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$speed=timedistanceβ$

Let the time taken from X to Y at a speed of $30Βkm/hr$ be $t_{1}$

Therefore,Β

$β30=t_{1}dβ$

$βd=30Βt_{1}$

And,Β Let the time taken from Y to X at a speed of $20Βkm/hr$ be $t_{2}$

Therefore,Β

$β20=t_{2}dβ$

$βd=20Β(t_{2})$

Therefore, $30Β(t_{1})=20Β(t_{2})$

$βt_{2}t_{1}β=3020β$

$βt_{1}=32βt_{2}$Β Β ..............(i)

Now, the total time taken is $10Βhrs$

Therefore, $10=t_{1}+t_{2}$Β Β .................(ii)

Substituting equation i in ii, we get

$β10=32βt_{2}+t_{2}$

$β5Βt_{2}=30$

$βt_{2}=6$

Substituting $t_{2}=6$ in $20=t_{2}dβ$ we get

$20=6dβ$

$d=120Βkm$

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