Proportionality allows us to relate two or more quantities. In the case of compound proportionality, we are relating three quantities at the same time. Compound proportionality is represented as y∝xz. If we use the constant k, we can form the equation y = kxz.
Here, we will look at a brief summary on how to solve compound proportion exercises. In addition, we will explore several examples with answers to understand the application of the process.
Summary of compound proportion
We can solve compound proportion problems by following these steps:
Step 1: We write the correct equation. Compound proportion problems are solved using the equation $latex y = kxz$. The variables y, x, and z can be changed to use variables that are more relevant to certain word problems.
In addition, we have to analyze the problem carefully to determine whether we need to modify the compound proportion equation by including squares, cubes, or square roots.
Step 2: We use the information given in the problem to find the value of k, called the constant of proportionality.
Step 3: We rewrite the equation from step 1 and substitute the value of k found in step 2.
Step 4: We use the equation found in step 3 and the rest of the information given in the problem to find the answer to the problem.
Compound proportion – Examples with answers
Use the following compound proportion examples to study the use of the above steps in detail. Each of the examples has a detailed solution that facilitates understanding.
EXAMPLE 1
If y varies jointly with x and z, and if we have $latex y=24$ when $latex x=18$ and $latex z=6$, find z when $latex y=12$ and $latex x=30$.
Solution
Step 1: We write the correct equation:
$latex y=kxz$
Step 2: We use the information given in the problem to find the value of k. In this case, we have $latex y=24$, $latex x=18$ and $latex z=6$:
$latex 24=k(18)(6)$
$latex 24=108k$
$latex \frac{2}{9}=k$
Step 3: We rewrite the equation from step 1 and substitute $latex k=\frac{2}{9}$:
$latex y=\frac{2}{9}xz$
Step 4: We use the equation found in step 3 and use $latex y=12$ and $latex x=30$:
$latex 12=\frac{2}{9}(30)z$
$latex 12=\frac{20}{3}z$
$latex 36=20z$
$latex \frac{9}{5}=z$
EXAMPLE 2
If a varies jointly with b and c squared and we have $latex a=225$ when $latex b=4$ and $latex c=3$, find the value of a when $latex b=6$ and $latex c=8$.
Solution
Step 1: We write the correct equation. In this case, we use the variables a, b and c and we square c :
$latex a=kb{{c}^2}$
Step 2: We use the information given in the problem to find the value of k. In this case, we have $latex a=225$, $latex b=4$ and $latex c=3$:
$latex 225=k(4)({{3}^2})$
$latex 225=36k$
$latex \frac{25}{4}=k$
Step 3: We rewrite the equation from step 1 and substitute $latex k=\frac{25}{4}$:
$latex a=\frac{25}{4}b{{c}^2}$
Step 4: We use the equation found in step 3 and substitute $latex b=6$ and $latex c=8$:
$latex a=\frac{25}{4}(6)({{8}^2})$
$latex a=2400$
EXAMPLE 3
If m varies jointly with n cube and o and if we have $latex m = 24$ when $latex n=8$ and $latex o=6$, find the value of m when $latex n=4$ and $latex o=12$.
Solution
Step 1: We write the correct equation. In this case, we have the variables m, n and o and we have the variable n cubed.
$latex m=k{{n}^3}o$
Step 2: We use the information given in the problem to find the value of k. In this case, we have $latex m=24$, $latex n=8$ and $latex o=6$:
$latex 24=k(8)(6)$
$latex 24=48k$
$latex \frac{1}{2}=k$
Step 3: We rewrite the equation from step 1 and substitute $latex k=\frac{1}{2}$:
$latex m=\frac{1}{2}{{n}^3}o$
Step 4: We use the equation found in step 3 and substitute $latex n=4$ and $latex o=12$:
$latex m=\frac{1}{2}({{4}^3})(12)$
$latex m=384$
EXAMPLE 4
The volume of a cone varies with its height and the square of its radius. A cone with a radius of 4 meters and a height of 9 meters has a volume of 48π cubic meters. Find the volume of a cube that has a radius of 8 meters and a height of 6 meters.
Solution
Step 1: We write the correct equation. In this case, we are going to use the variables v, h, and r to represent the volume, height, and radius respectively:
$latex v=kh{{r}^2}$
Step 2: We use the information given in the problem to find the value of k. In this case, we have $latex v=48\pi$, $latex h=9$ and $latex r=4$:
$latex 48\pi=k(9)({{4}^2})$
$latex 48\pi=144k$
$latex \frac{\pi}{3}=k$
Step 3: We rewrite the equation from step 1 and substitute $latex k=\frac{\pi}{3}$:
$latex v=\frac{\pi}{3}h{{r}^2}$
Step 4: We use the equation found in step 3 and substitute $latex r=8$ and $latex h=6$:
$latex v=\frac{\pi}{3}(6)({{8}^2})$
$latex v=128\pi$
EXAMPLE 5
Kinetic energy varies in conjunction with mass and the square of the velocity. A mass of 6 kilograms and a speed of 10 meters per second has a kinetic energy of 300 Joules. Find the kinetic energy for a mass of 5 kilograms and a speed of 6 meters per second.
Solution
Step 1: We write the correct equation. Here, we are going to use the variables e, m, and v to represent energy, mass, and velocity respectively.:
$latex e=km{{v}^2}$
Step 2: We use the information given in the problem to find the value of k. In this case, we have $latex e=300$, $latex m=6$ and $latex v=10$:
$latex 300=k(6)(10)$
$latex 300=600k$
$latex \frac{1}{2}=k$
Step 3: We rewrite the equation from step 1 and substitute $latex k=\frac{1}{2}$:
$latex e=\frac{1}{2}m{{v}^2}$
Step 4: We use the equation found in step 3 and substitute $latex m=5$ and $latex v=6$:
$latex e=\frac{1}{2}(5)({{6}^2})$
$latex e=90$
Compound proportion – Practice problems
Use the following problems to test your skills and understanding of compound proportion. Select an answer and check it to see if you chose the correct answer. Look at the solved examples above if you have problems with this topic.
See also
Interested in learning more about proportionality? Take a look at these pages:
Jefferson Huera Guzman
Jefferson is the lead author and administrator of Neurochispas.com. The interactive Mathematics and Physics content that I have created has helped many students.
