31. Percentage and ratio-based calculations problems

31. Percentage and ratio-based calculations problems - step by step solution procedure and detailed explanation for each problem. Give me 10 examples for each variety of problem.

I’ll provide a detailed explanation of percentage and ratio-based calculation problems, including step-by-step solution procedures for each type. I’ll cover 10 percentage problems and 10 ratio problems, each with unique examples, detailed explanations, and clear steps to ensure understanding. The problems will vary in complexity to demonstrate different applications of percentages and ratios.

Percentage-Based Problems

Percentage problems involve calculating a part of a whole, expressed as a fraction of 100. Common types include finding a percentage of a number, calculating percentage increase/decrease, finding what percentage one number is of another, and reverse percentage problems.

General Procedure for Percentage Problems

Understand the problem: Identify what is given (e.g., total amount, percentage, part) and what needs to be found.
Convert percentages: Convert the percentage to a decimal (divide by 100) or fraction for calculations.
Set up the equation: Use the formula or relationship (e.g., Part = Whole × Percentage, or Percentage = Part/Whole × 100).
Solve step-by-step: Perform calculations carefully, ensuring units align.
Verify: Check if the answer makes sense in the context of the problem.

Below are 10 percentage problems with step-by-step solutions and explanations.

Problem 1: Finding a Percentage of a Number

Question: What is 25% of 200?

Solution:

Step 1: Convert the percentage to a decimal: 25% = 25/100 = 0.25.
Step 2: Multiply the decimal by the number: 0.25 × 200 = 50.
Step 3: Verify: 25% is one-fourth, and one-fourth of 200 is 200 ÷ 4 = 50.
Answer: 50.

Explanation: This is a basic percentage calculation where the percentage is applied directly to the whole to find the part. Converting to a decimal simplifies multiplication.

Problem 2: Percentage Increase

Question: A shirt costs $40. Its price increases by 15%. What is the new price?

Solution:

Step 1: Calculate the increase: 15% of 40 = 0.15 × 40 = 6.
Step 2: Add the increase to the original price: 40 + 6 = 46.
Alternative Method: Use the multiplier: 100% + 15% = 115% = 1.15. Then, 1.15 × 40 = 46.
Step 3: Verify: The increase is $6, and 40 + 6 = 46.
Answer: $46.

Explanation: Percentage increase involves finding the additional amount (percentage of the original) and adding it to the original value. The multiplier method is efficient for such problems.

Problem 3: Percentage Decrease

Question: A car’s price is $25,000. It is discounted by 20%. What is the sale price?

Solution:

Step 1: Calculate the discount: 20% of 25,000 = 0.20 × 25,000 = 5,000.
Step 2: Subtract the discount from the original price: 25,000 − 5,000 = 20,000.
Alternative Method: Use the multiplier: 100% − 20% = 80% = 0.80. Then, 0.80 × 25,000 = 20,000.
Step 3: Verify: The discount is $5,000, and 25,000 − 5,000 = 20,000.
Answer: $20,000.

Explanation: Percentage decrease is similar to increase but involves subtraction. The multiplier method (using 100% minus the percentage) is often quicker.

Problem 4: Finding What Percentage One Number Is of Another

Question: A student scored 45 out of 60 on a test. What percentage is this?

Solution:

Step 1: Set up the percentage formula: Percentage = (Part/Whole) × 100.
Step 2: Substitute values: (45/60) × 100.
Step 3: Simplify: 45 ÷ 60 = 0.75, then 0.75 × 100 = 75.
Step 4: Verify: 75% of 60 = 0.75 × 60 = 45, which matches the score.
Answer: 75%.

Explanation: This problem requires expressing the part (score) as a percentage of the whole (total marks). The formula ensures the result is scaled to 100.

Problem 5: Reverse Percentage (Finding the Original Amount)

Question: After a 10% discount, a jacket costs $90. What was the original price?

