Ankita travels 14 ,km to her home partly by r | vedclass

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Question

(B) Let the speed of the rickshaw be $x \,km/h$ and the speed of the bus be $y \,km/h$.Time taken to travel $2 \,km$ by rickshaw is $t_1 = \frac{2}{x}

Vedclass · Accepted Answer

$10, 40$

Vedclass · Answer

(B) Let the speed of the rickshaw be $x \,km/h$ and the speed of the bus be $y \,km/h$.Time taken to travel $2 \,km$ by rickshaw is $t_1 = \frac{2}{x} \,h$.Time taken to travel the remaining distance $(14 - 2) = 12 \,km$ by bus is $t_2 = \frac{12}{y} \,h$.According to the first condition: $\frac{2}{x} + \frac{12}{y} = \frac{1}{2} \quad ...(i)$Time taken to travel $4 \,km$ by rickshaw is $t_3 = \frac{4}{x} \,h$.Time taken to travel the remaining distance $(14 - 4) = 10 \,km$ by bus is $t_4 = \frac{10}{y} \,h$.Since she takes $9 \,minutes$ longer,the total time is $\frac{1}{2} + \frac{9}{60} = \frac{1}{2} + \frac{3}{20} = \frac{13}{20} \,h$.According to the second condition: $\frac{4}{x} + \frac{10}{y} = \frac{13}{20} \quad ...(ii)$Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:$2u + 12v = \frac{1}{2} \quad ...(iii)$$4u + 10v = \frac{13}{20} \quad ...(iv)$Multiply equation $(iii)$ by $2$: $4u + 24v = 1 \quad ...(v)$Subtract $(iv)$ from $(v)$:$(4u + 24v) - (4u + 10v) = 1 - \frac{13}{20}$$14v = \frac{7}{20} \Rightarrow v = \frac{7}{20 	imes 14} = \frac{1}{40}$.Thus,$y = 40 \,km/h$.Substitute $v = \frac{1}{40}$ into $(iii)$:$2u + 12(\frac{1}{40}) = \frac{1}{2} \Rightarrow 2u + \frac{3}{10} = \frac{1}{2}$$2u = \frac{1}{2} - \frac{3}{10} = \frac{5-3}{10} = \frac{2}{10} = \frac{1}{5}$$u = \frac{1}{10}$. Thus,$x = 10 \,km/h$.The speed of the rickshaw is $10 \,km/h$ and the speed of the bus is $40 \,km/h$.

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