Question

Norachai traveled to the recycling plant and back. The trip back took 5 hours. He averaged 9mph faster on the trip there than on the return trip. What was norachai average speed on the outbound trip?

Answer 1

Answer:

$His\ average\ speed\ to\ trip\ there\ =45\ mph\\\\His\ average\ speed\ in\ return\ trip=45-9=36\ mph\\\\His\ average\ speed\ during\ whole\ trip=40\ mph$

Step-by-step explanation:

$Let\ distance\ from\ one\ side=d$

Trip:

$Let\ speed\ when\ he\ was\ going=x\ mph\\\\Time\ taken=4\ hours\\\\distance=d\\\\distance=speed\times time\\\\d=x\times 4\\\\d=4x\ ...................................eq(1)$

Return Trip:

$Speed=x-9\ \ \ \ \ (as\ it\ is\ 9\ mph\ slower)\\\\Time\ taken=5\ hours\\\\Distance=d\\\\Distance=speed\times time\\\\Distance=5\times (x-9)\\\\d=5x-45\ ........................................eq(2)$

$from\ eq(1)\ and\ eq(2)\\\\5x-45=4x\\\\5x-4x=45\\\\x=45\ mph\\\\from\ eq(1)\\\\d=4\times 45=180\ miles$

$Average\ speed\ during\ whole\ trip=\frac{total\ distance}{total\ time}=\frac{2\times 180}{5+4}=\frac{360}{9}=40\ mph$

$His\ average\ speed\ to\ trip\ there\ =45\ mph\\\\His\ average\ speed\ in\ return\ trip=45-9=36\ mph\\\\His\ average\ speed\ during\ whole\ trip=40\ mph$