💡 8th Grade Math (Algebra I): Create and Solve Linear Equations and Inequalities in One Variable with Real-Life Applications Practice Questions
8th Grade Math (Algebra I): Create and Solve Linear Equations and Inequalities in One Variable with Real-Life Applications Practice Questions
Sarah is saving money to buy a new bicycle that costs 250. She already has 70 saved. If she saves 15 each week, how many weeks will it take her to save enough money to buy the bicycle?
Let w represent the number of weeks Sarah needs to save.
- Step 1: Identify the knowns and unknowns.
- Cost of bicycle: 250
- Amount already saved: 70
- Amount saved per week: 15
- Unknown: Number of weeks (w)
- Step 2: Set up an equation to represent the situation.
- Step 3: Solve the equation for w.
- Subtract 70 from both sides:
- Divide both sides by 15:
The total amount saved will be the amount already saved plus the amount saved per week multiplied by the number of weeks. This total must be equal to the cost of the bicycle.
\[ 70 + 15w = 250 \]
\[ 15w = 250 - 70 \]
\[ 15w = 180 \]
\[ w = \frac{180}{15} \]
\[ w = 12 \]
Answer: It will take Sarah 12 weeks to save enough money for the bicycle. 💡
A rectangular garden has a perimeter of 50 meters. The length of the garden is 5 meters more than its width. What are the dimensions (length and width) of the garden?
Let w represent the width of the garden in meters.
Let l represent the length of the garden in meters.
- Step 1: Write down the formula for the perimeter of a rectangle.
- Step 2: Use the given information to set up equations.
- The perimeter is 50 meters:
- The length is 5 meters more than the width:
- Step 3: Substitute the second equation into the first equation.
- Step 4: Solve the equation for w.
- Distribute the 2:
- Combine like terms:
- Subtract 10 from both sides:
- Divide both sides by 4:
- Step 5: Find the length using the value of w.
\[ P = 2l + 2w \]
\[ 50 = 2l + 2w \]
\[ l = w + 5 \]
Replace l in the perimeter equation with (w + 5).
\[ 50 = 2(w + 5) + 2w \]
\[ 50 = 2w + 10 + 2w \]
\[ 50 = 4w + 10 \]
\[ 50 - 10 = 4w \]
\[ 40 = 4w \]
\[ w = \frac{40}{4} \]
\[ w = 10 \]
\[ l = w + 5 \]
\[ l = 10 + 5 \]
\[ l = 15 \]
Answer: The width of the garden is 10 meters and the length is 15 meters. ✅
Maria is planning a party. She has a budget of 300. She wants to spend 120 on decorations and food. She also wants to buy party favors for her guests. Each party favor costs 3. If she wants to have at least 15 guests, what is the maximum number of party favors she can buy?
Let f represent the number of party favors Maria can buy.
- Step 1: Determine the amount of money available for party favors.
- Total budget: 300
- Cost of decorations and food: 120
- Money remaining for party favors: 300 - 120 = 180
- Step 2: Set up an inequality to represent the cost of party favors.
- Step 3: Solve the inequality for f.
- Divide both sides by 3:
- Step 4: Consider the condition about the number of guests.
The cost of party favors is 3 per favor. The total cost must be less than or equal to the money available for favors.
\[ 3f \le 180 \]
\[ f \le \frac{180}{3} \]
\[ f \le 60 \]
Maria wants to have at least 15 guests. This means she needs to buy at least 15 party favors. The inequality \( f \le 60 \) already satisfies this condition, as 60 is greater than 15.
Answer: Maria can buy a maximum of 60 party favors. 👉
A phone plan costs 40 per month and includes 5 GB of data. For any data used over 5 GB, there is an additional charge of 5 per GB. If John's phone bill for a month was 65, how many GB of data did he use in total?
Let d represent the total number of GB of data John used.
- Step 1: Determine the cost incurred from extra data usage.
- Total bill: 65
- Base monthly cost: 40
- Cost from extra data: 65 - 40 = 25
- Step 2: Set up an equation to find the amount of extra data used.
- Step 3: Solve for the amount of extra GB used.
- Divide both sides by 5:
- Step 4: Calculate the total data used.
Each extra GB costs 5.
\[ 5 \times (\text{extra GB}) = 25 \]
\[ \text{extra GB} = \frac{25}{5} \]
\[ \text{extra GB} = 5 \]
John used the initial 5 GB plus the extra 5 GB.
\[ d = 5 \text{ GB} + 5 \text{ GB} \]
\[ d = 10 \text{ GB} \]
Answer: John used a total of 10 GB of data. 💡
Solve the inequality: \( 3x - 7 < 14 \)
We want to isolate the variable x.
