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absolute value ..plz help !

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absolute value ..plz help !

by charmaine » Mon Sep 15, 2008 11:43 pm
is lv-xl<8?
1) v and x are integers
2) lvl=4 and lxl =6

islxl+ lx-1l=1
1)x>=0
2)x<=1

if xyz cannot be 0, is x(y+z)>=0?
1)ly+zl= lyl+lzl
2) lx+yl = lxl+lyl

i have been trying to understand the answers, but i dont ..thanks for the help :) appreciated

[spoiler]
ans. 1. E
2. C
3. C[/spoiler/][spoiler][/spoiler]
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Re: absolute value ..plz help !

by parallel_chase » Tue Sep 16, 2008 1:54 am
charmaine wrote:is lv-xl<8?
1) v and x are integers
2) lvl=4 and lxl =6

Kindly post one question per post.

Most absolute value questions can be solved easily if you think of it as the distance on the number line.

lv-xl < 8

this means that the distance between x and v on the number is less than 8

Statement I
Nothing can be concluded out of this statement.

Statement II
lvl = 4
this can be written as lv-0l = 4 or l0-vl = 4
We just know the distance between 0 and v, v could be on the left side of 0 or v could be on the right side of 0

Similarly we dont know anything about lxl = 6.

if v is negative and x is positive lv-xl = 10 > 8
if v is negative and x is also negative lv-xl = 2 < 8

Insufficient.

Combining I & II
Nothing more can be concluded.

Hence E is the answer.
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Re: absolute value ..plz help !

by parallel_chase » Tue Sep 16, 2008 2:05 am
charmaine wrote: is lxl+ lx-1l=1
1)x>=0
2)x<=1
Using the number line approach.

lxl+ lx-1l=1

this means adding the distance between x and 0, and x and 1 = 1
Therefore we need to find the exact value of x or at least a range

Statement I
x is greater than equal to 0

we dont know the exact value of x except that x falls on the right side of 0

Insufficient.

Statement II

x is less than equal to 1

we dont know the value of x.

Insufficient.

Combining I & II

0 <= x <=1

For any value of x between 0 and 1 inclusive the expression lxl+ lx-1l will always result in 1

Sufficient.

Hence C.
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Re: absolute value ..plz help !

by parallel_chase » Tue Sep 16, 2008 2:13 am
charmaine wrote: if xyz cannot be 0, is x(y+z)>=0?
1)ly+zl= lyl+lzl
2) lx+yl = lxl+lyl
x(y+z)>=0 ?

x, y & z none of them is equal to 0

Statement I
ly+zl= lyl+lzl

We can conclude from the above that either y and z both are negative or both are positive. But we dont know anything about x. Insufficient.

Statement II

lx+yl = lxl+lyl

we can conclude the same as above for x and y. We dont know anything about z. Insufficient.

Combining I & II

either x, y & z all are negative or x, y & z all are positive.

if x y z are negative

-x(-y-z) = product of two negative numbers is positive. Hence greater than and equal to 0

if x y z are positive
x(y+z) = product of two positive numbers is positive. Hence greater than and equal to 0

Sufficient.

Hence C is the answer.

Let me know if you still have any doubts.

Hope this helps.
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by charmaine » Tue Sep 16, 2008 12:55 pm
sorry ...will post it individually nxt time ....thanks parallel..really appreciate it..i will try and evaluate the questions and ask if theres a question thanks again :)
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by charmaine » Tue Sep 16, 2008 3:14 pm
is lxl+ lx-1l=1
1)x>=0
2)x<=1


together.....0 <= x <=1

For any value of x between 0 and 1 inclusive the expression lxl+ lx-1l will always result in 1

Sufficient.

I Do not get it ....if lets say x=0.5 it would not result in 1 !

thanks parallel .....i detest absolute value questions :)
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by parallel_chase » Tue Sep 16, 2008 3:17 pm
charmaine wrote:is lxl+ lx-1l=1
1)x>=0
2)x<=1


together.....0 <= x <=1

For any value of x between 0 and 1 inclusive the expression lxl+ lx-1l will always result in 1

Sufficient.

I Do not get it ....if lets say x=0.5 it would not result in 1 !

thanks parallel .....i detest absolute value questions :)
if x = 0.5

l0.5l + l0.5-1l

0.5 + 0.5 = 1

you check for x = 0.3 , 0.23, 0.02 anything the result will always be 1.
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by vr4indian » Thu Sep 18, 2008 7:06 am
First Questions

we are given: |v-x| <8 => -8 < v-x < 8

1: |v| < 4 => -4 <v < 4 , cant get anything from this alone

2: |x| < 6 => -6 <x < 6 , cant get anything from this alone

take both together , take extreme values first

v = 4 and x = -6 => v-x = 4- (-)6 = 10 Not satisfy eqaution

v= 1 , x =1 => v-x = 1-1 = 0 satisfy equation

so E will be answer

Thanks
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by tendays2go » Thu Sep 18, 2008 1:15 pm
parallel_chase wrote:
charmaine wrote:is lxl+ lx-1l=1
1)x>=0
2)x<=1


together.....0 <= x <=1

For any value of x between 0 and 1 inclusive the expression lxl+ lx-1l will always result in 1

Sufficient.

I Do not get it ....if lets say x=0.5 it would not result in 1 !

thanks parallel .....i detest absolute value questions :)
if x = 0.5

l0.5l + l0.5-1l

0.5 + 0.5 = 1

you check for x = 0.3 , 0.23, 0.02 anything the result will always be 1.
we can always put values when in doubt as shown above, but here's my 2 penns:
by combining both, it's clear that 0<= x <= 1
thus, |x| = x (had x been less than 0, |x| = -x )
and |x-1| = - (x-1) = 1-x
=> |x| + |x-1| = x + 1-x = 1

nicely put by parallel_chase as a formula to remember ;-)
thus, (C) is the answer
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