Fractions in Disguise · Lesson 11 of 12
Tricky Percentages
“Reason through offers, successive discounts, markups, subgroup percentages and changing totals.”
• Compare percentage gains with absolute gains when making a choice. • Explain successive discounts using their changing bases. • Find the effect of a markup followed by a discount. • Interpret reverse comparisons, offers and percentages of subgroups. • Use algebra and diagrams to justify percentage reasoning.
Two offers promise different rewards. In one, you deposit ₹100 and receive ₹300 back. In the other, you deposit ₹1000 and receive ₹1500 back. The first multiplies your deposit more strongly, but the second gives you a larger extra amount of money. Which is better depends on what you are comparing and how much you can deposit. A large percentage is useful information, but it does not tell the whole story without its base and the practical conditions.
Many puzzling percentage questions work this way. Their arithmetic is manageable, but the reference quantity changes or the wording hides a comparison. A statement that sounds convincing can fail once you label the base at each stage.
We will use simple numbers, diagrams and equations to make those references visible. The aim is to understand the situations well enough to explain your answer, not merely pick a formula that contains a percent sign.
Would You Rather?
In the first reward offer, the gain is ₹300 − ₹100 = ₹200. Relative to the ₹100 deposited, the gain is 200%. In the second offer, the gain is ₹1500 − ₹1000 = ₹500. Relative to ₹1000, the gain is 50%.
The first offer gives a higher percentage gain, but the second gives a higher absolute gain. If each can be chosen only once and you can afford either deposit, the second adds more rupees. If you only have ₹100 available, it is not accessible. The conditions matter to the choice.
Do not compare the return with the deposit and then call the result a gain percentage. In the first offer, ₹300 is 300% of the deposit, while the extra ₹200 is a 200% gain. The original ₹100 must be separated from the extra money.
Problem
A store offers either 20% off or ₹50 off a qualifying purchase. Compare the choices for purchases of ₹180, ₹225 and ₹300.
- 1.For ₹180, 20% is ₹36. A ₹50 reduction is larger, giving a final price of ₹130 rather than ₹144.
- 2.For ₹225, 20% is ₹45. The ₹50 reduction is still larger, giving ₹175 rather than ₹180.
- 3.For ₹300, 20% is ₹60. The percentage discount is now larger, giving ₹240 rather than ₹250.
- 4.The offers are equal when 0.20W = 50, so W = ₹250.
- 5.Below ₹250, the fixed ₹50 reduction is better; above ₹250, 20% is better, provided the purchase meets the offer conditions.
Now compare “30% off, then an extra 20% off” with a single 50% discount. In ordinary arithmetic 30 + 20 = 50, but these discounts act on different prices. The second 20% reduction applies to the amount left after the first discount.
For a ₹200 cake, the first 30% discount removes ₹60 and leaves ₹140. The next 20% removes 20% of ₹140, which is ₹28. The combined reduction is ₹88, or 44% of the original ₹200. It is not a 50% reduction.
Problem
A ₹200 cake receives successive discounts of 30% and 20%. Another identical ₹200 cake receives a single 50% discount. Compare final prices.
- 1.After 30% off, the first cake costs 200 × 0.70 = ₹140.
- 2.After a further 20% off, it costs 140 × 0.80 = ₹112.
- 3.The second cake costs 200 × 0.50 = ₹100.
- 4.The single 50% discount is cheaper by ₹12.
- 5.The successive-discount multiplier is 0.70 × 0.80 = 0.56, meaning 56% is paid and 44% is saved.
A useful way to check successive discounts is to calculate the retained share. Keep 70% after the first discount, then 80% of that retained part. The product 0.7 × 0.8 gives the final share of the original.
In this pure percentage model, reversing the two discount rates gives the same product, 0.8 × 0.7. That does not turn the combined discount into their sum. Real offers may add fixed reductions or conditions, so use the precise terms of the problem.
A Mishap
Surbhi sets a selling price 50% above cost and later gives 50% off that selling price. She expects to recover the cost. But the 50% markup and the 50% discount use different bases. Start with a cost of ₹100: the marked price becomes ₹150, and halving it gives ₹75.
The ₹50 markup is followed by a ₹75 reduction, so the final selling price is ₹25 below cost. In general, if cost is x, the final price is 1.5 × 0.5 × x = 0.75x. The loss is 25% of cost.
To sell at cost after a 50% markup, the reduction must remove ₹50 from ₹150. Relative to the marked price, that is 50/150 = 1/3, or 33⅓%. A percentage that reverses a markup need not equal that markup.
Problem
Goods marked 50% above cost are sold at 50% off for ₹12,000. Find their cost, loss and the discount that would instead have allowed a sale at cost.
- 1.Let cost be x. Marked price is 1.5x, and the final sale is 0.5 × 1.5x = 0.75x.
- 2.Given 0.75x = 12,000, divide to obtain x = ₹16,000.
- 3.Loss = ₹16,000 − ₹12,000 = ₹4000, which is 25% of cost.
- 4.The marked price was 1.5 × ₹16,000 = ₹24,000.
- 5.To recover cost, the discount would need to be ₹24,000 − ₹16,000 = ₹8000. Its percentage of marked price is 8000/24,000 × 100 = 33⅓%.
Problem
A product is marked 35% above cost, then discounted by 30%. Does the seller make a profit or a loss?
- 1.Let cost be ₹100 to make the reference clear.
- 2.After the markup, marked price is ₹135.
