SOLUTION: Hello I'm having trouble setting this problem up. Any help would be appreciated. Thanks. A traveler having 18 miles to go, calculates that his usual rate would make him one-half

Question 1201248: Hello I'm having trouble setting this problem up. Any help would be appreciated. Thanks.
A traveler having 18 miles to go, calculates that his usual rate would make
him one-half hour late for an appointment; he finds that in order to arrive on
time he must travel at a rate one-half mile an hour faster. What is his usual
rate?
Found 3 solutions by ikleyn, josgarithmetic, MathTherapy:
Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.
Hello I'm having trouble setting this problem up. Any help would be appreciated. Thanks.
A traveler having 18 miles to go, calculates that his usual rate would make
him one-half hour late for an appointment; he finds that in order to arrive on
time he must travel at a rate one-half mile an hour faster. What is his usual rate?
~~~~~~~~~~~~

Let x be his usual rate, in miles per hour. The time to travel 18 miles at his usual rate is hours. The time to travel with the faster rate, (x+0.5) miles per hour, is hours. The longer time is of an hour longer: - = of an hour. (1) +----------------------------------------------+ | At this point, the setup is complete: | | you just have an equation to solve. | +----------------------------------------------+ To solve it, multiply both sides by 2x*(x+0.5) = 2x^2 + x. You will get 36(x+0.5) - 36x = x^2 + 0.5x 36x + 18 - 36x = x^2 + 0.5x x^2 + 0.5x - 18 = 0 = = = . Obviously, only positive root fits x = = = 4 miles per hour. Check it, by substituting into equation (1) = 4.5 - 4 = 0.5 = of an hour. ! correct ! ANSWER. The usual rate is 4 miles per hour.

Solved.

Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!

Using the description SPEED TIME DIST. USUAL r 18/r 18 FASTER r+1/2 18/(r+1/2) 18 DIFFERENCE 1/2

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

discrim,

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

Hello I'm having trouble setting this problem up. Any help would be appreciated. Thanks. A traveler having 18 miles to go, calculates that his usual rate would make him one-half hour late for an appointment; he finds that in order to arrive on time he must travel at a rate one-half mile an hour faster. What is his usual rate? Let usual/regular speed be S Then time taken to get to appointment = Increasing speed by , or .5 mph means that his new speed to get there on-time is: (S + .5) mph. So, time he'd take to get to the appointment, based on his new speed = At the new speed , he'll get there on-time by "shaving" , or .5 hour off of the previous late-time, or We then get the following TIME equation: ------ Multiplying by LCD, ----- Multiplying by 4 (S - 4)(2S + 9) = 0 ----- Factoring trinomial S - 4 = 0 or 2S + 9 = 0 Normal/Usual speed, or S = 4 mph or 2S = - 9 (ignore)

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