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Two small insects, which are x metres apart, take u minutes to pass each other when they are flying towards each other, and v minutes to meet each other when they are flying in the same direction. Then, the ratio of the speed of the slower insect to that of the faster insect is
Let the speed of two small insects be v1 and v2, where v1 > v2.
When they are travelling towards each other, then using the relative motion concept, their speeds will add up, and it takes u minutes to cover x metres distance
=> $$v1+v2=\dfrac{x}{u}$$
When they are travelling in the same direction, then using the relative motion concept, the difference of the speeds will be taken, and it takes v minutes to cover x metres distance
=> $$v1-v2=\dfrac{x}{v}$$
Dividing both equations -
$$\dfrac{v1+v2}{v1-v2}=\dfrac{v}{u}$$
Applying Componendo-Dividendo
$$\dfrac{\left(v1+v2\right)+\left(v1-v2\right)}{\left(v1+v2\right)-\left(v1-v2\right)}=\dfrac{v+u}{v-u}$$
$$\dfrac{2v1}{2v2}=\dfrac{v1}{v2}=\dfrac{v+u}{v-u}$$
The ratio of the speed of the slower insect to the faster insect is = $$\dfrac{v2}{v1}=\dfrac{v-u}{v+u}$$
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