SOLUTION: Tina and Joan averaged 6km/h walking from their house to school, and they arrived 15 min late. If they had averaged 9km/h, they would have arrived on time. How far is it from their

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Tina and Joan averaged 6km/h walking from their house to school, and they arrived 15 min late. If they had averaged 9km/h, they would have arrived on time. How far is it from their      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1145725: Tina and Joan averaged 6km/h walking from their house to school, and they arrived 15 min late. If they had averaged 9km/h, they would have arrived on time. How far is it from their house to the school?
Found 2 solutions by ikleyn, richwmiller:
Answer by ikleyn(52808) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let "d" be the distance to the school.


Then the time in the first scenario is  d%2F6  hours.


     The time in the second scenario is  d%2F9  hours.


The difference is 15 minutes, or 1%2F4 of an hour :


It gives you this "time" equation


    d%2F6 - d%2F9 = 1%2F4.


Multiply both sides by 36 to rid off the denominators.


You will get


    6d - 4d = 9

    2d      = 9

     d      = 9/2 = 4.5 miles.


ANSWER.  The distance to the school is  4.5 miles.


CHECK.  a)  4.5%2F6 = 3%2F4 of an hour;

        b)  4%2F5%2F9 = 1%2F2  of an hour.

        The difference is  3%2F4 - 1%2F2 = 1%2F4 of an hour.  ! Correct !

Solved.

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

Look into the lesson
    - How far do you live from school?
in this forum.

And find there the exact analogue (TWIN problem) to yours.



Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
6*(t+.25)=d
9*t=d
6*(t+.25)=9*t
6*t+1.5)=9*t
1.5=3t
.5=t
9*.5=4.5 km to school
check
6*(.5+.25)=d
6*.75=d
d=4.5
ok