Profit-Loss, SI & CI

Snapshot

• Profit and loss, discount, simple interest and compound interest are percentages with a story attached, so almost every error here is a wrong base rather than a wrong calculation.
• The golden rule: profit and loss are a percentage of the cost price, a discount is a percentage of the marked price, and interest grows on the principal (simple) or on the growing amount (compound).
• This page covers the reasoning behind each rule, a LaTeX formula sheet, shortcuts and where they fail, 15 worked examples, a traps table, 16 rapid-fire questions with answers and a one-week routine.
• Exam reality: +5 for a correct answer, −1 for a wrong one. Two or three of these in a paper are almost free marks if the bases are clear.

Part 1 — The core idea: ask "percent of what?"

Written and reviewed by Devendra Dubey, Mathematics & Commerce faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching). Prepared for the CUET UG General Test.

Why profit sits on the cost price

A shopkeeper thinks about what he paid (the cost price, CP) and what he received (the selling price, SP). Profit is the difference, so it is natural to compare it with what he put in. That is why profit% and loss% are always on CP. A discount, on the other hand, is a reduction from the price written on the tag (the marked price, MP), so discount% is on MP. Different bases, same percent symbol, and CUET builds traps from exactly this.

The anchor trick: let CP be 100

Because every figure is a percentage of CP, you can take CP = 100 and read everything as "100 plus profit". A 20% profit means SP = 120. A 15% loss means SP = 85. You only need real rupees at the very end, if at all.

Why compound interest curves

In simple interest the bank pays interest on the original principal only, so each year adds the same amount and the total climbs in a straight line. In compound interest each year's interest is added to the principal, so next year's interest is calculated on a bigger number. That makes growth multiplicative: each year the amount is multiplied by the same factor $\left(1 + \dfrac{R}{100}\right)$. A repeated multiplier is a geometric progression, which is why the curve bends upward.

Why CI beats SI only after the first year

In the first year there is nothing to compound yet, so SI and CI are equal. From year two, the interest earns interest, and the gap opens. For ₹10,000 at 10%, SI gives ₹14,000 after four years and CI gives ₹14,641.

Pause and predict: a trader sells two items at the same price, one at a 20% profit and one at a 20% loss. Does he gain, lose or break even overall?

Check your answer: he loses. The 20% profit is on a smaller cost price and the 20% loss is on a larger cost price, so the loss in rupees is bigger. With SP = ₹120 for each, the cost prices are ₹100 and ₹150. Total cost ₹250, total sale ₹240, a loss of ₹10, which is 4% of ₹250. The shortcut for equal selling prices is a loss of $\dfrac{x^2}{100}$ per cent, which is $\dfrac{20^2}{100} = 4\%$.

Part 2 — Formula sheet

Profit, loss and discount:

$$\text{Profit \%} = \dfrac{\text{SP} - \text{CP}}{\text{CP}} \times 100 \qquad \text{Discount \%} = \dfrac{\text{MP} - \text{SP}}{\text{MP}} \times 100$$

$$\text{SP} = \text{CP} \times \left(1 + \dfrac{p}{100}\right) \qquad \text{SP} = \text{MP} \times \left(1 - \dfrac{d}{100}\right)$$

Interest:

$$\text{SI} = \dfrac{P \times R \times T}{100} \qquad A = P\left(1 + \dfrac{R}{100}\right)^T \qquad \text{CI} = A - P$$

$$\text{Half-yearly compounding: } A = P\left(1 + \dfrac{R}{200}\right)^{2T}$$

Part 3 — Shortcut methods and when they fail

Fractional time in compound interest. For a time such as $2\tfrac{1}{2}$ years, the usual approach is to compound for the whole years and then apply simple interest for the remaining half year on the grown amount. Do not use a fractional exponent unless the question asks for it.

Direction check. A selling price after a discount must be below MP. A selling price with profit must be above CP. A compound amount must exceed the simple-interest amount for the same years once the time is above 1.

Part 4 — Worked examples: profit, loss and discount

Worked example 1: profit and loss percent. (a) A radio bought for ₹400 is sold for ₹500. (b) A table bought for ₹640 is sold for ₹560. Find the profit or loss percent.

(a) Profit = 100 and the base is CP = 400: $\dfrac{100}{400} \times 100 = 25\%$ profit. (b) Loss = 80, base CP = 640: $\dfrac{80}{640} \times 100 = 12.5\%$ loss.

Worked example 2: selling price and cost price. (a) CP ₹250, profit 18%. Find SP. (b) An item sold for ₹1,260 gives a 5% profit. Find CP.

(a) SP = 250 × 1.18 = ₹295. (b) 105% of CP is 1,260, so CP = 1,260 ÷ 1.05 = ₹1,200.

Common wrong step: in (b) subtracting 5% of 1,260 (₹63) to get ₹1,197. The 5% was on the cost price, not on the selling price.

