On a standard number line, point P is located at -17 and point Q is located at 31. Point R… — ACT

On a standard number line, point $P$ is located at $-17$ and point $Q$ is located at $31$ . Point $R$ lies between $P$ and $Q$ such that the ratio of the distance between $P$ and $R$ to the distance between $R$ and $Q$ is $3:5$ . What is the value of $|R - (-5)|$ ?

Answer: 6

Answer

The value of

|R - (-5)|

6

The total distance between point

P (-17)

and point

Q (31)

31 - (-17) = 48

. Since point

R

divides segment

PQ

in a

3:5

ratio, the distance from

P

R

\frac{3}{8} \times 48 = 18

. Adding this distance to

-17

gives

R = 1

. Substituting

R = 1

into

|R - (-5)|

yields

|1 - (-5)| = |6| = 6

Step-by-Step Solution

Calculate the total distance between points

P

and

Q

Distance

PQ = 31 - (-17) = 48

The distance between two points on a number line is found by taking the absolute difference of their coordinates.

Determine the coordinate of point

R

Coordinate of

R = 1

The ratio of

PR

RQ

3:5

, meaning

PR

\frac{3}{3+5} = \frac{3}{8}

of the total distance

PQ

. Adding

\frac{3}{8} \times 48 = 18

to the starting coordinate

-17

gives

R = 1

Evaluate the absolute value expression

|R - (-5)|

|1 - (-5)| = |6| = 6

Substitute

R = 1

into the target expression and simplify the double negative before taking the absolute value.

Key Concept

Calculating distance on a number line, segment ratio partitioning, and absolute value evaluation.

Estimated Time:1m 30s

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