Max has $125 consisting of bills each worth either $5 or $20

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 number of 5$ bills and \(y\) the number of 20$ bills, then we'd have: \(5x+20y=125\). The question asks to find the value of x.


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

From \(5x+20y=125\) we get \(y=\frac{125-5x}{20}=\frac{25-x}{4}\). Since given that \(x<5\) 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.

From \(5x+20y=125\) we get \(x=\frac{125-20y}{5}=25-4y\). Since given that \(y>5\) and \(x\) cannot be negative, then only possible value for \(y\) is 6, hence \(x=1\). Sufficient.

Answer: D.
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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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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

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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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.

Hi All,

We're told that Max has $125 consisting of bills worth $5 or $20 each. We're asked for the number of $5 bills that Max has. Based on the 'restrictions' in this question, there are only a handful of possible ways to get to $125. As such, I'm going to list them out first (before we deal with the two Facts):

$125:
Six 20s and one 5
Five 20s and five 5s
Four 20s and nine 5s
Three 20s and thirteen 5s
Two 20s and seventeen 5s
One 20 and twenty-one 5s
Zero 20s and twenty-five 5s

1) Max has fewer than 5 bills worth $5 each.

With this Fact, there's only one possible option: Six 20s and one 5.
Fact 1 is SUFFICIENT

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

With this Fact, there's only one possible option: Six 20s and one 5.
Fact 2 is SUFFICIENT

Final Answer:

GMAT assassins aren't born, they're made,
Rich

This type of question can also be solved by CHART METHOD with GREATER ACCURACY.

STEPS
1. Start with BIGGER MULTIPLE & CONTINUE up to MAXIMUM possible value of the BIGGER MULTIPLE.
2. Then Start COUNTING the REMAINING amount for the SMALLER MULTIPLE from the BOTTOM of the Chart. Keep the scenario, if the remaining number is divisible by the SMALLER number.

BACK to this QUESTION:
1. Start with 20, ( i.e. 20x0, 20x1.....) & Continue up to 20x6 [because 20x7 is GREATER than 125.
2. Then, CONTINUE from the BOTTOM of the Chart. REMAINING number is 5, which is ONE Multiple of FIVE. SINCE 20 is DIVISIBLE by 5 (for 4 tomes), just ADD 4 for with the bottom number of MULTIPLE of the SMALLER INTEGER to get SUBSEQUENTLY UPPER ONE

(here,
number of MULTIPLE of the SMALLER INTEGER in the BOTTOM: 1,
number of MULTIPLE of the SMALLER INTEGER in NEXT UPPER CELL: 1+4=5
number of MULTIPLE of the SMALLER INTEGER in NEXT UPPER CELL: 6+4=9
Thus, 1,5,9,13,17,21,25.).

STATEMENT 1: Max has fewer than 5 bills worth $5 each.
That means, ONLY SCENARIO (g) of my CHART is POSSIBLE.
SUFFICIENT.

STATEMENT 2: Max has more than 5 bills worth $20 each.
That means, ONLY SCENARIO (g) of my CHART is POSSIBLE.
SUFFICIENT.

CHART method will take little amount of time when one gets used to using it & provide greater accuracy.

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