Arithmetic Foundations — Numbers, Percentage, Ratio & Averages

The paper loves questions like "the unit digit of 7¹⁰⁵". You are not expected to compute 7¹⁰⁵ — you are expected to know that unit digits repeat in a cycle of at most 4. Divide the power by 4 and look at the remainder: remainder 1, 2, 3 → take the 1st, 2nd, 3rd entry of the cycle; remainder 0 → take the last (4th) entry, because a remainder of 0 means the cycle has just finished.

Worked example 1. Unit digit of 7¹⁰⁵. 105 ÷ 4 leaves remainder 1, so take the 1st entry of 7-cycle (7, 9, 3, 1) → 7. Worked example 2. Unit digit of 2⁵⁷: 57 ÷ 4 leaves remainder 1 → 1st entry of (2, 4, 8, 6) → 2. Worked example 3. Unit digit of 13⁴³ × 17²⁵. Only the last digits of the bases matter, so read it as 3⁴³ × 7²⁵. For 3⁴³: 43 ÷ 4 leaves 3 → 3rd entry of (3, 9, 7, 1) = 7. For 7²⁵: 25 ÷ 4 leaves 1 → 7. Now 7 × 7 = 49 → unit digit 9.

Trailing zeros in a factorial. A zero at the end comes from a factor of 10 = 2 × 5, and in any factorial there are always more 2s than 5s, so count the 5s. The count is n/5 + n/25 + n/125 + … (integer division each time). For 100!: 100 ÷ 5 = 20, 100 ÷ 25 = 4, 100 ÷ 125 = 0 → 20 + 4 = 24 zeros. For 50!: 10 + 2 = 12 zeros. For 1 × 2 × 3 × … × 25 (that is 25!): 5 + 1 = 6 zeros. For 125!: 25 + 5 + 1 = 31 zeros.

Remainder basics. Two ideas finish almost every remainder question. (i) Reduce the base first. Remainder of 17²³ ÷ 16: since 17 leaves remainder 1 on division by 16, 17²³ also leaves 1. (ii) Hunt for −1. Remainder of 7¹⁰⁰ ÷ 8: 7 = 8 − 1, so 7¹⁰⁰ behaves like (−1)¹⁰⁰ = +1 → remainder 1. Remainder of 5⁹⁹ ÷ 6: 5 = 6 − 1, and 99 is odd, so (−1)⁹⁹ = −1, which means remainder 6 − 1 = 5. (iii) Build a power that gives 1. Remainder of 2⁵¹ ÷ 7: 2³ = 8 leaves remainder 1, and 51 = 3 × 17, so 2⁵¹ = (2³)¹⁷ leaves 1.

HCF (highest common factor) is the biggest number that divides all of them; LCM (lowest common multiple) is the smallest number they all divide. The single most useful relation, true only for two numbers, is first × second = LCM × HCF.

Worked example. The HCF of two numbers is 12 and their LCM is 336. If one number is 84, find the other. Other = (12 × 336) ÷ 84 = 4032 ÷ 84 = 48. Verify independently: HCF(84, 48) = 12 ✓ and LCM(84, 48) = 336 ✓.

LCM and HCF of fractions. Two formulae that look backwards but are correct: LCM of fractions = LCM of numerators ÷ HCF of denominators, and HCF of fractions = HCF of numerators ÷ LCM of denominators. Worked on 2/3, 8/9 and 16/81: LCM = LCM(2, 8, 16) ÷ HCF(3, 9, 81) = 16 ÷ 3 = 16/3. HCF = HCF(2, 8, 16) ÷ LCM(3, 9, 81) = 2 ÷ 81 = 2/81. Sanity check: the LCM of fractions must be at least as large as the largest fraction (16/3 > 16/81 ✓) and the HCF at most the smallest (2/81 < 2/3 ✓).

