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(C) Let the distance to point $A$ be $d$ km and the time taken to reach at $2\, p.m.$ be $t$ hours.If speed is $10\, km/hr$,time taken is $t$,so $d =

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$12$

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(C) Let the distance to point $A$ be $d$ km and the time taken to reach at $2\, p.m.$ be $t$ hours.If speed is $10\, km/hr$,time taken is $t$,so $d = 10t$.If speed is $15\, km/hr$,he reaches at $12\, noon$,which is $2$ hours earlier than $2\, p.m.$,so $d = 15(t - 2)$.Equating the two expressions for $d$: $10t = 15(t - 2)$.$10t = 15t - 30$.$5t = 30$,so $t = 6$ hours.The distance $d = 10 	imes 6 = 60\, km$.To reach at $1\, p.m.$,he needs to reach $1$ hour earlier than $2\, p.m.$So,the required time is $t - 1 = 6 - 1 = 5$ hours.Required speed $= \frac{	ext{Distance}}{	ext{Time}} = \frac{60\, km}{5\, hr} = 12\, km/hr$.

Robert is travelling on his cycle and has calculated to reach point $A$ at $2\, p.m.$ if he travels at $10 \,km/hr$; he will reach there at $12\, noon$ if he travels at $15\, km/hr$. At what speed must he travel to reach $A$ at $1\, p.m.$? (in $km/hr$)

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