SOLUTION: I did this wrong twice and I am on my final chance to get it right; can someone please assist. Thank you. A car travels 560 mi averaging a certain speed. If the car had gone 10

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: I did this wrong twice and I am on my final chance to get it right; can someone please assist. Thank you. A car travels 560 mi averaging a certain speed. If the car had gone 10      Log On


   

Question 1128494: I did this wrong twice and I am on my final chance to get it right; can someone please assist. Thank you.
A car travels 560 mi averaging a certain speed. If the car had gone 10 mph​ faster, the trip would have taken 1 hr less. Find the​ car's average speed.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let x be the car average speed.

Theт the time for the trip is 560%2Fx hours.


Had the car moved 10 mph faster, its speed would be (x+10) mph and the time for the trip would be 560%2F%28x%2B10%29 hours.


The condition says that


    560%2Fx - 560%2F%28x%2B10%29 = 1  hour.     (1)


The number 560 has two remarcable divisors 70 and 80 that differ by 10.

It gives me an idea that x= 70.


Check.  560%2F70 - 560%2F80 = 8 - 7 = 1.


So, the answer is:  the car' speed under the question is  70 mph.


Surely, it is my guess ( ! the correct and checked guess, though !).


But you can solve the equation (1) algebraically, too.


For it, multiply both its sides by  x*(x+10).

You will get a quadratic equation


   560*(x+10) - 560x = x*(x+10),

    x^2 + 10x - 5600 = 0.


Factor the left side polynomial


    (x-70)*(x+80) = 0.


Only positive root  x= 70  is meaningful.


Thus we completed the algebraic solution, and got THE SAME answer:  the averaged speed of the car is 70 miles per hour.


Solved.

Is everything clear to you in my post ?

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

You may find other similar solved problems in the lessons
    - Selected Travel and Distance problems from the archive
    - Had a car move faster it would arrive sooner
in this site.


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


Apparently this question is from a course you are taking. Our giving you the answer won't help you learn anything; and presumably learning something is your objective in taking the course.

It would help us help you if, as you are asked to when you post your question, you told us what you tried to do on the problem.

Algebraically, if x is the car's speed, then the number of hours taken to drive 560 miles is 560/x. The number of hours to travel the same 560 miles at a speed 10mph faster would be 560/(x+10). And the second time is 1 hour less than the first:

560%2F%28x%2B10%29+=+560%2Fx-1

Multiply through by the least common denominator of all the fractions, x(x+10):

560x+=+560%28x%2B10%29-x%28x%2B10%29
560x+=+560x%2B5600-x%5E2%2B10x
x%5E2-10x-5600+=+0

Factor into the form (x+?)(x-?) to finish the problem.

I hope you have learned enough from the course that you can do that....

If you can't finish, then try finding the answer by logical guess-and-check. Presumably this is a computer based question, and only an answer is needed; it doesn't matter how you get the answer. (Well... if you want to learn from the course, it matters; but if you just need the right answer, you can guess it.)

You need two numbers -- a reasonable highway speed in mph and a reasonable number of hours -- whose product is 560. Then you need another speed 10mph faster and another number of hours that is 1 less, whose product is again 560.

A little playing with "nice" numbers should allow you to guess the answer to the problem.