A plane flies between two cites M and P, first from M to P and then f
[#permalink]
08 Feb 2025, 11:01
OE
$$
\({ We know that average speed }=\frac { Total Distance }{ Total Time }\)
$$
The distance between $\(M\)$ and $\(P\)$ is 500 miles and the speed while going from $\(M\)$ to $\(P\)$ is 500 miles per hour and that from $\(P\)$ to $\(M\)$ is 400 miles per hour, so the average speed is
$\(\frac{ Total Distance }{Total Time} =\frac{500+500}{\frac{500}{500}+\frac{500}{400}}=\frac{2}{\frac{4+5}{2000}}=\frac{4000}{9}=444.44 \mathrm{mph}\)$
$($ \(Using - Distance $=$ Speed $\times$ Time $)\)$
Note: - When two same distances are travelled at different speeds say $\(S_1 \& S_2\)$, the average speed for the entire journey is the harmonic mean of the speeds i.e. $\(\frac{2}{\frac{1}{S_1}+\frac{1}{S_2}}=\frac{2\left(\mathrm{~S}_1 \times \mathrm{S}_2\right)}{\mathrm{S}_1+\mathrm{S}_2}\)$
Hence column $\(B\)$ has higher quantity when compared with column $\(A\)$, so the answer is (B).