Answer
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Hint – To find the distance between two stations we consider the original speed to be a variable x, we then write the speed after fault in terms of x and establish respective equations using given data, then we solve that relation using the given data for x.
Complete step by step answer:
Let the original speed at which the train travels be x.
Therefore, $\dfrac{3}{5}$th of the original speed is $\dfrac{3}{5}$x = 0.6x
The time take to cover the distance after fault is $\dfrac{{50}}{{0.6{\text{x}}}}$ and time taken before fault is$\dfrac{{50}}{{\text{x}}}$, now we equate it to time. We get (distance = speed × time)
$\dfrac{{50}}{{0.6{\text{x}}}} - \dfrac{{50}}{{\text{x}}} = \dfrac{{40}}{{60}}$
$
\Rightarrow \dfrac{{\left( {250 - 150} \right)}}{{{\text{3x}}}} = \dfrac{2}{3} \\
\Rightarrow \dfrac{{100}}{{3{\text{x}}}} = \dfrac{2}{3} \\
\Rightarrow 100 = 2{\text{x}} \\
\Rightarrow {\text{x = 50}} \\
$
Therefore 0.6x = 0.6 × 50 = 30.
Hence speed reduced = 50 – 30 = 20.
Given, the train is late by two hours and by this time at a speed = 30, it would have gone a distance,
30 × 2 = 60 miles.
If the train had went at the original speed, the time it would have taken after the first hour is
Time = $\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$=$\dfrac{{60}}{{20}} = 3$, i.e. 3 hours after first hour
Hence total time = 3 + 1 = 4 hours.
Therefore total distance travelled = 4 × 50 = 200 miles.
Hence, the distance between the two stations is 200 miles.
Note – In order to solve this type of question the key is to write the sentences given the question in the form of equations and solve accordingly. Writing all the quantities in terms of one quantity simplifies the equation. While there are more than one values for the variable x after solving, we compare them with the data given in the question to verify.
Complete step by step answer:
Let the original speed at which the train travels be x.
Therefore, $\dfrac{3}{5}$th of the original speed is $\dfrac{3}{5}$x = 0.6x
The time take to cover the distance after fault is $\dfrac{{50}}{{0.6{\text{x}}}}$ and time taken before fault is$\dfrac{{50}}{{\text{x}}}$, now we equate it to time. We get (distance = speed × time)
$\dfrac{{50}}{{0.6{\text{x}}}} - \dfrac{{50}}{{\text{x}}} = \dfrac{{40}}{{60}}$
$
\Rightarrow \dfrac{{\left( {250 - 150} \right)}}{{{\text{3x}}}} = \dfrac{2}{3} \\
\Rightarrow \dfrac{{100}}{{3{\text{x}}}} = \dfrac{2}{3} \\
\Rightarrow 100 = 2{\text{x}} \\
\Rightarrow {\text{x = 50}} \\
$
Therefore 0.6x = 0.6 × 50 = 30.
Hence speed reduced = 50 – 30 = 20.
Given, the train is late by two hours and by this time at a speed = 30, it would have gone a distance,
30 × 2 = 60 miles.
If the train had went at the original speed, the time it would have taken after the first hour is
Time = $\dfrac{{{\text{distance}}}}{{{\text{speed}}}}$=$\dfrac{{60}}{{20}} = 3$, i.e. 3 hours after first hour
Hence total time = 3 + 1 = 4 hours.
Therefore total distance travelled = 4 × 50 = 200 miles.
Hence, the distance between the two stations is 200 miles.
Note – In order to solve this type of question the key is to write the sentences given the question in the form of equations and solve accordingly. Writing all the quantities in terms of one quantity simplifies the equation. While there are more than one values for the variable x after solving, we compare them with the data given in the question to verify.
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