
A train travels $360km$at uniform speed. If the speed had been $5km/hr$more, it would have taken $1hr$ less for the same journey. Find the speed of the train.
Answer
536.4k+ views
Hint: - Use, ${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{time}}}}$
Let the speed of the train be $x{\text{ }}km/hr$.
And the time to cover $360km $be $t{\text{ }} hr$.
As you know,${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{time}}}}$
$\therefore x = \dfrac{{360}}{t}km/hr............\left( 1 \right)$
It is also given that if speed is increased by $5km/hr$, the time to cover $360km $is $1hr$ less than the previous time.
$\therefore x + 5 = \dfrac{{360}}{{t - 1}}km/hr.............\left( 2 \right)$
From equation (1) and (2)
$
xt = \left( {x + 5} \right)\left( {t - 1} \right) \\
\therefore xt = xt - x + 5t - 5 \\
\therefore x - 5t + 5 = 0..............\left( 3 \right) \\
$
From equation (1) $t = \dfrac{{360}}{x}$
Substitute this value in equation (3)
$
x - 5\left( {\dfrac{{360}}{x}} \right) + 5 = 0 \\
{x^2} + 5x - 1800 = 0 \\
$
Now, factorize the above equation
$
{x^2} + 45x - 40x - 1800 = 0 \\
x\left( {x + 45} \right) - 40\left( {x + 45} \right) = 0 \\
\left( {x + 45} \right)\left( {x - 40} \right) = 0 \\
\left( {x + 45} \right) = 0,{\text{ }}\left( {x - 40} \right) = 0 \\
\therefore x = - 45,{\text{ }}x = 40 \\
$
Speed cannot be negative
So the required speed is $40{\text{ }}km/hr$
Note: -In such types of questions the key concept we have to remember is that always remember the formula of speed, time, and distance, then formulate the equations using this formula and the given conditions, then simplify the equations using substitution, then we will get the required answer.
Let the speed of the train be $x{\text{ }}km/hr$.
And the time to cover $360km $be $t{\text{ }} hr$.
As you know,${\text{Speed}} = \dfrac{{{\text{Distance}}}}{{{\text{time}}}}$
$\therefore x = \dfrac{{360}}{t}km/hr............\left( 1 \right)$
It is also given that if speed is increased by $5km/hr$, the time to cover $360km $is $1hr$ less than the previous time.
$\therefore x + 5 = \dfrac{{360}}{{t - 1}}km/hr.............\left( 2 \right)$
From equation (1) and (2)
$
xt = \left( {x + 5} \right)\left( {t - 1} \right) \\
\therefore xt = xt - x + 5t - 5 \\
\therefore x - 5t + 5 = 0..............\left( 3 \right) \\
$
From equation (1) $t = \dfrac{{360}}{x}$
Substitute this value in equation (3)
$
x - 5\left( {\dfrac{{360}}{x}} \right) + 5 = 0 \\
{x^2} + 5x - 1800 = 0 \\
$
Now, factorize the above equation
$
{x^2} + 45x - 40x - 1800 = 0 \\
x\left( {x + 45} \right) - 40\left( {x + 45} \right) = 0 \\
\left( {x + 45} \right)\left( {x - 40} \right) = 0 \\
\left( {x + 45} \right) = 0,{\text{ }}\left( {x - 40} \right) = 0 \\
\therefore x = - 45,{\text{ }}x = 40 \\
$
Speed cannot be negative
So the required speed is $40{\text{ }}km/hr$
Note: -In such types of questions the key concept we have to remember is that always remember the formula of speed, time, and distance, then formulate the equations using this formula and the given conditions, then simplify the equations using substitution, then we will get the required answer.
Recently Updated Pages
Master Class 9 General Knowledge: Engaging Questions & Answers for Success

Earth rotates from West to east ATrue BFalse class 6 social science CBSE

The easternmost longitude of India is A 97circ 25E class 6 social science CBSE

Write the given sentence in the passive voice Ann cant class 6 CBSE

Convert 1 foot into meters A030 meter B03048 meter-class-6-maths-CBSE

What is the LCM of 30 and 40 class 6 maths CBSE

Trending doubts
Which one is a true fish A Jellyfish B Starfish C Dogfish class 10 biology CBSE

Dr BR Ambedkars fathers name was Ramaji Sakpal and class 10 social science CBSE

A boat goes 24 km upstream and 28 km downstream in class 10 maths CBSE

Why is there a time difference of about 5 hours between class 10 social science CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

What is the full form of POSCO class 10 social science CBSE
