x = 1, no other combinations in that range are possible between X and Y. And since 1) makes it impossible for x = 5, x must be = 1.. So 1 is sufficient.

2) tells us that y > 5, so again we test if there are different possible values for y. For y = 6 we have x = 1.. And y can't actually be a higher value than 6 because then we break the restriction of 125 USD.. So clearly, y = 6 and x = 1, so 2 is sufficient.

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