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Number Line

by AkiB » Wed Jun 25, 2014 11:36 am
On the number line, if r<s, if p is halfway between rand s, and if t is halfway between p and r, then s- t/t-r =

(A) 1/4

(B) 1/3

(C) 4/3

(D) 3

(E) 4

[spoiler](OA) D[/spoiler]
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by GMATGuruNY » Wed Jun 25, 2014 11:41 am
AkiB wrote:On the number line, if r<s, if p is halfway between rand s, and if t is halfway between p and r, then s- t/t-r =

(A) 1/4

(B) 1/3

(C) 4/3

(D) 3

(E) 4
Let r=1 and p=3.
Since t is halfway between r=1 and p=3, t=2.
Since p=3 is halfway between r=1 and s, s=5.
Thus, (s-t)/(t-r) = (5-2)/(2-1) = 3.

The correct answer is D.
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by Mathsbuddy » Thu Jun 26, 2014 7:06 am
First put the letters in order to see what it looks like:
SMALLEST... r t p q s (where q is a space holder half-way between p and s) ...BIGGEST
Now use simple numbers, such as r = 0, t = 1, p = 2, s = 4:
(s-t)/(t-r) = (4-1)/(1-0) = 3.
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by GMATinsight » Thu Jun 26, 2014 7:25 am
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by ceilidh.erickson » Sat Jun 28, 2014 10:02 am
I agree with other posters that picking numbers is by far the easiest approach to this question. If you insist upon doing it algebraically, though, here is the solution:

p is halfway between rand s
We can infer that s - p = p - r

t is halfway between p and r
p - t = t - r

Solve for s in the first equation:
s - p = p - r
s = 2p - r

Solve for p in the 2nd equation:
p - t = t - r
p = 2t - r

Plug in for p in the 1st:
s = 2(2t - r) - r
s = 4t - 3r

Plug this value of s into the expression (s - t)/(t - r):
(4t - 3r - t)/(t - r)
(3t - 3r)/(t - r)

Factor a 3 out of the numerator, and the (t - r) terms will cancel. We're left with 3.
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