10. Charlie biked 15 minutes home from school and then walked 12 minutes to the pizza place near his house. This trip was 3.6 miles in total. His rate on his bike is 3 miles per hour less than 5 times his walking rate. In miles per hour, what is Charlie’s speed as he walks and what is his speed as he bikes?

GIVEN: Charlie biked [tex]15[/tex] minutes home from school and then walked [tex]12[/tex] minutes to the pizza place near his house. This trip was [tex]3.6[/tex] miles in total. His rate on his bike is [tex]3[/tex] miles per hour less than [tex]5[/tex] times his walking rate. In miles per hour.

TO FIND: Charlie’s speed as he walks and his speed as he bikes.

SOLUTION:

Let speed of charlie when he walks be [tex]x\text{ mph}[/tex]

Then speed of charlie as he biked [tex]=5x-3\text{ mph}[/tex]

Total length of trip [tex]=3.6\text{ miles}[/tex]

Total time taken[tex]=27\text{ minutes}=\frac{27}{60}\text{ hour}=\frac{9}{20}\text{hour}[/tex]

Now,

[tex]\text{Distance}=\text{speed}\times\text{time}[/tex]

distance traveled on bike [tex]=(5x-3)\times\frac{15}{60}=\frac{(5x-3)}{4}[/tex]

distance traveled by walking [tex]=x\times\frac{12}{60}=\frac{x}{5}[/tex]

As, [tex]\frac{(5x-3)}{4}+\frac{x}{5}=3.6[/tex]

[tex]\frac{29x-15}{20}=3.6\implies 29x=87[/tex]

[tex]\implies x=3\text{ mph}[/tex]

Speed of Charlie on bike [tex]=5x-3=12\text{ mph}[/tex]

Hence speed of charlie on bike and by walking is 12mph and 3mph respectively