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3(x – 7) ≥ 9

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3(x – 7) ≥ 9 [#permalink] New post 23 Jul 2018, 09:54
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Re: 3(x – 7) ≥ 9 [#permalink] New post 27 Jul 2018, 09:35
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Carcass wrote:
\(3(x – 7) \geq 9\)

\(0.25y – 3 \leq 1\)

Quantity A
Quantity B
\(x\)
\(y\)


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.

Kudos for R.A.E


The first equation can be solved algebraically,
\(3(x – 7) \geq 9\)

or\(x - 7 \geq 3\)

or \(x \geq 10\)

For 2nd equation:

\(0.25y – 3 \leq 1\)

or \(0.25y \leq 4\)

or \(y\leq 16\)

So we cannot reach to a definite solution for this because x can be of any value from 10 and y can be any value less than or equal to 16

So option is D
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Re: 3(x – 7) ≥ 9 [#permalink] New post 11 Apr 2020, 06:33
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Expert's post
Carcass wrote:
\(3(x – 7) \geq 9\)

\(0.25y – 3 \leq 1\)

Quantity A
Quantity B
\(x\)
\(y\)


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.

Kudos for R.A.E


Given: \(3(x – 7) \geq 9\)
Divide both sides of the inequality by \(3\) to get: \(x – 7 \geq 3\)
Add \(7\) to both sides to get: \(x \geq 10\)

Also given: \(0.25y – 3 \leq 1\)
Add \(3\) to both sides to get: \(0.25y \leq 4\)
Multiply both sides by \(4\) to get: \(y \leq 16\)

We now know that \(x \geq 10\) and \(y \leq 16\)
So, for example, it could be the case that x = 12 and y = 12, in which case the two quantities are equal
Or, it could be the case that x = 13 and y = 12, in which case Quantity A is greater

Answer: D

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
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Re: 3(x – 7) ≥ 9   [#permalink] 11 Apr 2020, 06:33
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