When variables are in the denominator...

In the post "TSD - Time Difference between cities" we came across the equation . Solving such equations theoretically can be slightly painful as most of them boil down to a quadratic equation. As seen in the post, the best way to solve them is by hit and trial, assuming that each of the fractions are going to be integers. One can take cues for the 'hit and trial' from two sources...
1. the denominators in both the fractions should be factors of 3000; and
2. Since the RHS is 11, the LHS should boil down to something of the type 10 + 1 or 9 + 2 or 8 + 3 or so on.

The purists would argue that the fractions need not necessarily turn out to be integers. While they 'need not', usually they do. So don't discount this simple aspect, it can save you a lot of time and trouble.

And such equations are very common in problems on boats and streams. A few of them for you to practice the 'hit and trial'...

1. A motor boat went down the river for 14 km and then up the river for 9 km. It took a total of 5 hours for the entire journey. Find the speed of the river flow if the speed of the boat in still water is 5 kmph.

The equation is a straightforward .

2. A man can cover 50 km upstream and 72 km downstream in 9 hours. He can also row 70 km upstream and 90 km downstream in 12 hours. Find the rate of current.

The equations are . Instead of plugging values for b and r individually, try pluggin values for (b - r) and (b + r).

The last question is a CAT question. Try it out theoretically and see how much of a trouble is avoided by just assuming that the fractions would be integers.