Struggle with Modulus questions

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abhishekswamy: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Mon May 23, 2011 3:30 am

by abhishekswamy » Sun Jul 10, 2011 6:33 am

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| â‰ 0

OA. C

Can anybody explain the solution for the above question in detail. M having a tough time solving modulus questions

Tnks in advance

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Sun Jul 10, 2011 6:40 am

hi,
decode statement 1 by taking two possible values of LHS of the eq's.

first take -x-1 = 2x-2.
which gives x = 1/3

then take x+1 = 2x-2
whih gives x = 3

now check if both these satisfy statement 1.

they do, hence statement 1 is not sufficient,

2 negates one of the values obtained above.

hence answer is c

x =1/3

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abhishekswamy: Junior | Next Rank: 30 Posts; Posts: 16; Joined: Mon May 23, 2011 3:30 am

by abhishekswamy » Sun Jul 10, 2011 6:47 am

bblast wrote:hi,
decode statement 1 by taking two possible values of LHS of the eq's.

first take -x-1 = 2x-2.
which gives x = 1/3

then take x+1 = 2x-2
whih gives x = 3

now check if both these satisfy statement 1.

they do, hence statement 1 is not sufficient,

2 negates one of the values obtained above.

hence answer is c

x =1/3

Bblast: can you please explain...you have taken two case for LHS i.e when |x-1| is positive and negative,y not 2 cases for RHS also??

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bblast: Legendary Member; Posts: 1077; Joined: Mon Dec 13, 2010 1:44 amFollowed by: 33 members; Thanked: 118 times; GMAT Score:710

by bblast » Sun Jul 10, 2011 7:07 am

Hi abhishek,

actually there are 4 cases :

lhs -ve rhs +ve case1
lhs -ve rhs -ve case2

lhs +ve rhs +ve case3
lhs +ve rhs -ve case4

As u can see case 2 and case 3 above are the same. similarly case 1 and case 4 are the same.
which are the actual 2 cases u need to consider.

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goalevan: Master | Next Rank: 500 Posts; Posts: 111; Joined: Tue Dec 30, 2008 1:25 pm; Location: USA; Thanked: 28 times; GMAT Score:770

by goalevan » Mon Jul 11, 2011 3:49 pm

-1 < x < 1?

Statement 1) |x + 1| = 2|x - 1| has two cases:

x + 1 = 2x - 2
x = 3

x + 1 = -(2x - 2)
3x = 1
x = 1/3

We always verify that both solutions satisfy the original equation:

|3 + 1| = 2|3 - 1|
|4| = 2|2|

|1/3 + 1| = 2|1/3 - 1|
|4/3| = 2|-2/3|

With these two possible values of x, statement 1 is not sufficient:

-1 < 1/3 < 1? yes
-1 < 3 < 1? no

Statement 2) |x - 3| â‰ 0

x - 3 â‰ 0
x â‰ 3

Note that no second case exists in this equation since -0 = 0.

With different possible values of x, statement 2 is not sufficient:

-1 < 0 < 1? yes
-1 < -500,000,000 < 1? no

Combined)
x = 3 OR x = 1/3
AND x â‰ 3
x = 1/3

-1 < 1/3 < 1

Sufficient.

C

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amit2k9: Master | Next Rank: 500 Posts; Posts: 461; Joined: Tue May 10, 2011 9:09 amFollowed by: 3 members; Location: pune; Thanked: 36 times

by amit2k9 » Mon Jul 11, 2011 9:10 pm

we need to check if -1<x<1

a for 0.3<x<0.4 the LHS approx = RHS.
for x=3 LHS= RHS. not sufficient.

b x # 3 not sufficient.

a+b

x=0.33. sufficient.
C it is.

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