2005 AMC 10A Problems/Problem 7

Revision as of 13:47, 17 February 2021 by Pathu05 game-2005 (talk | contribs) (Problem)

Problem

Wahida and Parthib live $13$ miles apart. Yesterday Wahida started to ride his bicycle toward Parthib's house. A little later Parthib started to ride his bicycle toward Wahida's house. When they met, Wahida had ridden for twice the length of time as Parthib and at four-fifths of Parthib's rate. How many miles had Parthib ridden when they met?

$\mathrm{(A) \ } 2\qquad \mathrm{(B) \ } 5\qquad \mathrm{(C) \ } 9\qquad \mathrm{(D) \ } 7\qquad \mathrm{(E) \ } 17$

Solution

Let $m$ be the distance in miles that Prthib rode.

Since Wahida rode for twice the length of time as Parthib and at four-fifths of Parthib's rate, he rode $2\cdot\frac{4}{5}\cdot m = \frac{8}{5}m$ miles.

Since their combined distance was $13$ miles,

$\frac{8}{5}m + m = 13$

$\frac{13}{5}m = 13$

$m = 5 \Longrightarrow \mathrm{(B)}$

Video Solution

-Check this amazing solution: https://youtu.be/WIR8yPLET9Y

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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