Arithmetic at CAT Depth: Percentages, Profit, Interest, Ratio, Mixtures & Averages

Arithmetic is the largest single block of CAT's QA section — typically eight to ten of the twenty-two questions, split between MCQs (+3/−1) and TITA questions with no negative marking. No other topic repays practice so reliably, because the paper draws on a small set of ideas — percentages, profit and loss, interest, ratio, mixtures, averages — and combines them in two or three steps rather than inventing anything new. The single habit that separates a fast CAT arithmetic solver from a school-level one is thinking in multipliers: a 20% rise is ×1.2, a 25% discount is ×0.75, two years at 10% compound is ×1.21, and a whole chain of changes is one product. This chapter builds that habit across every arithmetic topic, shows the alligation cross and the replacement formula that turn mixture questions into arithmetic, and closes with the averages traps — replaced members and harmonic-mean speeds — that CAT sets almost every year.

Percentages as Multipliers

A change of +x% is multiplication by (1 + x/100); a change of −x% is multiplication by (1 − x/100). Every percentage question in CAT reduces to writing the right multiplier chain and, usually, reading a fraction off it. Two successive changes of a% and b% give a net change of a + b + ab/100: +20% then −20% is 20 − 20 − 4 = −4%, and +10% twice is 10 + 10 + 1 = +21% — which is exactly why two years of 10% compound interest yield 21%. The order of the changes does not matter, because multiplication commutes.

Fraction–percentage equivalents to know cold
FractionPercentFractionPercent
1/250%1/812.5%
1/333⅓%3/837.5%
2/366⅔%5/862.5%
1/425%1/911.11%
1/520%1/119.09%
1/616⅔%1/128.33%
1/714.28%1/166.25%
  • Price up, consumption down. If price rises by r%, consumption must fall by r/(100 + r) × 100 to hold expenditure constant. Price up 25% → multiplier 5/4 → consumption ×4/5 → 20% fall. Price up 20% → ×6/5 → consumption ×5/6 → 16⅔% fall. Never answer the same percentage.
  • "A is 25% more than B" is not "B is 25% less than A". A = 1.25B, so B = A/1.25 = 0.8A: B is 20% less than A. In general "x% more" reverses to x/(100 + x) less, and "x% less" reverses to x/(100 − x) more: a 20% pay cut needs a 25% raise to restore.
  • Percentage points versus percent. A pass rate that moves from 40% to 50% has risen by 10 percentage points but by 25 percent (10/40). CAT phrases questions both ways; read which one is being asked before computing.
  • Reverse percentage. If a price after a 20% rise is ₹ 600, the original is 600/1.2 = ₹ 500, not 600 × 0.8 = 480. Always divide by the multiplier to go backwards. Worked: a candidate scoring 30% fails by 20 marks and one scoring 40% passes by 10 marks. Then 10% of the total = 30, so the total is 300 and the pass mark is 0.3 × 300 + 20 = 110.
  • Election example (two steps). 10% of votes cast are invalid; the winner takes 60% of the valid votes and wins by 1,800. The margin is 60% − 40% = 20% of valid votes, so valid votes = 1,800/0.2 = 9,000, and total votes cast = 9,000/0.9 = 10,000. Forgetting the invalid votes stops at 9,000; multiplying by 0.9 instead of dividing gives 8,100.
⚠️ Percent of what?
Every percentage has a base, and CAT changes the base mid-question. "Profit of 20%" is on cost unless stated; "discount of 20%" is on the marked price; "20% of the valid votes" is not 20% of all votes. Before computing, write the base beside each percentage. Most wrong options in this topic are correct percentages of the wrong base.

Profit, Loss and Discount — the Marked-Price Chain

Three prices sit in a line: cost price (CP) → marked price (MP) → selling price (SP). Mark-up is on CP, discount is on MP, and profit or loss is measured against CP: SP = CP × (1 + profit%) and SP = MP × (1 − discount%). Worked: CP ₹ 800, marked 25% above cost, sold at 10% discount. MP = 800 × 1.25 = 1,000; SP = 1,000 × 0.9 = 900; profit = 100 on 800 = 12.5%. In multipliers: 1.25 × 0.9 = 1.125 — the answer without ever computing the rupee amounts, which is how it should be done under the clock.

