y is directly proportional to the square of x - OCR - GCSE Maths - Question 11 - 2018 - Paper 1

Let the initial value of x be denoted as $x_0$. The initial value of y, therefore, will be:

$y_0 = kx_0^2$

An increase of 15% in x can be calculated as:

$x = x_0 + 0.15x_0 = 1.15x_0$

The new value of y after the increase in x will be:

$y = k(1.15x_0)^2 = k(1.3225)x_0^2$

Now, using the values of $y_0$ and $y$, we can find the percentage increase in y:

$ext{Percentage Increase} = \frac{y - y_0}{y_0} imes 100$

Substituting the values, we have:

$\text{Percentage Increase} = \frac{k(1.3225)x_0^2 - kx_0^2}{kx_0^2} \times 100$

This simplifies to:

$\text{Percentage Increase} = (1.3225 - 1) \times 100 = 32.25\%$