On a real number line, point A has coordinate -5 and point B has coordinate 11. Point C ha… — GRE General Test

On a real number line, point $A$ has coordinate $-5$ and point $B$ has coordinate $11$ . Point $C$ has coordinate $x$ such that the distance between $C$ and the midpoint of line segment $AB$ is equal to $\frac{1}{3}$ of the distance between $C$ and point $B$ . What is the maximum possible value of $x$ ?

Answer: 5

Answer

The maximum possible value of

x

5

The midpoint of

A(-5)

and

B(11)

M = 3

. Setting up the distance equation

|x - 3| = \frac{1}{3}|x - 11|

leads to

3|x - 3| = |x - 11|

. Evaluating the two cases

3(x - 3) = x - 11

and

3(x - 3) = -(x - 11)

yields solutions

x = -1

and

x = 5

. The maximum possible value among these is 5.

Step-by-Step Solution

Calculate the coordinate of the midpoint of line segment

AB

Midpoint coordinate

M = 3

The midpoint of two coordinates

a

and

b

on a number line is given by

\frac{a + b}{2} = \frac{-5 + 11}{2} = 3

Formulate the distance equation using absolute values.

3|x - 3| = |x - 11|

The distance between

x

and

3

|x - 3|

and the distance between

x

and

11

|x - 11|

. The given condition is

|x - 3| = \frac{1}{3}|x - 11|

Solve the absolute value equation for all possible values of

x

x = -1

and

x = 5

Splitting

3(x - 3) = x - 11

yields

x = -1

, and splitting

3(x - 3) = -(x - 11)

yields

x = 5

Select the maximum value among all valid solutions.

5

Comparing

x = -1

and

x = 5

, the maximum value is

5

Key Concept

Distance on a number line using absolute value and midpoint formula

Estimated Time:2m 0s

Question

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