algebra word problem...?

Question

During rush hour, Ella can drive 65 miles using the side roads in the same time that it takes to travel 15 miles on the freeway. If Ella's rate on the side roads is 10 mi/h faster than her rate on the freeway, find her rate on the side roads.

Answer 1

d = r * t

Let the rate on the side roads be s. Then the rate on the freeway is s - 10.

t = d / r, so

15 / (s - 10) = 65 / s
15s = 65s - 650
650 = 50s
s = 13 mph

Her rate on the freeway is 3 mph. She can drive the 15 miles on the freeway in five hours, same as 65 miles on side roads at 13 mph.

Answer 2

Let her rate on the freeway be x m/hour
Thus her rate on the side roads= (x+10) m/hour

Ella drives 65 miles in side roads and 15 miles on the freeway at same time.

We know that, time=distance/rate
thus,time on side roads=65/(x+10) hour
time on the freeway=15/x hour

Therefore,
65/(x+10)=15/x
=>65x=15x+150
=>50x=150
=>x=3
Thus
her rate on the freeway is x=3 m/hour
Thus her rate on the side roads= (x+10) m/hour=3+10=
13 m/hour

Answer 3

Time is the same, so you have two equations:

D(sideroads) / R(sideroads) = t
D(freeway) / R(freeway) = t

Equate these and substitute what you know:
65 / R = 15 / (R - 10)

Now cross multiply and solve
65(R - 10) = 15R
65R - 650 = 15R
50R = 650
R = 13

So she travels at 13 mph on the side roads and 3 mph on the freeway. Both trips take 5 hours.

Answer 4

We'll say the rate on the side road is r, while her rate on the freeway is r-10.
Side road: d=rt and t=d/r, so t=65/r
Freeway: d=(r-10)t and t=d/(r-10), so t=15/(r-10)
Since this is in the same amount of time and they both equal t, you set them equal to each other.
65/r=15/(r-10)
Now you cross multiply and solve
65(r-10)=15r
65r-650=15r
50r=650
r=13 mi/h
Ella was going 13 mi/h on the side roads.

Answer 5

Let

65 = miles driven on the side roads

15 = miles driven on the freeway

r = rate driven on the side roads

r - 10 = rate driven on the freeway

- - - - - - - - - - - - - - - - - - - -

Porportional problem

Multiply the means

Multiply the extremes

65/r = 15/r - 10

The means are 15r

The extremes are 65(r - 10)

15r = 65(r - 10

15r = 65r - 650

15r - 15r = 65r - 650 - 15r

0 = 50r - 650

0 + 650 = 50r - 650 + 650

650 = 50r

650 / 50 = 50r / 50

13 = r . . .rate on side road

The answer is r = 13

- - - - - - -s-

Answer 6

Use the relationship time = distance/rate.

Let

s = speed on side streets
f = speed on freeway

Given

65/s = 15/f
s = f + 10

65f = 15s
13f = 3s = 3(f + 10) = 3f + 30
10f = 30
f = 3 miles/hour
s = 3 + 10 = 13 miles/hour

The speed on side streets is 13 miles/hour.

That's one bad traffic jam.

Answer 7

If the scholars contemporary grade is a ninety% then we’re of direction assuming it’s out of a hundred% ninety/a hundred Now because of the fact the scholars grade more desirable 20% from the final grading era then you definately upload that to the previous a hundred% providing you with ninety/ one hundred twenty = .75 and because grades are in opportunities you multiply .75 via a hundred providing you with 75% because of the fact the scholars previous grade