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

thanks bunuel
i cud not express it in a correct manner , but my intention was same since i tried on no. line and i got it in a easier way ,however cudn't express that in my post(appologies for that), on a no. line it was clear that the position of m and r is not fix with bth the given information hence insuff....

IMO ... E..

Question: wat is r?

ST1: The distance between r and 0 is 3 times the distance between m and 0
Since the statement has the term 'distance' in it, it signifies that we are not consider the -ve or +ve possibility of the number position.
Hence ST1 can be written algebrically as:
|r-0| = 3|m-0| ---> |r| = 3|m|
Clearly NOT SUFF as m could be anything and even if m is constant, r could be -3m or 3m

ST2: 12 is halfway between m & r is clearly NOT SUFF as the same is true for (m=11,r = 13) , (m=10,r = 14)....

Both ST1 and ST2 together would give us:
m = 6 and r = 18, m=-12 & r = 36 ...etc..Hence NOT SUFF....

OA as D.... ... Not sure..

I somehow remembered the answer for this question.

Try m=6 and r=18. |r|=3|m| and 12 is midway
OR
m=-12 and r=36; |r|=3*|m| and 12 is midway

I don't remember any algebraic solution for this, but it would be great.

Ans: "E"

I used the number line and tested two cases.

Case 1. Assume m is negative and r is positive, each dashed segments ----- is 1x

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

m = -x
r = 3x (3 times the distance between m and 0)

If 12 is the midpoint, the graph becomes:

m ----- 0 ----- 12 ----- ----- r

That means x = 12 and m = -12 and r = 36


Case 2. Assume both m and r are positive

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

m = x
r = 3x

Add 12 as the mid point:

0 ----- m ----- 12 ----- r

Therefore m = 6 and r = 18


This shows that even if you combine the 2 statements, you still can't get a unique answer. Therefore the answer should be E.

*Edited the question. It should be r instead of n.

Stmnt 1 alone: Too many values possible. Say r = 3, m = 1 OR r= 6, m = 2 etc
Stmnt 2 alone: Again too many values possible. Think 12 is in the middle. m and r are equidistant from it so m = 11, r = 13 OR m = 10, r = 14 etc

Both together:
Focus on the logic behind it. You don't need to do any calculations.
We are looking for two values equidistant from 12. Let's say both m and r are at 12 initially. Their distance from 0 is the same i.e. 12 at this point. As they both start moving away from 12 simultaneously, the distance of m from 0 is reducing and that of r from 0 is increasing. There will be point when the distance of m from 0 will be a third of the distance of r from 0. This will be our first pair (shown in blue).
Let's say they keep moving. m will finally reach 0 when its distance from 0 is 0 while r will be at 24. Then m will move in the negative range and its distance from 0 will start increasing. Distance of r from 0 is continuing to increase. There will be a point again when distance of m from 0 is a third of the distance of r from 0 (shown in red).

\(|a-b|=c\) ---> Distance of \(a\) from \(b\) equals \(c\)

Question: r=?

Statement 1:

\(|r-0|=3*|m-0|\) ---> \(r=3*m\) or \(r=-3*m\),

\(r\) depends on \(m\), and since we don't know \(m\), Insufficient.

Statement 2:

Number line is like a set with consecutive numbers. Since this set is an evenly spaced set we know that median=average.

Because 12 is halfway of \(m\) and \(r\) :

\(12=\frac{m+r}{2}\) ---> \(m+r=24\), Insufficient.

