GMAT Prep DS Numberline

M and R might lie anywhere on the number line.

From [1]

M = 1, R =3
M -1, R = -3
M = 2, R = 6

So we have multiple answers and hence [1] is not sufficient

From [2]

M = 5, R =17
M = 6, R = 18
M = 7, R = 19

So again we have multiple answers and [2] is not sufficient.

Combining [1] and [2],

M = 6 and R = 18 is the only answer

So [1] and [2] together are sufficient.

Whats the OA?

Thanks

Ich think (1) and (2) are still not sufficient, in addition to chatekar's answer there's M = -12, R = 36. OA?

In strict mathematical terms, in case of M = -12 and R = 36,
The distance OM is -3 times distance OR.

Please correct me if I'm wrong.

Thanks

Thanks ,

the OA is E

Hi erdnah,

Is there a algebraic eqn /method for solving this . I also chose C as the answer assuming the values to be 6 and 18.

Of course there'll be a method to solve this, but I don't know it; if got a meeting in a min, perhaps I'll find some seconds to think about it
The important thing is, that it is asked for the distance, which can never be negative (move 0 to 12 and you should still get two - complete positive - solutions).

Statement 1 is obviously not sufficient and so is Statement# 2

Combining Statement1 and Statement2:

R HAS to lie on the positive side of zero, because 12 is in the middle of M & R.

Scenario A:
========
M lies on the positive size of 0(zero). Point P represents number 12. The points are in the following order: O M P R.

OM = x,
MP = PR = y, say. (from statement# 2)
OR = x+2y = 3(OM) (from statement #1)
x+2y = 3x, therefore x = y

Also, x+y = 12(from statement# 1). therefore x = 6 and R=18

Scenario B:
=======
M lies on the negative side of 0(zero). Point P represents number 12. The points are in the following order: M O P R

MO = x,
MP = PR = y, say (from statement# 2)
OR = 2y-x = 3OM
2y-x = 3x => y = 2x.

Also MP - OM = OP = 12
y - x = 12 => x = 12.
Therefore y = 24. Hence R is 36.

Hence the answer is E.