Lesson Objectives

Dear learner;

At the end of the lesson you will be able to:

1. solve linear equation based on the desired concentration and total volume of the solution.
2. Applying algebraic methods to ensure the final mixture of the concentration.
3. Find the precise volumes of the solutions required to produce some concentration of solution.

4.1. Applications of Equations

Applications of Equations: Using equations to model and solve real-world problems, such as calculating costs, determining quantities, predicting outcomes, and analyzing relationships between variables.

Question 1: Mixture Problem

A chemist needs to prepare 500 mL of a 20% alcohol solution. She has a 10% alcohol solution and a 30% alcohol solution. How much of each solution should she mix to obtain the desired solution?

Solution: Let x be the volume of the 10% solution, and 500 − x be the volume of the 30% solution.

The total amount of alcohol in the mixture should be 20% of 500 mL, which is 100 mL.

$0.10x + 0.30(500 − x) = 100$

$0.10x + 150 − 0.30x = 100$

$−0.20x + 150 = 100$

$−0.20x = −50$

$x = 250$

So, she should mix $250 mL$ of the 10% solution with $250 mL$ of the 30% solution.

Question 2: Motion Problem

A car travels from Town A to Town B at an average speed of 60 km/h. On the return trip, due to traffic, the average speed is reduced to 40 km/h. The total travel time is 5 hours. How far is Town B from Town A?

Solution: Let the distance between Town A and Town B be d km.

The time taken to travel from Town A to Town B: $\frac{d}{60}$​

The time taken to travel from Town B to Town A: $\frac{d}{40}$

The total travel time is 5 hours:

$\frac{d}{60} + \frac{d}{40} = 5$

To solve this equation, find a common denominator (120):

$\frac{2d}{120} + \frac{3d}{120} = 5$

$\frac{5d}{120} = 5$

$5d=600$

$d=120$

So, the distance between Town A and Town B is $120 km$.

Question 3: Investment Problem

A person invests a total of $10,000$ birr in two accounts. One account pays 5% annual interest, and the other pays 7% annual interest. If the total interest earned in one year is $620$ birr, how much was invested in each account?

Solution: Let $x$ be the amount invested at 5%, and $10000 – x$ be the amount invested at 7%.

The total interest earned is $620$ birr:

$0.05x + 0.07(10000 – x) = 620$  

$0.05x + 700 − 0.07x = 620$

$− 0.02x + 700 = 620$

$− 0.02x = − 80$ 

$x = 4000$

So, $4,000$ birr was invested at 5%, and $6,000$ birr was invested at 7%.

Question 4: Geometry Problem

The perimeter of a rectangle is $54 cm$. The length is $3 cm$ more than twice the width. Find the dimensions of the rectangle.

Solution: Let $w$ be the width, and $2w + 3$ be the length.

The perimeter of a rectangle is given by:

$2(length + width) = 54$  

$2((2w + 3) + w) = 54$

$2(3w + 3) = 54$

$6w + 6 = 54$

$6w = 48$

$w = 8$

The width is $8 cm$, and the length is: $2(8) + 3 = 16 + 3 = 19 cm$

So, the dimensions of the rectangle are $8 cm$ by $19 cm$.

Question 5: Work Problem

Two workers can complete a job in 6 hours if they work together. If one worker can complete the job alone in 10 hours, how long will it take the second worker to complete the job alone?

Solution: Let the time taken by the second worker to complete the job alone be x hours.

The work rates of the two workers are:

$\frac{1}{10} \quad \text{and} \quad \frac{1}{x}$​

Together, their combined work rate is:

$\frac{1}{10} + \frac{1}{x} = \frac{1}{6}$

Solve for $x$: $\frac{1}{10} + \frac{1}{x} = \frac{1}{6}$

$\frac{x + 10}{10x} = \frac{1}{6}$

$6(x+10) = 10x$

$6x + 60 = 10x$

$60 = 4x$

$x = 15$

So, it will take the second worker 15 hours to complete the job alone.

Managing Transportation Costs: Calculating Budget Surpluses and Deficits Question:

A transport company in Ethiopia needs to adjust its budget for fuel costs. The company finds that the difference between their planned and actual fuel costs for different routes must not exceed a certain limit. If the absolute value of the difference in costs between two routes is $x$, and the company wants this difference to be at most $50$ birr, how can they determine the range of acceptable values for $x$?

Answer: To ensure the difference in costs does not exceed $50$ birr, we set up the following absolute value inequality:
$|x| \leq 50$
This can be rewritten as:
$-50 \leq x \leq 50$
Therefore, the acceptable range for $x$ (the difference in fuel costs) is between $-50$ and $50$ birr.