Bunuel wrote:

TooLong150 wrote:

Bunuel wrote:

TIP:

On the GMAT we can often see such statement: \(k\) is halfway between \(m\) and \(n\) on the number line. Remember this statement can ALWAYS be expressed as:

\(\frac{m+n}{2}=k\).

Also on the GMAT when we see the distance between x and y, this can be expressed as \(|x-y|\).

Back to the question:

If m and r are two numbers on a number line, what is the value of r?

(1) The distance between r and zero is 3 times the distance between m and zero --> \(|r-0|=3|m-0|\) --> \(|r|=3|m|\) --> \(r=3m\) OR \(r=-3m\). Clearly insufficient.

(2) 12 is halfway between m and r --> \(\frac{r+m}{2}=12\) --> \(r+m=24\). Clearly insufficient.

(1)+(2) \(r=3m\) OR \(r=-3m\) and \(r+m=24\).

\(r=3m\) --> \(r+m=3m+m=24\) --> \(m=6\) and \(r=18\)

OR

\(r=-3m\) --> \(r+m=-3m+m=24\) --> \(m=-12\) and \(r=36\)

Two different values for \(r\). Not sufficient.

Answer: E.

Bunuel, can you explain how \(|r|=3|m|\) --> \(r=3m\) OR \(r=-3m\)?

\(|r|=3|m|\) means that the distance from r to 0 is thrice the distance from m to 0:

-----0--m-----r------

r-----m--0--------------

--m--0--------r------

r--------0--m------------

If r and m have the same sign (cases A and B), then r=3m but if r and m have different signs (cases C and D), then r=-3m.

Hope it's clear.

Hi Buenel,

I looked into your solution and noticed many are saying there is only 2 solutions hence insufficient. But just to fully understand this problem, isn't there more than 2 solutions?

For instance:

r could be - or +, m could be the opposite sign of r (- or +), or they both can be negative. Therefore multiple solutions? Or, am I missing something. I understand you wrote |r| = 3|m| -> r = 3m or r=-3m, however, could we also have -r = 3m?

Thank you