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Max has $125 consisting of bills each worth either$5 or $20 [#permalink] ### Show Tags 08 Oct 2012, 01:53 Expert's post 13 This post was BOOKMARKED 00:00 Difficulty: 55% (hard) Question Stats: 59% (02:20) correct 41% (01:17) wrong based on 975 sessions ### HideShow timer Statistics 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 _________________ Math Expert Joined: 02 Sep 2009 Posts: 36590 Followers: 7092 Kudos [?]: 93369 [0], given: 10557 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]

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08 Oct 2012, 01:53
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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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08 Oct 2012, 02:20
imo d...each alone is sufficient
let the number of bills of 5$b x and bills for 20$ be y
now we have only two situations where 5x+20y=125
either
x=1 and y=6 --> (true wen we use statement 1)
or
x=5 and y=5 --> (true wen we use statement 2)

using any of the above two statements we can find out the answer.

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08 Oct 2012, 04:07
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SOURH7WK wrote:
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.

A suggestion: always divide an equation by the GCD of the coefficients, it becomes easier to handle. In this case, 5x + 20y = 125, divide through by 5 and get:
x + 4y = 25. Smaller numbers, positive integers...isn't it easier to see the solutions?

(1): Another approach would be to look at 25 as being a M4+1 (remainder 1 when divided by 4). 4y is divisible by 4, therefore x must leave a remainder of 1 when divided by 4. Since x is less than 5, the only possibility is x = 1. Just to practice divisibility properties...:O)
(2): y > 5, then 4y > 20. Because 4*7 = 28 > 25, the only possible value for y is 6, and x must be 1.

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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] ### Show Tags 11 Oct 2012, 13:20 1 This post received KUDOS 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. Answer: D. Kudos points given to everyone with correct solution. Let me know if I missed someone. _________________ VP Joined: 08 Jun 2010 Posts: 1401 Followers: 3 Kudos [?]: 114 [0], given: 813 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]

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26 Feb 2013, 03:34
not easy at all, I want to follow to meet this again.
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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] ### Show Tags 10 Jan 2014, 07:45 1 This post received KUDOS 1 This post was BOOKMARKED 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. Our answer is D. GMAT Club Legend Joined: 09 Sep 2013 Posts: 13489 Followers: 576 Kudos [?]: 163 [0], given: 0 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]

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09 Aug 2015, 12:00
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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] ### Show Tags 17 Nov 2016, 00:33 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Max has$125 consisting of bills each worth either $5 or$20   [#permalink] 17 Nov 2016, 00:33
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