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Max has $125 consisting of bills each worth either $5 or [#permalink] New post 05 Nov 2009, 12:26
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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.

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Re: Q57, OG 12 Max bills [#permalink] New post 05 Nov 2009, 12:54
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pm4553 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

Discuss..


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

(1) 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.

(1) 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.
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Re: Q57, OG 12 Max bills [#permalink] New post 06 Nov 2009, 00:13
agree with Bunuel, i wud go by assuming numbers than forming equations, saves time..
in 1, take values less than 5 and only value satisying the result is 1. must be sufficient to get the answer.
in 2, take values more than 20, and only value satisfying the result is 6, giving us resultant value of 5$ bills to be 1. hence sufficient.

Ans must be D.
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Re: Q57, OG 12 Max bills [#permalink] New post 06 Nov 2009, 01:25
Bunuel wrote:
pm4553 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

Discuss..


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

(1) x<5 --> 5x+20y=125 --> y=\frac{125-5x}{20}=\frac{25-x}{4} as x<5 and y is an integer value only possible value for x is 1. Sufficient.

(1) y>5 --> 5x+20y=125 --> x=\frac{125-20y}{5}=25-4y as y>5 and x is not negative only possible value for y is 6, hence x=1. Sufficient.

Answer: D.


Good explanation, thanks!
OA is D
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Re: Max has $125 consisting of bills [#permalink] New post 15 Nov 2010, 01:14
monirjewel wrote:
Max has $125 consisting of bills each worth either $5 or $20. How many bills worth $5 Max have?
(1) Max has fewer than 5 bills worth $5 each
(2) Max has more than 5 bills worth $20 each


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

(1) 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.

(1) 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.
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Re: Max has $125 consisting of bills [#permalink] New post 15 Nov 2010, 08:10
monirjewel wrote:
Max has $125 consisting of bills each worth either $5 or $20. How many bills worth $5 Max have?
(1) Max has fewer than 5 bills worth $5 each
(2) Max has more than 5 bills worth $20 each


Bunuel has already discussed the algebraic way pretty elegantly. Let me just give you the non-algebra method here.

When we say we need $125, in $5 and $20 bills, first thing that comes to mind is that I need at least one $5 bill to get the 5 of 125. If I have only one $5 bill, I will need six $20 bills to make $120.

Also, if I have more than one $5 bill, I will need to reduce $20 bills. For every one $20 bill I reduce, I need four $5 bills to make up for it.
So number of $5 bills will be 1 or 5 or 9 or 13 and so on..

Stmnt 1: If $5 bills are fewer than 5, Max must have only one $5 bill. Sufficient.
Stmnt 2: If he has more than five $20 bills, he must have six $20 bills because that is the maximum number of $20 bills he can have. In that case he must have only one $5 bill. Sufficient.
Answer (D)
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Re: Max has $125 consisting of bills [#permalink] New post 15 Nov 2010, 19:42
The answer is D, and I didn't need to use the algebra method although it was helpful to see that as on a harder question it could be very useful.
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Re: Max has $125 consisting of bills [#permalink] New post 17 Nov 2010, 07:55
answer is D, without any algebra :)
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Re: Q57, OG 12 Max bills [#permalink] New post 19 Sep 2012, 18:55
Bunuel wrote:
pm4553 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

Discuss..


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

(1) 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.

(1) 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.


Dude, you're the **** best - thanks!
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Re: Max has $125 consisting of bills each worth either $5 or [#permalink] New post 19 Sep 2012, 22:13
pm4553 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.

Official Guide 12 Question

GMAT Official Guide 12

Question: 57
Page: 277
Difficulty: 600

Find All Official Guide Questions

Video Explanations:


125 = 5x + 20y
x = ?

the units digit of 125 is 5. From the equation 20y will always yield a value with 0 as it's units digit.
While 5x will then have to yield a value that ends with 5.

(1) if x is less than 5 then, we can choose from 0,1,2,3,4 but 5x should end with 5 so, we are left with 1 and 3 as possibilities.
Let x=1, the 20y should be 120, y=6
let x=3, then 5x=15 so 20y should be 125-15=110. But 110 is not a multiple of 20 so, not possible.
Thus, x=1. SUFFICIENT

(2) if y is greater than 5
Let us try y=6, this is possible!
Let us try y=7, 20y = 140, this is not possible or our eq.
thus, y=6. SUFFICIENT

Answer: D
Re: Max has $125 consisting of bills each worth either $5 or   [#permalink] 19 Sep 2012, 22:13
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