Kara and Anna are sisters that go to the same school. They started simultaneously moving from their school to their home at different rates on the same route. Kara took a bicycle and Anna a car, the rate of which was 5 times faster than Kara's bicycle. At the half-way point, Anna's car broke and she began to walk home at a rate 2 times slower than the bicycle's rate. Who reached home last? What was the percent increase in time spent versus the other sister?

Respuesta :

Answer:

[tex]Anna\ reached\ last.\\\\Percent\ increase=10\%[/tex]

Step-by-step explanation:

[tex]Let\ speed\ of\ bicycle=x\\\\Let\ total\ distance=d[/tex]

Time taken by Kara:

[tex]Kara\ took\ bicycle\\\\speed=x\\\\distance=d\\\\Time=\frac{distance}{speed}\\\\Time(T_{Kara})=\frac{d}{x}[/tex]

Time taken by Anna:

[tex]Half\ of\ the\ distance\ coverd\ by\ car\\\\Speed=5\times speed\ of\ bicycle=5\times x=5x\\\\distance=\frac{d}{2}\\\\Time\ taken\ to\ cover\ this\ half\ distance=\frac{\frac{d}{2}}{5x}\\\\Time\ taken\ to\ cover\ this\ half\ distance=\frac{d}{10x}\\\\Other\ half\ distance\ is\ covered\ by\ walking[/tex]

[tex]Speed=\frac{speed\ of\ bicycle}{2}= \frac{x}{2}\\\\distance=\frac{d}{2}\\\\Time\ taken\ to\ cover\ this\ half\ distance=\frac{\frac{d}{2}}{\frac{x}{2}}\\\\Time\ taken\ to\ cover\ this\ half\ distance=\frac{d}{x}\\\\Total\ time\ taken\ by\ Anna=T_{Anna}=\frac{d}{10x}+\frac{d}{x}\\\\T_{Anna}-T_{Kara}=\frac{d}{10x}+\frac{d}{x}-\frac{d}{x}=\frac{d}{10x}\\\\T_{Anna}-T_{Kara}>0\\\\T_{Anna}>T_{Kara}\\\\Hence\ Anna\ reached\ last\\\\Percent\ Increase=\frac{T_{Anna}-T_{Kara}}{T_{Kara}}\times 100[/tex]

[tex]Percent\ Increase=\frac{\frac{d}{10x}}{\frac{d}{x}}\times 100\\\\Percent\ Increase=\frac{100}{10}=10\%[/tex]