2.5 Successive Discounts

Figure: Successive discounts compound — a 20% then 10% discount is a net 28%, not 30%.

When two (or more) discounts are applied one after another (e.g., “20% off, plus an additional 10% off”), you cannot add them directly. The second discount applies to the already-reduced price, not the original MP.

For two successive discounts of a% and b%, the single equivalent discount is:

Effective Discount% = a + b − (ab/100)

This formula is derived directly by multiplying factors: SP = MP × (1 − a/100) × (1 − b/100). Expand and compare with SP = MP × (1 − d/100):

1 − d/100 = (1 − a/100)(1 − b/100) = 1 − a/100 − b/100 + ab/10000 d/100 = a/100 + b/100 − ab/10000 d = a + b − ab/100

For three successive discounts a%, b%, c%, first combine any two using the formula above to get an equivalent single discount, then combine that result with the third discount using the same formula again. Equivalently, just multiply all three factors directly: SP = MP(1−a/100)(1−b/100)(1−c/100).

Worked Example 10: Find the single discount equivalent to two successive discounts of 20% and 10% on a marked price of Rs 2500. Also find the SP.

Solution: Effective discount% = 20 + 10 − (20×10)/100 = 30 − 2 = 28%. SP = 2500 × 0.8 × 0.9 = 2500 × 0.72 = Rs 1800. (Check: 2500 × 0.72 = 1800 ✓)

Worked Example 11: A shopkeeper offers successive discounts of 10%, 20%, and 25% on an article marked at Rs 4000. Find the final selling price.

Solution: SP = 4000 × 0.9 × 0.8 × 0.75 = 4000 × 0.54 = Rs 2160. (0.9 × 0.8 = 0.72; 0.72 × 0.75 = 0.54)

Worked Example 12: Two successive discounts on an article together amount to an overall single discount of 32%. If the first discount given was 20%, find the second discount percentage.

Solution: Let MP = 100. After overall 32% discount, SP = 68. After the first 20% discount alone: 100 × 0.8 = 80. Let the second discount be d%: 80 × (1 − d/100) = 68 ⇒ 1 − d/100 = 0.85 ⇒ d = 15%.

Worked Example 12A (three equal successive discounts): A retailer offers three successive discounts of 10% each on an article marked at Rs 2000. Find the final selling price and the single equivalent discount percentage.

Solution: SP = 2000 × 0.9 × 0.9 × 0.9 = 2000 × 0.729 = Rs 1458. Equivalent single discount% = [(2000−1458)/2000] × 100 = (542/2000) × 100 = 27.1%. Notice this is not 30% (which a hasty aspirant might guess by adding 10+10+10) — three successive 10% cuts always compound to slightly less than a flat 30% cut, because each subsequent discount acts on a smaller base.

Worked Example 12B (three unequal successive discounts — comparison edge case): A trader allows three successive discounts of 15%, 10%, and 5% on an article marked at Rs 8000. Find the final selling price and the single equivalent discount percentage. A customer complains that a flat discount of 27% would have been fairer to him — determine whether the customer is right.

Solution: SP = 8000 × 0.85 × 0.90 × 0.95. Step-by-step: 0.85 × 0.90 = 0.765; 0.765 × 0.95 = 0.72675. SP = 8000 × 0.72675 = Rs 5814. Equivalent single discount% = (1 − 0.72675) × 100 = 27.325%. Since the actual combined discount (27.325%) is higher than the flat 27% the customer proposed, the three successive discounts actually work slightly in the customer’s favour here — he would have paid more (8000 × 0.73 = Rs 5840) under a flat 27% discount than the Rs 5814 he actually pays. This is a useful reminder that successive discounts are not automatically worse for the customer than they look; each case must be computed, not assumed.

Worked Example 12C (reverse — find the missing middle discount): Three successive discounts of 25%, x%, and 10% on an article are together equivalent to a single discount of 40.6%. Find x.

Solution: Combined factor = 1 − 0.406 = 0.594. So 0.75 × (1 − x/100) × 0.90 = 0.594. 0.75 × 0.90 = 0.675. So (1 − x/100) = 0.594/0.675 = 0.88 ⇒ x/100 = 0.12 ⇒ x = 12%. Check: 0.75 × 0.88 × 0.90 = 0.66 × 0.90 = 0.594 = 1 − 0.406 ✓. This variant — finding a missing discount buried in the middle of a three-discount chain rather than the last one (as in Worked Example 12) — is the kind of twist Tier-II examiners like to add.