- 3.The 30% discount leaves 70% of ₹135: 0.70 × 135 = ₹94.50.
- 4.The seller loses ₹5.50 on a ₹100 cost, a 5.5% loss.
- 5.For any positive cost x, the same conclusion follows from 1.35 × 0.70x = 0.945x.
Changing the direction of a comparison also changes its percentage. If Ariba has 120% as many marbles as Arun, write A = 1.2B. Solving for B gives B = A/1.2 = (5/6)A. Arun therefore has 83⅓% as many as Ariba, not 80%.
A concrete pair makes this easier to see: Arun has 100 and Ariba has 120. The difference of 20 is 20% of Arun’s 100 but 16⅔% of Ariba’s 120. Always ask whose quantity is serving as the hundred-per-cent reference.
Problem
Forty per cent of the students on an excursion belong to one class. Sixty per cent of that class group are girls. What percentage of all excursion students are girls from that class?
- 1.Imagine 100 excursion students. The class group contains 40 of them.
- 2.Girls from that class form 60% of the 40, giving 0.60 × 40 = 24.
- 3.Thus the required share of all excursion students is 24%. Algebraically, 0.40 × 0.60 = 0.24.
- 4.If the total is 160, the model count is 0.24 × 160 = 38.4. That is not a possible exact count of students.
- 5.Therefore the percentages and total 160 cannot all be exact counts. They can describe approximate survey shares, or an exact-count version could use 200 students, giving 48 girls from that class.
An appealing shortcut is that x% of y equals y% of x as a numerical calculation. Both equal xy/100. For instance, 5% of 40 equals 2 and 40% of 5 also equals 2. Choosing the easier calculation can save effort.
This symmetry does not make every percentage comparison reversible. “A is 120% of B” asks for a ratio in one direction, not for two interchangeable numbers in a product. Explain which relationship you are using before applying a shortcut.
Problem
Identical items normally cost ₹100 each. Compare “buy one, get one free”, “buy two, get one free”, and “buy three, get one free”.
- 1.First offer: pay ₹100 for 2 items. Effective price is ₹50 each; the free share is 1/2, so the discount is 50%.
- 2.Second offer: pay ₹200 for 3 items. Effective price is ₹200/3 each; discount is 1/3 = 33⅓%.
- 3.Third offer: pay ₹300 for 4 items. Effective price is ₹75 each; discount is 1/4 = 25%.
- 4.The denominator for the discount includes all items received, including the free one.
- 5.If exactly four items are needed and complete bundles may be repeated, the first offer costs ₹200. The second costs ₹300 for one bundle plus one ordinary item, and the third costs ₹300 for one bundle. State these bundle assumptions when comparing.
Problem
A room has 100 people, of whom 99 are left-handed. How many left-handed people must leave for exactly 98% of those remaining to be left-handed?
- 1.The one right-handed person stays. If 98% are left-handed, 2% of the remaining group must be right-handed.
- 2.Thus one person represents 2%, and the whole remaining group has 1 ÷ 0.02 = 50 people.
- 3.Of these, 49 are left-handed and one is right-handed.
- 4.Originally there were 99 left-handed people, so 99 − 49 = 50 must leave.
- 5.Check: 49/50 × 100 = 98%. Removing only one left-handed person gives 98/99, which is not 98%.
Quiz
Which offer gives the larger absolute gain: deposit ₹100 and receive ₹300, or deposit ₹1000 and receive ₹1500?
Successive discounts of 30% and 20% are equivalent to which single discount?
A 50% markup followed by a 50% discount produces what result on cost?
If A is 120% of B, what percentage of A is B?
For “buy two, get one free” with identical unit prices, what is the discount on the normal total value received?
Practice Problems
- Find 5% of 40 and 40% of 5. Explain why they agree. Solution: Both equal 2. Their expressions are 5 × 40/100 and 40 × 5/100, equal because multiplication can be reordered.
- Which is better on a ₹450 purchase: 12% off or ₹50 off? At what purchase value are they equal? Solution: 12% of ₹450 = ₹54, so the percentage offer saves ₹4 more. Equality occurs when 0.12W = 50, giving W = ₹416⅔, approximately ₹416.67.
- A product marked ₹1500 receives successive discounts of 20% and 10%. Find final price and effective discount. Solution: Final price = 1500 × 0.8 × 0.9 = ₹1080. The retained share is 72%, so the effective discount is 28%, or ₹420.
- A seller sets the marked price of goods 25% above cost. What discount on that price allows a sale at cost? Solution: Use cost ₹100, giving marked price ₹125. The required reduction is ₹25. Discount percentage = 25/125 × 100 = 20%. Check: 1.25 × 0.8 = 1.
- Of 200 excursion students, 40% belong to one class, and 60% of those are girls. Find the number and percentage of all students who are girls from that class. Explain why applying 60% to 200 directly is wrong. Solution: The class group is 0.4 × 200 = 80. Girls in it number 0.6 × 80 = 48. They form 48/200 × 100 = 24% of all students. Applying 60% to 200 would use all excursion students as the base, but the stated 60% applies only to the class subgroup.
Key Takeaways
• A larger percentage gain can accompany a smaller absolute gain. • Successive discounts act on successive prices, so multiply their retained factors. • A markup and a discount of equal percentage generally do not cancel. • Reversing a comparison requires changing the denominator carefully. • A percentage of a subgroup is found by multiplying the corresponding shares. • For free-item offers, compare savings with the normal value of all items received. • Check that exact percentage data lead to possible whole-number counts when counting people.