Statement 1+2:

\(r=3*m\) ---> \(r=3*(24-r)\) ---> \(r=18\)
\(r=-3*m\) ---> \(r=-3*(24-r)\) ---> \(r=36\)

Therefore Insufficient and the correct answer is E.

stmnt1:

let m = 4 then r = 12
let m = 6 then r = 18

Hence insuff

stmnt2:
We can have different combinations for this as well

m= 10 and r = 14
m = 6 and r = 18

Hence insuff

taking together when m = 6 then r = 18 and 12 is halfway between m and r

also for m = -12 r = +36, 12 is halfway of m and r and r = 3 times the distance from 0 and m (distance is +ve value)

Algebraically this can be solved as a system of two equations

|r| = 3|m|

\(\frac{m+r}{2}=12\)

1) m=6; r=18
2) m=-12; r=36

Not sufficient, so the answer is E

STAT1.
Firstly, we dont know the value of m. So r being 3times m will not help.Also, r can be positive or negative so we are not sure about the value of r.
So, INSUFFICIENT

STAT2.
12 is halfway between m and r.
Now. There are four cases.
1. Both m and r are positive.
2. Both m and r are negative
3. m is positive and r is negative
4. m is negative and r is positive
So, NOT SUFFICIENT

Taking both together:--
Still all the four cases mentioned in statement 2 are possible. So NOT SUFFICIENT.

Answer will be E

Hope it helps!

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.
tnx lot for this

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.

1) m=6; r=18
2) m=-12; r=36

Not sufficient,

E

r = 0?

1) Gives 4 scenarios:
------------0--m------r
----r----m--0
------r------0--m
---------m--0------r

r can have 4 values => Not Sufficient

2)
m--12--r
r--12--m

r can have 2 values => Not Sufficient

1+2)
Apply conditions from stmt 2 into stmt 1
------------0--m--12----r => Valid
----r----m--0 => Not valid since 12 has to be in the middle of m and r
------r------0-12-m => Not valid. When 12 is between m and r, since r would have to be 0 (because of stmt 2), but it would violate stmt 1.
---------m--0------r => Valid

r will have 2 values => Not sufficient

ANSWER: E

Spotting "Distance from 0" in the Statements immediately makes you think about Absolute Value:

Absolute Value of [R] = the Distance of the Number R from 0-Zero on the Number Line

[M] = the Distance of the Number M from 0-Zero on the Number Line



What is the Value of R?


S1: Distance b/w R and 0 = (3) * (Distance b/w M and 0)

[R] = 3 * [M]


if M = 1

R = +3 or R = -3 ------either Satisfy Statement 1. Not Sufficient.


S2: 12 is Halfway between M and R

shout out to IanStewart for his Note Collections. One section covers this type of problem (and all of Coordinate Geometry) very thoroughly and in detail.

Rule: the Average of 2 Numbers will always be the Mid-Point ("halfway") between those 2 Numbers

So statement 2 is telling us that the Average of M and R = 12

(M + R)/2 = 12

M + R = 24

M could = 10 ---------> R could = 14

M could = 8 ---------> R could = 16

Not Sufficient



The hard part - Together:

S1: [R] = 3[M]

S2: R + M = 24


Thinking in terms of (+)positive and (-)negative numbers, for the first case we can try when M < 0:

[M] = (-)M

Let R = (+)Positive

R = 3 * (-M)

R = -3M

---Substitute into Statement 2----

-3M + M = 24

-2M = 24

M = -12 ----------> Since we assumed R is Positive, R would be = +36, which is 3 Times the Distance from 0 that -12 is.

Is 12 the Halfway Point between M = -12 and R = +36?

M = -12 ------> 24 Units away from 12
R = + 36 ------> 24 Units away from 12

Yes, 12 is the Halfway Point:

M = -12 and R = +36 are Valid. R can = 36.


Case 2: Let M >0 and R > 0 for this Case

R = 3M

R + M = 24

---substituting again----

3M + M = 24

4M = 24

M = +6 -------> Since we assumed R>0, Positive R would have to be 3 Times as far from 0. R would have to be = 18

12 is the Halfway Point between M = +6 and R = +18 (Statement 2 Satisfied also)


M can = +6
and
R can = +18


Since we have 2 Possible Values for R: (18 or 36) : even together the Statements are NOT Sufficient


E

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