SOLUTION: Hi a student goes to school at 2.5km/hr and arrives 6 minutes late. If he travels at 3km/hr he arrives 10 minutes early. What is the distance to the school. Thanks

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Question 1142987: Hi
a student goes to school at 2.5km/hr and arrives 6 minutes late. If he travels at 3km/hr he arrives 10 minutes early.
What is the distance to the school.
Thanks

Found 3 solutions by josgarithmetic, josmiceli, ikleyn:
Answer by josgarithmetic(39618)   (Show Source): You can put this solution on YOUR website!
-----
a student goes to school at 2.5km/hr and arrives 6 minutes late. If he travels at 3km/hr he arrives 10 minutes early.
What is the distance to the school.
-------

-------
a student goes to school at 2.5km/hr and arrives late. If he travels at 3km/hr he arrives early.
What is the distance to the school.
--------

              SPEED       TIME        DISTANCE
SLOW          2.5         x+1/10       d
FAST          3.0         x-1/6        d



-





--------the usual travel time, hours





---------distance to school, km

Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
Let = time in hrs it takes when the
student arrives on time
Let = the distance in km to the school
---------------------------------------------------
(1)
(1)
and
(2)
(2)
------------------------------------



Plug this result back into (1) or (2)
(2)
(2)
(2)
The distance to school is 4 km
--------------------------------------
check:
(1)
(1)
(1)
(1)
and
(2)
(2)
(2)
OK

Answer by ikleyn(52794)   (Show Source): You can put this solution on YOUR website!
.
Let d be the distance to the school (in kilometers).


Going at the speed of  2.5 km/h, the student spends    hours.


Going at the speed  of  3 km/h, the student spends    hours.


The difference is 6 + 10 minutes = 16 minutes =  of an hour =   of an hour.



It gives you the "time" equation


     -  = .


At this point, the setup is just completed.


To find "d", multiply both sides of the equation (1)  by 30.  You will get


    12d - 10d = 8,

    2d        = 8

     d        = 8/2 = 4 kilometers.


ANSWER.  The distance to the school is 4 kilometers.


CHECK.    = 1.6 hours = 1 hour 36 minutes;    

          =  hours = 1 hour and 20 minutes.

         The difference is 16 minutes -- ! Correct !

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

Using  "time"  equation is the  STANDARD  method of solving such problems.

It is simple,  logical,  straightforward and economic.  Going in this way,  you will not make a mistake - the logic of the method
prevents you of making mistakes.

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
    - Had a car move faster it would arrive sooner
    - How far do you live from school?    (*)
    - Earthquake waves
    - Time equation: HOW TO use, HOW TO write and HOW TO solve it
in this site.


For the TWIN problem,  see the lesson  (*)  in the list.



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