27. Successive discounts problems

27. Successive discounts problems - step by step solution procedure and detailed explanation for each problem.    

Successive discounts involve applying multiple discounts one after another on an original price, where each discount is calculated on the reduced price after the previous discount. The key is to compute the price step-by-step or use the equivalent single discount formula for efficiency. Below, I’ll provide 10 successive discount problems, each with a step-by-step solution, detailed explanation, and, where applicable, the use of the equivalent discount formula. The problems will vary in complexity to cover different scenarios.
General Approach to Successive Discounts
  1. Step-by-Step Method: Apply each discount sequentially. After each discount, calculate the new price by multiplying the current price by
    (1 - \frac{\text{discount percentage}}{100})
    .
  2. Equivalent Discount Formula: For two successive discounts
    d_1\%
    and
    d_2\%
    , the single equivalent discount is:
    \text{Equivalent Discount} = d_1 + d_2 - \frac{d_1 \cdot d_2}{100}
    For more than two discounts, apply the formula iteratively or compute the final price directly.
  3. Final Price Formula: If the original price is ( P ), and discounts are
    d_1\%, d_2\%, \ldots
    , the final price is:
    P \cdot \left(1 - \frac{d_1}{100}\right) \cdot \left(1 - \frac{d_2}{100}\right) \cdot \ldots
Let’s dive into the 10 problems.
Problem 1
An item is priced at $100 and is offered two successive discounts of 10% and 20%. Find the final price.
Step-by-Step Solution:
  1. Original Price: $100.
  2. First Discount (10%):
    • Discount amount =
      100 \cdot \frac{10}{100} = 10
      .
    • Price after first discount =
      100 - 10 = 90
      .
  3. Second Discount (20%) on $90:
    • Discount amount =
      90 \cdot \frac{20}{100} = 18
      .
    • Final price =
      90 - 18 = 72
      .
Alternative: Equivalent Discount Formula:
  • Discounts:
    d_1 = 10
    ,
    d_2 = 20
    .
  • Equivalent discount =
    10 + 20 - \frac{10 \cdot 20}{100} = 30 - 2 = 28\%
    .
  • Final price =
    100 \cdot (1 - \frac{28}{100}) = 100 \cdot 0.72 = 72
    .
Explanation:
  • The 10% discount reduces the price to 90% of the original. The 20% discount is applied on this reduced price, not the original, so it’s 20% of $90, not $100.
  • The equivalent discount shows that two successive discounts don’t simply add up (10% + 20% ≠ 28%) due to the second discount being applied on a smaller amount.
Final Answer: $72.
Problem 2
A laptop costs $800 and has successive discounts of 15% and 25%. What is the final price?
Step-by-Step Solution:
  1. Original Price: $800.
  2. First Discount (15%):
    • Discount amount =
      800 \cdot \frac{15}{100} = 120
      .
    • Price after first discount =
      800 - 120 = 680
      .
  3. Second Discount (25%) on $680:
    • Discount amount =
      680 \cdot \frac{25}{100} = 170
      .
    • Final price =
      680 - 170 = 510
      .
Alternative: Final Price Formula:
  • Final price =
    800 \cdot (1 - \frac{15}{100}) \cdot (1 - \frac{25}{100}) = 800 \cdot 0.85 \cdot 0.75
    .
  • Compute:
    800 \cdot 0.85 = 680
    , then
    680 \cdot 0.75 = 510
    .
Explanation:
  • The step-by-step method ensures clarity by showing how each discount reduces the price.
  • The final price formula is more efficient for calculations, especially with multiple discounts, as it avoids intermediate steps.
Final Answer: $510.
Problem 3
A shirt is marked at $50 with three successive discounts of 10%, 10%, and 10%. Find the final price.
Step-by-Step Solution:
  1. Original Price: $50.
  2. First Discount (10%):
    • Price =
      50 \cdot (1 - 0.1) = 50 \cdot 0.9 = 45
      .
  3. Second Discount (10%) on $45:
    • Price =
      45 \cdot 0.9 = 40.5
      .
  4. Third Discount (10%) on $40.5:
    • Price =
      40.5 \cdot 0.9 = 36.45
      .
Alternative: Final Price Formula:
  • Final price =
    50 \cdot 0.9 \cdot 0.9 \cdot 0.9 = 50 \cdot (0.9)^3
    .
  • Compute:
    (0.9)^3 = 0.729
    , so
    50 \cdot 0.729 = 36.45
    .
Explanation:
  • Three 10% discounts don’t equal a 30% discount because each discount is applied on a progressively smaller amount.
  • The formula
    (0.9)^3
    reflects the compounding effect of the discounts.
Final Answer: $36.45.
Problem 4
A car is priced at $20,000 with successive discounts of 5%, 10%, and 15%. Find the final price.
Step-by-Step Solution:
  1. Original Price: $20,000.
  2. First Discount (5%):
    • Price =
      20,000 \cdot 0.95 = 19,000
      .
  3. Second Discount (10%) on $19,000:
    • Price =
      19,000 \cdot 0.9 = 17,100
      .
  4. Third Discount (15%) on $17,100:
    • Price =
      17,100 \cdot 0.85 = 14,535
      .
Alternative: Final Price Formula:
  • Final price =
    20,000 \cdot 0.95 \cdot 0.9 \cdot 0.85
    .
  • Compute:
    0.95 \cdot 0.9 = 0.855
    , then
    0.855 \cdot 0.85 = 0.72675
    .
  • Final price =
    20,000 \cdot 0.72675 = 14,535
    .
Explanation:
  • The discounts are applied sequentially, and the final price formula combines them into a single multiplication, which is computationally efficient.
Final Answer: $14,535.
Problem 5
A book costs $40. After two successive discounts, the final price is $28.80. If the first discount is 10%, what is the second discount?
