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a, b, c, and d are positive
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31 Dec 2021, 07:20
31 Dec 2021, 07:20
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a, b, c, and d are positive
\(\frac{ a}{2} + \frac{b}{6}=\frac{c}{3}+\frac{d}{9}\)
The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given
\(\frac{ a}{2} + \frac{b}{6}=\frac{c}{3}+\frac{d}{9}\)
Quantity A |
Quantity B |
\(9a+3b\) |
\(6c+3d\) |
The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given
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Re: a, b, c, and d are positive
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31 Dec 2021, 07:38
31 Dec 2021, 07:38
1
Carcass wrote:
a, b, c, and d are positive
\(\frac{ a}{2} + \frac{b}{6}=\frac{c}{3}+\frac{d}{9}\)
The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given
\(\frac{ a}{2} + \frac{b}{6}=\frac{c}{3}+\frac{d}{9}\)
Quantity A |
Quantity B |
\(9a+3b\) |
\(6c+3d\) |
The quantity in Column A is greater
The quantity in Column B is greater
The two quantities are equal
The relationship cannot be determined from the information given
Given: \(\frac{ a}{2} + \frac{b}{6}=\frac{c}{3}+\frac{d}{9}\)
We can eliminate all of the fractions by multiplying both sides of the equation by 18 (the least common multiple of the 4 denominators).
When we do so we get: \(\frac{ 18a}{2} + \frac{18b}{6}=\frac{18c}{3}+\frac{18d}{9}\)
Simplify: \(9a + 3b = 6c + 2d\)
In Quantity A, we can replace \(9a + 3b\) with \(6c + 2d\) to get:
QUANTITY A: \(6c + 2d\)
QUANTITY B: \(6c + 3d\)
Subtract \(6c\) from both quantities:
QUANTITY A: \(2d\)
QUANTITY B: \(3d\)
Subtract \(2d\) from both quantities:
QUANTITY A: \(0\)
QUANTITY B: \(d\)
Since we're told d is positive, the correct answer is B
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Re: a, b, c, and d are positive [#permalink]
31 Dec 2021, 07:38
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