From the total amount available, a man keeps 25,000$ for

There are two conventional algebraic ways to solve these types of problems. In the first, we just introduce an unknown for the amounts each brother received. We then use the fact that a ratio is just a fraction in order to translate each statement in the question into algebra:

If the elder brother initially got $e, and the younger brother initially got $y, then from the ratio given, we know that e/y = 3/2, or 2e = 3y. Further, if the younger brother is given $25,000, he will then have y + 25000 dollars. We know the ratio of e to y+25000 is 2 to 3, so e/(y + 25000) = 2/3, or 3e = 2y + 50000. We now have two equations in two unknowns:

3e = 2y + 50,000
2e = 3y

If we multiply the first equation by 3 and the second equation by 2, we can then subtract the second from the first:

9e = 6y + 150,000
4e = 6y
5e = 150,000

So e = 30,000.

It's faster to still to use a multiplier. If the ratio of the amounts given to the elder and younger brothers is 3 to 2, then for some number x, the elder brother got $3x and the younger brother got $2x. We want to find $3x. Since the ratio of 3x to 2x+25,000 is 2 to 3, we have

3x/(2x + 25,000) = 2/3
9x = 4x + 50,000
5x = 50,000
x = 10,000

And since we wanted to find 3x, the answer is 30,000.

Finally, you can solve this in a kind of conceptual way. If we rewrite each ratio so that the elder brother's amount is the same in each we have:

* before any money is transferred, ratio of elder's $ to younger's $: 6 to 4
* after the money is transferred, ratio of elder's $ to younger's $: 6 to 9

So the $25,000 transfer is equivalent to 5 parts in the ratio (the difference between 9 and 4). Since the amount the elder brother has is equivalent to 6 parts, his amount is $30,000.

And I suppose you could also backsolve the question fairly easily, though if the numbers were different, that could be a very bad approach.