Gmat_mission wrote: ↑Thu Dec 17, 2020 12:27 pm
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On the number line shown, the distance between \(0\) and \(a,\) \(a\) and \(b,\) and \(a\) and \(c\) is in the ratio of \(1:2:3.\) If the distance of point \(b\) from \(15\) is twice the distance of point a from \(15,\) what is the value of \(|c|?\)
A. 9
B. 27
C. 36
D. 45
E. 54
Answer:
C
Solution:
Since a < b, we see that if b < 15, then b would be closer to 15 than a to 15. Since the distance between b and 15 is twice that between a and 15, we see that b can’t be less than 15. If a > 15, then the distance between b and 15 would be more than twice that between a and 15 since the distance between b and 0 is twice that between a and 0. We see that a can’t be greater than 15. Therefore, 15 must be between a and b. Now, using the distance relationships between the points a and b and between either one and 15, we can create the equations.
(b - a)/a = 2/1
and
(b - 15)/(15 - a) = 2
Simplifying the first equation, we have b = 3a. Substituting this into the second equation, we have:
(3a - 15)/(15 - a) = 2
3a - 15 = 30 - 2a
5a = 45
a = 9
Therefore, b = 3(9) = 27. Since (c - b)/(b - a) = 3/2, we have:
(c - 27)/(27 - 9) = 3/2
2(c - 27) = 3(27 - 9)
2c - 54 = 54
2c = 108
c = 54
Since c is positive, |c| = c = 54.
Answer: E
