Find two positive fractions such that if we switch the denominators, the sum afterthe switch is 50 times the original sum.
To solve this problem, let's start by assigning variables to the fractions. Let's call the first fraction a/b and the second fraction c/d.
According to the problem, if we switch the denominators, the sum after the switch is 50 times the original sum. In other words, (a/d) + (c/b) = 50 * [(a/b) + (c/d)].
To simplify the equation, let's clear the fractions by multiplying both sides by b * d:
(b * d * (a/d) + b * d * (c/b)) = 50 * [(b * d * (a/b)) + (b * d * (c/d))].
This simplifies to:
(a * d) + (c * d) = 50 * (a * d + b * c).
Now, let's solve the equation by trial and error. We can start by assuming one fraction with a denominator of 2 and another fraction with a denominator of 3:
(a * 2) + (c * 3) = 50 * (a * 3 + 2 * c).
Simplifying further:
2a + 3c = 150a + 100c.
Rearranging the equation:
148a = 97c.
Now we need to find positive integers a and c that satisfy this equation. Let's start with a = 97 and c = 148:
148 * 97 = 97 * 148.
It turns out that these values satisfy the equation. Therefore, the fractions that meet the conditions are:
a/b = 97/2 and c/d = 148/3.