Mark-up m, discount d → profit as a multiplier (1 + m)(1 − d)
Mark-up on CPDiscount on MPMultiplierResult
25%20%1.25 × 0.80 = 1.00No profit, no loss
40%25%1.40 × 0.75 = 1.055% profit (not 15%)
50%30%1.50 × 0.70 = 1.055% profit (not 20%)
20%10%1.20 × 0.90 = 1.088% profit
100%50%2.00 × 0.50 = 1.00Break-even — "50% off" on a doubled price is no discount at all
  • Successive discounts. 20% then 10% is a single discount of 1 − 0.8 × 0.9 = 1 − 0.72 = 28%, not 30%. By the a + b + ab/100 rule with both negative: −20 − 10 + 2 = −28.
  • False weights. A trader who "sells at cost price" but gives 800 g for a kilogram collects the price of 1,000 g for goods costing 800 g: profit = 200/800 = 25%. Formula: error/(true weight − error) × 100. Combined with a price change: marks up 25%, discounts 20%, and gives 900 g per kg. The customer pays 1.25 × 0.8 = 1.00 × CP of a kilogram for goods that cost 0.9 × CP, so profit = 1.00/0.9 − 1 = 11.11%. Ignoring the weight says 0%; adding the percentages says 25 − 20 + 10 = 15%; both are options CAT will list.
  • Same selling price, x% profit and x% loss. Two items sold at the same SP, one at 10% profit and one at 10% loss, always yield an overall loss of x²/100 % = 1%. Reason: the CPs are SP/1.1 and SP/0.9, whose sum exceeds 2·SP. The "no profit no loss" instinct is wrong every time.
  • Profit on SP versus on CP. "Profit is 20% of the selling price" means SP − CP = 0.2 SP, so CP = 0.8 SP and profit on CP = 0.2/0.8 = 25%. A ₹ 5 profit on a ₹ 105 sale is 4.76% of SP but 5% of CP.
  • Break-even. With fixed cost F, selling price s and variable cost v per unit, break-even is at F/(s − v) units. Fixed cost ₹ 60,000, s = ₹ 50, v = ₹ 30 → 60,000/20 = 3,000 units; every unit beyond adds ₹ 20 profit, so 4,500 units earn ₹ 30,000.

Simple and Compound Interest — Growth as a Multiplier

Simple interest grows linearly: SI = P·r·t/100, the same rupee amount every year. Compound interest grows geometrically: amount A = P(1 + r/100)ⁿ, so each year's interest is r% of a larger base. Compounded half-yearly, use rate r/2 and 2n periods; quarterly, r/4 and 4n. Population growth, inflation, depreciation (multiplier 1 − r/100) and bacterial growth are all compound interest under another name, and CAT sets them with the same formula.

₹ 10,000 at 10% per annum — SI versus CI
YearsSimple interestCompound interestCI − SI
11,0001,0000
22,0002,100 (×1.21)100 = P(r/100)²
33,0003,310 (×1.331)310 = P(r/100)²(3 + r/100)
2½ (annual compounding)2,5002,705 (×1.21 × 1.05)205
  • Difference formulas. For two years CI − SI = P(r/100)²; for three years CI − SI = P(r/100)²(3 + r/100). So "the difference between CI and SI for 2 years at 10% is ₹ 50" gives P × 0.01 = 50, P = ₹ 5,000 in one line.
  • Fractional periods. With annual compounding, the whole years compound and the fraction earns simple interest on the accumulated amount: ₹ 8,000 for 1½ years at 10% → 8,000 × 1.1 × 1.05 = ₹ 9,240. If the question says compounded half-yearly instead, it is 8,000 × 1.05³ = 9,261. Read the compounding period before choosing.
  • Instalments (EMI logic). A loan is cleared by equal instalments X when the present values of the instalments add up to the loan: P = X/(1 + r) + X/(1 + r)² + … Worked: ₹ 3,310 borrowed at 10% compound, repaid in three equal annual instalments. Then X/1.1 + X/1.21 + X/1.331 = 3,310. Multiply through by 1.331: X(1.21 + 1.1 + 1) = 3,310 × 1.331, so 3.31 X = 4,405.61 and X = ₹ 1,331. Check by present values: 1,331/1.1 = 1,210; 1,331/1.21 = 1,100; 1,331/1.331 = 1,000; total 3,310 ✓. The same loan over two years needs 2,100 → instalments of 1,210.
  • Growth and depreciation. A town of 8,000 growing 5% a year has 8,000 × 1.05² = 8,820 people after two years; a ₹ 20,000 machine losing 10% a year is worth 20,000 × 0.9² = ₹ 16,200 after two. At 10% compound, money doubles in a little over 7 years (1.1⁷ ≈ 1.95, 1.1⁸ ≈ 2.14) — the "rule of 72" estimate 72/10 = 7.2 is good enough to eliminate options.
🧠 Memorise the powers, not the formula
1.05² = 1.1025; 1.1² = 1.21, 1.1³ = 1.331, 1.1⁴ = 1.4641; 1.2² = 1.44, 1.2³ = 1.728; 1.25² = 1.5625; 1.5² = 2.25. With these, "₹ 12,100 after two years at 10%" is instantly P = 12,100/1.21 = ₹ 10,000, and "₹ 1,728 after three years at 20%" is P = ₹ 1,000. The numbers CAT chooses are almost always built from this list.

