Question

An aeroplane started 30 minutes later than the schedule time from a place 1500 km away from its destination to reach the destination at the schedule time the pilot had to increase the speed by \[250{}^{km}/{}_{hr}\]. What was the speed of the aeroplane per hour during the journey?
A. \[100{}^{km}/{}_{hr}\]
B. \[1000{}^{km}/{}_{hr}\]
C. \[1050{}^{km}/{}_{hr}\]
D. \[900{}^{km}/{}_{hr}\]

Answer 1

- Hint: Take usual speed of plane as ‘x’. Find the speed of the plane to reach your destination. Use the formula of speed to get a relation. Solve the quadratic equation formed and find the speed of the plane per hour.

Complete step-by-step answer:
It is said that an aeroplane started 30 minutes later than scheduled time.
The total distance the plane has to cover = 1500 km.
Let us consider the usual speed of the plane as \[x{}^{km}/{}_{hr}\].
Now to reach the destination on time, the pilot has to increase speed to \[250{}^{km}/{}_{hr}\].
Thus the speed in which the plane has to go to reach its destination on time \[=\left( x+250 \right){}^{km}/{}_{hr}\].
\[\therefore \]Speed of plane \[=\left( x+250 \right){}^{km}/{}_{hr}\].
Now we know Speed \[=\dfrac{Distance}{Time}\].
\[\therefore \]Time \[=\dfrac{Distance}{Speed}\].
The distance to cover is the same 1500 km, in whichever speed the plane goes.
\[\dfrac{1500}{x}-\dfrac{1500}{x+250}=\dfrac{30}{60}......(1)\]
Time is given in minutes, i.e. 30 minutes, to convert it to hour, we divide it by 60, i.e. \[\dfrac{30}{60}\] hour.
Now let us simplify equation (1).
\[\begin{align}
  & 1500\left[ \dfrac{1}{x}-\dfrac{1}{x+250} \right]=\dfrac{1}{2} \\
 & 1500\left[ \dfrac{\left( x+250 \right)-x}{x\left( x+250 \right)} \right]=\dfrac{1}{2} \\
 & 2\times 1500\left[ x+250-x \right]=x\left( x+250 \right) \\
 & 3000\times 250={{x}^{2}}+250x \\
 & \Rightarrow {{x}^{2}}+250x-750000=0.......(2) \\
\end{align}\]
From the above expression, we can say that,
Sum of zeroes = 250
Product of zeroes = -750000
i.e., \[{{x}^{2}}+\left( sum\text{ }of\text{ }zeroes \right)x+\left( product\text{ }of\text{ }zeroes \right)=0\].
Now let us take 2 terms as a and b.
Sum of zeroes = a + b
Product of zeroes = ab = -750000
Let us consider a = 1000 and b = -750
\[\begin{align}
  & \therefore a+b=1000+\left( -750 \right)=250 \\
 & ab=1000\times \left( -750 \right)=-750000 \\
\end{align}\]
Thus we split the 2nd term of equation (2) as,
\[\begin{align}
  & {{x}^{2}}+\left( 1000-750 \right)x-750000=0 \\
 & \therefore {{x}^{2}}+1000x-750x-7500000=0 \\
 & \therefore x\left( x+1000 \right)-750\left( x+1000 \right)=0 \\
 & \left( x-750 \right)\left( x+1000 \right)=0 \\
 & \therefore x-750=0 \\
 & x=750{}^{km}/{}_{hr} \\
\end{align}\]
Thus we got the usual speed of the plane as, \[x=750{}^{km}/{}_{hr}\].
\[\therefore \]The speed of plane per hour during the journey = x + 250 = 750 + 250 = \[1000{}^{km}/{}_{hr}\].
\[\therefore \]Speed of the plane per hour of the journey = \[1000{}^{km}/{}_{hr}\].
Thus option B is the correct answer.

Note: We neglect the term \[\left( x+1000 \right)\] here as,
\[\begin{align}
  & x+1000=0 \\
 & x=-1000 \\
\end{align}\]
We know that ‘x’ denotes the speed of the plane. And the speed can never be negative. We got \[\left( -1000 \right)\] here, as it is negative, we neglect this value.