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

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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(52810)   (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    hours.


     The time in the second scenario is    hours.


The difference is 15 minutes, or  of an hour :


It gives you this "time" equation


     -  = .


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)   =  of an hour;

        b)   =   of an hour.

        The difference is   -  =  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)   (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

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