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Is y > -4?

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Is y > -4? [#permalink]  05 Mar 2012, 05:46
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Question Stats:

35% (02:24) correct 65% (01:51) wrong based on 249 sessions
Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14)

(2) 4y^2 + 12 y < 0
[Reveal] Spoiler: OA

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Posts: 28352
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Kudos [?]: 45493 [3] , given: 6762

Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12 [#permalink]  05 Mar 2012, 05:57
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Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

(2) 4y^2 + 12 y < 0 --> reduce by 4 and factor out $$y$$: $$y(y+3)<0$$ --> roots are -3 and 0, "<" sign indicates that the solution lies between the roots: $$-3<y<0$$, hence $$y>-4$$. Sufficient.

Solving inequalities (to understand the reasoning for second statement):
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
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Joined: 01 Sep 2010
Posts: 2634
Followers: 470

Kudos [?]: 3646 [0], given: 726

Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12 [#permalink]  05 Mar 2012, 06:22
Expert's post
Bunuel wrote:
Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

(2) 4y^2 + 12 y < 0 --> reduce by 4 and factor out $$y$$: $$y(y+3)<0$$ --> roots are -3 and 0, "<" sign indicates that the solution lies between the roots: $$-3<y<0$$, hence $$y>-4$$. Sufficient.

Solving inequalities (to understand the reasoning for second statement):
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

For 1) is the same (of course is subjective) to do: 7^-4y > 7^-8y+14.......in the end: y>7/2 ?????

For 2) divide by 4 we have : y(y+3)<0 the sign is LESS so, as consequence the value of x is between (not at the extreme as in case of sign > ) -3<x<0. In this case of course our variable is y but is the same, is only one unknown.

By the way OA is D but I have never seen a post from you wrong. Precious work.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 28352
Followers: 4487

Kudos [?]: 45493 [0], given: 6762

Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12 [#permalink]  05 Mar 2012, 06:28
Expert's post
carcass wrote:
Bunuel wrote:
Is y > -4?

(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

(2) 4y^2 + 12 y < 0 --> reduce by 4 and factor out $$y$$: $$y(y+3)<0$$ --> roots are -3 and 0, "<" sign indicates that the solution lies between the roots: $$-3<y<0$$, hence $$y>-4$$. Sufficient.

Solving inequalities (to understand the reasoning for second statement):
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

For 1) is the same (of course is subjective) to do: 7^-4y > 7^-8y+14.......in the end: y>7/2 ?????

For 2) divide by 4 we have : y(y+3)<0 the sign is LESS so, as consequence the value of x is between (not at the extreme as in case of sign > ) -3<x<0. In this case of course our variable is y but is the same, is only one unknown.

By the way OA is D but I have never seen a post from you wrong. Precious work.

The red part should be 7^(-4y) > 7^(-8y-14). Else is correct.
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Kudos [?]: 45493 [0], given: 6762

Re: Is y > -4? [#permalink]  17 Jun 2013, 04:03
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12 [#permalink]  28 Oct 2013, 23:15
Bunuel wrote:
Is y > -4?
(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

Hi Bunuel! I have read the theory for solving inequalities, but still I am not very comfortable with the subject. Could you please explain why while solving statement 1 we do not consider the sign for y? Generally while solving inequality for a variable we consider the cases when y>0 and y<0. Then in this particular case why the sign of y is not necessary? TIA.
Math Expert
Joined: 02 Sep 2009
Posts: 28352
Followers: 4487

Kudos [?]: 45493 [1] , given: 6762

Re: Is y > -4? (1) (1/7)^(4y) > (1/7)^(8y + 14) (2) 4y^2 + 12 [#permalink]  29 Oct 2013, 01:36
1
KUDOS
Expert's post
vjns wrote:
Bunuel wrote:
Is y > -4?
(1) (1/7)^(4y) > (1/7)^(8y + 14) --> $$\frac{1}{7^{4y}}>\frac{1}{7^{8y+14}}$$ --> $$7^{8y+14}>7^{4y}$$ --> $$8y+14>4y$$ --> $$y>-3.5$$, hence $$y>-4$$. Sufficient.

Hi Bunuel! I have read the theory for solving inequalities, but still I am not very comfortable with the subject. Could you please explain why while solving statement 1 we do not consider the sign for y? Generally while solving inequality for a variable we consider the cases when y>0 and y<0. Then in this particular case why the sign of y is not necessary? TIA.

The sign of y has nothing to do when solving $$7^{8y+14}>7^{4y}$$.

Consider another way. Since $$7^{4y}$$ is positive irrespective of value of y, then we can safely reduce both sides by it and write: $$\frac{7^{8y+14}}{7^{4y}}>1$$ --> $$7^{8y+14-4y}>1$$.

$$7^{8y+14-4y}>1$$ to hold true the exponent must be positive: $$8y+14-4y>0$$ --> $$y>-3.5$$.

Hope this helps.
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Re: Is y > -4? [#permalink]  07 May 2015, 16:03
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: Is y > -4?   [#permalink] 07 May 2015, 16:03
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