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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82
Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 57% (02:09) correct 43% (03:19) wrong based on 28 sessions

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[GMAT math practice question]

Two inequalities $$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}$$ and $$5x - 3 < \frac{(3x + a)}{2}$$ have same solution. What is the value of $$a$$?

A. $$\frac{3}{4}$$

B. $$\frac{1}{4}$$

C. $$0$$

D. $$\frac{- 3}{4 }$$

E. $$\frac{- 1}{4}$$

_________________

Originally posted by MathRevolution on 26 Nov 2019, 06:07.
Last edited by MathRevolution on 29 Nov 2019, 09:46, edited 1 time in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 8284
Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

Two inequalities $$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}$$ and $$5x - 3 > \frac{(3x + a)}{2}$$ have same solution. What is the value of $$a$$?

A. $$\frac{3}{4}$$

B. $$\frac{1}{4}$$

C. $$0$$

D. $$\frac{- 3}{4 }$$

E. $$\frac{- 1}{4}$$

$$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}.......6x-9>10x-12......4x<3...x<\frac{3}{4}$$

$$5x - 3 > \frac{(3x + a)}{2}.....10x-6>3x+a.....7x>a+6.....x>\frac{a+6}{7}$$

$$\frac{3}{4}+x>\frac{a+6}{7}+x...........\frac{3}{4}>\frac{a+6}{7}......21>4a+24....4a<-3......a<\frac{-3}{4}$$

Had your Inequality been $$\frac{(2x - 3)}{2} >< \frac{(5x - 6)}{3}$$, then $$x>\frac{3}{4}$$ would have meant $$\frac{3}{4}=\frac{a+6}{7}.....a=\frac{-3}{4}$$.

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Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82
Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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=>

Since we have $$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}$$, if we cross multiply we get $$3(2x - 3) > 2(5x - 6)$$ or $$-4x > -3.$$ Then, we have $$x < \frac{3}{4}.$$

Since we have $$5x - 3 < \frac{(3x + a)}{2}$$, if we cross multiply we get $$10x - 6 < 3x + a$$ or $$7x < a + 6.$$ Then we have $$x < \frac{(a + 6)}{7}.$$

Since those two inequalities are equivalent each other, we have $$\frac{(a + 6)}{7} = \frac{3}{4}$$ or $$4a + 24 = 21$$. Then we have $$a = \frac{-3}{4}.$$

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Originally posted by MathRevolution on 28 Nov 2019, 01:20.
Last edited by MathRevolution on 29 Nov 2019, 09:47, edited 1 time in total.
Math Expert V
Joined: 02 Aug 2009
Posts: 8284
Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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MathRevolution wrote:
=>

Since we have $$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}$$, if we cross multiply we get $$3(2x - 3) > 2(5x - 6)$$ or $$-4x > -3.$$ Then, we have $$x < \frac{3}{4}.$$

Since we have $$5x - 3 > \frac{(3x + a)}{2}$$, if we cross multiply we get $$10x - 6 > 3x + a$$ or $$7x > a + 6.$$ Then we have $$x < \frac{(a + 6)}{7}.$$

Since those two inequalities are equivalent each other, we have $$\frac{(a + 6)}{7} = \frac{3}{4}$$ or $$4a + 24 = 21$$. Then we have $$a = \frac{-3}{4}.$$

You are wrong in the highlighted portion.
You cannot change the inequality sign
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Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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Hello,

I do not understand the part where you say x<3/4 and x<(a+6)/7
Then you equated both of them 3/4=(a+6)/7. Can we equate this ? (a+6)/7 can be some value greater than 3/4?
According to me we can only add both sides.

chetan2u am I correct ?

Posted from my mobile device
Math Expert V
Joined: 02 Aug 2009
Posts: 8284
Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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Sidmehra wrote:
Hello,

I do not understand the part where you say x<3/4 and x<(a+6)/7
Then you equated both of them 3/4=(a+6)/7. Can we equate this ? (a+6)/7 can be some value greater than 3/4?
According to me we can only add both sides.

chetan2u am I correct ?

Posted from my mobile device

Yes, but here it is said that both solutions are same. Not that I have seen a question on GMAT and as I said that the question is flawed
So if we say x<5 and x<a and solutions are same then a=5.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82
Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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chetan2u wrote:
MathRevolution wrote:
=>

Since we have $$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}$$, if we cross multiply we get $$3(2x - 3) > 2(5x - 6)$$ or $$-4x > -3.$$ Then, we have $$x < \frac{3}{4}.$$

Since we have $$5x - 3 > \frac{(3x + a)}{2}$$, if we cross multiply we get $$10x - 6 > 3x + a$$ or $$7x > a + 6.$$ Then we have $$x < \frac{(a + 6)}{7}.$$

Since those two inequalities are equivalent each other, we have $$\frac{(a + 6)}{7} = \frac{3}{4}$$ or $$4a + 24 = 21$$. Then we have $$a = \frac{-3}{4}.$$

You are wrong in the highlighted portion.
You cannot change the inequality sign

Yes. There was a mistake.
The question and the solution are fixed.
_________________
Math Revolution GMAT Instructor V
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42 GPA: 3.82
Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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chetan2u wrote:
Sidmehra wrote:
Hello,

I do not understand the part where you say x<3/4 and x<(a+6)/7
Then you equated both of them 3/4=(a+6)/7. Can we equate this ? (a+6)/7 can be some value greater than 3/4?
According to me we can only add both sides.

chetan2u am I correct ?

Posted from my mobile device

Yes, but here it is said that both solutions are same. Not that I have seen a question on GMAT and as I said that the question is flawed
So if we say x<5 and x<a and solutions are same then a=5.

Yes. It is right, since two inequalities have the same solution set.
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Joined: 03 Jun 2019
Posts: 1875
Location: India
Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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MathRevolution wrote:
[GMAT math practice question]

Two inequalities $$\frac{(2x - 3)}{2} > \frac{(5x - 6)}{3}$$ and $$5x - 3 < \frac{(3x + a)}{2}$$ have same solution. What is the value of $$a$$?

A. $$\frac{3}{4}$$

B. $$\frac{1}{4}$$

C. $$0$$

D. $$\frac{- 3}{4 }$$

E. $$\frac{- 1}{4}$$

First inequality
x<3/4

Second inequality
x<(6+a)/7

(6+a)/7=3/4
24+4a=21
a=-3/4

IMO D

Posted from my mobile device Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have   [#permalink] 29 Nov 2019, 10:32
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# Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  