Worked example 3: two items at the same price. A man sells two articles for ₹1,000 each. On one he gains 25% and on the other he loses 20%. What is his overall result?

Common wrong step: using the shortcut $x^2/100$ here. It needs the same percentage of profit and loss. Here they differ, so work from the cost prices.

Worked example 4: marked price and discount. A shopkeeper marks goods 40% above cost and then gives a 25% discount. What is his profit percent?

Take CP = 100. MP = 140. SP = 75% of 140 = 105. Profit = 5%. The shortcut agrees: $40 - 25 - \dfrac{40 \times 25}{100} = 5$.

Worked example 5: successive discounts. A shop offers 10% off, followed by a further 20% off the reduced price. What is the single equivalent discount?

$10 + 20 - \dfrac{10 \times 20}{100} = 28\%$. In multipliers, $0.9 \times 0.8 = 0.72$, so the buyer pays 72% and the discount is 28%, not 30%.

Worked example 6: dishonest dealer. A grocer sells at cost price but uses a 900 g weight for a kilogram. Find his gain percent.

For every 900 g he gives, he is paid for 1000 g. Gain = $\dfrac{1000 - 900}{900} \times 100 = 11.11\%$.

Common wrong step: dividing by 1000 to get 10%. The cost is for the 900 g he actually hands over, so 900 is the base.

Worked example 7: article-count trick. The cost price of 20 articles equals the selling price of 16 articles. Find the profit percent.

Let each article cost ₹1. Then 20 articles cost ₹20, and that same ₹20 is what he gets for selling 16. Per article he gets $\dfrac{20}{16} = 1.25$. Profit = 25%.

Part 5 — Worked examples: simple and compound interest

Worked example 8: simple interest, finding the principal. The simple interest on a sum for 3 years at 8% a year is ₹1,200. Find the sum.

$$P = \dfrac{\text{SI} \times 100}{R \times T} = \dfrac{1200 \times 100}{8 \times 3} = 5000$$

The sum is ₹5,000. Check: 8% of 5,000 is 400 a year, and 3 years give 1,200.

Worked example 9: amount under simple interest. What does ₹12,000 become in 4 years at 5% simple interest?

Interest per year = 5% of 12,000 = 600. For 4 years: 2,400. Amount = 12,000 + 2,400 = ₹14,400.

Worked example 10: compound interest by the 2-year rate. Find the compound interest on ₹5,000 at 10% a year for 2 years.

Worked example 11: working back from the difference. The difference between compound and simple interest on a sum for 2 years at 10% is ₹450. Find the sum.

$$P\left(\dfrac{R}{100}\right)^2 = 450 \Rightarrow P \times 0.01 = 450 \Rightarrow P = 45000$$

The sum is ₹45,000.

Worked example 12: half-yearly compounding. Find the amount on ₹8,000 for 1 year at 10% a year, compounded half-yearly.

The rate per half-year is 5% and there are 2 half-years: $8000 \times 1.05 \times 1.05 = 8000 \times 1.1025 = 8820$, so the amount is ₹8,820. The CI is ₹820, which is more than the ₹800 you would get with yearly compounding.

Worked example 13: doubling time. In how many years does money double at 12.5% simple interest? And roughly how long at 8% compound interest?

Simple: $\dfrac{100}{12.5} = 8$ years. Compound by the rule of 72: $\dfrac{72}{8} = 9$ years. Check: $1.08^9 \approx 1.999$.

Worked example 14: depreciation. A machine worth ₹50,000 loses 10% of its value every year. What is it worth after 2 years?

It keeps 90% each year: $50000 \times 0.9 \times 0.9 = 50000 \times 0.81 = 40{,}500$. So the value is ₹40,500.

Common wrong step: subtracting 20% in one go to get ₹40,000. The second 10% is on the reduced value, not on the original.

Worked example 15: finding the rate. A sum of money becomes 1.44 times itself in 2 years at compound interest. Find the rate.

$\left(1 + \dfrac{R}{100}\right)^2 = 1.44$, so $1 + \dfrac{R}{100} = 1.2$ and $R = 20\%$.

Part 6 — Common traps

Part 7 — Rapid-fire practice with answers

Part 8 — A timed-practice routine

In the exam. Spend the first five seconds writing "CP = 100" or "P = 100" in the margin. Then convert every percentage into a multiplier and do the arithmetic. If an answer does not pass the direction check, you have swapped a base.

Part 9 — Where this connects

This topic is applied percentage, so keep Percentage, Ratio and Alligation close, particularly the successive-change rule and the multiplier habit. Compound growth is a geometric progression, covered in Sequences and Series. Percentage change and profit questions appear again in the chart sets of Data Interpretation, and ratio-based profit sharing returns in partnership problems on the Averages and Ages page.

One-line revision: profit and loss are on cost price, discount is on marked price, two changes combine as a + b − ab/100, equal-price sales at x% profit and loss lose x²/100 per cent, SI is linear and CI is multiplicative, and for 2 years CI − SI equals P(R/100)².