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.

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.

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.

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

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

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.

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

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