Ratio, Proportion, Variation and Partnership

To chain ratios, match the shared term: a : b = 2 : 3 and b : c = 4 : 5 give a : b : c = 8 : 12 : 15 (scale the first by 4 and the second by 3 so that b = 12 in both). If a/b = c/d then componendo–dividendo gives (a + b)/(a − b) = (c + d)/(c − d), which turns an equation into a ratio in one step. Worked: (3x + 4y)/(3x − 4y) = 7/3. Apply componendo–dividendo backwards: 3x/4y = (7 + 3)/(7 − 3) = 10/4 = 5/2, so x/y = (5/2) × (4/3) = 10/3. Check with x = 10, y = 3: (30 + 12)/(30 − 12) = 42/18 = 7/3 ✓.

  • Proportion vocabulary. In a : b :: c : d the product of extremes equals the product of means, ad = bc. The mean proportional between a and b is √(ab) (between 4 and 9 it is 6); the third proportional to a and b is b²/a (to 4 and 6 it is 9). Duplicate ratio of a : b is a² : b²; sub-duplicate is √a : √b.
  • Ages, the standard ratio problem. Present ages are in the ratio 3 : 5; in 8 years the ratio will be 5 : 7. Let the ages be 3x and 5x: (3x + 8)/(5x + 8) = 5/7 → 21x + 56 = 25x + 40 → 4x = 16 → x = 4. The ages are 12 and 20 (check: 20 : 28 = 5 : 7 ✓). The trap is adding 8 to the ratio terms rather than to the ages.
  • Partnership: profit follows capital × time. A invests ₹ 5,000 for 12 months and B invests ₹ 6,000 for 8 months. Their shares are 5,000 × 12 : 6,000 × 8 = 60,000 : 48,000 = 5 : 4, so a profit of ₹ 18,000 splits as ₹ 10,000 and ₹ 8,000. If a partner changes capital mid-year, add the products for each period separately: ₹ 4,000 for 6 months then ₹ 8,000 for 6 months counts as 24,000 + 48,000 = 72,000.
  • Variation. Direct: y = kx (double x, double y). Inverse: y = k/x (double x, halve y). Joint: y = k·x²/z, in which case doubling x and doubling z multiplies y by 4/2 = 2. Worked: the price of a diamond varies as the square of its weight. A 6-carat stone breaks into pieces in the ratio 1 : 2 : 3, i.e. 1, 2 and 3 carats. Value before: k × 36; after: k(1 + 4 + 9) = 14k. Loss = 22k/36k = 61.1% — and a fall in value is inevitable whenever the exponent exceeds 1.
🎯 Why ratios rather than actual values
CAT frequently gives only ratios and asks for a ratio or a percentage, so the actual numbers never exist. Assign the smallest convenient values (let the capital be 100, the speed be the LCM of the numbers given, the total votes be 100) and compute with those. Any answer that is itself a ratio or percentage is unchanged by the choice, and the arithmetic becomes trivial.

Mixtures, Alligation, Replacement and Averages

The alligation cross: when two ingredients of "strength" c₁ and c₂ (price, concentration, or average) are mixed to give a mean strength m lying between them, the quantities are in the ratio (c₂ − m) : (m − c₁) — each ingredient's share is the other ingredient's distance from the mean. Rice at ₹ 20/kg and ₹ 30/kg mixed to sell at ₹ 24/kg: cheaper : dearer = (30 − 24) : (24 − 20) = 6 : 4 = 3 : 2. Check: (3 × 20 + 2 × 30)/5 = 120/5 = 24 ✓. The same cross handles a class average from two groups, an average speed from two legs of equal time, and a profit percentage from two batches.