Solution:

Step 1: The sale price is 90% of the original price (100% − 10% = 90%).
Step 2: Let the original price be ( x ). Then, 0.90( x ) = 90.
Step 3: Solve for ( x ): ( x ) = 90 ÷ 0.90 = 90 × (10/9) = 100.
Step 4: Verify: 10% of 100 = 10, so 100 − 10 = 90, which matches.
Answer: $100.

Explanation: Reverse percentage problems involve working backward from the final amount (after increase or decrease) to find the original. The key is recognizing the final amount represents a percentage of the original.

Problem 6: Percentage of a Percentage

Question: 40% of a class are girls, and 25% of the girls have blue eyes. What percentage of the class are girls with blue eyes?

Solution:

Step 1: Let the class size be 100 for simplicity.
Step 2: Girls = 40% of 100 = 40.
Step 3: Girls with blue eyes = 25% of 40 = 0.25 × 40 = 10.
Step 4: Percentage of the class = (10/100) × 100 = 10%.
Alternative: Multiply percentages: 40% × 25% = 0.40 × 0.25 = 0.10 = 10%.
Step 5: Verify: The calculation is consistent across methods.
Answer: 10%.

Explanation: This involves sequential percentage calculations. Multiplying the percentages (as decimals) directly is a shortcut for finding the combined effect.

Problem 7: Percentage Profit

Question: A phone is bought for $300 and sold for $360. What is the percentage profit?

Solution:

Step 1: Calculate profit: 360 − 300 = 60.
Step 2: Percentage profit = (Profit/Cost Price) × 100 = (60/300) × 100.
Step 3: Simplify: 60 ÷ 300 = 0.2, then 0.2 × 100 = 20.
Step 4: Verify: 20% of 300 = 0.20 × 300 = 60, and 300 + 60 = 360.
Answer: 20%.

Explanation: Percentage profit is calculated relative to the cost price. The formula ensures the profit is expressed as a percentage of the initial investment.

Problem 8: Percentage Loss

Question: A bicycle is bought for $500 and sold for $400. What is the percentage loss?

Solution:

Step 1: Calculate loss: 500 − 400 = 100.
Step 2: Percentage loss = (Loss/Cost Price) × 100 = (100/500) × 100.
Step 3: Simplify: 100 ÷ 500 = 0.2, then 0.2 × 100 = 20.
Step 4: Verify: 20% of 500 = 0.20 × 500 = 100, and 500 − 100 = 400.
Answer: 20%.

Explanation: Similar to percentage profit, percentage loss uses the cost price as the base. The loss is the difference between cost and selling price.

Problem 9: Compound Percentage Increase

Question: A population of 1,000 grows by 10% each year for 2 years. What is the population after 2 years?

Solution:

Step 1: Use the compound growth formula: Final = Initial × (1 + Rate)^Years.
Step 2: Rate = 10% = 0.10, Years = 2, Initial = 1,000.
Step 3: Final = 1,000 × (1 + 0.10)^2 = 1,000 × 1.10^2 = 1,000 × 1.21 = 1,210.
Step 4: Verify: Year 1: 1,000 × 1.10 = 1,100. Year 2: 1,100 × 1.10 = 1,210.
Answer: 1,210.

Explanation: Compound percentage increase accounts for growth on the increased amount each period. The formula avoids repetitive calculations.

Problem 10: Percentage Error

Question: A stick is measured as 105 cm but its actual length is 100 cm. What is the percentage error?

Solution:

Step 1: Calculate error: 105 − 100 = 5.
Step 2: Percentage error = (Error/Actual Value) × 100 = (5/100) × 100.
Step 3: Simplify: 5 ÷ 100 = 0.05, then 0.05 × 100 = 5.
Step 4: Verify: 5% of 100 = 5, and 100 + 5 = 105.
Answer: 5%.

Explanation: Percentage error measures the deviation relative to the actual value, useful in scientific and technical contexts.

