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CAT (Common Admission Test) · Advanced Quantitative Aptitude

Percentages, Profit-Loss and Interest

Percentage change, profit and loss, discounts, simple and compound interest.

Six concepts covering the arithmetic CAT packs into one chapter: percentage change, the base a percent sits on, holding a bill constant, the equal-and-opposite selling trap, and the gap between simple and compound interest. Every question here is decided by which number is underneath.

  • CAT (Common Admission Test)
  • Medium level
  • 6 concepts
  • 21 practice questions

1Percentage change, and two changes in a row

A change from A to B is \frac{B-A}{A}\times 100\%. The numerator is the difference; the denominator is always the value you started from, never the one you arrived at and never the larger of the two.

Two successive changes of x\% and y\% do not add. The second acts on a quantity the first has already moved, so the net is x+y+\frac{xy}{100}\% — with increases positive and decreases negative. Convert awkward percentages to fractions before multiplying: 12.5\%=\frac{1}{8}, 16.\overline{6}\%=\frac{1}{6}, 37.5\%=\frac{3}{8}.

Figure. Three amounts on one scale. The middle bar is 20% above the dashed line; the right bar then drops 10% of that raised amount, not 10% of the original, and lands at 1080 — an 8% net rise, not a 10% one.

How it works

  1. Name the baseUnderline the value after "from", "more than" or "compared with" — that is the denominator.
  2. Sign each changeA fall of 10% is y=-10, not 10. The product term flips with the signs.
  3. Add, then add the product over 100Net = x+y+\frac{xy}{100}. Two falls make xy positive, so 20% then 10% off is 28%, not 30%.
Percentages worth converting first
PercentageFractionPercentageFraction
50%1/212.5%1/8
25%1/437.5%3/8
20%1/562.5%5/8
16⅔%1/68⅓%1/12

A 20% rise, then a 10% fall

A price rises by 20\% and then falls by 10\%. What is the net percentage change from the original price?

  • x = +20, y = −10; net = x + y + xy/10020 − 10 − 2 = +8%
  • Check on ₹1000: after +20%₹1200
  • Then −10% of 1200₹1080
  • 1080 against 10008% net rise

Pro tip. The −2 in row one is the whole content of the formula: the second change is charged on ₹1200, so 10% of 1200 is 120 rather than 100. Multiplying the factors 1.2 × 0.9 = 1.08 gives the same answer in one step and never has a sign to lose.

Successive discounts of 20% and 10% on a marked price are equivalent to a single discount of
  1. 30%
  2. 28%
  3. 18%

a = −20, b = −10 gives −20 − 10 + 2 = −28%, or 0.8 × 0.9 = 0.72. Adding to 30% overstates because the second discount acts on the already-reduced price. 18% is 20 − 10 + (−2) with a sign error on the product.

2Profit sits on cost; discount sits on marked price

Profit% and Loss% are always taken on the cost price. Discount% is always taken on the marked price. Mixing the two bases is the most common trap in store problems, and the question is written to invite exactly that mix-up.

Chain the two as multipliers and the arithmetic collapses: \text{SP}=\text{CP}\times\left(1+\frac{\text{markup}}{100}\right)\times\left(1-\frac{\text{discount}}{100}\right). Two successive discounts of a\% then b\% are equivalent to a single discount of a+b-\frac{ab}{100}\%, which is always less than a+b.

Figure. Cost, marked price and selling price on one scale. Markup lifts CP to MP; discount then pulls MP down to SP — and the 5 left above the dashed cost line is the profit, measured against cost, not against the marked price.

How it works

  1. Fix the cost at 100Assume CP = 100 so every later percentage reads off as a rupee figure and as a percent at once.
  2. Apply the markupMP = CP × (1 + markup/100). Discount never touches this step.
  3. Apply the discount to MPSP = MP × (1 − discount/100). Profit% is then (SP − CP)/CP × 100 — still on cost.

Markup with discount

A shopkeeper marks his goods 40\% above cost and then offers a 25\% discount. Find his overall profit percentage.

