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c < –1

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c < –1 [#permalink]  19 Jul 2019, 09:47
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Question Stats:

65% (00:20) correct 34% (00:44) wrong based on 23 sessions
$$c < –1$$

 Quantity A Quantity B $$6c-3$$ $$3c-5$$

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
[Reveal] Spoiler: OA

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Re: c < –1 [#permalink]  14 Jan 2020, 06:46
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Let's assume A > B
=> 6c - 3 > 3c -5
=> 6c - 3c > 3 -5
=> 3c > -2
=> C > -2/3
Which is false as c < -1
So, Quantity B > Quantity A
Hope it helps!
Carcass wrote:
$$c < –1$$

 Quantity A Quantity B $$6c-3$$ $$3c-5$$

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: c < –1 [#permalink]  14 Jan 2020, 23:20
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Qa: 6c............................Qb=3c-2

Subtract 3C from both quantities,
Qa: 3c............................Qb=-2

Even if c=-1, 3c=-3. Since c<-1, 3c must be less than -3.
So, Quantity b is greater.
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Re: c < –1 [#permalink]  15 Jan 2020, 03:07
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Addition and subtraction in inequality in any case is accepted:
(A) 6c-3 ? (B)3c-5 just solve it like algebraic expressions.
3C ? -2 Simple addition and subtraction on both sides.
C ? -2/3 Dividing by 3(being positive no.) on both sides is accepted.
Since c<–1, (B) is the answer.
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Re: c < –1 [#permalink]  15 Jan 2020, 16:12
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Expert's post
Carcass wrote:
$$c < –1$$

 Quantity A Quantity B $$6c-3$$ $$3c-5$$

We can solve this question using matching operations

Given:
Quantity A: $$6c-3$$
Quantity B: $$3c-5$$

Add $$3$$ to both quantities to get:
Quantity A: $$6c$$
Quantity B: $$3c-2$$

Subtract $$3c$$ from both quantities to get:
Quantity A: $$3c$$
Quantity B: $$-2$$

Divide both quantities by $$3$$ to get:
Quantity A: $$c$$
Quantity B: $$-\frac{2}{3}$$

Since $$-1 < -\frac{2}{3}$$, and since it's given that $$c < –1$$, we can combine both inequalities to get: $$c < -1 < -\frac{2}{3}$$
This means $$c < -\frac{2}{3}$$, so Quantity B is greater.

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Re: c < –1   [#permalink] 15 Jan 2020, 16:12
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