Ratio-Based Problems

Ratio problems involve comparing quantities or dividing a total in a given proportion. Common types include simplifying ratios, dividing amounts in a ratio, finding equivalent ratios, and solving problems involving proportional relationships.

General Procedure for Ratio Problems

Understand the problem: Identify the quantities and the ratio relationship.
Simplify the ratio: Reduce to lowest terms if needed.
Set up proportions: Use the ratio to form equations or divide quantities.
Solve step-by-step: Perform calculations, ensuring consistency in units.
Verify: Check that the solution respects the given ratio.

Below are 10 ratio problems with step-by-step solutions and explanations.

Problem 1: Simplifying a Ratio

Question: Simplify the ratio 24:36.

Solution:

Step 1: Find the greatest common divisor (GCD) of 24 and 36: GCD = 12.
Step 2: Divide both terms by the GCD: 24 ÷ 12 = 2, 36 ÷ 12 = 3.
Step 3: Write the simplified ratio: 2:3.
Step 4: Verify: 24/36 = 2/3, which is equivalent.
Answer: 2:3.

Explanation: Simplifying a ratio involves dividing both terms by their GCD to get the smallest equivalent ratio.

Problem 2: Dividing an Amount in a Ratio

Question: Divide $120 between Alice and Bob in the ratio 2:3.

Solution:

Step 1: Total parts = 2 + 3 = 5.
Step 2: Value of one part = 120 ÷ 5 = 24.
Step 3: Alice’s share = 2 parts = 2 × 24 = 48.
Step 4: Bob’s share = 3 parts = 3 × 24 = 72.
Step 5: Verify: 48 + 72 = 120, and 48:72 = 2:3 (simplifies correctly).
Answer: Alice: $48, Bob: $72.

Explanation: Dividing in a ratio requires calculating the value of one “part” and distributing according to the ratio’s terms.

Problem 3: Finding an Equivalent Ratio

Question: Find a ratio equivalent to 5:2 with a total of 35 parts.

Solution:

Step 1: Total parts in 5:2 = 5 + 2 = 7.
Step 2: Let the equivalent ratio be 5k:2k. Total parts = 5k + 2k = 7k.
Step 3: Set up equation: 7k = 35.
Step 4: Solve for k: k = 35 ÷ 7 = 5.
Step 5: Equivalent ratio = 5 × 5 : 2 × 5 = 25:10.
Step 6: Verify: 25 + 10 = 35, and 25:10 simplifies to 5:2.
Answer: 25:10.

Explanation: Equivalent ratios maintain the same proportion. Scaling by a constant (k) adjusts the total parts.

Problem 4: Ratio to Total Amount

Question: The ratio of boys to girls in a class is 3:2. There are 30 students. How many are boys?

Solution:

Step 1: Total parts = 3 + 2 = 5.
Step 2: Value of one part = 30 ÷ 5 = 6.
Step 3: Boys = 3 parts = 3 × 6 = 18.
Step 4: Verify: Girls = 2 parts = 2 × 6 = 12. Total = 18 + 12 = 30, and 18:12 = 3:2.
Answer: 18 boys.

Explanation: The ratio gives the relative proportions, and the total quantity allows calculation of each group’s size.

Problem 5: Combining Ratios

Question: The ratio of red to blue marbles is 2:3, and blue to green is 3:4. Find the ratio of red to green marbles.

Solution:

Step 1: Align the ratios by making the blue terms equal. Red:Blue = 2:3, Blue:Green = 3:4.
Step 2: The blue term is 3 in both ratios, so combine directly.
Step 3: Red:Blue:Green = 2:3:4.
Step 4: Red:Green = 2:4 = 1:2 (simplified).
Step 5: Verify: The ratios are consistent (e.g., Blue:Green = 3:4 holds).
Answer: 1:2.

Explanation: Combining ratios requires a common term (blue) to link the proportions, allowing derivation of the desired ratio.

Problem 6: Ratio with a Difference

Question: The ratio of two numbers is 5:3, and their difference is 16. Find the numbers.