  • Two solutions to a target concentration. Mix 30% and 60% alcohol to get 50%: ratio = (60 − 50) : (50 − 30) = 10 : 20 = 1 : 2 — one part of the weaker to two of the stronger. Check: (30 + 2 × 60)/3 = 150/3 = 50 ✓. With a fixed quantity: 60 L of 40% solution needs how much 90% solution to reach 60%? Ratio 40% : 90% = (90 − 60) : (60 − 40) = 30 : 20 = 3 : 2, so 60 L of the weaker pairs with 40 L of the stronger.
  • Adding water (concentration 0). 20 L of 25% acid contains 5 L of acid; to dilute to 20% we need 5/(20 + x) = 0.2, so 20 + x = 25 and x = 5 L. Track the quantity that does not change — here the acid — rather than the percentage.
  • Repeated replacement. From a vessel holding V litres of pure liquid, x litres are drawn and replaced by water, n times. Pure liquid left = V(1 − x/V)ⁿ. From 50 L of milk, 10 L replaced three times: 50 × (0.8)³ = 50 × 0.512 = 25.6 L milk, so 24.4 L water. Linear thinking says 30 L of water — wrong, because each later draw removes some water too. From 40 L with 8 L replaced thrice: 40 × 0.8³ = 20.48 L milk.
  • Weighted average. 30 students average 40 and 20 students average 60: overall (30 × 40 + 20 × 60)/50 = 2,400/50 = 48, not 50. By alligation: the mean divides 40 and 60 in the ratio 20 : 30 = 2 : 3, so it sits 8 above 40.
  • Replacing a member. If the average of n people rises by d when one person is replaced, the newcomer is heavier than the leaver by n × d. Ten people, average up 2 kg, leaver 50 kg → newcomer = 50 + 10 × 2 = 70 kg. Eight people, average up 2.5 kg, leaver 65 kg → 65 + 20 = 85 kg. Adding d once (52 or 67.5) is the standard slip.
  • Average speed. For equal distances at u and v, the average speed is the harmonic mean 2uv/(u + v): 40 and 60 give 4,800/100 = 48, not 50. For equal times it is the plain mean (u + v)/2 = 50. For unequal legs, always compute total distance ÷ total time: a quarter of a journey at 30 km/h and the rest at 60 km/h takes D/120 + 3D/240 = 5D/240 hours, so the average is 240/5 = 48 km/h.
⚠️ Average speed is never the average of the speeds
Unless the times are equal, (u + v)/2 is wrong — and it is always one of the four options. Equal distances → harmonic mean, which is always below the arithmetic mean. If a candidate's answer for a two-speed journey over equal distances is at or above the simple average, it is wrong before any checking.

Key takeaways

  • Every percentage change is a multiplier — ×(1 ± x/100) — and a chain of changes is one product; two successive changes of a% and b% net to a + b + ab/100.
  • Mark-up is on cost, discount is on marked price, profit is on cost: profit multiplier = (1 + m)(1 − d), and a false weight divides it further by the fraction actually delivered.
  • CI − SI over two years is P(r/100)²; a fractional year compounds the whole years and adds simple interest for the fraction; equal instalments are found by equating present values.
  • Alligation: quantities are in the ratio (c₂ − m) : (m − c₁), each share being the other ingredient's distance from the mean; repeated replacement leaves V(1 − x/V)ⁿ of the original liquid.
  • Replacing one of n members shifts the total by n × (change in average), and equal distances at two speeds average to the harmonic mean 2uv/(u + v), never (u + v)/2.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. The price of sugar rises by 25%. By what percentage must a household reduce its consumption so that its expenditure on sugar stays the same?

    1. 25%
    2. 20%
    3. 12.5%
    4. 33⅓%
    Show answer

    Answer: B — 20%

    A 25% rise is the multiplier 5/4, so to keep price × quantity fixed the quantity must be multiplied by 4/5, a fall of 1/5 = 20%. Equivalently r/(100 + r) × 100 = 25/125 × 100 = 20%. The value 25% is the trap of answering the same percentage in reverse, which ignores that the base has changed. The value 33⅓% is r/(100 − r), the reverse of a 25% fall, not a rise. The value 12.5% is half the rise and corresponds to no correct method.

  2. In an election between two candidates, 10% of the votes cast were invalid. The winner received 60% of the valid votes and won by 1,800 votes. How many votes were cast in total?

    1. 9,000
    2. 8,100
    3. 10,000
    4. 3,000
    Show answer

    Answer: C — 10,000

    The loser received 40% of the valid votes, so the margin 60% − 40% = 20% of valid votes equals 1,800. Valid votes = 1,800/0.2 = 9,000. These are 90% of the votes cast, so total cast = 9,000/0.9 = 10,000. The value 9,000 stops at the valid votes and forgets the invalid 10%. The value 8,100 multiplies 9,000 by 0.9 instead of dividing — going the wrong way through the multiplier. The value 3,000 is 1,800/0.6, treating the winning margin as if it were the winner's entire vote.

