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Here's a problem that I can't seem to solve. It was discussed to us last week, but I had a fever so I wasn't able to understand the answer and solution.

Due to inclement weather, the pilot of a plane slows down the plane’s regular flying rate by $25. This results to an additional $1.5$ hours in covering the 3,000-km distance to its regular time required for the trip. Find the regular rate of the trip.

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    Did you try? If yes, then add it here. And if you didn't try to solve then. – 2017-01-15
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    The `rational-functions` tag (which I just removed) has the description "*Rational functions are ratios of two polynomials...*". What in your question matched that description? P.S. Hint: $v\,t=0.75\,v\,(t+1.5)=\,$... – 2017-01-15

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The pilot is moving at $\frac {3v}{4}$ of its normal speed and it took the regular time $t+1.5hrs$ to arrive. Perhaps you can use the formula $vt=d$ and apply it here?

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Let regular rate be S.

Then reduced rate during trip = .75S

Time to cover 3,000 km at regular rate = $\frac{3000}{S}$

Time to cover 3,000 km at reduced rate = $\frac{3000}{.75S}$

Difference in time 1.5 hours.

$\frac{3000}{.75S} - \frac{3000}{S} = 1.5$

Solve it to find S that is the required answer.

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    If any doubt please let me know. – 2017-01-15
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    Is the answer 666.67 kph? Thank you for your time. – 2017-01-15
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    Mine pleasure.. – 2017-01-15