Informative line

# Representation of Constant of Proportionality (Unit Rate) Through Equation

• Two or more ratios are in proportion if their unit rates or constant of proportionality are same.
• We can represent the proportional relationship by an equation.
• Let the ordered pair of the equation be $$(x,y)$$, the equation will b:

$$y=k\,x$$                         ...(i)

Here, $$k=$$ constant of proportionality

$$k=\dfrac {y}{x}$$

In equation (i), $$x$$ and $$y$$ are the two values/ratios which are in proportion.

Now, we can say that the two values/ratios are said to be in proportion if one will always be the product of the other and a constant.

For example:  The cost of $$4$$ pens is $$12$$ and the cost of $$3$$ pens is $$9$$. If the total cost of pens is proportional to the number of pens purchased at a constant price, write an equation to represent their relationship.

To solve the given problem, we should find the constant of proportionality or the unit rate.

Let $$x$$ be the number of pens and $$y$$ be the total cost.

• When,

Number of pens $$(x)=4$$

Total cost $$(y)=12$$

So,  $$\dfrac {y}{x}=\dfrac {12}{4}=3$$

$$\dfrac {y}{x}=3$$

• When,

Number of pens $$(x)=3$$

Total cost $$(y)=9$$

So, $$\dfrac {y}{x}=\dfrac {9}{3}=3$$

$$\dfrac {y}{x}=3$$

• On observing both the situations, we can conclude:

The cost of pens varies according to the number of pens.

If $$\dfrac {y}{x}=3$$,

$$y=3x$$

Here $$3$$ is the constant of proportionality i.e. the cost is $$3$$ per pen.

#### Alex runs $$4$$ miles in $$2$$ hours. Keith runs $$6$$ miles in $$3$$ hours. The total distance $$(d)$$ is proportional to the total time $$(t)$$ taken at a constant speed. Which equation is expressing the given situation?

A $$t=3d$$

B $$d=2t$$

C $$2d=t$$

D $$3t=2d$$

×

To solve the given problem, we should find the constant of proportionality or the unit rate.

Here, distance is represented by $$d$$.

Time taken is represented by $$t$$.

Distance $$(d)$$ covered by Alex $$=4$$ miles

Time $$(t)$$ taken by Alex = $$2$$ hours

So,

$$\dfrac {d}{t}=\dfrac {4}{2}=2$$

$$\dfrac {d}{t}=2$$

Distance $$(d)$$ covered by Keith $$=6$$ miles

Time $$(t)$$ taken by Keith = $$3$$ hours

So,

$$\dfrac {d}{t}=\dfrac {6}{3}=2$$

$$\dfrac {d}{t}=2$$

On observing both the situation, we can conclude:

Distance covered varies according to the time taken.

If  $$\dfrac {d}{t}=2$$

$$d=2\,t$$

Here, $$2$$ is the constant of proportionality i.e. the distance covered is $$2$$ miles in one hour.

Hence, option (B) is correct.

### Alex runs $$4$$ miles in $$2$$ hours. Keith runs $$6$$ miles in $$3$$ hours. The total distance $$(d)$$ is proportional to the total time $$(t)$$ taken at a constant speed. Which equation is expressing the given situation?

A

$$t=3d$$

.

B

$$d=2t$$

C

$$2d=t$$

D

$$3t=2d$$

Option B is Correct

# Concept of Rates

• There are many times when we hear the word 'rate'.
• Till now, we have not learnt the cocept, rate.
• Let's learn it.

## Rate

• A rate is a quantity that describes a ratio relationship between two types of quantities.

For example:  $$1.75$$ miles/hour is a rate that describes a ratio relationship between hours and miles.

If an object travels at this rate, then after the end of 1st hour, it will cover $$1.75$$ miles, after 2 hours it will cover $$3.50$$ miles, after 3 hours it will cover $$5.25$$ miles, and so on.

• The rate can be identified by the keyword 'per'.

## Equivalent Rates:

When two units of measure, express the same relationship, they are called equivalent rates.

