Three friends, Joey, Chandler and Ross meet for breakfast at $9 : 00$ AM at Central Perk every morning. Joey walks, Chandler bikes, and Ross drives. Yesterday all three were exactly on time. Joey left his house at $8 : 30$ AM and walked at a rate of $3$ miles per hour. Chandler left his house at $8 : 15$ AM and biked at a rate of $14$ mph. Ross left her house at $8 : 45$ AM and drove an average speed of $35$ mph. If the person who traveled the farthest live $m$ miles from the restaurants, the what is the value of $2 m$ ?

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Solution

$D i s t a n c e$ Traveled By

$(i)$ As Joey Walked $3 m p h$ for $0.5 H o u r$ , $D_{J o e y} = 3 \times 0.5 = 1.5 m i l e$

$(i i)$ As Chandler biked $14 m p h$ for $0.75$ $H o u r$ , $D_{C h a n d l e r} = 14 \times 0.75 = 10.5 m i l e$

$(i)$ As Ross Drove $35 m p h$ for $0.25 H o u r$ , $D_{R o s s} = 35 \times 0.25 = 8.75 m i l e$

Therefore, $m = 10.5$
and $2 m = 21$

Hence, the answer is $21$

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Distance Traveled By(i) As Joey Walked 3mph for 0.5Hour , DJoey=3×0.5=1.5mile(ii) As Chandler biked 14mph for 0.75 Hour , DChandler=14×0.75=10.5mile(i) As Ross Drove 35mph for 0.25Hour , DRoss=35×0.25=8.75mileTherefore, m=10.5and 2m=21Hence, the answer is 21