Edit:-

Total time

$=\left(\frac{D}{2\times60}\right)+\left(\frac{D}{4\times30}\right)$ $+\left(\frac{D}{4\times10}\right)$

Total time

$=\left(\frac{D}{2\times60}\right)+\left(\frac{D}{4\times30}\right)$ $+\left(\frac{D}{4\times10}\right)$

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18 votes

Best answer

Let the total distance be $D$ then

Avg speed $=\dfrac{D}{\text{total time taken}}\qquad \to (1)$

Total time taken $=\left(\dfrac{D}{2} \times \dfrac{1}{60}\right)+ \left(\dfrac{D}{4}\times \dfrac{1}{30}\right)+ \left(\dfrac{D}{4}\times \dfrac{1}{10}\right) = \dfrac{5D}{120}\qquad \to (2)$

$\text{Avg speed}= D \div(2) =\dfrac{120}{5}=24.$

Correct Answer: $C$

Avg speed $=\dfrac{D}{\text{total time taken}}\qquad \to (1)$

Total time taken $=\left(\dfrac{D}{2} \times \dfrac{1}{60}\right)+ \left(\dfrac{D}{4}\times \dfrac{1}{30}\right)+ \left(\dfrac{D}{4}\times \dfrac{1}{10}\right) = \dfrac{5D}{120}\qquad \to (2)$

$\text{Avg speed}= D \div(2) =\dfrac{120}{5}=24.$

Correct Answer: $C$

0 votes

$\text{we can solve by using weighted average concept,here speed acts like average and time acts like weights}$

$time=\frac{distance}{speed}$

$\text{the ratio of times can by found by multiplying distance ratios and inverse of speeds ratios}$

$distance ~ratio= \frac12:\frac14:\frac14 = 2:1:1$

$inverse~ of ~speed ~ratio= \frac1{60}:\frac1{30}:\frac1{10} = 1:2:6$

$\text{times ratio = (2:1:1)*(1:2:6)=2:2:6=1:1:3} ~and~ speeds ~are~ 60,30,10$

$\text{now average speed=} \frac{1*60+1*30+3*10}{1+1+3}=\frac{120}5=24$

$time=\frac{distance}{speed}$

$\text{the ratio of times can by found by multiplying distance ratios and inverse of speeds ratios}$

$distance ~ratio= \frac12:\frac14:\frac14 = 2:1:1$

$inverse~ of ~speed ~ratio= \frac1{60}:\frac1{30}:\frac1{10} = 1:2:6$

$\text{times ratio = (2:1:1)*(1:2:6)=2:2:6=1:1:3} ~and~ speeds ~are~ 60,30,10$

$\text{now average speed=} \frac{1*60+1*30+3*10}{1+1+3}=\frac{120}5=24$