Question

A plane left 30 minutes later than the scheduled time in order to reach its destination, 1500 km away it has to increase its speed by 250km/hr than his usual speed. Find its usual speed.

Answer 1

Hint: We use the formula for determining Time in terms of Distance and Speed which is,
\[T=\dfrac{D}{S}\], where T is the time, D is the distance and S is the speed of the plane.

Complete step-by-step answer:
Given Distance D = 1500km and the Speed increase by S= 250km/h also that the plane left 30 minutes later than the scheduled time.

We have to determine the usual speed of the plane
Let the usual speed of the plane be x.

We have the formula to determine Time in terms of Distance and Speed which is,
\[T=\dfrac{D}{S}\], where T is the time, D is the distance and S is the speed of the plane.
Substituting the values of Distance D and the Speed S (as assumed x) in above equation we get,
Time \[T=\dfrac{1500}{x}\] …………(i)

Now given that the speed is increased by 250km/hr.
Thus, we obtain a new speed adding with the previous one x, which gives the new speed as \[S'=x+250\]
Similarly, the new time T’ becomes \[T'=\dfrac{1500}{x+250}\] ………….(ii)

Now because we are doing all calculations in km/hour so we convert 30 mins into hour.
We have, I hour= 60 mins
Implies 60 mins= 1 hour
Implies 1 min= \[\dfrac{1}{60}\]hour
Implies 30 mins=\[\dfrac{30}{60}=\dfrac{1}{2}\]hour
That means the plane left \[\dfrac{1}{2}\]hour later than the scheduled time, thus adding the Time T to \[\dfrac{1}{2}\]hour gives us the new Time T’
Thus, \[T\text{ }+\dfrac{1}{2}=\text{ }T\]
Now substituting the values of T and T’ obtained by the equation (i) and (ii) we get
\[T\text{ }+\dfrac{1}{2}=\text{ }T\]

\[\Rightarrow \dfrac{1500}{x}+\dfrac{1}{2}=\dfrac{1500}{x+250}\]

\[\Rightarrow \dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{1}{2}\]

 \[\Rightarrow \dfrac{1500(x+250)-1500x}{x(x+250)}=\dfrac{1}{2}\]

\[\Rightarrow \dfrac{1500x+375000-1500x}{x(x+250)}=\dfrac{1}{2}\]

\[\Rightarrow \dfrac{375000}{x(x+250)}=\dfrac{1}{2}\]

Cross multiplying the above equation we get,

\[750000={{x}^{2}}+250x\]

Rearranging the equation we get,
\[{{x}^{2}}+250x-750000=0\]

Now we find two numbers whose sum=+250 and multiple =-750000, we get possible numbers as 1000 and -750
Splitting the above equation, we get,

\[x{}^\text{2}+1000x-750x-750000=0\]

\[\Rightarrow x\left( x+1000 \right)-750\left( x+1000 \right)=0\]

\[\Rightarrow \left( x+1000 \right)\left( x-750 \right)=0\]

\[\Rightarrow x+1000=0\] or \[\left( x-750 \right)=0\]

implies \[x=-1000\] or \[x=750\]

Speed can't be in negative so \[x=-1000\] is not possible, then only possible value of x is \[x=750\]

So, the usual speed of the plane is \[x=750\]km/hour

Note: The possibility of error is that you can forget to convert 30 min in terms of hour before doing the calculation, which is wrong. We need to convert all the terms in one unit and then proceed accordingly.