For example: 1 kg of sugar for $$2$$ kg of flour and $$3$$ kg of sugar for $$6$$ kg of flour. Consider the given steps to show them as equivalent rates:

Step 1: Write each rate in fraction form.

$$1$$ kg of sugar for $$2$$ kg of flour $$=\dfrac {1}{2}$$

$$3$$ kg of sugar for $$6$$ kg of flour $$=\dfrac {3}{6}$$

Step 2:  Write the proportion using the $$\stackrel{?}{=}$$ symbol.

$$\dfrac {1}{2}\stackrel{?}{=}\dfrac {3}{6}$$

Step 3: Solve the problem by using cross product.

$$6×1\stackrel{?}{=}3×2$$

$$6\stackrel{?}{=}6$$

Hence, $$\dfrac {1}{2}$$ and $$\dfrac {3}{6}$$ rates are equivalent.

#### Which one of the following options represents the equivalent rates?

A $$5$$ miles in $$30$$ minutes and $$4$$ miles in $$20$$ minutes.

B $$5$$ miles in $$10$$ minutes and $$7$$ miles in $$15$$ minutes.

C $$8$$ miles in $$40$$ minutes and $$5$$ miles in $$30$$ minutes.

D $$8$$ miles in $$48$$ minutes and $$5$$ miles in $$30$$ minutes.

×

For option (A),

Writing the rates in fraction form.

$$5$$ miles in $$30$$ minutes $$=\dfrac {5}{30}$$

$$4$$ miles in $$20$$ minutes $$=\dfrac {4}{20}$$

Writing the proportion using the $$\stackrel{?}{=}$$ symbol.

$$\dfrac {5}{30} \stackrel{?}{=} \dfrac {4}{20}$$

Applying the cross-multiplication method:

$$5×20\stackrel{?}{=}4×30$$

$$100\stackrel{?}{=}120$$

$$100\neq120$$

The cross products are not equal.

So, the given rates are not equivalent.

Hence, option (A) is incorrect.

For option (B)

Writing the rates in fraction form.

$$5$$ miles in $$10$$ minutes $$=\dfrac {5}{10}$$

$$7$$ miles in $$15$$ minutes  $$=\dfrac {7}{15}$$

Writing the proportion using the $$\stackrel{?}{=}$$  symbol.

$$\dfrac {5}{10}\stackrel{?}{=}\dfrac {7}{15}$$

Applying the cross-multiplication method:

$$5×15\stackrel{?}{=}7×10$$

$$75\stackrel{?}{=}70$$

$$75\neq70$$

The cross products are not equal.

So, the given rates are not equivalent.

Hence, option (B) is incorrect.

For option (C),

Writing the rates in fraction form.

$$8$$ miles in $$40$$ minutes $$=\dfrac {8}{40}$$

$$5$$ miles in $$30$$ minutes $$=\dfrac {5}{30}$$

Writing the proportion using the $$\stackrel{?}{=}$$ symbol.

$$\dfrac {8}{40}\stackrel{?}{=}\dfrac {5}{30}$$

Applying the cross multiplication method:

$$8×30\stackrel{?}{=}5×40$$

$$240\stackrel{?}{=}200$$

$$240\neq200$$

The cross products are not equal.

So, the given rates are not equivalent.

Hence, option (C) is incorrect.

For option (D),

Writing the rates in fraction form.

$$8$$ miles in $$48$$ minutes $$=\dfrac {8}{48}$$

$$5$$ miles in $$30$$ minutes $$=\dfrac {5}{30}$$

Writing the proportion using the $$\stackrel{?}{=}$$ symbol.

$$\dfrac {8}{48}\stackrel{?}{=}\dfrac {5}{30}$$

Applying the cross multiplication method:

$$8×30\stackrel{?}{=}5×48$$

$$240\stackrel{?}{=}240$$

$$240=240$$

The cross products are equal.

So, the given rates are equivalent.

Hence, option (D) is correct.

### Which one of the following options represents the equivalent rates?

A

$$5$$ miles in $$30$$ minutes and $$4$$ miles in $$20$$ minutes.

.

B

$$5$$ miles in $$10$$ minutes and $$7$$ miles in $$15$$ minutes.

C

$$8$$ miles in $$40$$ minutes and $$5$$ miles in $$30$$ minutes.

D

$$8$$ miles in $$48$$ minutes and $$5$$ miles in $$30$$ minutes.

