Proportion equations questions with answers

Questions

1. Question 1Find the equation

   $y$ is inversely proportional to the square of $x$. Given that $y = 2$ when $x = 6$, express $y$ in terms of $x$.

   Show the worked solution

   1. Step 1: Write the equation connecting y and x, with a constant k$y = \dfrac{k}{x^2}$
   2. Step 2: Substitute the given values to find k$2 = \dfrac{k}{(6)^2}$$2 = \dfrac{k}{36}$$k = 2 \times 36$$= 72$
   3. Step 3: Put the value of k into the equation$y = \dfrac{72}{x^2}$
2. Question 2Find k, then a value of y

   $y$ is directly proportional to the square of $x$. Given that $y = 12$ when $x = 4$, find the value of $y$ when $x = 2$.

   Show the worked solution

   1. Step 1: Write the equation connecting y and x, with a constant k$y = kx^2$
   2. Step 2: Substitute the given values to find k$12 = k(4)^2$$12 = 16k$$k = \dfrac{12}{16}$$= \dfrac{3}{4}$
   3. Step 3: Substitute the new value of x$y = \dfrac{3}{4}x^2$$y = \dfrac{3}{4}(2)^2$$= 3$
3. Question 3Find k, then a value of x

   $y$ is directly proportional to the square of $x$. Given that $y = 12$ when $x = 2$, find the positive value of $x$ when $y = 75$.

   Show the worked solution

   1. Step 1: Write the equation connecting y and x, with a constant k$y = kx^2$
   2. Step 2: Substitute the given values to find k$12 = k(2)^2$$12 = 4k$$k = \dfrac{12}{4}$$= 3$
   3. Step 3: Substitute the new value of y$y = 3x^2$$75 = 3x^2$$x^2 = \dfrac{75}{3}$$= 25$
   4. Step 4: Take the positive square root$x = \sqrt{25}$$= 5$
4. Question 4The effect of changing x (a factor)

   $y$ is inversely proportional to the square of $x$. When $x$ is multiplied by $4$, $y$ is multiplied by a number. Find this number.

   Show the worked solution

   1. Step 1: Write the equation connecting y and x, with a constant k$y = \dfrac{k}{x^2}$
   2. Step 2: Replace x by the new x$\text{New } y = \dfrac{k}{(4x)^2}$$= \dfrac{1}{16} \times \dfrac{k}{x^2}$
   3. Step 3: Compare the new y with the old y$y$ is multiplied by $\dfrac{1}{16}$.
5. Question 5The percentage change in y

   $y$ is directly proportional to the square of $x$. Find the percentage increase in $y$ when $x$ is increased by $20\%$.

   Show the worked solution

   1. Step 1: Write the equation connecting y and x, with a constant k$y = kx^2$
   2. Step 2: The increase in x: what is x multiplied by?$\dfrac{100 + 20}{100} = 1.2$
   3. Step 3: Replace x by the new x$\text{New } y = k(1.2x)^2$$= 1.44kx^2$
   4. Step 4: Find the percentage increase in y$\text{Percentage increase} = (1.44 - 1) \times 100\%$$= 44\%$
6. Question 6A missing value in a table

   $y$ is directly proportional to the cube of $x$. The table shows two pairs of values of $x$ and $y$. Find the value of $p$.

   $x$	$3$	$p$
   $y$	$108$	$864$

   Show the worked solution

   1. Step 1: Write the equation connecting y and x, with a constant k$y = kx^3$
   2. Step 2: Substitute the given values to find k$108 = k(3)^3$$108 = 27k$$k = \dfrac{108}{27}$$= 4$
   3. Step 3: Substitute the new value of y$y = 4x^3$$864 = 4x^3$$x^3 = \dfrac{864}{4}$$= 216$
   4. Step 4: Take the cube root$x = \sqrt[3]{216}$$= 6$So $p = 6$.
7. Question 7Find k from a change (or a sum) in y

   $y$ is directly proportional to the square root of $x$. The values of $y$ when $x = 25$ and when $x = 100$ add up to $180$. Find the value of $y$ when $x = 16$.

   Show the worked solution

   1. Step 1: Write the equation with a constant k$y = k\sqrt{x}$
   2. Step 2: Write each value of y in terms of kWhen $x = 25$, $y = k\sqrt{25}$$y = 5k$.When $x = 100$, $y = k\sqrt{100}$$y = 10k$.
   3. Step 3: The two values add up to 180: solve for k$5k + 10k = 180$$15k = 180$$k = \dfrac{180}{15}$$= 12$
   4. Step 4: Put x = 16 into the equation$y = 12\sqrt{x}$$y = 12\sqrt{16}$$= 48$