Question 50

A train after travelling 150 km meets with an accident and then proceeds at $$\frac{3}{5}$$ km of its former speed arrives at its destination 8 hours late. Had the accident occurred 360 km further, it would have reached the destination 4 hours late. What is the total distance travelled by the train?


Let initial speed of train be $$5x$$ km/hr and total distance travelled be $$d$$ km

According to ques, => $$\frac{150}{5x}+\frac{(d-150)}{3x}=\frac{d}{5x}+8$$

=> $$\frac{30}{x}-\frac{50}{x}+\frac{d}{3x}-\frac{d}{5x}=8$$

=> $$\frac{2d}{15}-8x=20$$ --------------(i)

Similarly, $$\frac{510}{5x}+\frac{(d-510)}{3x}=\frac{d}{5x}+4$$

=> $$\frac{102}{x}-\frac{170}{x}+\frac{d}{3x}-\frac{d}{5x}=4$$

=> $$\frac{2d}{15}-4x=68$$ --------------(ii)

Subtracting equation (i) from (ii), we get : $$4x=48$$

=> $$x=12$$

Substituting above value in equation (i), => $$\frac{2d}{15}=20+96$$

=> $$d=58\times15=870$$ km

=> Ans - (B)

SHORTCUT METHOD (to find original speed) = $$\frac{360}{3x}-\frac{360}{5x}=4$$ => $$x=12$$

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