QUESTION

# In a flight of 2800 Km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 100 km/hr and time increased by 30 min. Find the original duration of the flight.A.$3\dfrac{1}{2}\text{ hours}$ B.$4\dfrac{1}{2}\text{ hours}$C.$5\dfrac{1}{2}\text{ hours}$D.$2\dfrac{1}{2}\text{ hours}$

Hint: Think of the basic definition of speed and focus on the point that the speed mentioned here is the average speed. Use the formula $\text{distance covered}={{v}_{avg}}\times \text{time taken}$ to convert the conditions given in the question to the form of the equation. Solve the equations to get the answer.

Let’s start with what is speed. Speed is a scalar quantity defined as the distance travelled by a particle or object per unit time.
Generally, we deal with two kinds of speeds. One is instantaneous, and the other is the average speed. For uniform motion, both are identical.
Average speed is defined as the total distance covered by a body divided by the time taken by the body to cover it.
$\therefore {{v}_{avg}}=\dfrac{\text{distance covered}}{\text{time taken}}$
$\Rightarrow {{v}_{avg}}\times \text{time taken}=\text{distance covered}$
Now, starting with the solution to the above question.
It is given in the question that the total distance to travel is 2800 Km.
Let the original average speed be x Km/hr, and the original duration be t hours.
Therefore, we can say that the new average speed is (x – 100) km/hr, and the new duration of the flight is (t hours + 30 min), which can be written as $\left( t+\dfrac{1}{2} \right)$ hours.
Now, it is given that the distance covered in both cases is 2800 Km, and we know that distance covered is the product of average speed and time. So, we get,
Original distance covered = 2800km
$\Rightarrow xt=2800$
$x=\dfrac{2800}{t}.............(i)$
Also, new distance covered = 2800km
$\Rightarrow \left( x-100 \right)\left( t+\dfrac{1}{2} \right)=2800$
Now we will substitute the value of x from equation (i), so our equation becomes:
$\left( \dfrac{2800}{t}-100 \right)\left( t+\dfrac{1}{2} \right)=2800$
$\Rightarrow \dfrac{2800}{t}\times t+\dfrac{2800}{t}\times \dfrac{1}{2}-100t-100\times \dfrac{1}{2}=2800$
$\Rightarrow 2800+\dfrac{1400}{t}-100t-50=2800$
$\Rightarrow 50\left( \dfrac{28}{t}-2t-1 \right)=0$
Now we will take the LCM to be t for solving it further.
$\Rightarrow 50\dfrac{\left( 28-2{{t}^{2}}-t \right)}{t}=0$
Now, we know t cannot be zero as the time cannot be zero. So, sending $\dfrac{50}{t}$ to the other side of the equation, we get,
$28-2{{t}^{2}}-t=0\times \dfrac{t}{50}$
$\Rightarrow -2{{t}^{2}}-t+28=0$
Now using the formula of roots of the quadratic equation, we get
$t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{1\pm \sqrt{1-4\times (-2)\times 28}}{2\times (-2)}=\dfrac{1\pm \sqrt{225}}{-4}$
We know that time is always positive. Therefore, the possible value of t is:
$t=\dfrac{1-\sqrt{225}}{-4}=\dfrac{1-15}{-4}=\dfrac{14}{4}=\dfrac{7}{2}$
And we know we can write $\dfrac{7}{2}$ as $3\dfrac{1}{2}$ . Hence, the answer is option (a) $3\dfrac{1}{2}$ hours.

Note: Always try to keep the quantities according to a standardised unit system; this helps you to solve the question in an error-free manner. Also, it is prescribed to write each and every statement given in the question in mathematical form as it ensures that you are not missing any information given in the question.