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|2x + 3| > |7x - 2|

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|2x + 3| > |7x - 2|

by sanju09 » Tue Feb 24, 2009 4:26 am
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1
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by DanaJ » Tue Feb 24, 2009 5:24 am
Break it down into pieces, after noticing that |2x + 3| > |7x - 2| is equivalent to |2x + 3| - |7x - 2| > 0.
a. x is smaller than -3/2 then you get that:
|2x + 3| = -2x - 3
|7x - 2| = 2 - 7x
This means that the initial equation is equivalent to:
5x - 5 > 0 or 5x > 5 or x > 5. This is inconsistent with x < -3/2

b. x is between -3/2 and 2/7. You have:
|2x + 3| = 2x + 3
|7x - 2| = 2 - 7x
Initial equation will be:
9x + 1 > 0 or x > -1/9. So for this one the solution will be (-1/9, 2/7]

c. x is greater than 2/7. You get:
|2x + 3| = 2x + 3
|7x - 2| = 7x - 2
Initial euqation:
-5x + 5 > 0 or -5x > -5 or x < 1. Solution: (2/7, 1)


Put it all together and you get -1/9 < x < 1.
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Re: |2x + 3| > |7x - 2|

by sureshbala » Tue Feb 24, 2009 5:29 am
sanju09 wrote:What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1
A quick way is to eliminate the options....

Clearly, 0 satisfies the given expression.
So choices A and E are out.

1 does not satisfy the equation.
So choice D is out.

1/2 satisfies the equation.
So choice B is out.

Hence C is the answer
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Re: |2x + 3| > |7x - 2|

by ven4gmat » Tue Feb 24, 2009 5:46 am
sureshbala wrote:
sanju09 wrote:What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1
A quick way is to eliminate the options....

Clearly, 0 satisfies the given expression.
So choices A and E are out.

1 does not satisfy the equation.
So choice D is out.

1/2 satisfies the equation.
So choice B is out.

Hence C is the answer
One day I must get to this way of handling GMAT Quant...and the day is not so far if i stick to BTG
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by willbeatthegmat » Tue Feb 24, 2009 6:16 am
these kinda quest tortures me.....can someone tell me the technique to tackle such questions?????????

Thanks...
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Re: |2x + 3| > |7x - 2|

by x2suresh » Tue Feb 24, 2009 12:35 pm
My approach is similar to Suresh Bala's approach.


x=2 satisfies give equation.

eliminate A,D,E

x=1/2 eliminate B.

C is the answer
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Re: |2x + 3| > |7x - 2|

by sudi760mba » Tue Feb 24, 2009 1:28 pm
sanju09 wrote:What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1
I'm going to take a stab at it and would like some validation:

2x + 3 > 7x - 2
=-5x > -5
= x< 1

The only one that seems to work is C.
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Re: |2x + 3| > |7x - 2|

by sanju09 » Wed Feb 25, 2009 1:51 am
sudi760mba wrote:
sanju09 wrote:What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < (-1/9) or x > 5

B. -1 < x < (1/9)

C. (-1/9) < x < 1

D. (-1/9) < x < 5

E. x < (-1/9) or x > 1
I'm going to take a stab at it and would like some validation:

2x + 3 > 7x - 2
=-5x > -5
= x< 1

The only one that seems to work is C.
More than one of the given choices could lead you to wonder that x < 1 works for those too. Yours is not a holistic treatment of the problem :) stab at its heart now :lol:
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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