A) 50km
B) 300km
C) 600km
D) 12km
Correct Answer: C
Solution :
Let the actual speed of the train be x km/h and actual time taken be y h. |
We know that, |
Distance = Speed \[\times \] Time |
\[\therefore \] Distance = xy km |
According to the question, |
\[xy=\left( x+10 \right)\left( y-2 \right)\] |
\[\Rightarrow \,\,xy=xy-2x+10y-20\] |
\[\Rightarrow \,2x-10y+20=0\] |
\[\Rightarrow \,x-5y+10=0\] |
[dividing both sides by 2] ...(i) |
and \[xy=\left( x-10 \right)\left( y+3 \right)\] |
\[\Rightarrow \,\,xy=xy+3x-10y-30\] |
\[\Rightarrow \,\,\,\,3x-10y-30=0\] (ii) |
On multiplying Eq. (i) by 3 and then subtracting Eq. (ii) from it, we get |
\[3\times \left( x-5y+10 \right)-\left( 3x-10y-30 \right)=0\] |
\[\Rightarrow \,\,\,-5y=-60\] |
\[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,y=12\] |
On putting y = 12 in Eq. (i), we get |
\[x-5\times 12+10=0\] |
\[\Rightarrow \,\,\,\,x-60+10=0\] |
\[\Rightarrow \,\,\,\,x=50\] |
Hence, the distance covered by train |
\[xy=50\times 12=600\,km\] |
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