Grade 8 Word Problem Set 2 Worksheet Answer Key

Grade 8 Word Problem Set 2 Worksheet Answer Key: Complete Solutions and Study Tips

If you’re looking for a reliable grade 8 word problem set 2 worksheet answer key that not only provides correct answers but also explains the reasoning behind each step, you’ve landed on the right page. Even so, this complete walkthrough walks you through every problem in the typical Grade 8 Word Problem Set 2, offering clear, step‑by‑step solutions, helpful hints, and common pitfalls to avoid. Whether you’re a student needing quick verification, a parent supporting homework, or a teacher preparing supplementary material, this article serves as a one‑stop resource for mastering the concepts covered in this worksheet.

How to Use This Answer Key Effectively

• Start with the problem statement. Read the entire word problem carefully before attempting a solution. Identify the key information, variables, and what is being asked.
• Work through the solution step by step. Each answer below includes the reasoning process, formulas used, and final result. Follow the same order to build confidence.
• Check your work. After you solve a problem on your own, compare your steps with the detailed solution. Highlight any differences and discuss why they matter.
• Practice independently. Use the answer key as a feedback tool, not a shortcut. Attempt problems without looking at the solutions first, then verify your answers later.

Detailed Solutions for Grade 8 Word Problem Set 2

Below are the problems commonly found in a Grade 8 Word Problem Set 2, followed by complete solutions. The problems are grouped by topic to help you focus on specific skills Simple, but easy to overlook..

1. Fractions and Mixed Numbers

Problem: A recipe calls for 3 ½ cups of flour and 1 ¼ cups of sugar. If you want to make half of the recipe, how many cups of flour and sugar do you need in total?

Solution:

1. Convert mixed numbers to improper fractions.
   • Flour: 3 ½ = 7⁄2
   • Sugar: 1 ¼ = 5⁄4
2. Find half of each quantity.
   • Flour: (7⁄2) × ½ = 7⁄4 = 1 ¾ cups
   • Sugar: (5⁄4) × ½ = 5⁄8 cups
3. Add the two amounts.
   • 7⁄4 + 5⁄8 = (14⁄8 + 5⁄8) = 19⁄8 = 2 ³⁄₈ cups

Answer: You need 2 ³⁄₈ cups of flour and sugar combined Surprisingly effective..

2. Decimals and Money

Problem: A store offers a 15 % discount on a $48 jacket. After the discount, a sales tax of 8 % is added. What is the final price of the jacket?

Solution:

1. Calculate the discount amount.
   • Discount = 15 % of $48 = 0.15 × 48 = $7.20
2. Subtract the discount from the original price.
   • Price after discount = $48 – $7.20 = $40.80
3. Compute the sales tax on the discounted price.
   • Tax = 8 % of $40.80 = 0.08 × 40.80 = $3.264
4. Add the tax to the discounted price.
   • Final price = $40.80 + $3.264 = $43. (Rounded to the nearest cent: $43.06)

Answer: The final price of the jacket is $43.06 Small thing, real impact. Took long enough..

3. Percentages – Increase and Decrease

Problem: The population of a town increased from 12,000 to 13,800 over a year. What is the percentage increase?

Solution:

1. Find the change in population.
   • Increase = 13,800 – 12,000 = 1,800
2. Divide the increase by the original population.
   • Ratio = 1,800 ÷ 12,000 = 0.15
3. Convert to a percentage.
   • Percentage increase = 0.15 × 100 = 15 %

Answer: The population increased by 15 % That's the part that actually makes a difference..

4. Geometry – Area of Composite Shapes

Problem: A garden consists of a rectangle (10 m by 6 m) attached to a semicircle whose diameter equals the width of the rectangle. Find the total area of the garden.

Solution:

1. Area of the rectangle.
   • A_rect = length × width = 10 m × 6 m = 60 m²
2. Radius of the semicircle.
   • Diameter = 6 m → radius = 3 m
3. Area of the full circle (πr²) then halve it.
   • A_circle = π × 3² = 9π m²
   • A_semicircle = ½ × 9π = 4.5π m²
4. Total area.
   • A_total = 60 + 4.5π ≈ 60 + 14.137 = 74.14 m² (rounded to two decimal places)

Answer: The garden’s total area is approximately 74.14 m².

5

5. Algebra – Linear Equations in Context

Problem: A plumber charges a flat fee of $75 for a service call plus $60 per hour of labor. If a customer’s bill totals $315, how many hours did the plumber work?

Solution:

1. Define the variable.
   • Let ( h ) = number of hours worked.
2. Set up the equation based on the pricing structure.
   • ( 75 + 60h = 315 )
3. Solve for ( h ).
   • Subtract 75 from both sides: ( 60h = 240 )
   • Divide by 60: ( h = 4 )

Answer: The plumber worked for 4 hours That's the part that actually makes a difference. Surprisingly effective..

6. Ratios and Proportions

Problem: To make a specific shade of green paint, blue and yellow paint must be mixed in a ratio of 2:5. If a painter has 14 liters of blue paint, how many liters of yellow paint are needed to maintain the correct shade, and what is the total volume of green paint produced?

Solution:

1. Set up the proportion for the yellow paint (( y )).
   • ( \frac{2}{5} = \frac{14}{y} )
2. Cross-multiply and solve.
   • ( 2y = 70 )
   • ( y = 35 ) liters of yellow paint.
3. Calculate total volume.
   • Total = Blue + Yellow = ( 14 + 35 = 49 ) liters.

Answer: 35 liters of yellow paint are needed, producing 49 liters of green paint total.

7. Geometry – Volume and Surface Area

Problem: A cylindrical water tank has a radius of 1.5 meters and a height of 4 meters. If the tank is filled to 80% of its capacity, how many liters of water does it hold? (Note: 1 m³ = 1,000 liters)

Solution:

1. Calculate the total volume of the cylinder (( V = \pi r^2 h )).
   • ( V = \pi \times (1.5)^2 \times 4 )
   • ( V = \pi \times 2.25 \times 4 = 9\pi \approx 28.274 \text{ m}^3 )
2. Find 80% of the total volume.
   • ( 0.80 \times 28.274 \approx 22.619 \text{ m}^3 )
3. Convert cubic meters to liters.
   • ( 22.619 \times 1,000 = 22,619 ) liters.

Answer: The tank holds approximately 22,619 liters of water It's one of those things that adds up. Nothing fancy..

8. Statistics – Mean, Median, and Outliers

Problem: A student’s test scores for the semester are: 82, 76, 91, 88, 45, and 85. The teacher allows the lowest score to be dropped. By how many points does the mean (average) increase after the lowest score is removed?

Solution:

1. Original Mean (6 scores):
   • Sum = ( 82 + 76 + 91 + 88 + 45 + 85 = 467 )
   • Mean = ( 467 \div 6 \approx 77.83 )
2. New Mean (5 scores, dropping 45):
   • New Sum = ( 467 - 45 = 422 )
   • New Mean = ( 422 \div 5 = 84.4 )
3. Calculate the increase.
   • Increase = ( 84.4 - 77.83 = 6.57 )

Answer: The mean increases by approximately 6.57 points.

9. Probability – Compound Events

Problem: A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. Two marbles are drawn sequentially without replacement. What is the probability that both marbles are red?

Solution:

1. Probability the first marble is red.
   • ( P(\text{Red}_1) = \frac{5}{10} = \frac{1}{2} )
2. Probability the second marble is red (given the first was red).
   • Remaining: 4 red, 9 total.
   • ( P(\text{Red}_2 | \text{Red}_1) = \frac{4}{9} ) 3