  • Let CP = 100, so MP = 100 × 1.40140
  • SP = 140 × (1 − 0.25) = 140 × 0.75105
  • Profit = 105 − 1005
  • Profit% on CP = 5/100 × 1005%

Pro tip. Assume CP = 100 and treat markup and discount as multipliers; the final number is the profit percentage directly. Writing SP = CP × 1.40 × 0.75 = 1.05·CP reaches the same 5% in one line.

Goods marked 50% above cost are sold at a 20% discount on the marked price. The profit percent is
  1. 30%
  2. 20%
  3. 25%

SP = CP × 1.50 × 0.80 = 1.20·CP, a 20% profit. 30% adds the markup and discount as if they shared a base; 25% is half the markup with the discount ignored.

3Holding the bill constant

Expenditure is price times consumption. If the price rises by x\% and the bill must stay unchanged, consumption has to fall by \frac{x}{100+x}\times 100\% — the reduction is measured against the new, higher price, so the denominator is the raised price plus 100, not 100 alone.

A 25\% rise in price needs a 20\% cut in consumption; a 20\% fall in price allows a 25\% rise. The pairs are reciprocal, and cutting consumption by the same percentage the price rose by overshoots and leaves you spending less than before.

Figure. Every point on this curve spends the same money. Walking from 100×100 to 125×80 raises the price by a quarter and lowers consumption by a fifth — unequal percentages, equal products. In the drawn axis frame, (x − 0.08)×(0.92 − y) is constant across the sampled points.

How it works

  1. Name the productWrite down what is fixed — the sugar bill, the fuel spend. Everything follows from it.
  2. Use the new baseThe cut is \frac{x}{100+x}\times 100\%, because you divide by the raised price.
  3. Check by multiplying125 × 80 = 10000 = 100 × 100. If the two products differ, the percentage is wrong.

Constant-expenditure adjustment

The price of sugar rises by 25\%. By what percentage must a family cut consumption to keep its sugar bill unchanged?

  • Required cut = x/(100 + x) × 100 with x = 2525/125 × 100
  • 25/1251/5
  • 1/5 × 10020%
  • Check: 1.25 × 0.801.00, bill unchanged

Pro tip. A 25% rise equals a factor of 5/4, so consumption must scale by 4/5 — a drop of exactly 20%. The concrete route (old price ₹40, bill ₹1000 → 25 kg; new price ₹50 → 20 kg) lands on the same cut and is worth doing once so the formula is believed.

The price of a commodity falls by 20%. To keep expenditure the same, consumption must rise by
  1. 20%
  2. 25%
  3. 16⅔%

The price is now 80, so the rise is measured against 80: 20/80 × 100 = 25%. Answering 20% keeps the base at the old price. 16⅔% is 20/120 × 100, the answer when the price had risen by 20% instead.

4Same selling price, equal percent gain and loss

Sell two items at the same selling price, one at a profit of x\% and the other at a loss of x\%, and the net result is always a loss of \left(\frac{x}{10}\right)^2\%. Equal percentages on opposite sides do not cancel, because the two cost prices are different — the loss item cost more than the profit item brought in.

The square is why a 10% pair loses 1% overall and a 20% pair loses 4%. The order of the two sales does not matter; the two cost prices and one common selling price fix the result.

Figure. The dashed line is the shared selling price. The profit item sits 90 below it and the loss item sits 110 above it — unequal gaps on unequal costs — so the 90 gained does not cover the 110 lost.

How it works

  1. Fix the common SPPick a selling price that divides cleanly by both (1 + x/100) and (1 − x/100).
  2. Back out each CPCP₁ = SP / (1 + x/100) and CP₂ = SP / (1 − x/100). The loss item has the larger cost.
  3. Compare totalsTotal CP exceeds 2·SP by exactly (x/10)² percent of total CP.

Two sales at ₹990, ±10%

Two articles are sold at ₹990 each. One is sold at a 10% profit and the other at a 10% loss. Find the overall profit or loss percent.

