A is faster than B. A and B each walk 24 , km | vedclass

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(B) Let the speed of $A$ be $v_A$ and the speed of $B$ be $v_B$ in $km/hr$.Given: $v_A + v_B = 7$ and $v_A > v_B$.The time taken by $A$ is $t_A = \fra

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$4$

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(B) Let the speed of $A$ be $v_A$ and the speed of $B$ be $v_B$ in $km/hr$.Given: $v_A + v_B = 7$ and $v_A > v_B$.The time taken by $A$ is $t_A = \frac{24}{v_A}$ and the time taken by $B$ is $t_B = \frac{24}{v_B}$.Given: $t_A + t_B = 14$,so $\frac{24}{v_A} + \frac{24}{v_B} = 14$.$\frac{24(v_A + v_B)}{v_A v_B} = 14$.Substituting $v_A + v_B = 7$: $\frac{24 	imes 7}{v_A v_B} = 14$.$v_A v_B = \frac{24 	imes 7}{14} = 12$.We have $v_A + v_B = 7$ and $v_A v_B = 12$. The quadratic equation $x^2 - 7x + 12 = 0$ gives the speeds.$(x - 4)(x - 3) = 0$,so $x = 4$ or $x = 3$.Since $A$ is faster than $B$,$v_A = 4 \, km/hr$ and $v_B = 3 \, km/hr$.

$A$ is faster than $B$. $A$ and $B$ each walk $24 \, km$. The sum of their speeds is $7 \, km/hr$,and the sum of the time taken by them is $14 \, hours$. Then $A$'s speed is equal to (in $km/hr$)

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