SOLUTION: Jennifer drove 300 miles to see her friend. On the return trip, she drove 10 mph faster, so she made the return trip in 1 hour less time. What was her speed on the way to friend's

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Question 1163114: Jennifer drove 300 miles to see her friend. On the return trip, she drove 10 mph faster, so she made the return trip in 1 hour less time. What was her speed on the way to friend's home.
Found 4 solutions by josgarithmetic, ikleyn, greenestamps, MathTherapy:
Answer by josgarithmetic(39618)   (Show Source):
You can put this solution on YOUR website!
Going to:

Trip back:

-------------A Different Way------------

SPEED TIME DISTANCE GOING r 300/r 300 BACK r+10 300/(r+10) 300 DIFF 1

------------simplify and solve this.

Answer by ikleyn(52797)   (Show Source):
You can put this solution on YOUR website!
.

Let x be the rate (in miles per hour) going to friend's home. Then the speed going back is (x+10) mph. The time going to there is hours. The time going back is hours. The "time" equation is - = 1 hour. The solution/(the answer) is just seen from here: x = 50 mph. To get formal algebra solution, multiply both sides by x*(x+10). You will get 300*(x+10) - 300x = x*(x+10) 300x + 3000 - 300x = x^2 + 10x x^2 + 10x - 3000 = 0 (x-50)*(x+60) = 0 and x= 50 is the only meaningful positive solution. ANSWER. The rate going to friend's home was 50 mph. CHECK. - = 6 - 5 = 1 hour. ! Correct !

Solved.

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Using "time" equation is a STANDARD method of solving such problems.
From this lesson, learn on how to write, how to use and how to solve a "time" equation.

To see many other similar solved problems, look into the lessons
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in this site.

Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

You should certainly know how to set up and solve this problem using formal algebra, as shown by the other tutors.

However, if getting the answer as fast as possible (for example, on a timed test) is important, and a formal algebraic solution is not required, then a bit of mental arithmetic can get you there in almost no time.

The "nice" numbers in this problem make a quick mental solution easy.

Considering different combinations of reasonable speeds and driving times that produce a 300 mile trip, you should be able to quickly come up with 5 hours at 60mph and 6 hours at 50mph.

And since those numbers satisfy the condition of the problem (10mph faster --> 1 hour less), they provide the answer to the problem.

ANSWER: Her speed on the way was 50mph.

Answer by MathTherapy(10552)   (Show Source):
You can put this solution on YOUR website!

Jennifer drove 300 miles to see her friend. On the return trip, she drove 10 mph faster, so she made the return trip in 1 hour less time. What was her speed on the way to friend's home.

Let speed on outbound leg be S
Then speed on return leg = S + 10
We then get the following TIME equation:
300(S + 10) = 300S + S(S + 10) ------ Multiplying by LCD, S(S + 10)

0 = (S - 50)(S + 60)
S - 50 = 0 OR S + 60 = 0
Speed on outbound leg, or OR S = - 60 (ignore)