Question

In flight of $600\text{ km}$ an aircraft was slowed down due to bad weather, its average speed for the trip was reduced by $200\text{ km/hr}\text{.}$ and the time of flight increased by $30$ minutes. The duration of the flight is:
(a) $1$ hour
(b) $3$ hour
(c) $2$ hour
(d) $4$hour

Answer 1

Hint: To calculate the duration of flight time we will need to calculate the original speed of aircraft and we can calculate the original speed with help of duration of flight at original speed and duration of flight at reduced speed with increased time.
Formula used:
$\text{time=}\dfrac{\text{Distance}}{\text{Speed}}$

Complete Step by step Solution:
As per the data given in the question,
We have,
Distance $=600\text{ km}$
Let's assume,
Original speed of Aircraft be $x\text{ km/hr}\text{.}$
And new reduced speed $=\left( x-200 \right)\text{ km/hr}\text{.}$
So, Duration of Aircraft at original speed will be
$T=\dfrac{600}{x}\text{ hr}\text{.}$
And duration of Aircraft at reduced speed will be
${{T}_{r}}=\dfrac{600}{x-200}hr.$
And time increased of flight is $30$ minus $=\dfrac{1}{2}\text{ hr}\text{.}$
So,
$\dfrac{600}{x-200}-\dfrac{600}{x}=\dfrac{1}{2}$
$\dfrac{600\left( x-x+200 \right)}{x\left( x-200 \right)}=\dfrac{1}{2}$
$600\times 200\times 2={{x}^{2}}-200x$
${{x}^{2}}-200x-240000=0$
${{x}^{2}}-600x+4100x-240000=0$
$x\left( x-600 \right)+400\left( x-600 \right)=0$
$\left( x+400 \right)\left( x-600 \right)=0$
$x=600$ or $X=-400$
Since, speed cannot be in Negative then speed of Aircraft is $600\text{ km/hr}\text{.}$
Now, after obtaining the original speed of Aircraft. We can calculate direction of flight.
So, duration of flight $(T)=\dfrac{\text{distance}}{\text{speed}\left( x \right)}$
$T=\dfrac{600}{600}$
$T=1\text{ hour}\text{.}$.
Duration of flight is $1$ hour.

Hence, the option (a) is the correct answer.

Note:
Here increased time is given in minutes and other parameters are in hour units. So, we have to convert minutes to hours.