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(39630) About Me  (Show Source):
You can put this solution on YOUR website!
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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.
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a student goes to school at 2.5km/hr and arrives 1%2F10hour late. If he travels at 3km/hr he arrives 1%2F6hour 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


2.5%28x%2B1%2F10%29=3%28x-1%2F6%29=d
-
30%2A2.5%28x%2B1%2F10%29=30%2A3%28x-1%2F6%29
75%28x%2B1%2F10%29=90%28x-1%2F6%29
75x%2B7.5=90x-15
15x=22.5
30x=45
x=3%2F2--------the usual travel time, hours

d=3%28x-1%2F6%29
d=3%283%2F2-1%2F6%29
d=3%289%2F6-1%2F6%29=3%288%2F6%29
d=8%2F2
highlight%28d=4%29---------distance to school, km

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +t+ = time in hrs it takes when the
student arrives on time
Let +d+ = the distance in km to the school
---------------------------------------------------
(1) +d+=+2.5%2A%28+t+%2B+6%2F60+%29+
(1) +d+=+2.5t+%2B+.25+
and
(2) +d+=+3%2A%28+t+-+10%2F60+%29+
(2) +d+=+3t+-+.5+
------------------------------------
+2.5t+%2B+.25+=+3t+-+.5+
+.5t+=+.75+
+t+=+1.5+
Plug this result back into (1) or (2)
(2) +d+=+3%2A1.5+-+.5+
(2) +d+=+4.5+-+.5+
(2) +d+=+4+
The distance to school is 4 km
--------------------------------------
check:
(1) +d+=+2.5t+%2B+.25+
(1) +4+=+2.5%2A1.5+%2B+.25+
(1) +4+=+3.75+%2B+.25+
(1) +4+=+4+
and
(2) +4+=+3%2A1.5+-+.5+
(2) +4+=+4.5+-+.5+
(2) +4+=+4+
OK

Answer by ikleyn(52878) About Me  (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  d%2F2.5  hours.


Going at the speed  of  3 km/h, the student spends  d%2F3  hours.


The difference is 6 + 10 minutes = 16 minutes = 16%2F60 of an hour = 4%2F15  of an hour.



It gives you the "time" equation


    d%2F2.5 - d%2F3 = 4%2F15.


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.   4%2F2.5 = 1.6 hours = 1 hour 36 minutes;    

         4%2F3 = 11%2F3 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.