Step-by-Step Solution:
  1. Original Price: $40.
  2. First Discount (10%):
    • Price =
      40 \cdot 0.9 = 36
      .
  3. Second Discount:
    • Let the second discount be
      d\%
      .
    • Final price =
      36 \cdot (1 - \frac{d}{100}) = 28.80
      .
    • Solve:
      36 - 0.36d = 28.80
      .
    • 0.36d = 36 - 28.80 = 7.20
      .
    • d = \frac{7.20}{0.36} = 20
      .
    • Second discount = 20%.
Verification:
  • Price after 20% discount on $36 =
    36 \cdot 0.8 = 28.80
    , which matches.
Explanation:
  • The problem requires working backward from the final price to find the unknown discount. The equation is set up based on the final price formula.
Final Answer: 20%.
Problem 6
A TV is sold for $900 after two successive discounts of 20% and 25% on the marked price. Find the marked price.
Step-by-Step Solution:
  1. Let the marked price be ( P ).
  2. First Discount (20%):
    • Price =
      P \cdot 0.8
      .
  3. Second Discount (25%):
    • Final price =
      P \cdot 0.8 \cdot 0.75 = 900
      .
    • P \cdot 0.6 = 900
      .
    • P = \frac{900}{0.6} = 1500
      .
Verification:
  • First discount:
    1500 \cdot 0.8 = 1200
    .
  • Second discount:
    1200 \cdot 0.75 = 900
    , which matches.
Explanation:
  • The marked price is found by reversing the discount process, dividing the final price by the product of the retention factors (0.8 and 0.75).
Final Answer: $1500.
Problem 7
A store offers successive discounts of 30%, 20%, and 10%. If the final price is $336, find the original price.
Step-by-Step Solution:
  1. Let the original price be ( P ).
  2. Apply Discounts:
    • After 30%:
      P \cdot 0.7
      .
    • After 20%:
      P \cdot 0.7 \cdot 0.8
      .
    • After 10%:
      P \cdot 0.7 \cdot 0.8 \cdot 0.9 = 336
      .
  3. Compute:
    • 0.7 \cdot 0.8 = 0.56
      .
    • 0.56 \cdot 0.9 = 0.504
      .
    • P \cdot 0.504 = 336
      .
    • P = \frac{336}{0.504} \approx 666.67
      .
Verification:
  • First:
    666.67 \cdot 0.7 \approx 466.67
    .
  • Second:
    466.67 \cdot 0.8 \approx 373.34
    .
  • Third:
    373.34 \cdot 0.9 \approx 336
    , which matches.
Explanation:
  • The original price is calculated by dividing the final price by the product of the retention factors. Rounding may occur due to decimal approximations.
Final Answer: $666.67 (approx).
Problem 8
A jacket has a marked price of $120. After two successive discounts, the price is $81.60. If the second discount is 15%, find the first discount.
Step-by-Step Solution:
  1. Original Price: $120.
  2. Second Discount (15%):
    • Let the price after the first discount be
      P_1
      .
    • Final price =
      P_1 \cdot 0.85 = 81.60
      .
    • P_1 = \frac{81.60}{0.85} = 96
      .
  3. First Discount:
    • 120 \cdot (1 - \frac{d}{100}) = 96
      .
    • 1 - \frac{d}{100} = \frac{96}{120} = 0.8
      .
    • \frac{d}{100} = 0.2
      , so
      d = 20\%
      .
Verification:
  • First:
    120 \cdot 0.8 = 96
    .
  • Second:
    96 \cdot 0.85 = 81.60
    , which matches.
Explanation:
  • The first discount is found by working backward from the price after the second discount to the original price.
Final Answer: 20%.
Problem 9
An item has two successive discounts where the equivalent single discount is 40%. If the first discount is 25%, find the second discount.
Step-by-Step Solution:
  1. Equivalent Discount Formula:
    • Let the second discount be
      d_2\%
      .
    • 25 + d_2 - \frac{25 \cdot d_2}{100} = 40
      .
    • d_2 - 0.25d_2 = 40 - 25
      .
    • 0.75d_2 = 15
      .
    • d_2 = \frac{15}{0.75} = 20
      .
Verification:
  • Equivalent discount =
    25 + 20 - \frac{25 \cdot 20}{100} = 45 - 5 = 40\%
    , which matches.
Explanation:
  • The equivalent discount formula directly relates the two discounts to the single discount, allowing us to solve for the unknown.
Final Answer: 20%.
Problem 10
A phone is priced at $1000. After three successive discounts, the price is $729. Find the equal successive discounts that produce the same final price.
Step-by-Step Solution:
  1. Final Price Ratio:
    • 1000 \cdot (1 - \frac{d}{100})^3 = 729
      .
    • (1 - \frac{d}{100})^3 = \frac{729}{1000} = 0.729
      .
    • 1 - \frac{d}{100} = \sqrt[3]{0.729} = 0.9
      .
    • \frac{d}{100} = 0.1
      , so
      d = 10\%
      .
Verification:
  • Three 10% discounts:
    1000 \cdot 0.9 \cdot 0.9 \cdot 0.9 = 1000 \cdot 0.729 = 729
    .
Explanation:
  • The problem assumes three equal discounts. The cube root is used because the discounts are applied three times, and each discount is the same.
Final Answer: 10% each.
Summary
  • Key Concept: Successive discounts are applied sequentially, and the equivalent discount is less than the sum of individual discounts due to the reduced base price for each subsequent discount.
  • Methods: Step-by-step calculations are intuitive, while the final price formula or equivalent discount formula is efficient.
  • Applications: These problems test understanding of percentages, sequential operations, and algebraic manipulation for unknowns.
Each problem was solved with both a step-by-step approach and an alternative method to reinforce understanding. 

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