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On the number line shown, the distance between 0 and a, a

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On the number line shown, the distance between 0 and a, a

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Source: e-GMAT



In 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

The OA is C.

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BTGmoderatorLU wrote:
Source: e-GMAT



In 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

The OA is C.
We see that b > a. Given that the distance of point b from 15 is twice the distance of point a from 15, we can deduce that 15 must lie of the left of b. If were at the right of b, the condition "the distance of point b from 15 is twice the distance of point a from 15," cannot be met.

Given that the distance between 0 and a, a and b, and a and c is in the ratio of 1 : 2 : 3, let's assume that the coordinate of a = x; thus, ab = 2x, and the coordinate of b = 3x and that of c is 4x.

There are two scenarios:

Scenario 1: 15 lies between 0 and a.

Thus, b - 15 = 2(a - 15)

b - 15 = 2a - 30

2a - b = 15

=> 2x - 3x = 15

x = -15

Not possible since a = x and the coordinate of a is positive.

Scenario 2: 15 lies between a and b.

Thus, b - 15 = 2(15 - a )

b - 15 = 30 - 2a

2a + b = 45

=> 2x + 3x = 45

x = 9 (possible)

Thus, the coordinate of c = 4x = 4*9 = 36

The correct answer: C

Hope this helps!

I think there was no need to ask for the absolute value of c. Given c is on the right of 0 on the number line, it is positive.

-Jay
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BTGmoderatorLU wrote:
Source: e-GMAT



In 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
Using info given (figure included) we have DATA and FOCUS as below:
\[0 < a < b < c\,\,\left( * \right)\]
\[{\text{dist}}\left( {0,a} \right) = a\]
\[\left. \begin{gathered}
{\text{dist}}\left( {a,b} \right) = 2a\, \hfill \\
{\text{dist}}\left( {a,c} \right) = 3a \hfill \\
\end{gathered} \right\}\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,{\text{dist}}\left( {b,c} \right) = 3a - 2a = a\]
\[? = c = {\text{dist}}\left( {0,c} \right) = 4a\]


From the fact that
\[{\text{dist}}\left( {b,15} \right) = 2 \cdot {\text{dist}}\left( {a,15} \right)\,\,\,\,\,\]
We have three possibilities:
\[b < 15:\,\]
\[15 - b = 2\left( {15 - a} \right)\,\,\,\, \Rightarrow \,\,\,\,15 = 2a - b = 2a - 3a = - a\,\,\,\, \Rightarrow \,\,\,a < 0\,\,\,\,{\text{impossible}}\,\,\]
\[a < 15 \leqslant b:\]
\[b - 15 = 2\left( {15 - a} \right)\,\,\,\, \Rightarrow \,\,\,\,3 \cdot 15 = 2a + b = 2a + 3a = 5a\,\,\,\, \Rightarrow \,\,? = 4a = 4 \cdot 9 = 36\]
From the fact that we have already found (one viable) correct alternative choice, no need to evaluate the third scenario!


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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BTGmoderatorLU wrote:
Source: e-GMAT



In 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
In 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.
To determine the ratio of a to b to c, TEST AN EASY CASE.
Let a=1.
Since a is ONE PLACE to the right of 0, b must be TWO PLACES to the right of a (implying that b=3), while c must be THREE PLACES to the right of a (implying that c=4).
Thus:
a:b:c = 1:3:4.

We can PLUG IN THE ANSWERS, which represent the value of c.
When the correct answer is plugged in, the distance between b and 15 will be twice the distance between a and 15.
The ratio above implies that the value of c will almost certainly be a MULTIPLE OF 4.
Of the five answer choices, only C is a multiple of 4.

C: 36
Since c=36 and a:b:c = 1:3:4 = 9:27:36, a=9 and b=27.
Distance between b and 15 = 27-15 = 12.
Distance between a and 15 = 15-9 = 6.
Success!
The distance between b and 15 is twice the distance between a and 15.

The correct answer is C.

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