
The bus covers a distance of 240 km at a uniform speed. Due to heavy rain, its speed gets reduced by 10 km/h and as such it takes 2 hours longer to cover the total distance. Assuming the uniform speed to be x km/h, form an equation and solve it to evaluate the value of x(in km/h).
Answer
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Hint: To solve the above question we will calculate the time taken by the bus to cover the distance in both the cases that are in case 1 when the bus travel at a uniform speed and in case 2 when its speed gets reduced by 10km/h so it will take more time than usual and from the question, we know that taken time is in case 2 is 2 hours more than that taken in case 1. So, we can say that:
Time is taken in case 2 – Time taken in case 1 = 2 hours.
And, to calculate the time we will use the formula \[Distance=Speed\times Time\]
Complete step-by-step solution:
We can see from the question that the bus is traveling a distance of 240 km and the uniform speed of the bus is x km/h.
We know from the relation of Distance, Speed, and Time that \[Distance=Speed\times Time\].
So, we can say that $\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$
So, we can say that time taken by the bus to travel the distance of 240 km at the uniform speed of x km/h =$\dfrac{240}{x}hour$.
And, again according to the question when it starts raining, the speed of the bus is reduced by 10 km/h. So, the speed of the bus after its speed reduction = (x - 10) km/h.
So, time is taken by bus to travel the distance of 240 km when it is raining = $\dfrac{240}{x-10}hour$
And, according to the question when it starts raining, the bus takes 2 hours more than what it takes when it travels at a uniform speed. So, we can say that:
Time is taken by bus to travel the distance of 240 km when it is raining - Time is taken by the bus to travel the distance of 240 km at the uniform speed of x km/h = 2 hours
$\Rightarrow \dfrac{240}{x-10}-\dfrac{240}{x}=2$
$\Rightarrow \dfrac{240\left( x-x+10 \right)}{x\left( x-10 \right)}=2$
$\Rightarrow \dfrac{240\times 10}{x\left( x-10 \right)}=2$
$\Rightarrow \dfrac{120\times 10}{x\left( x-10 \right)}=1$
$\Rightarrow {{x}^{2}}-10x-1200=0$
This is our required equation.
Now, we will solve the above quadratic equation.
Now, after splitting the middle term we will get:
$\Rightarrow {{x}^{2}}-40x+30x-1200=0$
$\Rightarrow x\left( x-40 \right)+30\left( x-40 \right)=0$
$\therefore \left( x+30 \right)\left( x-40 \right)=0$
Hence, x = -30 km/h, 40km/h
But we know that bus speed cannot be negative so bus uniform speed must be 40 km/h.
Hence, 40 km/h is the uniform speed of the bus.
This is our required answer.
Note: Students are required to write the equation correctly and they should not make any calculation mistakes. Students are required to note that speed is a scalar quantity and it practically bus speed can’t be negative. So, we should reject the negative value obtained after calculation.
Time is taken in case 2 – Time taken in case 1 = 2 hours.
And, to calculate the time we will use the formula \[Distance=Speed\times Time\]
Complete step-by-step solution:
We can see from the question that the bus is traveling a distance of 240 km and the uniform speed of the bus is x km/h.
We know from the relation of Distance, Speed, and Time that \[Distance=Speed\times Time\].
So, we can say that $\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$
So, we can say that time taken by the bus to travel the distance of 240 km at the uniform speed of x km/h =$\dfrac{240}{x}hour$.
And, again according to the question when it starts raining, the speed of the bus is reduced by 10 km/h. So, the speed of the bus after its speed reduction = (x - 10) km/h.
So, time is taken by bus to travel the distance of 240 km when it is raining = $\dfrac{240}{x-10}hour$
And, according to the question when it starts raining, the bus takes 2 hours more than what it takes when it travels at a uniform speed. So, we can say that:
Time is taken by bus to travel the distance of 240 km when it is raining - Time is taken by the bus to travel the distance of 240 km at the uniform speed of x km/h = 2 hours
$\Rightarrow \dfrac{240}{x-10}-\dfrac{240}{x}=2$
$\Rightarrow \dfrac{240\left( x-x+10 \right)}{x\left( x-10 \right)}=2$
$\Rightarrow \dfrac{240\times 10}{x\left( x-10 \right)}=2$
$\Rightarrow \dfrac{120\times 10}{x\left( x-10 \right)}=1$
$\Rightarrow {{x}^{2}}-10x-1200=0$
This is our required equation.
Now, we will solve the above quadratic equation.
Now, after splitting the middle term we will get:
$\Rightarrow {{x}^{2}}-40x+30x-1200=0$
$\Rightarrow x\left( x-40 \right)+30\left( x-40 \right)=0$
$\therefore \left( x+30 \right)\left( x-40 \right)=0$
Hence, x = -30 km/h, 40km/h
But we know that bus speed cannot be negative so bus uniform speed must be 40 km/h.
Hence, 40 km/h is the uniform speed of the bus.
This is our required answer.
Note: Students are required to write the equation correctly and they should not make any calculation mistakes. Students are required to note that speed is a scalar quantity and it practically bus speed can’t be negative. So, we should reject the negative value obtained after calculation.
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