Option D is Correct

# Constant of Proportionality

## Unit rate

• When rates are expressed a single quantity, they are called unit rates.

For example: $$1$$ mile distance, $$1$$ pound of vegetables, etc.

• A unit rate is the ratio of two quantities in which the second term is always $$1$$.

Example:

2 miles per hour

$$2$$ miles : $$1$$ hour

$$2:1$$

• The unit rate is also called constant of proportionality.
• The constant ratio between two quantities is called as the constant of proportionality.

For example: $$27.0$$ for $$3$$ dozen oranges and $$18.0$$ for $$2$$ dozen oranges.

Price of $$3$$ dozen oranges $$=27.0$$

$$\because$$ Price of $$1$$ dozen oranges $$=\dfrac {27}{3}=\dfrac {27\div3}{3\div3}$$

$$\therefore$$ Unit Rate $$=\dfrac {9}{1}$$

Here, Unit Rate of the constant of Proportionality equals to $$9$$.

Price of $$2$$ dozen oranges $$=18.0$$

$$\because$$ Price of $$1$$ dozen oranges $$=\dfrac {18.0}{2}=\dfrac {18\div2}{2\div2}$$

$$\therefore$$ Unit Rate $$=\dfrac {9}{1}$$

Here, Unit Rate or constant of proportionality equals to $$9$$.

#### Which represents a unit rate?

A $$3.5$$ miles per $$2$$ hours

B $$4$$ miles per hour

C $$20$$ miles per $$3.5$$

D $$2$$ pizza per $$6.0$$

×

A unit rate is the ratio of two quantities in which the second term is always $$1$$.

In option (A),

$$3.5$$ miles per $$2$$ hours.

$$\Rightarrow\;3.5$$ miles : $$2$$ hours

$$\therefore$$  Ratio $$=3.5:2$$

Here, the second term is not equal to $$1$$.

Hence, option (A) is incorrect.

In option (B),

$$4$$ miles per hour.

$$\Rightarrow\;4$$ miles : Per hour

$$\therefore$$ Ratio =  $$4:1$$

Here, the second term of the ratio is $$1$$.

Hence, option (B) is correct.

In option (C),

$$20$$ miles per $$3.5$$

$$\Rightarrow\;20$$ miles : Per hour $$3.5$$

$$\therefore$$ Ratio =  $$20:3.5$$

Here, the second term is not equal to $$1$$.

Hence, option (C) is incorrect.

In option (D),

$$2$$ Pizza per $$6.0$$

$$\Rightarrow\;2$$ Pizza : $$6.0$$

$$\therefore$$ Ratio =  $$2:6.0$$

Here, the second term is not equal to $$1$$.

Hence, option (D) is incorrect.

### Which represents a unit rate?

A

$$3.5$$ miles per $$2$$ hours

.

B

$$4$$ miles per hour

C

$$20$$ miles per $$3.5$$

D

$$2$$ pizza per $$6.0$$

Option B is Correct

# Solving Proportions Using Tables

• Some problems have tables that show the relation between two quantities and we need to find the constant of proportionality.
• Consider an example to understand these types of problems:

Example:  A table shows the proportional relationship between distance covered and time taken.

 Distance Covered (in miles) Time taken (in hours) 80 2 160 4 200 5 320 8

Find the constant of proportionality in the above table.

To solve this problem, we should calculate the distance for each hour.

For row 1,

Distance covered in $$2$$ hours = $$80$$ miles

Distance covered in  $$1$$ hour = $$\dfrac {80}{2}=\dfrac {80\div2}{2\div2}=40$$ miles

For row 2,

Distance covered in $$4$$ hours = $$160$$ miles

Distance covered in $$1$$ hour = $$\dfrac {160}{4}=\dfrac {160\div4}{4\div4}=40$$ miles

For row 3,

Distance covered in $$5$$ hours = $$200$$ miles

Distance covered in $$1$$ hour = $$\dfrac {200}{5}=\dfrac {200\div5}{5\div5}=40$$ miles

For row 4,

Distance covered in $$8$$ hours = $$320$$ miles

Distance covered in $$1$$ hour = $$\dfrac {320}{8}=\dfrac {320\div8}{8\div8}=40$$ miles

Each row has the same unit rate.

