SOLUTION: A train runs 224km between 2 cities. If the speed on the way back is 8km/h slower, the run takes half an hour longer. Determine the speed on the way back.

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Question 691629: A train runs 224km between 2 cities. If the speed on the way back is 8km/h slower, the run takes half an hour longer. Determine the speed on the way back.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +t+ = time in hrs for 1st trip
+t+%2B+.5+ = time in hrs for 2nd trip
Let +s+ = speed in km/hr for 1st trip
+s+-+8+ = speed in km/hr for 2nd trip
-----------------------------------
1st trip:
(1) +224+=+s%2At+
2nd trip:
(2) +224+=+%28+s+-+8+%29%2A%28+t+%2B+.5+%29+
---------------------------
(2) +224+=+s%2At+-+8t+%2B+.5s+-+4+
and
(1) +t+=+224%2Fs+
Substitute (1) into (2)
(2) +224+=+s%2A%28224%2Fs%29+-+8%2A%28224%2Fs%29+%2B+.5s+-+4+
(2) +224+=+224+-+8%2A%28224%2Fs%29+%2B+.5s+-+4+
(2) +4+=++-+8%2A%28224%2Fs%29+%2B+.5s+
Multiply both sides by +2s+
(2) +8s+=+-16%2A224+%2B+s%5E2+
(2) +s%5E2+-+8s+-+3584+=+0+
I'll complete the square
(2) +s%5E2+-+8s+=+3584++
(2) +s%5E2+-+8s+%2B+%28-8%2F2%29%5E2+=+3584+%2B+%28-8%2F2%29%5E2++
(2) +s%5E2+-+8s+%2B+16+=+3584+%2B+16++
(2) +%28+s+-+4+%29%5E2+=+3600+
(2) +%28+s+-+4+%29%5E2+=+60%5E2+
(2) +s+-+4+=+60+
(2) +s+=+64+ ( ignore the negative square root of 60 )
and
+s+-+8+=+56+
The train's speed on the way back is 56 km/hr
check answer:
(2) +224+=+56%2A%28+t+%2B+.5+%29+
(2) +t+%2B+.5+=+224%2F56+
(2) +t+%2B+.5+=+4+
(2) +t+=+3.5+
and
(1) +224+=+s%2At+
(1) +224+=+64t+
(1) +t+=+224%2F64+
(1) +t+=+3.5+
OK