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# M26-12

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Manager
Joined: 06 Jun 2010
Posts: 159

Kudos [?]: 19 [0], given: 151

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28 Mar 2013, 04:15
If −3<x<5 and −7<y<9, which of the following represents the range of all possible values of y−x?
a. −4<y−x<4
b. −2<y−x<4
c. −12<y−x<4
d. −12<y−x<12
e. 4<y−x<12

In the solution provided,the max value of y is taken as 9 whereas the min value of x is taken as -3.Is this a mistake in the question or we can ignore < and consider it as <= ?

Kudos [?]: 19 [0], given: 151

VP
Status: Far, far away!
Joined: 02 Sep 2012
Posts: 1122

Kudos [?]: 2294 [0], given: 219

Location: Italy
Concentration: Finance, Entrepreneurship
GPA: 3.8

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28 Mar 2013, 04:26
We can solve it using this method:
−3<x<5 all *-1 = 3>-x>-5 = $$-5<-x<3$$
$$-7<y<9$$
We can sum the inequalities : $$-7-5<y-x<9+3 = -12<y-x<12$$

shreerajp99 wrote:
In the solution provided,the max value of y is taken as 9 whereas the min value of x is taken as -3.Is this a mistake in the question or we can ignore < and consider it as <= ?

To answer your question, you can ignore the "<" and consider it "<=": it makes no difference.
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Kudos [?]: 2294 [0], given: 219

MBA Section Director
Status: Back to work...
Affiliations: GMAT Club
Joined: 22 Feb 2012
Posts: 4635

Kudos [?]: 3622 [0], given: 2404

Location: India
City: Pune
GMAT 1: 680 Q49 V34
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28 Mar 2013, 04:44
shreerajp99 wrote:
If −3<x<5 and −7<y<9, which of the following represents the range of all possible values of y−x?
a. −4<y−x<4
b. −2<y−x<4
c. −12<y−x<4
d. −12<y−x<12
e. 4<y−x<12

In the solution provided,the max value of y is taken as 9 whereas the min value of x is taken as -3.Is this a mistake in the question or we can ignore < and consider it as <= ?

Range of y-x will be defined as least value of y-x, and greatest value of y-x

y-x will least when y has its smallest value and x has its greatest value. so y - x = -7 - 5 = -12
y-x will greatest when y has its greatest value and x has its least value. so y - x = 9 - (-3) = 9+3 = 12

So range y-x = -12 < y-x < 12 Choice D

Regards,

Abhijit
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Kudos [?]: 3622 [0], given: 2404

Re: M26-12   [#permalink] 28 Mar 2013, 04:44
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# M26-12

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