Questions & Answers

Question

Answers

Answer

Verified

146.1k+ views

Hint: Use Distance = Speed $\times $ Time. Consider the original speed of the train to be x km/hr and apply the data given in the question. Use a cross multiplication method to solve the equations.

Complete step-by-step Solution:

Cross multiplication method is the method where the numerator from the left-hand side is multiplied with the denominator from the right-hand side and numerator from the right-hand side is multiplied with the denominator from the left-hand side.

Let, the original speed of the train be x km/hour.

Using the formula for distance = Speed $\times $ Time we will calculate the time as Time = distance/speed.

$\therefore$ Time $=\dfrac{90}{x}$

If the speed has been 15 km/hour more, new speed becomes x + 15 , new time becomes \[\dfrac{90}{x+15}\] , when the speed becomes 15 km/hour more, it would take 30minutes less for the journey.

Convert the minutes into the hour, therefore 30 minutes is converted into $\dfrac{1}{2}$ an hour as the speed has been given in an hour.

According to the question,

$\dfrac{90}{x}-\dfrac{90}{x+15}=\dfrac{1}{2}$

Taking 90 as common we get,

$\begin{align}

& \Rightarrow 90\left( \dfrac{1}{x}-\dfrac{1}{x+15} \right)=\dfrac{1}{2} \\

& \\

\end{align}$

$\Rightarrow 90\left( \dfrac{x+15-x}{x\left( x+15 \right)} \right)=\dfrac{1}{2}$

$\Rightarrow \dfrac{90\times 15}{x\left( x+15 \right)}=\dfrac{1}{2}$

By cross-multiplication we get,

$\Rightarrow 90\times 15\times 2={{x}^{2}}+15x$

$\Rightarrow {{x}^{2}}+15x-2700=0$

We will use the middle term splitting method to find the factors of this quadratic equation.

We will find the factors of 2700 through which we get the sum or difference of their factors as 15. These factors are 60 and 45 whose difference is 15.

$\therefore {{x}^{2}}+60x-45x-2700=0$

$\Rightarrow x\left( x+60 \right)-45\left( x+60 \right)=0$

$\Rightarrow \left( x+60 \right)\left( x-45 \right)=0$

The values of x are -60 and 45.

But the speed cannot be negative. Therefore, negative values are excluded.

Hence, the speed of the train is 45 km/hour.

NOTE: While forming the equation, one must be very careful in converting the data from question to equation form. The student commit a mistake by forming the equation as \[\dfrac{90}{x}+\dfrac{90}{\left( x+15 \right)}=\dfrac{1}{2}\] instead of \[\dfrac{90}{x}-\dfrac{90}{\left( x+15 \right)}=\dfrac{1}{2}\] . Do not forget to convert minutes into the hour. Be careful about the negative values as speed, distance and time can never be negative.

Complete step-by-step Solution:

Cross multiplication method is the method where the numerator from the left-hand side is multiplied with the denominator from the right-hand side and numerator from the right-hand side is multiplied with the denominator from the left-hand side.

Let, the original speed of the train be x km/hour.

Using the formula for distance = Speed $\times $ Time we will calculate the time as Time = distance/speed.

$\therefore$ Time $=\dfrac{90}{x}$

If the speed has been 15 km/hour more, new speed becomes x + 15 , new time becomes \[\dfrac{90}{x+15}\] , when the speed becomes 15 km/hour more, it would take 30minutes less for the journey.

Convert the minutes into the hour, therefore 30 minutes is converted into $\dfrac{1}{2}$ an hour as the speed has been given in an hour.

According to the question,

$\dfrac{90}{x}-\dfrac{90}{x+15}=\dfrac{1}{2}$

Taking 90 as common we get,

$\begin{align}

& \Rightarrow 90\left( \dfrac{1}{x}-\dfrac{1}{x+15} \right)=\dfrac{1}{2} \\

& \\

\end{align}$

$\Rightarrow 90\left( \dfrac{x+15-x}{x\left( x+15 \right)} \right)=\dfrac{1}{2}$

$\Rightarrow \dfrac{90\times 15}{x\left( x+15 \right)}=\dfrac{1}{2}$

By cross-multiplication we get,

$\Rightarrow 90\times 15\times 2={{x}^{2}}+15x$

$\Rightarrow {{x}^{2}}+15x-2700=0$

We will use the middle term splitting method to find the factors of this quadratic equation.

We will find the factors of 2700 through which we get the sum or difference of their factors as 15. These factors are 60 and 45 whose difference is 15.

$\therefore {{x}^{2}}+60x-45x-2700=0$

$\Rightarrow x\left( x+60 \right)-45\left( x+60 \right)=0$

$\Rightarrow \left( x+60 \right)\left( x-45 \right)=0$

The values of x are -60 and 45.

But the speed cannot be negative. Therefore, negative values are excluded.

Hence, the speed of the train is 45 km/hour.

NOTE: While forming the equation, one must be very careful in converting the data from question to equation form. The student commit a mistake by forming the equation as \[\dfrac{90}{x}+\dfrac{90}{\left( x+15 \right)}=\dfrac{1}{2}\] instead of \[\dfrac{90}{x}-\dfrac{90}{\left( x+15 \right)}=\dfrac{1}{2}\] . Do not forget to convert minutes into the hour. Be careful about the negative values as speed, distance and time can never be negative.

Students Also Read