# A train covers a distance at a uniform speed . If the train could have been \[{\text{10 kmph}}\] faster , it would have taken two hours less than the scheduled time. And if the train were slower by \[{\text{10 kmph}}\], it would have taken \[{\text{3}}\] hours more than the scheduled time. Find the distance covered by the train.

Last updated date: 27th Mar 2023

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Answer

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Hint – Use the formula of speed and make equations according to the question .

Let the distance covered by the train \[ = d\;km\]

Speed \[ = s\;kmph\]

\[t{\text{ }}\]be the scheduled time

Distance = Speed X Time = constant in this case

\[

st = (s + 10)(t - 2)\; = d{\text{ }}\;{\text{ }}\;{\text{ }}......\left( i \right) \\

st = (s - 10)(t + 3) = d\;{\text{ }}\;{\text{ }}\;{\text{ }}\;.....\left( {ii} \right) \\

\]

Since the distance travelled is same in all the cases

Simplifying the \[\left( i \right)\],

\[

st = st - 2s + 10t - 20 \\

\Rightarrow 2s - 10t = - 20\;{\text{ }}\;{\text{ }}.....\left( {iii} \right) \\

\]

Simplifying the \[\left( {ii} \right)\] we get ,

\[

st = st + 3s + 10t - 30\; \\

\Rightarrow 3s - 10t = 30\;{\text{ }}\;............\left( {iv} \right) \\

\]

Multiplying \[\left( {iii} \right)\] by \[3\;\& {\text{ }}\left( {iv} \right)\] by \[\;2\], we get,

\[

6s - 30t = - 60\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}..........\left( v \right) \\

6s - 20t = 60\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;..........{\text{ }}\left( {vi} \right) \\

\]

Subtracting equation \[\left( {vi} \right)\] from \[\left( v \right)\], we get,

\[

- 10t = - 120 \\

t = 12hrs \\

\]

Putting the value of \[t{\text{ }}\] in equation \[\left( {iii} \right)\],

Then we get speed as,

\[s = 50\,kmph\,\]

And we know Distance = Speed x Time

So,

\[d = st = 50 \times 12 = 600\;km\;\]

Hence the distance is \[600\;km\].

Note – Whenever you struck with this type of problem always try to make equations of various variables according to question , then solve it and get the asked parameter. During solving this problem we kept in mind that the distance travelled is always same whatever the speed may be.

Let the distance covered by the train \[ = d\;km\]

Speed \[ = s\;kmph\]

\[t{\text{ }}\]be the scheduled time

Distance = Speed X Time = constant in this case

\[

st = (s + 10)(t - 2)\; = d{\text{ }}\;{\text{ }}\;{\text{ }}......\left( i \right) \\

st = (s - 10)(t + 3) = d\;{\text{ }}\;{\text{ }}\;{\text{ }}\;.....\left( {ii} \right) \\

\]

Since the distance travelled is same in all the cases

Simplifying the \[\left( i \right)\],

\[

st = st - 2s + 10t - 20 \\

\Rightarrow 2s - 10t = - 20\;{\text{ }}\;{\text{ }}.....\left( {iii} \right) \\

\]

Simplifying the \[\left( {ii} \right)\] we get ,

\[

st = st + 3s + 10t - 30\; \\

\Rightarrow 3s - 10t = 30\;{\text{ }}\;............\left( {iv} \right) \\

\]

Multiplying \[\left( {iii} \right)\] by \[3\;\& {\text{ }}\left( {iv} \right)\] by \[\;2\], we get,

\[

6s - 30t = - 60\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}..........\left( v \right) \\

6s - 20t = 60\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;{\text{ }}\;..........{\text{ }}\left( {vi} \right) \\

\]

Subtracting equation \[\left( {vi} \right)\] from \[\left( v \right)\], we get,

\[

- 10t = - 120 \\

t = 12hrs \\

\]

Putting the value of \[t{\text{ }}\] in equation \[\left( {iii} \right)\],

Then we get speed as,

\[s = 50\,kmph\,\]

And we know Distance = Speed x Time

So,

\[d = st = 50 \times 12 = 600\;km\;\]

Hence the distance is \[600\;km\].

Note – Whenever you struck with this type of problem always try to make equations of various variables according to question , then solve it and get the asked parameter. During solving this problem we kept in mind that the distance travelled is always same whatever the speed may be.

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