  3. A trader marks his goods 25% above cost price, offers a 20% discount on the marked price, and uses a weight of 900 g while claiming to sell 1 kg. His actual profit percentage is:

    1. 0%
    2. 15%
    3. 10%
    4. 11.11%
    Show answer

    Answer: D — 11.11%

    Let the cost of 1 kg be ₹ 100. Marked price = 125; after the 20% discount the customer pays 125 × 0.8 = ₹ 100. But the customer receives only 900 g, which cost the trader ₹ 90. Profit = 10 on a cost of 90 = 10/90 = 11.11%. The value 0% comes from the price chain alone (1.25 × 0.8 = 1) while ignoring the false weight. The value 15% adds and subtracts the percentages (25 − 20 + 10). The value 10% treats the 100 g shortfall as a profit on 1,000 g instead of on the 900 g actually supplied.

  4. What amount does ₹ 8,000 become in 1½ years at 10% per annum, interest compounded annually?

    1. ₹ 9,240
    2. ₹ 9,200
    3. ₹ 9,261
    4. ₹ 9,680
    Show answer

    Answer: A — ₹ 9,240

    With annual compounding, the first full year compounds and the remaining half-year earns simple interest at half the annual rate on the accumulated amount: 8,000 × 1.1 = 8,800, then 8,800 × 1.05 = ₹ 9,240. The value ₹ 9,200 is simple interest throughout (8,000 × 10 × 1.5/100 = 1,200 added). The value ₹ 9,261 = 8,000 × 1.05³ is the answer if interest were compounded half-yearly, which the question rules out. The value ₹ 9,680 = 8,000 × 1.21 is two full years of compounding.

  5. In what ratio must tea costing ₹ 15/kg be mixed with tea costing ₹ 20/kg so that the mixture costs ₹ 16.50/kg?

    1. 3 : 7
    2. 1 : 1
    3. 7 : 3
    4. 2 : 3
    Show answer

    Answer: C — 7 : 3

    By alligation, cheaper : dearer = (20 − 16.5) : (16.5 − 15) = 3.5 : 1.5 = 7 : 3. Check: (7 × 15 + 3 × 20)/10 = (105 + 60)/10 = 16.5 ✓. The value 3 : 7 is the cross read the wrong way round — each ingredient's share must be the other ingredient's distance from the mean, and since 16.50 is closer to 15, the cheaper tea dominates. The value 1 : 1 would give a mean of 17.50, and 2 : 3 would give 18.

  6. A vessel contains 50 litres of pure milk. 10 litres are drawn out and replaced with water, and this operation is performed three times in all. The quantity of water in the vessel at the end is:

    1. 30 L
    2. 25.6 L
    3. 18 L
    4. 24.4 L
    Show answer

    Answer: D — 24.4 L

    Each operation keeps (1 − 10/50) = 0.8 of whatever milk is present, so after three operations the milk is 50 × 0.8³ = 50 × 0.512 = 25.6 L, and the water is 50 − 25.6 = 24.4 L. The value 25.6 L is the milk, not the water — the question asks for water. The value 30 L is the linear slip 3 × 10, which ignores that the second and third draws remove some water along with the milk. The value 18 L is the water after only two operations (50 − 50 × 0.64).

  7. How many litres of a 90% alcohol solution must be added to 60 litres of a 40% alcohol solution so that the resulting mixture is 60% alcohol?

    1. 90 L
    2. 40 L
    3. 60 L
    4. 30 L
    Show answer

    Answer: B — 40 L

    By alligation the 40% and 90% solutions mix in the ratio (90 − 60) : (60 − 40) = 30 : 20 = 3 : 2. The 60 L of 40% solution is the "3" part, so the 90% solution is 60 × 2/3 = 40 L. Check: alcohol = 0.4 × 60 + 0.9 × 40 = 24 + 36 = 60 L in 100 L = 60% ✓. The value 90 L inverts the ratio (60 × 3/2). The value 60 L assumes equal quantities, which would give 65%. The value 30 L comes from misreading 3 : 2 as 2 : 1.

  8. A car covers the first quarter of a journey at 30 km/h and the remaining three-quarters at 60 km/h. Its average speed for the whole journey is:

    1. 40 km/h
    2. 48 km/h
    3. 45 km/h
    4. 52.5 km/h
    Show answer

    Answer: B — 48 km/h

    Let the journey be D km. Time for the first quarter = (D/4)/30 = D/120 h; time for the rest = (3D/4)/60 = D/80 h. Total time = D/120 + D/80 = (2D + 3D)/240 = D/48 h, so the average speed is D ÷ (D/48) = 48 km/h. The value 40 km/h is the harmonic mean 2 × 30 × 60/90, valid only for equal distances. The value 52.5 km/h weights the speeds by distance, (30 + 3 × 60)/4, which is not how average speed works. The value 45 km/h is the plain mean of the two speeds, valid only for equal times.