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

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(B) Let $A$'s speed be $x \, km/hr$.Since the sum of their speeds is $7 \, km/hr$,$B$'s speed is $(7 - x) \, km/hr$.Given that both walk $24 \, km$,th

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$4$

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(B) Let $A$'s speed be $x \, km/hr$.Since the sum of their speeds is $7 \, km/hr$,$B$'s speed is $(7 - x) \, km/hr$.Given that both walk $24 \, km$,the time taken by $A$ is $t_A = \frac{24}{x}$ and the time taken by $B$ is $t_B = \frac{24}{7-x}$.The sum of the times taken is $14 \, hours$,so $\frac{24}{x} + \frac{24}{7-x} = 14$.Dividing by $2$,we get $\frac{12}{x} + \frac{12}{7-x} = 7$.Multiplying by $x(7-x)$,we get $12(7-x) + 12x = 7x(7-x)$.$84 - 12x + 12x = 49x - 7x^2$.$7x^2 - 49x + 84 = 0$.Dividing by $7$,we get $x^2 - 7x + 12 = 0$.Factoring the quadratic equation: $(x - 4)(x - 3) = 0$.Thus,$x = 4$ or $x = 3$.Since $A$ is faster than $B$,$A$'s speed must be $4 \, km/hr$ and $B$'s speed must be $3 \, km/hr$.

$A$ 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 times taken by them is $14 \, hours$. Then $A$'s speed is equal to (in $km/hr$):

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