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# Max has $125 consisting of bills each worth either$5 or $20  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Math Expert Joined: 02 Sep 2009 Posts: 23538 Followers: 3483 Kudos [?]: 26116 [0], given: 2705 Max has$125 consisting of bills each worth either $5 or$20 [#permalink]  08 Oct 2012, 01:53
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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] 08 Oct 2012, 01:53 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. _________________ Director Joined: 24 Aug 2009 Posts: 509 Schools: Harvard, Columbia, Stern, Booth, LSB, Followers: 9 Kudos [?]: 371 [2] , given: 241 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]  08 Oct 2012, 02:16
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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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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] 08 Oct 2012, 02:20 imo d...each alone is sufficient let the number of bills of 5$ 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. statement1: Senior Manager Joined: 15 Jun 2010 Posts: 363 Schools: IE'14, ISB'14, Kellogg'15 WE 1: 7 Yrs in Automobile (Commercial Vehicle industry) Followers: 7 Kudos [?]: 160 [0], given: 50 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]  08 Oct 2012, 02:23
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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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] 08 Oct 2012, 04:07 4 This post received KUDOS 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. Hence Answer D. 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. Nonetheless, your answer is absolutely correct: D. _________________ PhD in Applied Mathematics Love GMAT Quant questions and running. Math Expert Joined: 02 Sep 2009 Posts: 23538 Followers: 3483 Kudos [?]: 26116 [0], given: 2705 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]  11 Oct 2012, 13:20
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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Re: Max has $125 consisting of bills each worth either$5 or $20 [#permalink] 26 Feb 2013, 03:34 not easy at all, I want to follow to meet this again. Manager Joined: 12 Jan 2013 Posts: 249 Followers: 0 Kudos [?]: 20 [0], given: 47 Re: Max has$125 consisting of bills each worth either $5 or$20 [#permalink]  10 Jan 2014, 07:45
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.