In this chapter we will learn method to solve direct variation problems utilizing the concept of ratio and proportion.
Let us first review the basic concept of direct variation.
What is direct variation?
Direct variation is represented by following expression;
y = k. x
Where y and x are variables and k is constant.
The expression is called direct variable as;
⟹ If value of x increases the value of y will also increase proportionally.
⟹ If value of x decrease, the value of y will also decrease.
Hence, the value of x & y are in direct relationship with each other.
Significance of k in direct variation
In direct variation the value of constant k will always remains the same.
Rewriting the above expression;
y = k. x
\mathtt{\frac{y}{x} \ =\ k}
It means that in direct variation, the value of y & x will always maintain the same ratio.
Even if the value of y & x change, the ratio value k will always be the same.
Let there are two set of values x1, y1 and x2, y2 for direct variation.
As the values maintain the same ratio, the expression can be written as;
\mathtt{\frac{y1}{x1} \ =\ \frac{y2}{x2} \ =\ k}
Remember this expression as it would help us to solve different problems related to direct variation.
I hope the concept is clear, let us now move to solve some problems.
Direct Variation Problem using Proportion method
Example 01
Cost of 6 kg mangoes is 36$. Find the cost of 15kg mangoes.
Solution
It’s a questions of direct variation as the number of mangoes increases the cost will also increase.
Let;
x1 = 6 kg
y1 = 36$
x2 = 15 kg
y2 = ?
Using the direct variation proportion formula;
\mathtt{\frac{y1}{x1} \ =\ \frac{y2}{x2} \ }
Putting the values;
\mathtt{\frac{36}{6} \ =\ \frac{y2}{15}}\\\ \\ \mathtt{y2\ =\ \frac{36\ \times 15}{6}}\\\ \\ \mathtt{y2\ =\ 90\ \$}
Hence, the cost of 15 kg mangoes is 90$.
Example 02
A blue collar worker earns 240$ in 4 days. In how many days will he earn 540$.
Solution
This is a question of direct variation because as number of working days increases, the earning will also increase.
Let;
x1 = 4
y1 = 240
x2 = ?
y2 = 540
Putting the values in direct variation expression.
\mathtt{\ \frac{y1}{x1} \ =\ \frac{y2}{x2} \ }\\\ \\ \mathtt{\frac{240}{4} \ =\ \frac{540}{x2}}\\\ \\ \mathtt{x2\ =\ \frac{540\ \times 4}{240}}\\\ \\ \mathtt{y2\ =\ 9\ days}
Hence, the worker will earn 540$ in 9 days.
Example 03
A car travels 90 kilometers in 15 liter diesel. Find the distance travelled by car on 22 liters diesel.
Solution
Distance travelled and diesel usage are in direct variation. As the diesel consumption increase, the distance travelled by car will also increase.
Let;
x1 = 15
y1 = 90
x2 = 22
y2 = ?
Putting the value in direct variation expression;
\mathtt{\frac{y1}{x1} \ =\ \frac{y2}{x2} \ }\\\ \\ \mathtt{\frac{90}{15} \ =\ \frac{y2}{22}}\\\ \\ \mathtt{y2\ =\ \frac{90\ \times 22}{15}}\\\ \\ \mathtt{y2\ =\ 132}
Hence in 22 liters oil, the car will travel 132 kilometers.
Example 04
A writer can write 4500 words in 3 hours. How many words can he type in 11 hours.
Solution
Words typed and time taken are in direct variation. As the types word increases, the time of work will also increase.
Let;
x1 = 3
y1 = 4500
x2 = 11
y2 = ?
Putting the numbers in direct variation equation.
\mathtt{\frac{y1}{x1} \ =\ \frac{y2}{x2} \ }\\\ \\ \mathtt{\frac{4500}{3} \ =\ \frac{y2}{11}}\\\ \\ \mathtt{y2\ =\ \frac{4500\ \times 11}{3}}\\\ \\ \mathtt{y2\ =\ 16500}
Hence in 11 hours, the writer can type 16500 words.
Example 05
A fan consumes electricity of 900 watt hour in 12 hours. Find the electricity consumed by fan in 36 hours.
Solution
Electricity consumed and time are in direct variation. As the time of usage increase, the electricity consumed will also increase.
Let;
x1 = 12
y1 = 900
x2 = 36
y2 = ?
Putting the values in direct variation ratios.
\mathtt{\frac{y1}{x1} \ =\ \frac{y2}{x2} \ }\\\ \\ \mathtt{\frac{900}{12} \ =\ \frac{y2}{36}}\\\ \\ \mathtt{y2\ =\ \frac{900\ \times 36}{12}}\\\ \\ \mathtt{y2=\ 2700}
Hence in 36 hours, 2700 watt hour of electricity is consumed.