A train covers a distance of 90km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for the journey. The original speed of the train is____km/hr.
Answer
379.8k+ 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.
Recently Updated Pages
Which of the following would not be a valid reason class 11 biology CBSE

What is meant by monosporic development of female class 11 biology CBSE

Draw labelled diagram of the following i Gram seed class 11 biology CBSE

Explain with the suitable examples the different types class 11 biology CBSE

How is pinnately compound leaf different from palmately class 11 biology CBSE

Match the following Column I Column I A Chlamydomonas class 11 biology CBSE

Trending doubts
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

Write a letter to the principal requesting him to grant class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
