if the absolute value of a-b = 5 and the absolute value of a-c=5 which could be the value of absolute value of b-c
I 2
II 4
III 8
A) I only
B) II Only
C) III only
D) I and II only
E) I and III only
@avneety Do you mean |a-b| = 5 and |a-c| = 5?
If so, the correct answer isn’t in the answer choices. Did you type the entire question correctly (maybe “5” should be “2”)? Note that |a-b| represents the distance between a and b when plotted on a number line.
i meant | a - c | = 3
@avneety So |a-b| = 5 and |a-c| = 3?
Hint: As I said above, |a-b| represents the distance between a and b on a number line. Same for |a-c|. So if the distance between a and b is 5 and the distance between a and c is 3, what are all possible distances between b and c?
Just to expand on the @MITer94 approach:
It might help if you pick some value for a (say, a=10), mark it on the number line, and see which numbers are 5 away from 10 (and mark them as b’s), and which numbers are 3 away from 10 (and mark them as c’s).
|a - b| = 5
|a - c| = 3
b - c =?
So remember, when you have an absolute value:
|a - b| = 5
-> a - b = 5
-> a - b = -5
We can apply the same thing to
|a - c| = 5
-> a - c = 5
-> a - c = -5
Let’s deal with the equations that equal to 5 first…
-> a - b = 5
-> a - c = 5
and we know 5 = 5, therefore
a - b = 5 = a - c
a - b = a - c
b - c = 0 <- let’s start this one it is important*****
Let’s deal with the equations that equal to -5 …
a - b = -5
a - c = -5
and we know -5 = -5, therefore
a - b = -5 = a - c
a - b = a - c
b - c = 0 <- let’s start this one it is important*****
Since b - c is 0 for both cases, |b - c| = 0, which isn’t the answer choices you provided.
You can do the same concept for |a - c| = 3
This is a tedious way of doing things- but really helps breaking things down!
@Perfect1600 you missed the two cases where a-b = 5, a-c = -5, etc. This leads to |b-c| = 10 as well.
I think the more intuitive (and faster!) approach is to visualize it using a number line, as @gcf101 and I pointed out.
I see it like this :
The a is consistent between A-b and A-c. The difference between the real values is either 2(5-3) or 8 (5+3). So, the addition of b and deduction of c (to change a-b into a-c ) has caused a difference of 2 or 8.
@perfect1600 – not trying to nitpick, just for the sake of accuracy: you stated first the corrected question
and then proceeded to work it out in its original (with a typo) form
missing on the way, as @MITer94 noted, a couple of cases to boot.
KISS principle is preached so heavily on this forum not accidentally. 