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(B) Let the speeds of the two cyclists be $v_1$ and $v_2$ km/h,and the total distance be $D$ km. Let $v_1 > v_2$.They meet after $t = 3$ hours $20$ mi

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) Let the speeds of the two cyclists be $v_1$ and $v_2$ km/h,and the total distance be $D$ km. Let $v_1 > v_2$.They meet after $t = 3$ hours $20$ minutes $= 3 + 1/3 = 10/3$ hours.Thus,$D = (v_1 + v_2) 	imes (10/3)$.The time taken by the first cyclist to cover the distance is $T_1 = D/v_1$ and by the second is $T_2 = D/v_2$.We are given $T_1 - T_2 = 5$ hours.Substituting $D$,we get $(10/3) 	imes (v_1 + v_2) / v_1 - (10/3) 	imes (v_1 + v_2) / v_2 = 5$.Let $x = v_1/v_2$. Then $(10/3) 	imes (x + 1) / x - (10/3) 	imes (x + 1) = 5$ is incorrect; rather,$(10/3)(1 + v_2/v_1) = T_1$ and $(10/3)(1 + v_1/v_2) = T_2$.Using $T_1 = D/v_1$ and $T_2 = D/v_2$,we have $D = v_1 T_1 = v_2 T_2$.From $D = (v_1 + v_2)(10/3)$,we get $v_1/v_2 + 1 = D / (v_2 	imes 10/3) = (3/10) T_2$.Also $v_2/v_1 + 1 = (3/10) T_1$.Since $T_2 - T_1 = 5$,we solve the quadratic equation to find $T_2 = 10$ hours.

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