Answer
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Hint: We solve this question by assuming the average speed of the bus will be \[x\] km/hr. As per question, average speed of a car can be taken as (20+x) km/hr. Then, we will use the basic concept of speed-distance formula, which is $ \text{speed=}\dfrac{\text{distance}}{\text{time}} $ . We will substitute the time, distance and speed and then we will get a quadratic equation in x. Solving for x using the factorisation method will give us the value of x and then adding it to 20 km/hr will give us the answer.
Complete step-by-step answer:
We have given that the total distance covered $ =300\text{ km} $
Also, we have given that the average speed of a car is $ 20 $ km/hr more than the average speed of a bus.
Let the average speed of the bus be \[x\] km/hr. So, the average speed of the car will be $ x+20 $ km/hr.
Also, we have given that the car takes 4 hours less than the bus to cover the same distance.
Now, the time taken by the bus will be $ \text{Time=}\dfrac{\text{distance}}{\text{speed}} $
$ \Rightarrow \dfrac{300}{x} $
Now, time taken by a car to cover the same distance will be $ \dfrac{300}{x}-4 $
Now, the average speed of the car will be $ \text{speed=}\dfrac{\text{distance}}{\text{time}} $
When we put values we get $ x+20=\dfrac{300}{\dfrac{300}{x}-4} $
After cross multiplying we get $ \left( x+20 \right)\left( \dfrac{300}{x}-4 \right)=300 $
When we solve further we get
$ \begin{align}
& \Rightarrow \left( x+20 \right)\left( \dfrac{300-4x}{x} \right)=300 \\
& \Rightarrow \left( x+20 \right)\left( 300-4x \right)=300x \\
& \Rightarrow 300x-4{{x}^{2}}+6000-80x=300x \\
& \Rightarrow 4{{x}^{2}}+80x=6000 \\
& \Rightarrow 4{{x}^{2}}+80x-6000=0 \\
\end{align} $
When we divide the whole equation by 4, we get
$ \begin{align}
& \Rightarrow \dfrac{4{{x}^{2}}}{4}+\dfrac{80x}{4}-\dfrac{6000}{4}=0 \\
& \Rightarrow {{x}^{2}}+20x-1500=0 \\
\end{align} $
Now, we have a quadratic equation of the form $ a{{x}^{2}}+bx+c=0 $ . Now, we solve this equation by using the factorization method.
$ \begin{align}
& {{x}^{2}}+20x-1500=0 \\
& {{x}^{2}}+50x-30x-1500=0 \\
& x\left( x+50 \right)-30\left( x+50 \right)=0 \\
& \left( x+50 \right)\left( x-30 \right)=0 \\
& \left( x+50 \right)=0\text{ or }\left( x-30 \right)=0 \\
& x=-50\text{ or }x=30 \\
\end{align} $
We have two values of $ x $ after solving the equation. As $ x $ is the average speed and it doesn’t have negative value. So, we take $ x=30 $
So, the average speed of the bus is $ 30 $ km/hr and the average speed of the car is $ x+20=30+20=50 $ km/hr.
Note: The main point to remember in these types of questions is to not forget to mention the units. If units are not mentioned, marks will be deducted. The possibility of mistake is in assuming the average speed of car and bus. As we have given that the average speed of a car is 20 km/hr more than the average speed of a bus, we have to add 20 to the average speed of the bus and not deduct it by mistake.
Complete step-by-step answer:
We have given that the total distance covered $ =300\text{ km} $
Also, we have given that the average speed of a car is $ 20 $ km/hr more than the average speed of a bus.
Let the average speed of the bus be \[x\] km/hr. So, the average speed of the car will be $ x+20 $ km/hr.
Also, we have given that the car takes 4 hours less than the bus to cover the same distance.
Now, the time taken by the bus will be $ \text{Time=}\dfrac{\text{distance}}{\text{speed}} $
$ \Rightarrow \dfrac{300}{x} $
Now, time taken by a car to cover the same distance will be $ \dfrac{300}{x}-4 $
Now, the average speed of the car will be $ \text{speed=}\dfrac{\text{distance}}{\text{time}} $
When we put values we get $ x+20=\dfrac{300}{\dfrac{300}{x}-4} $
After cross multiplying we get $ \left( x+20 \right)\left( \dfrac{300}{x}-4 \right)=300 $
When we solve further we get
$ \begin{align}
& \Rightarrow \left( x+20 \right)\left( \dfrac{300-4x}{x} \right)=300 \\
& \Rightarrow \left( x+20 \right)\left( 300-4x \right)=300x \\
& \Rightarrow 300x-4{{x}^{2}}+6000-80x=300x \\
& \Rightarrow 4{{x}^{2}}+80x=6000 \\
& \Rightarrow 4{{x}^{2}}+80x-6000=0 \\
\end{align} $
When we divide the whole equation by 4, we get
$ \begin{align}
& \Rightarrow \dfrac{4{{x}^{2}}}{4}+\dfrac{80x}{4}-\dfrac{6000}{4}=0 \\
& \Rightarrow {{x}^{2}}+20x-1500=0 \\
\end{align} $
Now, we have a quadratic equation of the form $ a{{x}^{2}}+bx+c=0 $ . Now, we solve this equation by using the factorization method.
$ \begin{align}
& {{x}^{2}}+20x-1500=0 \\
& {{x}^{2}}+50x-30x-1500=0 \\
& x\left( x+50 \right)-30\left( x+50 \right)=0 \\
& \left( x+50 \right)\left( x-30 \right)=0 \\
& \left( x+50 \right)=0\text{ or }\left( x-30 \right)=0 \\
& x=-50\text{ or }x=30 \\
\end{align} $
We have two values of $ x $ after solving the equation. As $ x $ is the average speed and it doesn’t have negative value. So, we take $ x=30 $
So, the average speed of the bus is $ 30 $ km/hr and the average speed of the car is $ x+20=30+20=50 $ km/hr.
Note: The main point to remember in these types of questions is to not forget to mention the units. If units are not mentioned, marks will be deducted. The possibility of mistake is in assuming the average speed of car and bus. As we have given that the average speed of a car is 20 km/hr more than the average speed of a bus, we have to add 20 to the average speed of the bus and not deduct it by mistake.
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