Flight A usually takes 1 , text{hour} more th | vedclass

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(A) Let the speed of Flight $A$ be $v_A$ and the speed of Flight $B$ be $v_B$ in $	ext{km/hr}$.Distance $D = 7200 \, 	ext{km}$.Let $T_A$ and $T_B$ b

Vedclass · Accepted Answer

$800$

Vedclass · Answer

(A) Let the speed of Flight $A$ be $v_A$ and the speed of Flight $B$ be $v_B$ in $	ext{km/hr}$.Distance $D = 7200 \, 	ext{km}$.Let $T_A$ and $T_B$ be the usual times taken by Flight $A$ and Flight $B$ respectively.From the first condition: $T_A = T_B + 1$,where $T_A = \frac{7200}{v_A}$ and $T_B = \frac{7200}{v_B}$.When the speed of Flight $B$ decreases by $\frac{1}{6}$,its new speed $v_B' = v_B(1 - \frac{1}{6}) = \frac{5}{6}v_B$.The new time taken by Flight $B$ is $T_B' = \frac{D}{v_B'} = \frac{7200}{\frac{5}{6}v_B} = \frac{6}{5} 	imes \frac{7200}{v_B} = 1.2 T_B$.From the second condition,$T_B' = T_A + 0.6$ (since $36 \, 	ext{minutes} = 0.6 \, 	ext{hours}$).Substitute $T_A = T_B + 1$ into the equation: $1.2 T_B = (T_B + 1) + 0.6$.$1.2 T_B = T_B + 1.6$.$0.2 T_B = 1.6 \implies T_B = 8 \, 	ext{hours}$.Then $T_A = 8 + 1 = 9 \, 	ext{hours}$.Speed of Flight $A = \frac{7200}{9} = 800 \, 	ext{km/hr}$.

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