A man travels 360 , km in 4 , hrs,partly by a | vedclass

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(B) Let the time spent in the air be $x \, hrs$ and the time spent in the train be $y \, hrs$.Given,$x + y = 4 \, hrs$ (Equation $1$).If he travelled

Vedclass · Accepted Answer

$90$

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(B) Let the time spent in the air be $x \, hrs$ and the time spent in the train be $y \, hrs$.Given,$x + y = 4 \, hrs$ (Equation $1$).If he travelled the entire distance by air,he would have saved $\frac{4}{5}$ of the time spent in the train $(y)$.Given that this time saved is $2 \, hrs$,we have $\frac{4}{5}y = 2$.Solving for $y$: $y = 2 	imes \frac{5}{4} = 2.5 \, hrs$.Substituting $y = 2.5$ into Equation $1$: $x + 2.5 = 4 \Rightarrow x = 1.5 \, hrs$.Let $v_a$ be the speed of the aeroplane and $v_t$ be the speed of the train.The total distance is $v_a x + v_t y = 360$.If he travelled the whole way by air,the time taken would be $\frac{360}{v_a}$.The time saved is $4 - \frac{360}{v_a} = 2 \Rightarrow \frac{360}{v_a} = 2 \Rightarrow v_a = 180 \, km/h$.Distance covered by air $= v_a 	imes x = 180 	imes 1.5 = 270 \, km$.Distance covered by train $= 360 - 270 = 90 \, km$.

$A$ man travels $360 \, km$ in $4 \, hrs$,partly by air and partly by train. If he had travelled all the way by air,he would have saved $\frac{4}{5}$ of the time he was in train and would have arrived at his destination $2 \, hrs$ early. Find the distance he travelled by train (in $km$).

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