Seeta and Geeta cycled towards each other,one | vedclass

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(B) Let the speeds of Seeta and Geeta be $v_S$ and $v_G$ respectively. Let $t$ be the time taken by Geeta to reach the meeting point $M$ from $B$. The

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$6$

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(B) Let the speeds of Seeta and Geeta be $v_S$ and $v_G$ respectively. Let $t$ be the time taken by Geeta to reach the meeting point $M$ from $B$. Then Seeta takes $(t-6)$ hours to reach $M$ from $A$.Let $d_S$ and $d_G$ be the distances travelled by Seeta and Geeta respectively until they meet at $M$. We have $d_S = v_S(t-6)$ and $d_G = v_G t$.Given $d_G - d_S = 12$,so $v_G t - v_S(t-6) = 12$.After meeting,Seeta travels the remaining distance to $B$ in $8$ hours,so $d_G = v_S 	imes 8$. Geeta travels the remaining distance to $A$ in $9$ hours,so $d_S = v_G 	imes 9$.Substituting these into the distance equations: $v_S(t-6) = 9 v_G$ and $v_G t = 8 v_S$.From $v_G t = 8 v_S$,we get $v_S/v_G = t/8$. Also from $v_S(t-6) = 9 v_G$,we get $v_S/v_G = 9/(t-6)$.Equating the two: $t/8 = 9/(t-6) \Rightarrow t(t-6) = 72 \Rightarrow t^2 - 6t - 72 = 0$.Solving the quadratic: $(t-12)(t+6) = 0$. Since $t > 0$,$t = 12$.Then $v_S/v_G = 12/8 = 3/2$. Let $v_S = 3k$ and $v_G = 2k$.Substitute into $v_G t - v_S(t-6) = 12$: $(2k)(12) - (3k)(12-6) = 12 \Rightarrow 24k - 18k = 12 \Rightarrow 6k = 12 \Rightarrow k = 2$.Thus,$v_S = 3(2) = 6 \, km/hr$ and $v_G = 2(2) = 4 \, km/hr$.The speed of the faster cyclist is $6 \, km/hr$.

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