
A passenger train takes one hour less for a journey of 150km if its speed is increased by 5km/hr from its usual speed. The usual speed of the train is …………… km/hr.
Answer
615.3k+ views
Hint- For solving such types of questions we need to let the speed of the vehicle and form an equation and then find the value of the quantity.
Complete step by step answer:
Let the usual speed of train be x km/hr, then
Time taken to cover the journey with usual speed=$\dfrac{{{\text{distance}}}}{{speed}}$=$\dfrac{{150}}{x}$ hr
On increasing the speed by 5km/hr, new speed=(x+5)km/hr
Therefore, time taken to complete the journey=$\dfrac{{{\text{distance}}}}{{speed}}$= $\dfrac{{150}}{{x + 5}}$ hr
According to question,
(Time taken at usual speed)- (Time taken at new speed)=1 hr
$
\dfrac{{150}}{x} - \dfrac{{150}}{{x + 5}} = 1 \\
150(\dfrac{1}{x} - \dfrac{1}{{x + 5}}) = 1 \\
150(\dfrac{{(x + 5) - (x)}}{{x.(x + 5)}}) = 1 \\
150(\dfrac{5}{{x.(x + 5)}}) = 1 \\
150 \times 5 = x.(x + 5) \\
{x^2} + 5x - 750 = 0 \\
{x^2} + 30x - 25x - 750 = 0 \\
x(x + 30) - 25(x + 30) = 0 \\
(x - 25)(x + 30) = 0 \\
x = 25, - 30 \\
$
The usual speed of the train is 25km/hr. (As speed can never be negative)
Hence the answer to this question is 25 km/hr.
Note- For this question we need to let the speed of the train and find the usual conditions and the new conditions after change in the speed of the train and then solve the relation to get the desired quantity.
Complete step by step answer:
Let the usual speed of train be x km/hr, then
Time taken to cover the journey with usual speed=$\dfrac{{{\text{distance}}}}{{speed}}$=$\dfrac{{150}}{x}$ hr
On increasing the speed by 5km/hr, new speed=(x+5)km/hr
Therefore, time taken to complete the journey=$\dfrac{{{\text{distance}}}}{{speed}}$= $\dfrac{{150}}{{x + 5}}$ hr
According to question,
(Time taken at usual speed)- (Time taken at new speed)=1 hr
$
\dfrac{{150}}{x} - \dfrac{{150}}{{x + 5}} = 1 \\
150(\dfrac{1}{x} - \dfrac{1}{{x + 5}}) = 1 \\
150(\dfrac{{(x + 5) - (x)}}{{x.(x + 5)}}) = 1 \\
150(\dfrac{5}{{x.(x + 5)}}) = 1 \\
150 \times 5 = x.(x + 5) \\
{x^2} + 5x - 750 = 0 \\
{x^2} + 30x - 25x - 750 = 0 \\
x(x + 30) - 25(x + 30) = 0 \\
(x - 25)(x + 30) = 0 \\
x = 25, - 30 \\
$
The usual speed of the train is 25km/hr. (As speed can never be negative)
Hence the answer to this question is 25 km/hr.
Note- For this question we need to let the speed of the train and find the usual conditions and the new conditions after change in the speed of the train and then solve the relation to get the desired quantity.
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