- Step 1: Add 7 to both sides of the inequality.
- Step 2: Divide both sides by 3.
\[ 3x - 7 + 7 < 14 + 7 \]
\[ 3x < 21 \]
Since we are dividing by a positive number, the inequality sign remains the same.
\[ \frac{3x}{3} < \frac{21}{3} \]
\[ x < 7 \]
Answer: The solution to the inequality is x < 7. 📌
A baker uses 3 cups of flour for every 2 dozen cookies. If the baker wants to make 10 dozen cookies, how many cups of flour will be needed?
Let f represent the number of cups of flour needed.
Let d represent the number of dozens of cookies.
- Step 1: Set up a proportion based on the given ratio.
- Step 2: Solve the proportion for f.
- Cross-multiply:
- Divide both sides by 2:
The ratio of flour to dozens of cookies is 3 cups : 2 dozen.
\[ \frac{3 \text{ cups}}{2 \text{ dozen}} = \frac{f \text{ cups}}{10 \text{ dozen}} \]
\[ 3 \times 10 = 2 \times f \]
\[ 30 = 2f \]
\[ f = \frac{30}{2} \]
\[ f = 15 \]
Answer: The baker will need 15 cups of flour. ✅
David is buying notebooks and pens. Each notebook costs 2. Each pen costs 1. He wants to buy at least 5 notebooks and at most 10 pens. If he spends a total of 24, how many notebooks and pens could he have bought?
Let n be the number of notebooks and p be the number of pens.
- Step 1: Write down the given information as equations and inequalities.
- Cost equation: \( 2n + 1p = 24 \)
- Notebook constraint: \( n \ge 5 \)
- Pen constraint: \( p \le 10 \)
- Step 2: Use the cost equation and the notebook constraint to find possible values for p.
- If \( n = 5 \): \( p = 24 - 2(5) = 24 - 10 = 14 \). This violates \( p \le 10 \).
- If \( n = 6 \): \( p = 24 - 2(6) = 24 - 12 = 12 \). This violates \( p \le 10 \).
- If \( n = 7 \): \( p = 24 - 2(7) = 24 - 14 = 10 \). This satisfies \( p \le 10 \). So, (n=7, p=10) is a possible solution.
- If \( n = 8 \): \( p = 24 - 2(8) = 24 - 16 = 8 \). This satisfies \( p \le 10 \). So, (n=8, p=8) is a possible solution.
- If \( n = 9 \): \( p = 24 - 2(9) = 24 - 18 = 6 \). This satisfies \( p \le 10 \). So, (n=9, p=6) is a possible solution.
- If \( n = 10 \): \( p = 24 - 2(10) = 24 - 20 = 4 \). This satisfies \( p \le 10 \). So, (n=10, p=4) is a possible solution.
- If \( n = 11 \): \( p = 24 - 2(11) = 24 - 22 = 2 \). This satisfies \( p \le 10 \). So, (n=11, p=2) is a possible solution.
- If \( n = 12 \): \( p = 24 - 2(12) = 24 - 24 = 0 \). This satisfies \( p \le 10 \). So, (n=12, p=0) is a possible solution.
From the cost equation, \( p = 24 - 2n \).
Since \( n \ge 5 \), let's test values for n starting from 5.
Answer: David could have bought combinations like 7 notebooks and 10 pens, or 8 notebooks and 8 pens, or 9 notebooks and 6 pens, or 10 notebooks and 4 pens, or 11 notebooks and 2 pens, or 12 notebooks and 0 pens. 💡
A local charity is holding a fundraising event. They aim to raise at least 5000. They have already received 1500 in donations. They plan to sell raffle tickets for 5 each. What is the minimum number of raffle tickets they need to sell to reach their goal?
Let t represent the number of raffle tickets they need to sell.
- Step 1: Determine the remaining amount of money to be raised.
- Fundraising goal: 5000
- Donations received: 1500
- Amount still needed: 5000 - 1500 = 3500
- Step 2: Set up an inequality to represent the earnings from raffle tickets.
- Step 3: Solve the inequality for t.
- Divide both sides by 5:
Each ticket sells for 5. The total earnings from tickets must be greater than or equal to the amount still needed.
\[ 5t \ge 3500 \]
\[ t \ge \frac{3500}{5} \]
\[ t \ge 700 \]
Answer: They need to sell a minimum of 700 raffle tickets. 📌
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