M and B leave their campsite and walk in opposite directions around a lake. If the shoreline is 15 miles long, M walks 0.5 miles per hour faster than B and they meet in 2 hours...how fast does each walk?

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M and B leave their campsite and walk in opposite directions around a lake. If the shoreline is 15 miles long, M walks 0.5 miles per hour faster than B and they meet in 2 hours...how fast does each walk?

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Answer 1 · 2016-06-23T11:41:04

M walks at 4mph, B walks at 3.5mph

Explanation:

$S_{x}$ $S_x$ denotes speed of person x

$S_{M} = S_{B} + 0.5$ $S_M = S_B + 0.5$ as M is walking 0.5 mph faster than B

$D = S_{M} t$ $D= S_M t$ t being the amount of times passed (in hours)

$D = 15 - (S_{B} t)$ $D=15 - (S_Bt)$ we know since M is walking faster B must meet at some location minus from the max location (as continues walking round)

$15 - (S_{B} t) = S_{M} t$ $15-(S_Bt) = S_Mt$ since D = D

$t = 2$ $t = 2$ as 2 hours - substitute in

$15 - S_{B} (2) = S_{M} (2)$ $15-S_B(2) = S_M(2)$

$S_{M} = S_{B} + 0.5$ $S_M = S_B+0.5$ so (as travelling faster) - substitute in

$15 - 2 S_{B} = 2 (S_{B} + 0.5)$ $15-2S_B = 2(S_B+0.5)$ expand and simplify

$S_{B} = 3.5$ $S_B = 3.5$ Speed of B = 3.5mph

$S_{M} = S_{B} + 0.5$ $S_M = S_B + 0.5$
$S_{M} = 3.5 + 0.5 = 4$ $S_M = 3.5 + 0.5 = 4$ Speed of M = 4mph

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M and B leave their campsite and walk in opposite directions around a lake. If the shoreline is 15 miles long, M walks 0.5 miles per hour faster than B and they meet in 2 hours...how fast does each walk?

1 Answer

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