$A$ man starts from a place $P$ and reaches the place $Q$ in $7 \, hours$. He travels $\frac{1}{4}^{th}$ of the distance at $10 \, km/h$ and the remaining distance at $12 \, km/h$. The distance,in $km$,between $P$ and $Q$ is

  • A
    $72$
  • B
    $90$
  • C
    $70$
  • D
    $80$

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$A$ train,after travelling $50 \, km$,meets with an accident and then proceeds at $\frac{3}{4}$ of its former speed and arrives at its destination $35 \, minutes$ late. Had the accident occurred $72 \, km$ further,it would have reached the destination only $15 \, minutes$ late. The normal speed of the train (in $km/hr$) is:

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$A$ and $B$ are $20 \text{ km}$ apart. $A$ can walk at an average speed of $4 \text{ km/hr}$ and $B$ at $6 \text{ km/hr}$. If they start walking towards each other at $7 \text{ a.m.}$,when will they meet? (in $\text{a.m.}$)

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