Word Problem - MAth?

Question

During rush hour, Fernando can drive 35 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando's rate on the side roads is 9 mi/h faster than his rate on the freeway, find his rate on the side roads.

Answer 1

Let

s = speed on side streets
f = speed on freeway

Given

35/s = 30/f
s = f + 9

Find s.

35/s = 30/f
f = (30/35)s = (6/7)s

s = f + 9
s = (6/7)s + 9
(1/7)s = 9
s = 63 miles per hour

This gives new meaning to the term "rush hour."

Answer 2

This is a distance, speed time problem.
Let x represent the speed on the side roads in mph.
Side road: d = 35 s = x t = d /s = 35/x
Free way d = 30 s= x - 9 t = 30/(x - 9)

Since time is equal: 35/x = 30/(x-9)
Cross multiply: 35(x - 9) = 30 x
Expand: 35x - 315 = 30x
Isolate x: 35x - 30x = 315
5x = 315
x = 63

Fernando's speed on the side road is 63 mph.

Answer 3

Okay. Here's what you do.
(miles/hour)/miles = the number of hours it took.
So, if it took the same amount of time, you can set the two distances and speeds equal to each other.

(x + 9 miles per hour on side roads)/(35 mi on side roads) = (x miles per hour on freeway)/30 mi on freeway

Multiply both sides by 30.
x miles per hour on freeway = 30 mi(x+9 mi/hr)/(35mi)
Distributive property with the 30.
x = (30x+270)/35

35x= 30x + 270
5x = 270
270/5 = 54 mi/hr on the freeway
If he's going 9 mi/hr faster on the side roads than the freeway, 54 + 9 is 63 mi/hr on the side roads.

Answer 4

3 miles on the side road

Answer 5

Let f = his rate on the freeway and s = his rate on the side roads

s = f + 9
35/s = 30/f (since d=rt, t=d/r and the two times are equal)

First write the related cross product equation for the second equation, then substitute. I'll leave it to you to solve...