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sree

asked 2016-08-25 03:22:11

A train left station A for station B at a certain speed. After traveling for $100$ km, the train meets with an accident and could travel at $4/5\text{th}$ of the original speed and reaches $45$ minutes late at station B. Had the accident taken place $50$ km further on, it would have reached $30$ minutes late at station B. What is the distance between station A and station B?

Javed Khan

answered 2016-08-28 18:38:16

Ans: $250$ km

**Solution 1**

See the difference of the two cases.

If $50$ km was travelled with $\dfrac{4}{5}$ of the original speed, $45-30=15$ more minutes was needed.

Travelling with $\dfrac{4}{5}$ of the original speed needs $\dfrac{5}{4}$ of the original time.

$\dfrac{5}{4}-1=\dfrac{1}{4}$

i.e., $\dfrac{1}{4}$ of the original time to travel $50$ km $=15$ minutes

⇒ Original time to travel $50$ km $=60$ minutes $=1$ hour

⇒ Original speed $=50$ km/hr.

After travelling $100$ km, train travels with $\dfrac{4}{5}$ of the speed and hence takes $45$ minutes more.

Travelling with $\dfrac{4}{5}$ of the original speed needs $\dfrac{5}{4}$ of the original time (i.e., $\dfrac{1}{4}$ of the original time in excess).

i.e., After travelling $100$ km, $\dfrac{1}{4}$ of the original time to cover remaining distance is $45$ minutes. (or original time is $45×4=180$ minutes $=3$ hour).

i.e., after travelling $100$ km, train takes $3$ hour for its journey and therefore covers a distance of $3×50=150$ km.

Therefore distance between A and B $=100+150=250$ km.

**Solution 2**

Let original speed $=s$ km/hr

total distance $=d$ km

From first case,

$\dfrac{100}{s}+\dfrac{d-100}{\left(\dfrac{4s}{5}\right)}=\dfrac{d}{s}+\dfrac{45}{60}\\

\Rightarrow \dfrac{100}{s}+\dfrac{5d-500}{4s}=\dfrac{d}{s}+\dfrac{3}{4}\\

\Rightarrow 400+5d-500=4d+3s\\

\Rightarrow d-100=3s\cdots(1)$

$\dfrac{150}{s}+\dfrac{d-150}{\left(\dfrac{4s}{5}\right)}=\dfrac{d}{s}+\dfrac{30}{60}\\

\Rightarrow \dfrac{150}{s}+\dfrac{5d-750}{4s}=\dfrac{d}{s}+\dfrac{1}{2}\\

\Rightarrow 600+5d-750=4d+2s\\

\Rightarrow d-150=2s~~\cdots(2)$

$(1) \implies 2d-200=6s\\

(2) \implies 3d-450=6s$

Therefore

$2d-200=3d-450\\

\Rightarrow d=250$

See the difference of the two cases.

If $50$ km was travelled with $\dfrac{4}{5}$ of the original speed, $45-30=15$ more minutes was needed.

Travelling with $\dfrac{4}{5}$ of the original speed needs $\dfrac{5}{4}$ of the original time.

$\dfrac{5}{4}-1=\dfrac{1}{4}$

i.e., $\dfrac{1}{4}$ of the original time to travel $50$ km $=15$ minutes

⇒ Original time to travel $50$ km $=60$ minutes $=1$ hour

⇒ Original speed $=50$ km/hr.

After travelling $100$ km, train travels with $\dfrac{4}{5}$ of the speed and hence takes $45$ minutes more.

Travelling with $\dfrac{4}{5}$ of the original speed needs $\dfrac{5}{4}$ of the original time (i.e., $\dfrac{1}{4}$ of the original time in excess).

i.e., After travelling $100$ km, $\dfrac{1}{4}$ of the original time to cover remaining distance is $45$ minutes. (or original time is $45×4=180$ minutes $=3$ hour).

i.e., after travelling $100$ km, train takes $3$ hour for its journey and therefore covers a distance of $3×50=150$ km.

Therefore distance between A and B $=100+150=250$ km.

Let original speed $=s$ km/hr

total distance $=d$ km

From first case,

$\dfrac{100}{s}+\dfrac{d-100}{\left(\dfrac{4s}{5}\right)}=\dfrac{d}{s}+\dfrac{45}{60}\\

\Rightarrow \dfrac{100}{s}+\dfrac{5d-500}{4s}=\dfrac{d}{s}+\dfrac{3}{4}\\

\Rightarrow 400+5d-500=4d+3s\\

\Rightarrow d-100=3s\cdots(1)$

$\dfrac{150}{s}+\dfrac{d-150}{\left(\dfrac{4s}{5}\right)}=\dfrac{d}{s}+\dfrac{30}{60}\\

\Rightarrow \dfrac{150}{s}+\dfrac{5d-750}{4s}=\dfrac{d}{s}+\dfrac{1}{2}\\

\Rightarrow 600+5d-750=4d+2s\\

\Rightarrow d-150=2s~~\cdots(2)$

$(1) \implies 2d-200=6s\\

(2) \implies 3d-450=6s$

Therefore

$2d-200=3d-450\\

\Rightarrow d=250$

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