Question

Two hikers are 55 miles apart and walking toward each other. They meet in 10 hours. Find the rate of each hiker iſ one hiker walks 3.3 mph faster than the other.

Answer 1

Answer:

Speed of hikers are 4.4 mph and 7.7 mph

Step-by-step explanation:

Two hikers are 55 miles apart and walking towards each other.

They meet after 10 hours.

Let the speeds of hikers are S and S' mph.

One hiker walks 3.3 mph faster than the other,

S = 3.3 + S'

S' = S - 3.3

Distance traveled by the hiker with higher speed S = Speed × Time

$d_1=S\times 10$

$d_1=10S$

Distance traveled by the slower hiker $d_2$ = (S - 3.3) × 10

Total distance between them = 55 miles

$d_1+d_2=55$

10S + 10(S - 3.3) = 55

20S - 33 = 55

20S = 88

S = 4.4 mph

Speed of first hiker = 4.4 mph

Speed of second hiker = 4.4 + 3.3 = 7.7 mph