So, the constant of proportionality or unit rate is $$10$$.

#### The table shows the proportional relationship between the number of trays and ice-cubes. Number of trays Number of Ice-cubes 2 12 3 18 4 24 5 30 6 36 Find the constant of proportionality in the given table.

A $$7$$

B $$6$$

C $$3$$

D $$5$$

×

For the given table, we need to calculate the number of ice-cubes in each tray.

For first row,

Number of Ice-cubes in $$2$$ trays $$=12$$

Number of Ice-cubes in $$1$$ tray $$=\dfrac {12}{2}$$

$$\Rightarrow\dfrac {12\div2}{2\div2}=6$$ ice -cubes

For second row,

Number of Ice-cubes in $$3$$ trays $$=18$$

Number of Ice-cubes in $$1$$ tray $$=\dfrac {18}{3}$$

$$\Rightarrow\dfrac {18\div3}{3\div3}=6$$ ice -cubes

For third row,

Number of Ice-cubes in $$4$$ trays $$=24$$

Number of Ice-cubes in $$1$$ tray $$=\dfrac {24}{4}$$

$$\Rightarrow\dfrac {24\div4}{4\div4}=6$$ ice -cubes

For fourth row,

Number of Ice-cubes in $$5$$ trays $$=30$$

Number of Ice-cubes in $$1$$ tray $$=\dfrac {30}{5}$$

$$\Rightarrow\dfrac {30\div5}{5\div5}=6$$ ice -cubes

For fifth row,

Number of Ice-cubes in $$6$$ trays $$=36$$

Number of Ice-cubes in $$1$$ tray $$=\dfrac {36}{6}$$

$$\Rightarrow\dfrac {36\div6}{6\div6}=6$$ ice -cubes

Each row has the same unit rate.

So, the constant of proportionality or unit rate is $$6$$.

Hence, option (B) is correct.

### The table shows the proportional relationship between the number of trays and ice-cubes. Number of trays Number of Ice-cubes 2 12 3 18 4 24 5 30 6 36 Find the constant of proportionality in the given table.

A

$$7$$

.

B

$$6$$

C

$$3$$

D

$$5$$

Option B is Correct

# Solving Proportions Using Graphs

• Some problems have graphs that show the relationship between the quantities, and we need to find the unit rate.
• A unit rate is calculated for a single quantity i.e., $$1$$.
• In a graph, the unit rate refers to the point $$(1,\;r)$$,  i.e.,  $$(x,\;y)=(1,\;r)$$
• On a coordinate plane, a point whose $$x$$-coordinate is $$1$$, the value of $$y$$ coordinate is the unit rate.
• To calculate the unit rate for a given graph, use: $$\dfrac {y}{x}$$

For Example:  Mr. Gibson is enquiring for the cost per grade $$7$$ Maths book in a book store. The graph of cost and number of books is shown:

What is the cost per book?

To solve this problem, we follow the procedure given below.

We know that point $$(1,\;r)$$ represents the unit rate on a coordinate plane.

• In $$(1,\;r)$$:

$$1$$ is at $$x$$-axis and  $$r$$ is at $$y$$ - axis.

• In the given graph, the ordered pairs represent:

$$(2,\;6);2$$ books will cost him $$6$$,

$$(4,\;12);4$$ books will cost him $$12$$,

$$(6,\;18);6$$ books will cost him $$18$$,

and so on.

So, $$(1,\;r)$$ represents that $$1$$ book will cost him $$r$$.

• Unit rate $$=\dfrac {y}{x}$$

Let the point $$(2,\;6)$$, where $$y=6,\;x=2$$

unit rate $$=\dfrac {6}{2}=\dfrac {3}{1}$$

So, $$(1,\;r)=(1,\;3)$$, i.e., $$1$$ book will cost him $$3$$.

Thus, the cost per book $$=3$$

#### The given graph shows the proportional relationship between the number of color pages xeroxed and the total price​​ (¢)​ for pages at a xerox shop. What is the unit price rate at the xerox shop?

A $$¢\,2.5$$

B $$¢\;5.4$$

C $$¢\,2.1$$

D $$¢\,3.2$$

×

We know that the point $$(1,\;r)$$ represents the unit rate on a coordinate plane.

