SOLUTION: Red Riding Hood drives the 432 miles to go Grandmother’s house in 1 hour less than it takes the Wolf to drive the same route. Her average speed is 6 mph faster than the Wolf’s ave

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Question 632407: Red Riding Hood drives the 432 miles to go Grandmother’s house in 1 hour less than it takes the Wolf to drive the same route. Her average speed is 6 mph faster than the Wolf’s average speed. How fast does Red Riding Hood drive?
Found 2 solutions by josmiceli, solver91311:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +t+ = the Wolf's time to get there in hours
Let +s+ = Wolf's average speed in mi/hr
given:
RRH's equation:
(1) +432+=+%28+s+%2B+6+%29%2A%28+t+-+1+%29+
Wolf's equation:
(2) +432+=+s%2At+
--------------
(1) +432+=+s%2At+%2B+6t+-+s+-+6+
and
(2) t = 432/s }}}
-------------
Substitute (2) into (1)
(1) +432+=+s%2A%28432%2Fs%29+%2B+6%2A%28432%2Fs%29+-+s+-+6+
Multiply both sides by +s+
(1) +432s+=+432s+%2B+2592-+s%5E2+-+6s+
(1) +s%5E2+%2B+6s+-+2592+=+0+
Use quadratic equation
s+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
+a+=+1+
+b+=+6+
+c+=+-2592+
s+=+%28-6+%2B-+sqrt%28+6%5E2-4%2A1%2A%28-2592%29+%29%29%2F%282%2A1%29+
s+=+%28+-6+%2B-+sqrt%28+36+%2B+10368+%29%29%2F2+
s+=+%28+-6+%2B-+sqrt%28+10404+%29%29%2F2+
+s+=+%28+-6+%2B+102+%29+%2F2+ ( ignore the negative square root )
+s+=+96%2F2+
+s+=+48+
RRH's speed was 48 mi/hr
check answer:
(2) +432+=+s%2At+
(2) +432+=+48%2At+
(2) +t+=+9+ hrs
and
(1) +432+=+%28+s+%2B+6+%29%2A%28+t+-+1+%29+
(1) +432+=+%28+s+%2B+6+%29%2A%28+9+-+1+%29+
(1) +432+=+8%2A%28+s+%2B+6+%29+
(1) +s+%2B+6+=+54+
(1) +s+=+48+
OK

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let represent RRH's rate, so must represent BBW's rate. Let represent the time it took RRH to make the trip and must then represent BBW's time.

Starting with Distance Equals Rate Times Time, we can easily derive that Distance divided by Rate equals Time, so RRH's trip is described by:



And BBW's trip:



A little algebra music, Sammy (and I leave it as an exercise for the student to verify)



Now that we have two different expressions in that are both equal to , set them equal to each other:



Cross-multiply:



Simplify to standard quadratic form:



Solve for the positive root, then calculate

John

My calculator said it, I believe it, that settles it
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