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Mod Problem

by gmat009 » Tue Oct 07, 2008 4:06 pm
Is |x| + |x -1| = 1
(1) x >= 0
(2) x <= 1
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA is C, IMO E
Plz. explain
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Source: — Data Sufficiency |
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by cubicle_bound_misfit » Tue Oct 07, 2008 5:54 pm
i try to solve this as great 'guru' Ian Stewart taught me long time back .

try to draw a numbe line.

now we have to prove if

distance of x from 0 + distance of x from 1 is equal to 1

try to draw all the different scenarios where this is possible.

if x <0 you will find in the number line however near x is to the 0 (left side of the zero) it will always be greater than 1 try with -.99 , -1/2 etc.

So, this equation to be true x should be >0

now think of two ranges in the number line

0<= x<=1 and x>1 for any value of x > 1 the mod itself will be greater than 1 so only possible range is [0,1] which we get by C.

hope this will help.
please let me know if this approach is wrong.
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by gmat009 » Tue Oct 07, 2008 6:21 pm
cubicle_bound_misfit wrote:i try to solve this as great 'guru' Ian Stewart taught me long time back .

try to draw a numbe line.

now we have to prove if

distance of x from 0 + distance of x from 1 is equal to 1

try to draw all the different scenarios where this is possible.

if x <0 you will find in the number line however near x is to the 0 (left side of the zero) it will always be greater than 1 try with -.99 , -1/2 etc.

So, this equation to be true x should be >0

now think of two ranges in the number line

0<= x<=1 and x>1 for any value of x > 1 the mod itself will be greater than 1 so only possible range is [0,1] which we get by C.

hope this will help.
please let me know if this approach is wrong.
I have not understood this solution, so I am not sure this approach is right or wrong but there must be more easy way to solve this problem
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by pre-gmat » Tue Oct 07, 2008 7:07 pm
I think there's an easy solution to this.

a and b all by itself is not sufficient.

Combining both X>=0 and X<=1 we get

0<=x<=1

so x is basically anywhere from 0 to 1.

Try any value between 0 and 1 (try 0 and 1 as well) you will get the equation

|x| + |x-1| = 1
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by lilweezy » Tue Oct 07, 2008 8:15 pm
you have to look at each statement on it's own for data sufficiency. i think the answer is A.

|x| + |x-1| = 1

(1) x>=0
(2) x<=1

for option (1), any value you put for x that satisfies the equation works in the formula. for option (2), you get different answers for x=1 or x=-1. therefore statement (2) is not sufficient, but statement (1) is.

Answer is A.
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by schakiiieee » Wed Oct 08, 2008 3:47 pm
lilweezy wrote:you have to look at each statement on it's own for data sufficiency. i think the answer is A.

|x| + |x-1| = 1

(1) x>=0
(2) x<=1

for option (1), any value you put for x that satisfies the equation works in the formula. for option (2), you get different answers for x=1 or x=-1. therefore statement (2) is not sufficient, but statement (1) is.

Answer is A.
No it does not. Take x=5. --> 5 + 4 = 9 which is not 1.
Any value that is between 0 and 1:
e.g. x=1/4
1/4 + |1/4-1| = 1/4 + |-3/4| = 1.
Works.
(as it is continuous between 0 and 1 --> ANS C)
For any x > 1 it fails!
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by namitrajiv » Wed Oct 08, 2008 9:28 pm
should be C

|X| = X when x>0

|X-1| = 1-X when x<1

combine x+ 1-x = 1 , thats what we need

first statement says x>=0 therefore we can conclude that |x| = x
but we are not sure about |x-1|
second statemnet says x<=1 thus we knw |x-1| =1-x
thus combining the two we can have the ans...
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by Gmatss » Thu Oct 09, 2008 8:41 am
IMO C

Combining both statment

0<=x<=1

Any positive fraction between 0 and 1 inclusive satifies the condition
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by lilweezy » Thu Oct 09, 2008 2:08 pm
schakiiieee wrote:
lilweezy wrote:you have to look at each statement on it's own for data sufficiency. i think the answer is A.

|x| + |x-1| = 1

(1) x>=0
(2) x<=1

for option (1), any value you put for x that satisfies the equation works in the formula. for option (2), you get different answers for x=1 or x=-1. therefore statement (2) is not sufficient, but statement (1) is.

Answer is A.
No it does not. Take x=5. --> 5 + 4 = 9 which is not 1.
Any value that is between 0 and 1:
e.g. x=1/4
1/4 + |1/4-1| = 1/4 + |-3/4| = 1.
Works.
(as it is continuous between 0 and 1 --> ANS C)
For any x > 1 it fails!
you're right, i was totally off on this one.
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by Ian Stewart » Thu Oct 09, 2008 2:47 pm
gmat009 wrote:
I have not understood this solution, so I am not sure this approach is right or wrong but there must be more easy way to solve this problem
The approach is very easy, and it is the most useful approach to know for the more difficult absolute value questions on the GMAT. It relies on the following:

|x| = the distance between x and zero on the number line
|x - y| = the distance between x and y on the number line

So, when the question asks if |x| + |x-1| = 1, it's just asking "Is the distance from x to zero plus the distance from x to 1 equal to 1?" And if you draw a number line, it's easy to see that this will be true if (and only if) x is between zero and one. So C.

I do think it takes a bit of time to get used to thinking this way about absolute value, but this question literally takes ten seconds if you can approach it as a distance question, not as an algebra question.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
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by maihuna » Wed Apr 15, 2009 7:58 am
my algebra approach will be:

Using 1: x + |x-1| =1
if x<1 x + (-(x-1)) = 1 or 1=1 true
if x>=1 x+x-1 =1 or x=1 but we dont know this

Using 2: If x<=1 we know we will get part 1 as above and so the answer C
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by cubicle_bound_misfit » Wed Apr 15, 2009 10:01 am
nother blast from the past
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by sanjay_dce » Wed Apr 15, 2009 10:46 am
Ian Stewart wrote:
gmat009 wrote:
I have not understood this solution, so I am not sure this approach is right or wrong but there must be more easy way to solve this problem
The approach is very easy, and it is the most useful approach to know for the more difficult absolute value questions on the GMAT. It relies on the following:

|x| = the distance between x and zero on the number line
|x - y| = the distance between x and y on the number line

So, when the question asks if |x| + |x-1| = 1, it's just asking "Is the distance from x to zero plus the distance from x to 1 equal to 1?" And if you draw a number line, it's easy to see that this will be true if (and only if) x is between zero and one. So C.

I do think it takes a bit of time to get used to thinking this way about absolute value, but this question literally takes ten seconds if you can approach it as a distance question, not as an algebra question.
Thank you Ian you Rock!
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by vittalgmat » Wed Apr 15, 2009 10:59 am
Thanks Ian ,
as u said it takes a little time to get used to looking at absolute as a distance and imagine in terms of number line.
But the idea is paying off.
Many thanks for that.

Needless to say my ans is a C
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