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Dave bought a few pencils and a few erasers. Each pencil costs 12 cent
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16 Oct 2023, 09:34
16 Oct 2023, 09:34
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Dave bought a few pencils and a few erasers. Each pencil costs 12 cents and each eraser costs 20 cents. He bought at least one of both the items and spent a total of 108 cents.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
Number of pencils |
Number of erasers |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Dave bought a few pencils and a few erasers. Each pencil costs 12 cent
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04 Dec 2023, 23:13
04 Dec 2023, 23:13
1
Right off the bat, we know that the numbers of pencils and erasers must be integers greater than or equal to 1.
Interpret into an equation, where x = number of pencils and y = number of erasers:
\(12x + 20y = 108\)
We can see that x must be a multiple of 4, because it's the only way that \(y=\frac{108-12x}{20}\) would also be an integer (i.e., have a remainder of 0).
Let's try \(x=4\):
\(12(4)+20y=108\)
\(20y=60\)
\(y=3\)
4>3, so A>B.
Interpret into an equation, where x = number of pencils and y = number of erasers:
\(12x + 20y = 108\)
We can see that x must be a multiple of 4, because it's the only way that \(y=\frac{108-12x}{20}\) would also be an integer (i.e., have a remainder of 0).
Let's try \(x=4\):
\(12(4)+20y=108\)
\(20y=60\)
\(y=3\)
4>3, so A>B.
gmatclubot
Dave bought a few pencils and a few erasers. Each pencil costs 12 cent [#permalink]
04 Dec 2023, 23:13
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