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abcd are different from zero
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15 Jul 2019, 02:26
15 Jul 2019, 02:26
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78% (01:17) correct
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\(abcd \neq 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{a^2b}{cd^2} \times \frac{d^3}{abc}\) |
\(\frac{d^2}{bc} \times \frac{ab^2}{bd}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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Re: abcd are different from zero
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15 Jul 2019, 07:06
15 Jul 2019, 07:06
3
Carcass wrote:
\(abcd \neq 0\)
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
\(\frac{a^2b}{cd^2} \times \frac{d^3}{abc}\) |
\(\frac{d^2}{bc} \times \frac{ab^2}{bd}\) |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Given:
Quantity A: \(\frac{a^2b}{cd^2} \times \frac{d^3}{abc}=\frac{a^2bd^3}{abc^2d^2}=\frac{ad}{c^2}\)
Quantity B: \(\frac{d^2}{bc} \times \frac{ab^2}{bd}=\frac{ab^2d^2}{b^2cd}=\frac{ad}{c}\)
From here, we can quickly see the answer is D by testing some possible values.
case i: a = c = d = 1
We get:
Quantity A: \(\frac{(1)(1)}{1^2}=1\)
Quantity B: \(\frac{(1)(1)}{1}=1\)
In this case, the quantities are equal
case ii: a = d = 1, and c = 2
We get:
Quantity A: \(\frac{(1)(1)}{2^2}=\frac{1}{4}\)
Quantity B: \(\frac{(1)(1)}{2}=\frac{1}{2}\)
In this case, Quantity B is greater
Answer: D
Cheers,
Brent
abcd are different from zero
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08 Mar 2024, 19:03
08 Mar 2024, 19:03
1
Quantity A
\(\frac{a^2b}{cd^2} \times \frac{d^3}{abc} = \frac{ad}{c^2} \)
Quantity B
\(\frac{d^2}{bc} \times \frac{ab^2}{bd} = \frac{ad}{c}\)
If the values of a, c, d are positive Quantity B is greater.
If the values of a, c, d are negative, Quantity A is greater.
The answer is thus D.
\(\frac{a^2b}{cd^2} \times \frac{d^3}{abc} = \frac{ad}{c^2} \)
Quantity B
\(\frac{d^2}{bc} \times \frac{ab^2}{bd} = \frac{ad}{c}\)
If the values of a, c, d are positive Quantity B is greater.
If the values of a, c, d are negative, Quantity A is greater.
The answer is thus D.
