Prep question

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Prep question

by ankit1383 » Wed Oct 01, 2008 8:20 am
If m and r are two numbers on the number line,what is the value of r ?

1 The distance between r and 0 is 3 times the distance between m and 0

2 12 is halfway between m and r


Ans E

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please explain

by ankit1383 » Wed Oct 01, 2008 6:18 pm
Please explain.....................

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by raju232007 » Fri Oct 03, 2008 5:19 am
Lets assume that the distance between r and 0 is x and the distance between m & 0 be y

As per the first statement we get x=3y

But pls note that r can be located on either side of the no line..
Lets consider the case when r is located on the positive site

let y=1,then we get x=3...so r=3

Now when r is located on the negative side

Let=1,then we get x=3...here x represents the distance(i.e the absolute value)...therefore r=-3

the value of r cannot be determined from this statement...

As per statement 2
its observed that 12 is the midpoint between m & r. But both m& r can be located on the positive side or negative side nd moreover we cant interpret much about the coordinates here...

Even when both the statements are combined we cant interpret r..

So E is the ans

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Re: Prep question

by stop@800 » Tue Oct 07, 2008 12:28 am
ankit1383 wrote:If m and r are two numbers on the number line,what is the value of r ?

1 The distance between r and 0 is 3 times the distance between m and 0
Incomplete

2 12 is halfway between m and r


Ans E

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by 4meonly » Tue Oct 07, 2008 8:58 am
If m and r are two numbers on the number line,what is the value of r ?
1 The distance between r and 0 is 3 times the distance between m and 0
2 12 is halfway between m and r

I used absolute value approach. I used it for the first time, so, somebody please correct if I am wrong.

(1)
The distance between r and 0 is 3 times the distance between m and 0
|r|=3|m|
m=?, r=?
INSUFF

(2)
12 is halfway between m and r
r-12=12-m if r>m
m-12=12-r if r<m
If 12 is halfway between m and r, then either r or m is positive
r=?
INSUFF

(1)(2)
from (1) |r|>|m|
12 (positive) is halfway between m and r, so r>0. So,
r-12=12-m

if m>0
r-12=12-m, 3m-12=12-m, 4m=24, m=6, r=18

if m<0
r-12=12-(-m), -3m-12=12+m, -4m=24, m=-6
if m=-6, |r|=3|m|, then either r=18, but 12 will not be on the halfway between r and m, or r=-18, but 12 will not be on the halfway between r and m.
So m=-6 is not possibe

Thus, the only way: m=6, r=18
SUFF

C

Somebody, please, find the flaw! I know that I am wrong but where?
Are there any negative m and positive r so that 12 is halfway between them and 3|m|=|r|?

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Re: Prep question

by Ian Stewart » Tue Oct 07, 2008 10:15 am
ankit1383 wrote:If m and r are two numbers on the number line,what is the value of r ?

1 The distance between r and 0 is 3 times the distance between m and 0

2 12 is halfway between m and r


Ans E
The answer is E because we have two (and only two) possibilities:

r = 18, m = 6
r = 36, m = -12

I think it's clear that neither statement is sufficient alone. When you consider the possibility that m is positive along with the possibility that m is negative, you can find each solution. 4meonly did that above, with one small error:
if m<0
r-12=12-(-m), -3m-12=12+m, -4m=24, m=-6
I've highlighted the mistake above. I'm not sure how that additional negative sign got in there, but if m is negative, you simply want to solve:

r - 12 = 12 - m

and since r = -3m, we find that m = -12, r = 36.
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by cubicle_bound_misfit » Tue Oct 07, 2008 10:45 am
wow Ian,

I was plotting on number lines and yet I could not find it.

thanks a lot.
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by aj5105 » Fri Jan 16, 2009 9:10 am
The answer is E because we have two (and only two) possibilities:

r = 18, m = 6
r = 36, m = -12


nice piece of info !

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Re: Prep question

by logitech » Fri Jan 16, 2009 11:26 am
Ans E

m = = r=0= r== m
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