Solve.
-John and Tony start from GraysLake at the same time and head for a town 10 miles away. John walks twice as fast as Tony and arrives 3 hours before Tony. Find how fast each walks.
A) Tony's speed $= \frac { 3 } { 5 } \mathrm {~m} / \mathrm { h }$ and John's $= \frac { 6 } { 5 } \mathrm {~m} / \mathrm { h }$.
B) Tony's speed $= \frac { 5 } { 3 } \mathrm {~m} / \mathrm { h }$ and John's $= \frac { 10 } { 3 } \mathrm {~m} / \mathrm { h }$.
C) Tony's speed $= 3 \mathrm {~m} / \mathrm { h }$ and John's $= 6 \mathrm {~m} / \mathrm { h }$.
D) Cannot be determined with information given

Question

Quizplus · Accepted Answer

Let Tony's speed be $x$ miles per hour. Then John's speed is $2x$ miles per hour. Let t1 be Tony's time and t2 be John's time. We know that $t2 = t1 - 3$. Using the formula $distance = rate \cdot time$, we have $10 = xt1$ and $10 = 2xt2$. Substituting for $t2$, we get $10 = 2x(t1-3)$. Simplifying gives $t1 = 6$ and $x=5/3$. Therefore, Tony's speed is $5/3$ miles per hour and John's speed is $2\cdot(5/3) = 10/3$ miles per hour. Thus, choice B is correct.

Solve -John and Tony Start from GraysLake at the Same =35 m/h= \frac { 3 } { 5 } \mathrm {~m} / \mathrm { h }=53​ m/h

Solve
-John and Tony Start from GraysLake at the Same $= \frac { 3 } { 5 } \mathrm {~m} / \mathrm { h }$