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Bunuel
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AryaSwagat

meanup
Concept:

\(A\)--------\({t_1}\)----------\(X\)---------\({t_2}\)---------\(B\)

A and B start walking towards each other. After meeting A covers his distance in \(t_1\) time and B covers his distance in \(t_2\) time. The ratio of their speed is;

\(\frac{S_A}{S_B}=\sqrt{\frac{t_2}{t_1}}\)

Based on this concept,

\(\frac{S_A}{S_B}=\sqrt{\frac{16}{25}} = \frac{4}{5}\)

\(S_B=5--->15\)
\(S_B--->3\)

\(S_A=4*3=12\) km/hr

Ans B
­If anyone is curious how this formula is derived.
Let's assume that speed of A is a, Speed of B is b. They meet each other after time 't' has been elapsed after starting.
Total distance is = a*t + b*t 
individually A takes t+16 hours and B takes t+25 hours to complete the total disrance. 
So the total distance (in terms of speed of A) is = a*(t+16) and that by B is = b*(t+25)
Now we have two equations 
a*t + b*t = a*(t+16) = b*(t+25)
solving this we get 16*a^2= 25*b^2. i.e. \(\frac{a}{b}=\sqrt{\frac{{t_2}}{{t_1}}}\)
from here we can get the answer b=12
Ans B­
­Hi,

How do you derive this from the above equation? Much appreciated.
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Let a be P1's speed, b be P2's speed, \(t_0\) be the time it takes to meet each other.
Then,

\(at_0 = 25b\)
\(bt_0 = 16a\)

By cross multiplying, \(16a^2t_0 = 25b^2t_0\). Simplify and solve the rest.­
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