Question

A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
(A) 1000 km/hr
(B) 1600 km/hr
(C) 800 km/hr
(D) 600 km/hr

Answer 1

Hint: Assume that the usual speed of the plane is x km/hr. The distance to be covered by the plane to reach its destination is equal to 1600 km. Now, use the formula, \[\text{Time=}\dfrac{\text{Distance}}{\text{Speed}}\] and calculate the usual time. The speed of the plane after being 40 minutes late is \[\left( x+400 \right)km/hr\] . The plane has to cover 1600 km to reach the destination. Now, use the formula, \[\text{Time=}\dfrac{\text{Distance}}{\text{Speed}}\] and calculate the time taken. Since the plane has to reach the destination on time, the time taken by plane after being 40 minutes late should be 40 minutes less than the usual time taken by the plane. Now, convert 40 minutes into hour using the relation, \[\text{60minutes=1hour}\]Therefore, the plane should take time = \[\dfrac{1600}{x}\text{hrs-}\dfrac{2}{3}\text{hrs}\] . We have also calculated time by using the formula \[\text{Time=}\dfrac{\text{Distance}}{\text{Speed}}=\dfrac{1600}{x+400}\text{hrs}\] . Therefore, \[\dfrac{1600}{x}\text{hrs-}\dfrac{2}{3}\text{hrs=}\dfrac{1600}{x+400}\] . Now, solve it further and get the value of x.

Complete step by step answer:
According to the question, it is given that A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. We have to find the usual speed of the plane
First of all, let us assume that the usual speed of the plane is x km/hr.
The usual speed of the plane = x km/hr ………………………………………(1)
The distance to be covered by the plane to reach its destination = 1600 km …………………………….(2)
We know the formula, \[\text{Time=}\dfrac{\text{Distance}}{\text{Speed}}\] …………………………………….(3)
Now, from equation (1), equation (2) and equation (3), we get
\[\text{Time=}\dfrac{1600}{x}\]
The usual time taken by the plane to reach its destination is \[\dfrac{1600}{x}\] hr ………………………………..(4)
It is given that the plane is 40 minutes late. The plane has to increase its speed by 400 km/hr to reach the destination at the usual time. It means the new speed of the plane is 400 more than that of the usual speed.
The new speed of the plane = \[\left( x+400 \right)km/hr\] ……………………………..(5)
The distance to be covered by the plane to reach its destination = 1600 km …………………………….(6)
Now, from equation (3), equation (5) and equation (6), we get
\[\text{Time=}\dfrac{1600}{x+400}\]
The time taken by plane after being 40 minutes late is \[\dfrac{1600}{x+400}\] hr ……………………………..(7)
Since the plane left 40 minutes late and it had to increase its speed by 400 km/hr from its usual speed. So, after being 40 minutes late, to reach the destination on time, the plane would travel faster. Therefore, the time taken by plane after being 40 minutes late should be 40 minutes less than the usual time taken by the plane.
We know that \[\text{60minutes=1hour}\] .
\[\Rightarrow \text{60minutes=1hour}\]
\[\begin{align}
  & \text{1 minute=}\dfrac{1}{60}\text{hour} \\
 & \Rightarrow 40\,\text{minutes=}\dfrac{40}{60}\text{hour} \\
\end{align}\]
\[\Rightarrow 40\,\text{minutes=}\dfrac{2}{3}\text{hour}\] …………………………………..(8)
Here, we have the required time taken by plane after being 40 minutes late = \[\dfrac{1600}{x}\text{hrs-}\dfrac{2}{3}\text{hrs}\] ………………………………………..(9)
From equation (4), and equation (9), we have the usual time taken by the plane and the time taken by plane after being 40 minutes late.
\[\begin{align}
 & \Rightarrow \dfrac{1600}{x}-\dfrac{2}{3}=\dfrac{1600}{x+400} \\
 & \Rightarrow \dfrac{1600}{x}-\dfrac{1600}{x+400}=\dfrac{2}{3} \\
 & \Rightarrow 1600\left( \dfrac{1}{x}-\dfrac{1}{x+400} \right)=\dfrac{2}{3} \\
 & \Rightarrow 1600\left( \dfrac{x+400-x}{x\left( x+400 \right)} \right)=\dfrac{2}{3} \\
 & \Rightarrow \dfrac{1600\times 400}{{{x}^{2}}+400x}=\dfrac{2}{3} \\
 & \Rightarrow 3\times 320000={{x}^{2}}+400x \\
 & \Rightarrow {{x}^{2}}+400x-960000=0 \\
 & \Rightarrow {{x}^{2}}+1200x-800x-960000=0 \\
 & \Rightarrow x\left( x+1200 \right)-800\left( x+1200 \right)=0 \\
 & \therefore \left( x-800 \right)\left( x+1200 \right)=0 \\
\end{align}\]
So, \[x=800\] or \[x=-1200\] .
Since the speed can not be negative so, \[x=-1200\] is not possible.
Therefore, \[x=800\] is possible. So, the speed of the plane is 800 km/hr.
So, the correct answer is “Option C”.

Note:
In this question, one might make a silly mistake, and take the speed of the plane equal to 400 km/hr after being late by 40 minutes. This is wrong because it is given that the plane has to increase its speed by 400 which means that the speed of the plane after being late is 400 more than the usual speed of the speed.