The ratio of the ages of A and B, four years ago, was 4 ∶ 5. Eight years from now the ratio of the ages of A and B will be 11 ∶ 13. What is the sum of their present ages?
The question asks us to find the sum of the present ages of two people, A and B. We are given information about the ratio of their ages at two different points in time: four years ago and eight years from now.
Let's denote the present age of A as $A$ and the present age of B as $B$. We can translate the information given into algebraic equations:
This gives us the equation: $\frac{A - 4}{B - 4} = \frac{4}{5}$
This gives us the equation: $\frac{A + 8}{B + 8} = \frac{11}{13}$
From the first condition:
$\frac{A - 4}{B - 4} = \frac{4}{5}$
Cross-multiplying gives:
$5(A - 4) = 4(B - 4)$
$5A - 20 = 4B - 16$
Rearranging the terms to form a standard linear equation:
$5A - 4B = 20 - 16$
$5A - 4B = 4$ (Equation 1)
From the second condition:
$\frac{A + 8}{B + 8} = \frac{11}{13}$
Cross-multiplying gives:
$13(A + 8) = 11(B + 8)$
$13A + 104 = 11B + 88$
Rearranging the terms:
$13A - 11B = 88 - 104$
$13A - 11B = -16$ (Equation 2)
Now we need to solve the system of two linear equations:
We can use the method of elimination. Let's eliminate B. Multiply Equation 1 by 11 and Equation 2 by 4:
| $11 \times (5A - 4B = 4)$ | Result: $55A - 44B = 44$ |
| $4 \times (13A - 11B = -16)$ | Result: $52A - 44B = -64$ |
Subtract the second resulting equation from the first:
$(55A - 44B) - (52A - 44B) = 44 - (-64)$
$55A - 52A - 44B + 44B = 44 + 64$
$3A = 108$
Now, solve for A:
$A = \frac{108}{3}$
$A = 36$
Substitute the value of $A$ back into Equation 1 to find B:
$5(36) - 4B = 4$
$180 - 4B = 4$
$180 - 4 = 4B$
$176 = 4B$
$B = \frac{176}{4}$
$B = 44$
The present age of A is 36 years, and the present age of B is 44 years.
The sum of their present ages is:
Sum = $A + B$
Sum = $36 + 44$
Sum = 80 years
Let's check if these ages satisfy the conditions:
The calculated ages are correct.
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