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Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have

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Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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New post Updated on: 29 Nov 2019, 09:46
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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. \(\frac{- 1}{4}\)

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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.
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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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New post 26 Nov 2019, 07:06
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. \(\frac{- 1}{4}\)



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

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

Add the two inequalities
\(\frac{3}{4}+x>\frac{a+6}{7}+x...........\frac{3}{4}>\frac{a+6}{7}......21>4a+24....4a<-3......a<\frac{-3}{4}\)

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

Please correct your question.
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Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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New post Updated on: 29 Nov 2019, 09:47
=>

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 \(a = \frac{-3}{4}.\)

Therefore, D is the answer.
Answer: D
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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.
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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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New post 28 Nov 2019, 01:46
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 \(a = \frac{-3}{4}.\)

Therefore, D is the answer.
Answer: D



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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New post 28 Nov 2019, 02:11
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 ?

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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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New post 28 Nov 2019, 05:13
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.
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Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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New post 29 Nov 2019, 09:48
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 \(a = \frac{-3}{4}.\)

Therefore, D is the answer.
Answer: D



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.
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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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New post 29 Nov 2019, 09:50
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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Re: Two inequalities (2x - 3)/2 > (5x - 6)/3 and 5x - 3 > (3x + a)/2 have  [#permalink]

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New post 29 Nov 2019, 10:32
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

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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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