Solution:

Step 1: Let the numbers be 5x and 3x.
Step 2: Set up the difference equation: 5x − 3x = 16.
Step 3: Simplify: 2x = 16.
Step 4: Solve for x: x = 16 ÷ 2 = 8.
Step 5: Numbers are 5 × 8 = 40 and 3 × 8 = 24.
Step 6: Verify: 40:24 = 5:3, and 40 − 24 = 16.
Answer: 40 and 24.

Explanation: The ratio provides a proportional relationship, and the difference gives a concrete value to solve for the scaling factor.

Problem 7: Ratio in Mixtures

Question: A mixture contains water and juice in the ratio 4:1. If there are 20 liters of mixture, how much is juice?

Solution:

Step 1: Total parts = 4 + 1 = 5.
Step 2: Value of one part = 20 ÷ 5 = 4 liters.
Step 3: Juice = 1 part = 1 × 4 = 4 liters.
Step 4: Verify: Water = 4 parts = 4 × 4 = 16 liters. Total = 16 + 4 = 20, and 16:4 = 4:1.
Answer: 4 liters.

Explanation: Ratios in mixtures divide the total quantity into parts, with each component proportional to its ratio term.

Problem 8: Proportional Division with Multiple Ratios

Question: Three friends share $200 in the ratio 1:2:3. How much does each get?

Solution:

Step 1: Total parts = 1 + 2 + 3 = 6.
Step 2: Value of one part = 200 ÷ 6 ≈ 33.33.
Step 3: First friend = 1 × 33.33 ≈ 33.33.
Step 4: Second friend = 2 × 33.33 ≈ 66.67.
Step 5: Third friend = 3 × 33.33 ≈ 100.
Step 6: Verify: 33.33 + 66.67 + 100 = 200, and ratios hold approximately.
Answer: $33.33, $66.67, $100 (rounded).

Explanation: Multiple ratios extend the division process to more parts, with each share proportional to the ratio.

Problem 9: Ratio and Proportion

Question: If 4 workers complete a job in 12 days, how many days will 6 workers take?

grounding context: This is a ratio problem because the work rate is inversely proportional to the number of workers.

Solution:

Step 1: Total work = Worker-days = 4 workers × 12 days = 48 worker-days.
Step 2: With 6 workers, days required = 48 ÷ 6 = 8 days.
Alternative: Use inverse ratio of workers: 4:6 = 2:3. Days ratio is inverse: 3:2. So, (2/3) × 12 = 8.
Step 3: Verify: 6 workers × 8 days = 48 worker-days, matching the work.
Answer: 8 days.

Explanation: This is an inverse proportion problem where the ratio of workers affects the time inversely.

Problem 10: Scaling Ratios

Question: A recipe uses flour and sugar in the ratio 3:2. If 15 cups of flour are used, how much sugar is needed?

Solution:

Step 1: Ratio of flour to sugar = 3:2.
Step 2: Let flour = 3x, sugar = 2x. Given flour = 15 cups.
Step 3: Solve for x: 3x = 15, so x = 15 ÷ 3 = 5.
Step 4: Sugar = 2x = 2 × 5 = 10 cups.
Step 5: Verify: Flour:Sugar = 15:10 = 3:2.
Answer: 10 cups.

Explanation: Scaling a ratio involves finding the multiplier (x) for one quantity and applying it to the other.

Summary

Percentage Problems: Covered finding percentages, increase/decrease, reverse percentages, profit/loss, compound growth, and error calculations. The key is converting percentages to decimals or fractions and using appropriate formulas.
Ratio Problems: Covered simplifying, dividing amounts, equivalent ratios, combining ratios, and proportional relationships. Ratios involve setting up proportions and solving for unknowns, often with a focus on parts of a whole.

Each problem included a step-by-step solution, verification, and explanation to clarify the concept and method.

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31. Percentage and ratio-based calculations problems

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