  • CP of profit item = 990 / 1.1₹900
  • CP of loss item = 990 / 0.9₹1100
  • Total CP = 900 + 1100; total SP = 2 × 990₹2000 against ₹1980
  • Loss = 20 on 2000; also (10/10)²1% loss

Pro tip. Memorise (x/10)²% as the net loss and skip the two cost prices when the question only asks for the overall percent. Checking forwards once — as in the ledger — is what makes the shortcut safe to trust.

Two items sold at the same price, one at +20% and one at −20%, produce an overall
  1. no profit or loss
  2. 4% loss
  3. 4% profit

(20/10)² = 4% loss. The equal percentages cancel only if they share a cost base; here they share a selling price, so the loss item was the more expensive of the two and the net is short.

5Simple interest grows in a straight line

Simple interest is \text{SI}=\frac{PRT}{100}. Interest is charged on the original principal every year, so the same amount is added each year and the total grows linearly. Nothing the interest earns ever earns anything itself.

A sum doubles under simple interest when SI equals the principal, so RT=100; it trebles when SI is twice the principal, so RT=200. The principal cancels and only the product of rate and time survives.

Figure. ₹5000 at 10% simple interest. Equal steps of ₹500 a year, forever, because each year's interest is charged on the original ₹5000 and never on anything earlier years produced. A straight line is the signature of simple interest.

How it works

  1. Interest per year is fixedPR/100, the same every year. Multiply by T for the total interest.
  2. Amount is principal plus interestA = P(1 + RT/100). The principal is never rewritten.
  3. For multiples, set SI against PDoubling needs SI = P; trebling needs SI = 2P. Only RT remains.

Interest on ₹5000 at 10% for 2 years

Find the simple interest on ₹5000 at 10% per annum for 2 years, and the amount.

  • SI = PRT/100 = 5000 × 10 × 2 / 100₹1000
  • Interest per year = 5000 × 10 / 100₹500
  • Amount = 5000 + 1000₹6000
  • As a check: A = P(1 + RT/100) = 5000 × 1.20₹6000

Pro tip. Keep this ₹5000 / 10% / 2-year case in mind — the next concept reuses the same numbers under compounding, and the ₹50 gap between the two regimes is exactly interest on the first year's interest.

A sum trebles itself in 20 years at simple interest. The annual rate is
  1. 10%
  2. 15%
  3. 5%

Trebling means the interest equals twice the principal, so RT = 200 and R = 200/20 = 10%. Reading "trebles" as SI = 3P gives RT = 300 and 15% — the amount is three times P, but the interest is only twice it.

6Compound interest, and the gap from simple

Compound amount is A=P\left(1+\frac{R}{100}\right)^{T}, and CI = A-P. Each year's interest joins the principal and earns in its turn, so the yearly interest grows. Over two years at rate R, the gap between compound and simple interest is exactly P\left(\frac{R}{100}\right)^{2} — one year's interest on the first year's interest.

For 2-year CI, add the two simple-interest legs plus that interest-on-interest. The same gap, given in a question, is the fastest route back to the principal: divide by (R/100)^{2}.

Figure. ₹5000 at 10%. The two curves leave together and stay together for a full year — nothing has compounded yet. They separate at two years by ₹50 (6000 simple against 6050 compound) and the gap then widens, because the compound curve bends up while the simple one stays straight.

How it works

  1. Grow by the factor each yearMultiply by (1 + R/100) once per year. Two years at 10% is ×1.21, not ×1.20.
  2. Read the gap at two yearsCI − SI = P(R/100)². Never compute both interests separately when the difference is what is asked.
  3. Invert if the gap is givenP = (CI − SI) / (R/100)². That is the usual CAT wording.

Finding principal from CI − SI

The difference between compound and simple interest on a sum for 2 years at 10% per annum is Rs 50. Find the sum.

  • For 2 years: CI − SI = P(R/100)²P × (0.10)²
  • 50 = P × 0.01P = 50 / 0.01
  • P₹5000
  • Check: year-1 interest = 500; 10% of 500₹50 gap

Pro tip. The 2-year CI − SI gap is exactly one year's interest earned on the first year's interest — memorise P(R/100)². Checking forwards (10% of 5000 is 500; 10% of that 500 is 50) catches a missed square faster than re-deriving the formula.