In $$(1,\;r)$$:

$$1$$ is at x-axis and $$r$$ is at y-axis.

In the given graph the ordered pairs represent:

$$(2,\;5);$$ The cost of 2 color pages xeroxed is ¢ $$5$$,

$$(4,\;10);$$ The cost of 4 color pages xeroxed is  ¢ $$10$$,

$$(6,\;15);$$ The cost of 6 color pages xeroxed is ¢ $$15$$,

and so on.

So, $$(1,\;r)$$ represents that cost of $$1$$color page xeroxed is ¢$$r$$.

Thus, unit rate $$=\dfrac {y}{x}$$

For the point $$(2,\;5)$$, where $$y=5,\;x=2$$,

Unit rate $$=\dfrac {5}{2}=\dfrac {2.5}{1}$$

So, $$(1,\;r)=(1,\;2.5)$$

i.e., cost of $$1$$ color page xeroxed is $$2.5$$.

Thus, the cost per color page xerox$$=¢2.5$$

Hence, option (A) is correct.

### The given graph shows the proportional relationship between the number of color pages xeroxed and the total price​​ (¢)​ for pages at a xerox shop. What is the unit price rate at the xerox shop?

A

$$¢\,2.5$$

.

B

$$¢\;5.4$$

C

$$¢\,2.1$$

D

$$¢\,3.2$$

Option A is Correct

# Calculation of Unit Rates

• A unit rate is the ratio of two quantities in which the second term is always $$1$$.
• We can find the unit rate by any of the two methods:
• By making the denominator $$1$$.
• Dividing the first term by the second term.
• Consider an example to understand both the methods:

Example: James spends $$20.0$$ for buying $$2$$ large sized cheese pizzas. What is the cost of each Pizza?

Method $$I$$By making the denominator $$1$$:

James spends $$20.0$$ for $$2$$ Pizzas.

Step 1: Write the ratio in fraction form, i.e. $$20.0$$ for $$2$$ Pizzas $$=\dfrac {20.0}{2}=\dfrac {20}{2}$$

Step 2: Make the denominator $$1$$.

$$\dfrac {20}{2}=\dfrac {20\div2}{2\div2}=\dfrac {10}{1}$$

Thus, the cost of each Pizza is $$10.0$$

Method $$II$$: Dividing the first term by the second term:

James spends $$20.0$$ for $$2$$ pizzas.

Step 1: Write the ratio in fraction form, i.e. $$20.0$$ for $$2$$ Pizzas $$=\dfrac {20.0}{2}=\dfrac {20}{2}$$

Step 2: Divide the numerator by the denominator.

$$20\div2$$

Thus, the cost of each pizza is $$10.0$$

#### $$8$$ boxes can hold $$72$$ books. How many books can each box hold?

A $$6$$

B $$3$$

C $$9$$

D $$7$$

×

According to the statement, $$8$$ boxes can hold $$72$$ books.

Writing the ratio in fraction form i.e.

$$\dfrac {72\,\text {books}}{8\,\text{boxes}}=\dfrac {72}{8}$$

Dividing the numerator by the denominator,

$$72\div8$$

Thus, each box can hold $$9$$ books.

Hence, option (C) is correct.

### $$8$$ boxes can hold $$72$$ books. How many books can each box hold?

A

$$6$$

.

B

$$3$$

C

$$9$$

D

$$7$$

Option C is Correct

# Comparison of Two Quantities by using Unit Rates

When rates are expressed for a single quantity, they are called unit rates.

For example : 1 mile distance, 1 pound of vegetables etc.

Now we can easily understand the comparison of two quantities using unit rates.

Let us consider an example:

Kara buys $$4$$ liters of milk for $$4.80$$ . If Marc buys $$5$$ liters of milk for $$5.75$$, who buys cheaper?

In this example, we need to compare the cost and quantity of milk and find the unit rate.

