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Old 2009 November 7th, 07:28 PM   #1 (permalink)
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algebra

If 4<7-x/3, which of the following must be true? [/font]
I 5<x
II |x+3|>2
III -(x+5) is positive

a) II and III only
b) I, II and III
c)II only;
d)III only;
e)I and II only


can somebody explain this please.

Last edited by gmat.raj : 2009 November 7th at 07:31 PM. Reason: junk characters came in
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Old 2009 November 7th, 07:46 PM   #2 (permalink)
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4<7-x/3
or 12-7<-x
or -5<-x
or 5>x
1. No
2.|5+3|>2 yes
3.-(4+5) is positive ...No
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Old 2009 November 7th, 10:09 PM   #3 (permalink)
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my pick for D
i hope that it is 4<(7-x)/3
12<7-x
x<-5
(1) x<5 not for sure
(2) |x+3|>2 it true for x>-1 and x<-5 as we have both yes and no answer not must be
(3) nor compulsory to check as we have no answer none but for training purposes
-(x+5)>0
-x-5>0
-x>5
x<-5 must be true
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Old 2009 November 8th, 05:01 AM   #4 (permalink)
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Yes , its D. thanks clock60.
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Old 2009 November 8th, 09:29 AM   #5 (permalink)
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Hi,

Thanks for the replies. OG answer is D

I am bit confused with the following
12<7-x
5<-x
is X> -5 Flip the sign when you divide by negative sign? Correct me if I am wrong.


Clock and touche have arrived to x<5.

This confused me with the option 1.

Please clarify.

Thanks
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Old 2009 November 8th, 11:08 AM   #6 (permalink)
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12<7-x
5<-x
is X> -5 Flip the sign when you divide by negative sign? Correct me if I am wrong.

-x>5
x<-5
x is less than -5, but not 5
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Old 2009 November 12th, 08:30 AM   #7 (permalink)
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Quote:
Originally Posted by clock60 View Post
my pick for D
i hope that it is 4<(7-x)/3
12<7-x
x<-5
(1) x<5 not for sure
(2) |x+3|>2 it true for x>-1 and x<-5 as we have both yes and no answer not must be
(3) nor compulsory to check as we have no answer none but for training purposes
-(x+5)>0
-x-5>0
-x>5
x<-5 must be true
I am not sure how did you discarded II

if X<-5

then X lies on a number line which is less than -5

-8...-7...-6...-5...-4....

so tha value of X should be -5.#?

|x+3| = | -5.#?+3| = |-2.#?| =+2.#?

as it is modulus

and 2.#? > 2

so IMO A
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Old 2009 November 12th, 10:13 PM   #8 (permalink)
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Quote:
Originally Posted by arindambanerji View Post
I am not sure how did you discarded II

if X<-5

then X lies on a number line which is less than -5

-8...-7...-6...-5...-4....

so tha value of X should be -5.#?

|x+3| = | -5.#?+3| = |-2.#?| =+2.#?

as it is modulus

and 2.#? > 2

so IMO A
my bad i agree with
you i missed that according problem x<-5 so st 2 valid
thank you for correcting me
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Old 2009 November 13th, 07:04 PM   #9 (permalink)
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IMO (D)
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http://mbachase.co/index.php?option=com_content&view=article&id=72&It emid=62
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