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 yxz. 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.

ALGEBRA
examples of compound proportion

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Exploring examples with answers of compound proportion.

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ALGEBRA
examples of compound proportion

Relevant for

Exploring examples with answers of compound proportion.

See examples

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 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 y=24 when x=18 and z=6, find z when y=12 and x=30.

Step 1: We write the correct equation:

y=kxz

Step 2: We use the information given in the problem to find the value of k. In this case, we have y=24, x=18 and z=6:

24=k(18)(6)

24=108k

 \frac{2}{9}=k

Step 3: We rewrite the equation from step 1 and substitute k=\frac{2}{9}:

y=\frac{2}{9}xz

Step 4: We use the equation found in step 3 and use y=12 and x=30:

12=\frac{2}{9}(30)z

12=\frac{20}{3}z

36=20z

 \frac{9}{5}=z

EXAMPLE 2

If a varies jointly with b and c squared and we have a=225 when b=4 and c=3, find the value of a when b=6 and c=8.

Step 1: We write the correct equation. In this case, we use the variables ab and c and we square c :

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 a=225, b=4 and c=3:

225=k(4)({{3}^2})

225=36k

 \frac{25}{4}=k

Step 3: We rewrite the equation from step 1 and substitute k=\frac{25}{4}:

a=\frac{25}{4}b{{c}^2}

Step 4: We use the equation found in step 3 and substitute b=6 and c=8:

a=\frac{25}{4}(6)({{8}^2})

a=2400

EXAMPLE 3

If m varies jointly with n cube and o and if we have m = 24 when n=8 and o=6, find the value of m when n=4 and o=12.

Step 1: We write the correct equation. In this case, we have the variables mn and o and we have the variable n cubed.

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 m=24, n=8 and o=6:

24=k(8)(6)

24=48k

 \frac{1}{2}=k

Step 3: We rewrite the equation from step 1 and substitute k=\frac{1}{2}:

m=\frac{1}{2}{{n}^3}o

Step 4: We use the equation found in step 3 and substitute n=4 and o=12:

m=\frac{1}{2}({{4}^3})(12)

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.

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:

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 v=48\pi, h=9 and r=4:

48\pi=k(9)({{4}^2})

48\pi=144k

 \frac{\pi}{3}=k

Step 3: We rewrite the equation from step 1 and substitute k=\frac{\pi}{3}:

v=\frac{\pi}{3}h{{r}^2}

Step 4: We use the equation found in step 3 and substitute r=8 and h=6:

v=\frac{\pi}{3}(6)({{8}^2})

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.

Step 1: We write the correct equation. Here, we are going to use the variables em, and v to represent energy, mass, and velocity respectively.:

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 e=300, m=6 and v=10:

300=k(6)(10)

300=600k

 \frac{1}{2}=k

Step 3: We rewrite the equation from step 1 and substitute k=\frac{1}{2}:

e=\frac{1}{2}m{{v}^2}

Step 4: We use the equation found in step 3 and substitute m=5 and v=6:

e=\frac{1}{2}(5)({{6}^2})

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.

If y varies jointly with x and z, and we have y=12, x=9 y z=3, find z when y=6 and x=15.

Choose an answer






If a varies jointly with b cubed and c, and we have a=36 when b=4 and c=6, find a when b=2 and c=14.

Choose an answer






The volume of a square pyramid varies together with the height and the square of one side of the base. A pyramid with a height of 6 and a side of 3 has a volume of 18. What is the volume of the pyramid that has a side of 8 and a height of 9?

Choose an answer







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