For Kara,

Cost of $$4$$ liters milk = $$4.80$$

Cost of 1 liter milk $$=\dfrac {4.80}{4}=\dfrac {4.80}{4}$$

$$=1.20$$

Dividing by ignoring the decimals

For Marc,

Cost of $$5$$ liters milk $$= 5.75$$

Cost of $$1$$ liter milk $$=\dfrac {5.75}{5}=\dfrac {5.75}{5}$$

Dividing by ignoring the decimal

$$=1.15$$

$$\because\;1.15<1.20$$

$$\therefore$$ Marc pays less than Kara for 1 liter of milk.

Thus, Marc buys cheaper than Kara.

#### Toby buys $$15$$ kg of rice for $$39.0$$ and Jose buys $$11$$ kg of rice for $$33.0$$ . Who buys cheaper?

C Toby and Jose buy at equal rate

×

For Toby,

Cost of $$15$$ kg of rice =? $$39.0$$

Cost of $$1$$ kg of rice = $$\dfrac {39.0}{15}=\dfrac {39}{15}$$

$$\therefore$$ Cost of 1 kg of rice = $2.6 For Jose, Cost of $$11$$ kg of rice$$= \; 33.0$$ Cost of $$1$$ kg of rice = $$\dfrac {\,33.0}{11}=\dfrac {33}{11}$$ Cost of 1 kg of rice =$3.0.

$$\because\;\;2.6<3.0$$

$$\therefore$$  Toby pays less than Jose for $$1$$ kg of rice.

Thus, Toby buys cheaper than Jose.

Hence, option (A) is correct.

### Toby buys $$15$$ kg of rice for $$39.0$$ and Jose buys $$11$$ kg of rice for $$33.0$$ . Who buys cheaper?

A

.

B

C

Toby and Jose buy at equal rate

D

Option A is Correct

# Real-World Problems Involving Proportions

To understand the real world problems involving proportions, consider an example:

Example: Mr. Alex wants to make a cheese pizza. He needs to buy an $$8$$ounce pack of cheese. If the cost of a $$12$$ ounce pack of cheese is $$3.36$$,

(i) How much will he pay for an $$8$$ounce pack?

(ii) How much does the cost differ between the two packs?

Step 1: To solve this problem, we should calculate the unit rate of $$12$$ ounce pack of cheese.

Cost of $$12$$ ounce pack of cheese $$=3.36$$

Cost of $$1$$ ounce pack of cheese $$=\dfrac {3.36}{12}$$

Step 2: To find the unit rate, make the denominator $$1$$, i.e.

$$\dfrac {3.36}{12}=\dfrac {3.36\div12}{12\div12}$$

The unit rate $$=\dfrac {0.28}{1}$$

$$=0.28$$ /ounce

Step 3:

Thus, the price of $$8$$-ounce pack of cheese $$=0.28×8$$

$$=2.24$$

$$\because$$ Difference between the two packs of cheese $$=3.36-2.24$$

$$=1.12$$

The result is:

(i) He pays $$2.24$$ for $$8$$-ounce pack of cheese.

(ii) The cost differs by $$1.12$$

#### In a building of $$8$$ floors, there are $$32$$ flats in total. The owner calls the painter to paint all the flats till $$5^{th}$$ floor, find: (a) how many flats get painted? (b) how many flats remain unpainted?

A $$20,\;4$$

B $$10,\,12$$

C $$20,\;10$$

D $$20,\;12$$

×

To solve the problem, we should calculate the number of flats painted in each floor.

Number of flats in $$8$$ floors $$=32$$

Number of flats in each floor $$=\dfrac {32}{8}$$

Finding the unit rate by making the denominator $$1$$, i.e.

$$\Rightarrow \dfrac {32}{8}=\dfrac {32\div8}{8\div8}=\dfrac {4}{1}$$

The number of flats in each floor $$=\dfrac {4}{1}$$

$$=4$$

(a) Number of painted flats in $$5$$ floors $$=4×5=20$$

(b) Number of flats remain unpainted

$$32$$ flats – $$20$$ flats

$$=12$$

Hence, option (D) is correct.

### In a building of $$8$$ floors, there are $$32$$ flats in total. The owner calls the painter to paint all the flats till $$5^{th}$$ floor, find: (a) how many flats get painted? (b) how many flats remain unpainted?

A

$$20,\;4$$

.

B

$$10,\,12$$

C

$$20,\;10$$

D

$$20,\;12$$

Option D is Correct