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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.

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