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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, 03:16

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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 Answer D
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08 Oct 2012, 03:20

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.

WE 1: 7 Yrs in Automobile (Commercial Vehicle industry)

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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.

Hence Answer D.
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08 Oct 2012, 05:07

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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.

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.
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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.
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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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