On ₹5000 at 10% for 2 years, compound interest exceeds simple interest by
  1. ₹50
  2. ₹100
  3. ₹500

P(R/100)² = 5000 × 0.01 = ₹50. ₹100 is two years of that gap counted twice; ₹500 is the first year's interest itself, which is the quantity the gap is interest on — not the gap.

Notes

  • Percentage change: A change from A to B is \frac{B-A}{A}\times100\%; two successive changes of x\% and y\% combine into a net change of x+y+\frac{xy}{100}\%.
  • Base of computation: Profit% and Loss% are always taken on the cost price, whereas discount% is taken on the marked price — mixing the two bases is the most common trap.
  • Simple vs compound interest: SI grows linearly as \frac{PRT}{100}, while CI grows multiplicatively as P\left(1+\frac{R}{100}\right)^T-P, so the gap widens each year.
  • Successive discounts: Two discounts of a\% then b\% give a single equivalent discount of a+b-\frac{ab}{100}\%, which is always less than a+b.
  • Price-consumption link: If price rises by x\%, consumption must fall by \frac{x}{100+x}\times100\% to keep total expenditure unchanged.

Formulas

  • Selling price: \text{SP}=\text{CP}\left(1+\frac{\text{Profit\%}}{100}\right)
  • Successive change: net=x+y+\frac{xy}{100}
  • Compound amount: A=P\left(1+\frac{R}{100}\right)^{T}
  • SI-CI gap over 2 years: \text{CI}-\text{SI}=P\left(\frac{R}{100}\right)^2
  • Equivalent discount: D=a+b-\frac{ab}{100}
  • Simple interest: \text{SI}=\frac{PRT}{100}

Exam traps & shortcuts

  • Convert percentages to fractions: 12.5\%=\frac18, 16.\overline{6}\%=\frac16, 37.5\%=\frac38 — fraction arithmetic is far faster than decimals.
  • For 2-year CI, add the two simple-interest legs plus interest-on-interest (R\% of R\% of P), which also equals the CI-SI difference.
  • Selling two items at the same price, one at +x\% and one at -x\%, always yields a net loss of \left(\frac{x}{10}\right)^2\%.
  • Chain markup and discount as multipliers: \text{SP}=\text{CP}\times(1+\text{markup})\times(1-\text{discount}) solves most store problems in one line.

Reference tables

Every line should be reconstructible from the concepts above, not merely recalled.

Formula sheet
QuantityRelationWatch for
Percentage change(B − A)/A × 100Base is always A, the start
Successive changesx + y + xy/100Signs first; discounts are negative
Selling price after markupSP = CP(1 + Profit%/100)Profit% is on CP
DiscountSP = MP(1 − d/100)Discount% is on MP
Equivalent discounta + b − ab/100Always less than a + b
Constant expenditurecut = x/(100 + x) × 100Denominator is the new price
Equal ±x% on same SPnet loss (x/10)²%Costs differ; result is never zero
Simple interestSI = PRT/100Same addition every year
Compound amountA = P(1 + R/100)^TCI = A − P
CI − SI over 2 yearsP(R/100)²Interest on the first year's interest

Recap

Read only this the night before.

The base
Percentage change divides by where you came from. Profit% divides by CP; discount% divides by MP. Name the base before computing.
Successive
x + y + xy/100, or multiply the factors. Two discounts always undershoot a + b.
Constant bill
Price up x% means consumption down x/(100 + x) × 100. Multiply back to check.
±x% pair
Same SP, +x% and −x%, nets a loss of (x/10)²%. Equal percents do not cancel on a shared selling price.
SI vs CI
SI is linear on P; CI is successive percentage change. Over two years the gap is P(R/100)².
Fractions
12.5% = 1/8, 16⅔% = 1/6, 37.5% = 3/8. Convert before multiplying.

Practise Percentages, Profit-Loss and Interest

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