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08 Oct 2012, 02:53
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D
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30%
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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.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
(1) Max has fewer than 5 bills worth $5 each.
(2) Max has more than 5 bills worth $20 each.
(A) Statement 1 alone is sufficient, but statement 2 alone is not.
(B) Statement 2 alone is sufficient, but statement 1 alone is not.
(C) Both statements together are sufficient, but neither alone is sufficient.
(D) Each statement alone is sufficient.
(E) Neither statement is sufficient, even when combined.
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08 Oct 2012, 02:53
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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 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.
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.
08 Oct 2012, 05:07
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SOURH7WK
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.
