Two places P and Q are 162 , km apart. A trai | vedclass

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(D) Let the speed of the train from $Q$ be $x \, km/h$.Since the train from $P$ is $8 \, km/h$ faster,its speed is $(x + 8) \, km/h$.When two objects

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$9 \frac{1}{2}$

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(D) Let the speed of the train from $Q$ be $x \, km/h$.Since the train from $P$ is $8 \, km/h$ faster,its speed is $(x + 8) \, km/h$.When two objects move towards each other,their relative speed is the sum of their individual speeds: $(x + 8 + x) = (2x + 8) \, km/h$.The distance between $P$ and $Q$ is $162 \, km$,and they meet in $6 \, hours$.Using the formula $	ext{Distance} = 	ext{Relative Speed} 	imes 	ext{Time}$:$162 = (2x + 8) 	imes 6$Divide both sides by $6$:$27 = 2x + 8$$2x = 27 - 8$$2x = 19$$x = 9.5 \, km/h$ or $9 \frac{1}{2} \, km/h$.

Two places $P$ and $Q$ are $162 \, km$ apart. $A$ train leaves $P$ for $Q$ and simultaneously another train leaves $Q$ for $P$. They meet at the end of $6 \, hours$. If the former train travels $8 \, km/h$ faster than the other,then the speed of the train from $Q$ is (in $km/h$):

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