Answer
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Hint: Assume the distance travelled by the delivery boy for one side of the trip to be D. Then calculate the time \[{{T}_{1}}\] & \[{{T}_{2}}\], where \[{{T}_{1}}\] is the time taken on onward journey & \[{{T}_{2}}\] is the time taken on return journey. Then work on the speed (Return), and calculate the average speed.
Complete step by step solution:D = distance travelled by the bike in each direction.
\[{{t}_{1}}\] = time spent on onward journey.
\[{{t}_{2}}\]= time spent on the return journey.
Thus, the total distance = D + D = 2D
And by the formula,
$Speed=\dfrac{Dist.}{Time}$
Or, \[{{S}_{1}}=\dfrac{{{D}_{1}}}{{{t}_{1}}}\] ; \[{{S}_{2}}=\dfrac{{{D}_{2}}}{{{t}_{2}}}\]
Rearranging,
\[{{t}_{1}}=\dfrac{{{D}_{1}}}{{{S}_{1}}}\] ; \[{{t}_{2}}=\dfrac{{{D}_{2}}}{{{S}_{2}}}\]
Now, we know that,
\[\text{Average Speed=}\dfrac{\text{Total Distance}}{\text{Total Time Taken}}\]
We got, Total Distance = 2D
& Total Time
\[\begin{align}
& =\dfrac{{{D}_{1}}}{{{S}_{1}}}+\dfrac{{{D}_{2}}}{{{S}_{2}}} \\
& =\dfrac{({{S}_{1}}+{{S}_{2}})D}{{{S}_{1}}{{S}_{2}}} \\
\end{align}\] \[\because {{D}_{1}}={{D}_{2}}\]
\[\begin{align}
& =\dfrac{2D}{\dfrac{({{S}_{1}}+{{S}_{2}})D}{{{S}_{1}}{{S}_{2}}}}=\dfrac{2{{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})} \\
& =94mph \\
& =24-\dfrac{12.5}{100}\times 24 \\
& =24-3=21mph \\
& =\dfrac{2\times {{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})} \\
& =2\times \dfrac{21\times 24}{(24+21)} \\
& =2\times \dfrac{21\times 24}{45} \\
& =22.4mph \\
\end{align}\]
Substituting these values in the formula, we get,
\[=\dfrac{2D}{\dfrac{({{S}_{1}}+{{S}_{2}})D}{{{S}_{1}}{{S}_{2}}}}=\dfrac{2{{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})}\] ………….(1)
Try to remember this general formula for a total average speed problem.
Step 2: Now let’s calculate the decreased speed of the bike.
So the original speed was \[=94mph\]
Now let’s calculate the decreased speed,
Decreased Speed = Original Speed – 12.5% of original speed
Putting values,
\[\begin{align}
& =24-\dfrac{12.5}{100}\times 24 \\
& =24-3=21mph \\
\end{align}\]
Step 3:
Putting values in the equation (1),
We get,
\[\begin{align}
& =\dfrac{2\times {{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})} \\
& =2\times \dfrac{21\times 24}{(24+21)} \\
& =2\times \dfrac{21\times 24}{45} \\
& =22.4mph \\
\end{align}\]
So, the average speed of the whole journey is 22.4 mph.
Hence, the correct answer is C.
Note: These questions come in the category where either you solve them using the formula directly, or derive the same formula and solve. SO, it is advised to remember the small formulas that can help you solve your problem easier and faster. Also, make sure to use the right units.
Complete step by step solution:D = distance travelled by the bike in each direction.
\[{{t}_{1}}\] = time spent on onward journey.
\[{{t}_{2}}\]= time spent on the return journey.
Thus, the total distance = D + D = 2D
And by the formula,
$Speed=\dfrac{Dist.}{Time}$
Or, \[{{S}_{1}}=\dfrac{{{D}_{1}}}{{{t}_{1}}}\] ; \[{{S}_{2}}=\dfrac{{{D}_{2}}}{{{t}_{2}}}\]
Rearranging,
\[{{t}_{1}}=\dfrac{{{D}_{1}}}{{{S}_{1}}}\] ; \[{{t}_{2}}=\dfrac{{{D}_{2}}}{{{S}_{2}}}\]
Now, we know that,
\[\text{Average Speed=}\dfrac{\text{Total Distance}}{\text{Total Time Taken}}\]
We got, Total Distance = 2D
& Total Time
\[\begin{align}
& =\dfrac{{{D}_{1}}}{{{S}_{1}}}+\dfrac{{{D}_{2}}}{{{S}_{2}}} \\
& =\dfrac{({{S}_{1}}+{{S}_{2}})D}{{{S}_{1}}{{S}_{2}}} \\
\end{align}\] \[\because {{D}_{1}}={{D}_{2}}\]
\[\begin{align}
& =\dfrac{2D}{\dfrac{({{S}_{1}}+{{S}_{2}})D}{{{S}_{1}}{{S}_{2}}}}=\dfrac{2{{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})} \\
& =94mph \\
& =24-\dfrac{12.5}{100}\times 24 \\
& =24-3=21mph \\
& =\dfrac{2\times {{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})} \\
& =2\times \dfrac{21\times 24}{(24+21)} \\
& =2\times \dfrac{21\times 24}{45} \\
& =22.4mph \\
\end{align}\]
Substituting these values in the formula, we get,
\[=\dfrac{2D}{\dfrac{({{S}_{1}}+{{S}_{2}})D}{{{S}_{1}}{{S}_{2}}}}=\dfrac{2{{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})}\] ………….(1)
Try to remember this general formula for a total average speed problem.
Step 2: Now let’s calculate the decreased speed of the bike.
So the original speed was \[=94mph\]
Now let’s calculate the decreased speed,
Decreased Speed = Original Speed – 12.5% of original speed
Putting values,
\[\begin{align}
& =24-\dfrac{12.5}{100}\times 24 \\
& =24-3=21mph \\
\end{align}\]
Step 3:
Putting values in the equation (1),
We get,
\[\begin{align}
& =\dfrac{2\times {{S}_{1}}{{S}_{2}}}{({{S}_{1}}+{{S}_{2}})} \\
& =2\times \dfrac{21\times 24}{(24+21)} \\
& =2\times \dfrac{21\times 24}{45} \\
& =22.4mph \\
\end{align}\]
So, the average speed of the whole journey is 22.4 mph.
Hence, the correct answer is C.
Note: These questions come in the category where either you solve them using the formula directly, or derive the same formula and solve. SO, it is advised to remember the small formulas that can help you solve your problem easier and faster. Also, make sure to use the right units.
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