Question

Yoona runs at a steady rate of 1 yard per second. Jessica runs 4 times as fast. If Jessica gives Yoona a head starts of 30 yards in a race, how much yards must Jessica run to catch up to Yoona?

Answer 1

Hint: We first find the speed of Jessica as Jessica runs 4 times as fast as Yoona. Then we assume the distance in yards which is required for Jessica to catch up to Yoona. We have the fixed time span as Jessica started running only when Yonna is already 30 yards ahead of her. From that relation, we solve the linear equation to find the solution.

Complete step by step answer:
Yoona runs at a steady rate of 1 yard per second. Jessica runs 4 times as fast as Yonna.
So, the speed of Jessica is $ 4\times 1=4 $ yards per second.
Now both of them are involved in a race. Jessica gives Yoona a head start of 30 yards.
This means Jessica is 30 yards behind Yonna.
Let’s assume that Jessica has to cover x yards to catch up to Yoona.
Also, we understand that because of the head start Jessica started running when Yonna is already 30 yards ahead of her.
If the time required for Jessica to cover x yards and catch up to Yoona is t seconds, then in that t seconds Jessica covers x yards whereas Yonna covers only $ \left( x-30 \right) $ yards.
For Yonna the value of t will be the distance covered divided by speed which is $ t=\dfrac{x-30}{1} $ .
For Jessica, the value of t will be the distance covered divided by speed which is $ t=\dfrac{x}{4} $ .
We equate these two to get $ t=\dfrac{x-30}{1}=\dfrac{x}{4} $ . Solving the equation, we get
 $ \begin{align}
  & \dfrac{x-30}{1}=\dfrac{x}{4} \\
 & \Rightarrow 4x-120=x \\
 & \Rightarrow x=\dfrac{120}{3}=40 \\
\end{align} $
Therefore, Jessica needs to run 40 yards to catch up to Yoona.

Note:
We can also find the distance difference for two people covering at the same time. The difference would be equal to 30 yards cause of the head start. Jessica covers further than Yoona because of greater speed but they meet up at the same spot as Yoona has the head start.