সমানুপাত ও ব্যস্তানুপাত: ডাকের অঙ্ক

(Videos) Variation and Proportionality Word Problems

by mathman1024

This video by mathman1024 goes over a word problem based on variation.

$\large{\text{Summary}}$

The problem says, Johnny earns $\$594$ when he works twenty-two hours. How much would he earn if he were to work thirty-seven hours?

We know that this is a direct variation problem because it deals with hours worked and pay. If you work more hours, you get paid more.

Once we know the variation we're using we can define our variables. $$P=\text{amount of paycheck}$$ $$h=\text{number of hours}$$

So we can now define our paycheck to be: $$P=kh$$

The goal for these problems is to find the value of $k$ (the constant of proportionality) and use it to solve for the desired value. To find $k$, take the specific example the problem has given. The given problem states Johnny earns $\$594$ when he works twenty-two hours. So we can plug those into our equation. $$\begin{align} P&=kh\\ \color{#8925AE}{(594)}&=k\color{#8925AE}{(22)}\\ \frac{594}{\color{#8925AE}{22}}&=\frac{22k}{\color{#8925AE}{22}}\\ 27&=k \end{align}$$

So what does this mean in terms of this problem? Well, it means that you make $\$27$ per hour. Thus, $k$ is now a known quantity, and our equation becomes: $$P=27h$$

Now we can answer the question they ask, How much would he earn if he were to work thirty-seven hours? $$\begin{align} P&=27h\\ P&=27\color{#8925AE}{(37)}\\ P&=999 \end{align}$$

We state, $$\text{When he works }37\text{ hours}\\ \text{he earns }999\text{ dollars.}$$

Direct Proportionality in Science

Directly proportional variables are used in the sciences. They come up most in the math-y sciences like chemistry and physics (i.e., the sciences that use variables).

That's where we're going to get our examples from today.

If you need a refresher on direct proportionality in general, click here.

There's also a page on inverse proportionality here.

Chemistry: Gas Laws

Chemical gas laws are a great example of proportionality. There are several of these, but they all involve the same $3$ variables:

• $P$ (pressure)
• $V$ (volume)
• $T$ (temperature).

Let's look at Charles' Law, which relates the volume and temperature of a gas.

$$\text{Charles' Law}$$ $$\frac{V_1}{T_1}=\frac{V_2}{T_2}$$ $$\begin{align} V_1 &= \text{ original volume}\\ V_2 &= \text{ new volume}\\ T_1 &= \text{ original temperature}\\ T_2 &= \text{ new temperature}\\ \end{align}$$

In this case, volume and temperature are directly proportional. If volume goes up, temperature does too. The ratio between temperature and volume is constant.

Let's practice.

Remember, no matter what the actual values of our volume and temperature are, the ratio between them will remain the same. This is what it means to be directly proportional.

$$\begin{align} \frac{V_1}{T_1}&=\frac{V_2}{T_2} \text{ (Charles' Law Equation)}\\ \frac{20L}{100K}&=\frac{100L}{T_2} \text{ (Plug in values we know)}\\ (20L)(T_2)&=(100L)(100K) \text{ (Cross-multiply to eliminate fractions)}\\ T_2 &= \frac{10000LK}{20L} \text{ (Solve for } T_2)\\ T_2 &= 500K \end{align}$$

Proportionality: Word Problems

Real quick, before we jump into problems. Here are some links in case you need refreshers on any of the topics involved:

• Direct proportionality: general explanation here and science examples here

• Inverse proportionality: general explanation here

If you're ready to test out some word problems, let's get started!

The best approach to word problems is to translate it from words to numbers.

Here's a hint for translating problems directly proportional variables:

Let's try this method out.

If the apples and days are directly proportional, this ratio will never change. No matter how many apples Anya eats, the ratio of apples to days will always be $3:1$.

Let's take a quick look at how we can translate the word problem into a numerical expression to solve.

Notice how our new ratio of apples to days is still $3:1$.

Now let's take a look at more complicated variation equations and problems.

Example: Jessie wants to buy a brand new pair of ice skates and a new skating suit. Jessie's mom said that Jessie could buy the skates and the suit, but only if he can figure out the following riddle:

We have $$1000$ to spend on ice skates and a skate suit. The cost of the ice skates varies inversely as the square of $5$. The cost of the skating suit varies directly as the cube of the cost of the ice skates. The constant of proportionality is the same for both equations, $5$. What are the costs of the ice skates and the skating suit?

In order to find out the cost of the skating suit, we must first find the cost of the ice skates. Let's define our variables. $$x=\text{cost of ice skates}$$ $$y=\text{cost of the suit}$$ Now, we'll write the equation for the cost of the ice skates from the word problem. $$x=\frac{k}{5^2}$$ $$x=\frac{5}{25}$$ $$x=\frac{1}{5}$$ Now let's tackle the second part of the word problem: The cost of the skating suit varies directly as the cube of the cost of the ice skates. $$y=kx^3$$ $$y=(5)(\frac{1}{5})^3$$ $$y=\frac{1}{25}$$

Now that we've figured out the fractions, we want to figure out what part of the 1000 * \frac{1}{5}$$$$=200$$So, we will spend$$20$on the ice skates. Now, how much will we spend on the skating suit?$$1000 * \frac{1}{25}$$$$=40$$We will spend$$40$ on the skating suit.

As you can see, this question tested your ability to remember the direct and inverse variation equations.

(Video) Algebraic Inverse Proportion

by HEGARTYMATHS

Thies video by HEGARTYMATHS goes over some word problems based on inverse proportionality.

$\large{\text{Summary}}$

Word problems involving proportionality will usually give you one set of points and ask you to find part of a second set given only some of the information. Using the first set of points you solve for the constant $k$. Then, with $k$ and part of the second set, you can find the final part of the second set.

Let's work through the first problem together.

$$\begin{align}&\text{y is inversely proportional to x. If y=10 when x=2, find:}\\ &\text{(i) A Formula for y in terms of x}\\ &\text{(ii) y when x=5}\\ &\text{(iii) x when y=8}\end{align}$$

The first step is to write our basic formula. We know that basic inverse formula has the form $y=k\cdot \frac1{x}$, or $y=\frac{k}{x}$.

We know that $y=10$ when $x=2$. Let's plug those in then solve for $k$. $$\begin{align}y&=\frac{k}{x}\\ 10&=\frac{k}{2}\\ 20&=k \end{align}$$

Our constant, $k$, is $20$. We're going to use $k=20$ and $x=5$ to solve for $y$ in the second part. $$\begin{align}y&=\frac{k}{x}\\ y&=\frac{20}{5}\\ y&=4 \end{align}$$

When $x=5$, $y=4$.

We use the constant $k=20$ to solve the second part as well. $$\begin{align}y&=\frac{k}{x}\\ 8&=\frac{20}{x}\\ 8x&=20\\ x&=\frac{20}{8}\\ x&=\frac52 \text{ or } 2.5\end{align}$$

Here's a graphic with another example!

Solution

First, let's set up the base variation. $$\begin{align}y&=\frac{k}{x}\\ 14&=\frac{k}7\end{align}$$

We can solve this for $k$ to find our constant. $$\begin{align}14&=\frac{k}{7}\\ 14\cdot7 &= k\\ k&=98 \end{align}$$

Now we can use the second set of data to solve. $$\begin{align}y&=\frac{k}{x}\\ 16&=\frac{98}{x}\\ 16x&=98\\ x&=\frac{98}{16}\\ x&=\frac{49}{8} \text{ or }6.125\end{align}$$