\(y=k\,x\) ...(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.
Number of pens \((x)=4\)
Total cost \((y)=$12\)
So, \(\dfrac {y}{x}=\dfrac {12}{4}=3\)
\(\dfrac {y}{x}=3\)
Number of pens \((x)=3\)
Total cost \((y)=$9\)
So, \(\dfrac {y}{x}=\dfrac {9}{3}=3\)
\(\dfrac {y}{x}=3\)
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.
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 1^{st }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.
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.
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 example: \(1\) mile distance, \(1\) pound of vegetables, etc.
Example:
2 miles per hour
\(2\) miles : \(1\) hour
\(2:1\)
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 \(9\).
A \(3.5\) miles per \(2\) hours
B \(4\) miles per hour
C \(20\) miles per \(3.5\)
D \(2\) pizza per \($6.0\)
Example: A table shows the proportional relationship between distance covered and time taken.
Distance Covered |
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 \(10\).
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.
\(1\) is at \(x\)-axis and \(r\) is at \(y\) - axis.
\((2,\;6);2\) 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 \($r\).
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 \(=$3\)
A \(¢\,2.5\)
B \(¢\;5.4\)
C \(¢\,2.1\)
D \(¢\,3.2\)
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 \($10.0\)
A \(6\)
B \(3\)
C \(9\)
D \(7\)
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.
A Toby buys cheaper
B Jose buys cheaper
C Toby and Jose buy at equal rate
D Toby buys costlier
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 \($1.12\)
A \(20,\;4\)
B \(10,\,12\)
C \(20,\;10\)
D \(20,\;12\)