When x is added to each of 13, 19, 16 and 23, then the numbers so obtained, in this order, are in proportion. Then, if $5x : y :: y : (8x-4)$, and $y > 0$, what is the value of y?
The problem states that when a number $x$ is added to each of the numbers 13, 19, 16, and 23, the resulting numbers are in proportion. This means:
$ \frac{13+x}{19+x} = \frac{16+x}{23+x} $To solve for $x$, we cross-multiply:
$ (13+x)(23+x) = (19+x)(16+x) $Expand both sides:
$ 13 \times 23 + 13x + 23x + x^2 = 19 \times 16 + 19x + 16x + x^2 $ $ 299 + 36x + x^2 = 304 + 35x + x^2 $Now, simplify the equation by canceling $x^2$ from both sides and rearranging terms:
$ 299 + 36x = 304 + 35x $ $ 36x - 35x = 304 - 299 $ $ x = 5 $So, the value of $x$ is 5.
We are given a second proportion involving $x$ and $y$: $5x : y :: y : (8x-4)$. We are also told that $y > 0$.
This proportion can be written as:
$ \frac{5x}{y} = \frac{y}{8x-4} $Cross-multiplying gives:
$ y^2 = 5x(8x-4) $Now, substitute the value of $x=5$ into this equation:
$ y^2 = 5(5)(8(5)-4) $ $ y^2 = 25(40-4) $ $ y^2 = 25(36) $To find $y$, we take the square root of both sides:
$ y = \sqrt{25 \times 36} $ $ y = \sqrt{25} \times \sqrt{36} $ $ y = 5 \times 6 $ $ y = 30 $Since the condition is $y > 0$, the value $y=30$ is the correct solution.