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3(x – 7) ≥ 9 [#permalink]
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Expert's post 00:00

Question Stats: 82% (00:44) correct 17% (00:34) wrong based on 142 sessions
$$3(x – 7) \geq 9$$

$$0.25y – 3 \leq 1$$

 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
[Reveal] Spoiler: OA

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Re: 3(x – 7) ≥ 9 [#permalink]
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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]
1
KUDOS
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

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