SOLUTION: An aeroplane flies from A to B at a speed of 750km/h. On the returnn trip there was a strong tail wind and airplane flew at 900km/h. The return trip took 20min less than original.
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Question 954012: An aeroplane flies from A to B at a speed of 750km/h. On the returnn trip there was a strong tail wind and airplane flew at 900km/h. The return trip took 20min less than original. How many hours did the return trip take? Found 3 solutions by josgarithmetic, hemu_da, MathTherapy:Answer by josgarithmetic(39625) (Show Source):
You can put this solution on YOUR website! R=900
r=750
t, unknown time length going A to B.
k=1/3 hour, how much time less on returning, B to A
Question asks for t-1/3;
Understand that 20 minutes is of an hour.
You can put this solution on YOUR website! Suppose t is the normal time to travel from A to B.
And normal speed is 750 km/hr.
Speed on return trip is 900 km/hr.
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Time taken is 20 min less than the original ( which is t here).
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I.e. hr
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In both the trip distance is same that is...
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5t = 6t - 2
t = 2 hr.
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Returning trip take 20 min less than the original, hence time is--
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2 hr - 20 min = 1 hr 40 min.
.....
You can put this solution on YOUR website!
An aeroplane flies from A to B at a speed of 750km/h. On the returnn trip there was a strong tail wind and airplane flew at 900km/h. The return trip took 20min less than original. How many hours did the return trip take?
Let time taken on return trip (B to A), be T
Then time taken on outbound trip (A to B) = , or
We then get:
750T + 250 = 900T
750T – 900T = - 250
- 150T = - 250
T, or time taken on return trip = , or hours
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