Question

the difference between present ages of a man and son is 25. If the product of their ages 10 years back was 350 find their present ages​

Answer 1

Step-by-step explanation:

We can either solve this intuitively, or set up an equation and solve algebraically. If we want to do this intuitively, we can simply think about whether or not the product of the ages between father and son is already greater (which we assume it is) than the 116 years.

When we find that it is, we just subtract one from each of of the years. It’s very important to note that we cannot simply subtract one year from only the father, or only the son, they have to be subtracted by the same amount, as time is both linear and unidirectional. So if we take a year off the son’s age, then we must also take a year off the fathers age.

Let’s try doing this and multiplying the ages together. We get 31 * 6 which equals 186. Let’s try subtracting 1 from both the father’s and son’s age again, then multiplying. We get 30 * 5 which equals 150. Repeating the process once more gets us 29 * 4 which equals 116.

We’ve done this process 3 times, subtracting one year from both father and son and multiplying their ages together. We can also subtract the final age of either father/son from their initial ages. Regardless we arrive at 3 years.

Now to go back to the first paragraph, you’ll notice that I mentioned an algebraic method. The equation is (32 - x)(7 - x) = 116. We can have x as the same variable because of the we’re subtracting the same amount from both ages of father and son. If you want to go about solving this problem, just foil the two quantities, and you’ll get 224 - 39x + x^2 = 116. From here, subtract 116 from both sides, so you can solve for x. 108 - 39x + x^2 = 0. From here you can use the quadratic formula, x = (-b +- sqrt(a^2 - 4ac))/2a. You’ll get two solutions, 3, and 36. This means that there are two solutions. We can subtract 3 from both ages, or we can subtract 36 from both equations, but if we subtract 36, then we will get negative ages for both. This is not possible in this scenario and neither father nor son can have negative ages, so 36 is what we call an extraneous solution. Therefor our only valid answer is 3.

I hope that by providing both of these scenarios, you can think intuitively about a mathematical problem to help you possibly save time during a test, or often find quick solutions to problems in an intuitive manner, rather than a mathematically formal manner.