Two numbers are in the ratio $7:11$. If the first number is increased by 10 and the second number is decreased by 20, the ratio becomes $2:3$. Find the sum of the original numbers.

1. Problem Description

Two numbers are in the ratio $7:11$. If the first number is increased by 10 and the second number is decreased by 20, the ratio becomes $2:3$. Find the sum of the original numbers.

2. Solution Steps

Let the two numbers be $7x$ and $11x$.

According to the problem, when the first number is increased by 10 and the second number is decreased by 20, the ratio becomes $2:3$. Therefore, we can write the equation:

$$\frac{7x+10}{11x-20} = \frac{2}{3}$$

Cross-multiply:

$$3(7x+10) = 2(11x-20)$$

$$21x+30 = 22x-40$$

$$22x - 21x = 30 + 40$$

$$x = 70$$

The original numbers are $7x$ and $11x$.

The first number is $7x = 7(70) = 490$.

The second number is $11x = 11(70) = 770$.

The sum of the original numbers is $490 + 770 = 1260$.

3. Final Answer

The sum of the original numbers is 1260.