Question

Abigail was skateboarding home when the wheel axle of her skateboard broke. She had already traveled two thirds of the way home and had to walk the rest of the way. Walking the rest of the way home took her twice as long as it took her to ride her skateboard. How many times faster is Abigail on her skateboard than she is walking?

Answer 1

Answer:

The number of times Abigail is faster on her skateboard than she is walking is 2 times faster

Step-by-step explanation:

Let the distance to Abigail's home = D

The distance Abigail had traveled before hr skateboard broke = 2/3·D

The distance from home where the skateboard broke = D - 2/3·D = 1/3·D

The time it took Abigail to walk the 1/3·D home = Twice the time it will take her if she was on her skateboard

Let the time  it will take Abigail to travel the 1/3·D home on her skateboard = t

Therefore;

The time it took Abigail to walk the 1/3·D home = 2 × t

The speed of Abigail walking, $s_w$= Distance/Time = 1/3·D/(2 × t) = D/(2·t)

The speed of Abigail skateboarding, $s_s$= Distance/Time =D/t

The ratio of the speed of Abigail skateboarding to the speed of Abigail walking = $\dfrac{s_s}{y_w} = \dfrac{\dfrac{D}{t} }{\dfrac{D}{2 \times t} } = \dfrac{D}{t} \times \dfrac{2 \times t}{D} = 2$

Therefore, Abigail is two times faster on her skateboard than she is walking.