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.

Illustration Questions

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.

Illustration Questions

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:

image

\(5×20\stackrel{?}{=}4×30\)

\(100\stackrel{?}{=}120\)

\(100\neq120\)

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:

image

\(5×15\stackrel{?}{=}7×10\)

\(75\stackrel{?}{=}70\)

\(75\neq70\)

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:

image

\(8×30\stackrel{?}{=}5×40\)

\(240\stackrel{?}{=}200\)

\(240\neq200\)

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:

image

\(8×30\stackrel{?}{=}5×48\)

\(240\stackrel{?}{=}240\)

\(240=240\)

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 \(9\).

Illustration Questions

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 \(10\).

Illustration Questions

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 \(=$3\)

Illustration Questions

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?

image
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 \($10.0\) 

Illustration Questions

\(8\) 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,

\(72\div8\)

image

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.

Illustration Questions

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

A Toby buys cheaper

B Jose buys cheaper

C Toby and Jose buy at equal rate

D Toby buys costlier

×

For Toby,

Cost of \(15\) kg of rice =? \( $ 39.0\)

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

 

image

\(\therefore\) Cost of 1 kg of rice = $2.6

For Jose, 

Cost of \(11\) kg of rice\(= $\; 33.0\)

Cost of \(1\) kg of rice = \(\dfrac {$\,33.0}{11}=\dfrac {33}{11} \)

image

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

Toby buys cheaper

.

B

Jose buys cheaper

C

Toby and Jose buy at equal rate

D

Toby buys costlier

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 \($1.12\)

Illustration Questions

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

Practice Now