If a, b, and c are three numbers on the number line shown above, is c between a and b?

1) b<0

Doesn't provide any information about the number c. Hence, insufficient to answer the question.

2) a-b > c

Case 1: a<0<b, then c<a<0<b. For e.g. a = -1, b = 3 then c<-4(a-b)
Case 2: 0<a<b, then c<a<b. For e.g. a = 2, b = 5 then c<-3(a-b)
Case 3: a<b<0, then we get two cases
For e.g. a = -4, b = -2 c<-2(a-b) or
If c = -6(<-2(a-b)), c<a<b<0.
If c = -3(<-2(a-b)), a<c<b<0.
So, we the number c may or may not be between a and b. Statement 2 is insufficient to answer the question.

1) + 2)

a<b<0, then we get two cases
For e.g. a = -4, b = -2 c<-2(a-b) or
If c = -6(<-2(a-b)), c<a<b<0.
If c = -3(<-2(a-b)), a<c<b<0.
So, we the number c may or may not be between a and b. Statement 1+2 combined is insufficient to answer the question.
Answer E

massi2884 wrote:If a, b, and c are three numbers on the number line shown above, is c between a and b?

1) b<0
2) a-b > c

OA E

Hi! Picking numbers is a great way to solve this (and many other) DS questions.

We quickly decide that (1) is insufficient, since it gives us no info about c. So, let's eliminate A and D and jump to (2).

(2) a-b>c

We can see that b is to the right of a on the number line, so let's pick nice simple numbers. We're not constrained by statement (1), since we're testing (2) by itself, but let's build in (1) in anticipation of having to combine.

Let's pick a = -2 and b = -1

So, we now know that:

-2 - (-1) > c
-2 + 1 > c
-1 > c

Well, if c = -1.5, then c IS between a and b.
If c = -1000, then c is NOT between a and b.

Accordingly, (2) is also insufficient alone - eliminate B.

Combining: since the numbers we chose for (2) also satisfy the rule in (1), no extra work is required: insufficient, choose E!