SOLUTION: If a boy walks from his house to school at 4 km/h, he reaches the school 10 minutes earlier than the scheduled time. However if he walks the rate of 3 km/h he reaches 10 minutes la

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Question 775609: If a boy walks from his house to school at 4 km/h, he reaches the school 10 minutes earlier than the scheduled time. However if he walks the rate of 3 km/h he reaches 10 minutes late. The distance of the school from his house is
Found 2 solutions by josgarithmetic, josmiceli:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
The required time to make his trip is x, in hours. When he is early using 4 km/hour, time is x-1/6. When he is late at 3 km/hour time is x+1/6. Whether early or late, the trip is the same distance.

Traveling to school, rate*time=distance
10 minutes = 1/6 hour
Early: 4%28x-1%2F6%29
Late: 3%28x%2B1%2F6%29
Equal Distance: highlight%284%28x-1%2F6%29=3%28x%2B1%2F6%29%29

4x-2%2F3=3x%2B1%2F2
x-2%2F3=1%2F2
x=1%2F2%2B2%2F3=%283%2B4%29%2F6=highlight%287%2F6%29 hours
x=1%261%2F6 hour
1 hour 10 minutes if he were to be "on-time".

HOW FAR FROM HOME TO SCHOOL?
Pick either early or late expression and compute the value.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +t+ = the time in hours for his trip
if he is to arrive at the scheduled time
Let +d+ = distance from school to house in km
--------------
Equation for leaving early:
+d+=+4%2A%28+t++-+10%2F60+%29+
---------------------
Equation for leaving late:
+d+=+3%2A%28+t+%2B+10%2F60+%29+
---------------------
By substitution:
+4%2A%28+t+-+1%2F6+%29+=+3%2A%28+t+%2B+1%2F6+%29+
+4t+-+2%2F3+=+3t+%2B+1%2F2+
+t+=+1%2F2+%2B+2%2F3+
Multiply both sides by +6+
+6t+=+3+%2B+4+
+t+=+7%2F6+
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By substitution:
+d+=+4%2A%28+70%2F60+-+10%2F60+%29+
+d+=+4+ km
The distance of the school from his house is 4 km
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check:
+d+=+3%2A%28+t+%2B+10%2F60+%29+
+d+=+3%2A%28+t70%2F60%2B+10%2F60+%29+
+d+=+3%2A%284%2F3%29+
+d+=+4+ km
OK