The distance from \(5\) = \(2x\)

and the distance from \(15 = x\)

The distance between \(15\) and\(5\)= \(15 - 5 = 10\)

That means \(2x + x = 10\)

or \(3x = 10\)

or \(x = \frac{10}{3}\)

or \(2x = \frac{20}{3}\)

Now we can calulate the value of \(x\) in \(2\)ways

i.e.\(5 + 2x = 5 + \frac{20}{3} = \frac{35}{3} = 11{\frac{2}{3}}\)

or \(15 - x = 15 - \frac{10}{3} = \frac{35}{3}= 11{\frac{2}{3}}\)
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Since x is BETWEEN 5 and 15, we know that:
15 - x = the distance from x to 15
x - 5 = the distance from x to 5

GIVEN: x is twice as far from 5 as x is from 15
In other words, the distance from x to 5 is TWICE the distance from x to 15.
So, we can write: x - 5 = 2(15 - x)
Expand: x - 5 = 30 - 2x
Rearrange values to get: 3x = 35
Solve: x = 35/3 = 11 2/3

Answer: C

Cheers,
Brent
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Another method you can use is plugging in numbers.

The question states that a number 'x' must be TWICE as far from 5 than from 15. To translate this into mathematical terms: 15-x = 'some distance' and 5-x = '2*some distance'
Note that the positive or negative aspect doesn't matter in this case but just plug in the answer choices as 'x' and see which of them works.

When plugging in C: 11.67 you get: 15-11.67 = 3.33 and 5-11.67 = -6.67 (6.67 is 2 times 3.33, so it works!)

The answer is C.