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x is an integer and 1<x<9
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15 Oct 2018, 09:28
15 Oct 2018, 09:28
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x is an integer.
\(1<x<9\)
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
\(1<x<9\)
Quantity A |
Quantity B |
\((\sqrt{x}+ \sqrt{x} )^2\) |
\(x + x \sqrt{x}\) |
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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Re: x is an integer and 1<x<9
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15 Oct 2018, 10:07
15 Oct 2018, 10:07
1
Carcass wrote:
x is an integer.
\(1<x<9\)
\(1<x<9\)
Quantity A |
Quantity B |
\((\sqrt{x}+ \sqrt{x} )^2\) |
\(x + x \sqrt{x}\) |
Given:
Quantity A: (√x + √x)²
Quantity B: x + x√x
Simplify:
Quantity A: (2√x)²
Quantity B: x + x√x
Simplify again:
Quantity A: 4x
Quantity B: x + x√x
Subtract x from both quantities to get:
Quantity A: 3x
Quantity B: x√x
Since x is POSITIVE, we can safely divide both quantities by x to get:
Quantity A: 3
Quantity B: √x
We're told that x = 2, 3, 4, 5, 6, 7, or 8
For EVERY possible x-value, √x is ALWAYS less than 3 (i.e., √2 < 3, and √3 < 3, and √4 < 3 . . . and √8 < 3,
Answer: A
Cheers,
Brent
Re: x is an integer and 1<x<9
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02 Nov 2018, 21:49
02 Nov 2018, 21:49
GreenlightTestPrep wrote:
Carcass wrote:
x is an integer.
\(1<x<9\)
\(1<x<9\)
Quantity A |
Quantity B |
\((\sqrt{x}+ \sqrt{x} )^2\) |
\(x + x \sqrt{x}\) |
Given:
Quantity A: (√x + √x)²
Quantity B: x√x
Simplify:
Quantity A: (2√x)²
Quantity B: x√x
Simplify again:
Quantity A: 4x
Quantity B: x√x
Since x is POSITIVE, we can safely divide both quantities by x to get:
Quantity A: 4
Quantity B: √x
We're told that x = 2, 3, 4, 5, 6, 7, or 8
For EVERY possible x-value, √x is ALWAYS less than 4
Answer: A
Cheers,
Brent
The quantity B is x+sqrt(x) and not xsqrtx
However the answer doesn't change