Remainder family, case 1 — the same remainder every time. "Find the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case." Take LCM(5, 6, 7, 8) = 840 and add the remainder: 843. Check: 843 = 840 + 3, so each of 5, 6, 7, 8 leaves exactly 3. ✓ The general form is N = k × LCM + r, so the next such numbers are 1683, 2523, …

Case 2 — a constant shortfall. "Find the least number which leaves remainder 4 on division by 5, remainder 5 on division by 6 and remainder 6 on division by 7." Notice each remainder is exactly 1 less than its divisor. So the number is 1 short of a common multiple: N = LCM(5, 6, 7) − 1 = 210 − 1 = 209. Check: 209 ÷ 5 leaves 4, ÷ 6 leaves 5, ÷ 7 leaves 6. ✓

Case 3 — the greatest number, so use HCF of differences. "Find the greatest number which divides 43, 91 and 183 leaving the same remainder in each case." Take the differences: 91 − 43 = 48, 183 − 91 = 92, 183 − 43 = 140. HCF(48, 92, 140) = 4. Check: 43, 91 and 183 all leave remainder 3 on division by 4. ✓ Variant with stated remainders: "the greatest number that divides 2011 and 2623 leaving remainders 9 and 5" — subtract the remainders first, then take the HCF: HCF(2011 − 9, 2623 − 5) = HCF(2002, 2618) = 154. Check: 154 × 13 = 2002 so 2011 leaves 9 ✓, and 154 × 17 = 2618 so 2623 leaves 5 ✓.

Bells tolling together. Six bells toll at intervals of 2, 4, 6, 8, 10 and 12 seconds. They start together; how often do they toll together again? LCM(2, 4, 6, 8, 10, 12) = 120 seconds, i.e. every 2 minutes. In 30 minutes = 1800 seconds, they coincide 1800 ÷ 120 = 15 times after the start, or 16 times if you also count the very first toll at zero seconds. Read the wording carefully — The paper prints both 15 and 16 as options, and the phrase "in 30 minutes" normally means the 15 later coincidences.

BODMAS fixes the order: Brackets, Of, Division, Multiplication, Addition, Subtraction — with division and multiplication ranked equally and worked left to right, and the same for addition and subtraction. Worked: 12 + 6 ÷ 2 × 3 − 4 = 12 + 3 × 3 − 4 = 12 + 9 − 4 = 17. Worked with "of": 1/2 of 60 + 20 ÷ 4 × 2 = 30 + 5 × 2 = 30 + 10 = 40. Note that "of" is done before division, which is exactly why it has its own letter.

Comparing fractions in seconds. Cross-multiply: for a/b versus c/d, compare a × d with c × b. Order 5/7, 7/9 and 9/11. 5/7 vs 7/9 → 5 × 9 = 45 against 7 × 7 = 49, so 7/9 is bigger. 7/9 vs 9/11 → 7 × 11 = 77 against 9 × 9 = 81, so 9/11 is bigger. Hence 5/7 < 7/9 < 9/11. The pattern is worth remembering: for a proper fraction, adding the same number to numerator and denominator increases it; for an improper fraction it decreases it.

Laws of indices, all one idea: aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, (aᵐ)ⁿ = aᵐⁿ, a⁰ = 1, a⁻ⁿ = 1/aⁿ, and a^(1/n) = ⁿ√a. Worked: (243)^0.16 × (243)^0.04 = (243)^0.20 = (3⁵)^(1/5) = 3. Worked: if 2^(x+1) = 32 = 2⁵ then x + 1 = 5, so x = 4. The technique in both is the same — force both sides onto the same base, then equate the powers.

Rationalisation. To clear a surd from the denominator, multiply above and below by the conjugate — change the middle sign. Worked: (√5 + √3)/(√5 − √3) × (√5 + √3)/(√5 + √3) = (√5 + √3)²/(5 − 3) = (5 + 3 + 2√15)/2 = (8 + 2√15)/2 = 4 + √15. Notice the denominator became a plain number because (a − b)(a + b) = a² − b².

Square roots and cube roots without long division. For a perfect square, the last digit of the number fixes the last digit of the root (1 → 1 or 9; 4 → 2 or 8; 5 → 5; 6 → 4 or 6; 9 → 3 or 7; 0 → 0), and the leading digits fix the tens digit. √1024: it ends in 4 so the root ends in 2 or 8; 1024 lies between 900 and 1600, so the root is in the thirties → 32 or 38; since 1024 is nearer 900, take 32 (32² = 1024 ✓). Also note that a perfect square never ends in 2, 3, 7 or 8 — an instant elimination.

Cube roots are even friendlier, because the last digit of a cube determines the last digit of the root uniquely: 1 → 1, 8 → 2, 7 → 3, 4 → 4, 5 → 5, 6 → 6, 3 → 7, 2 → 8, 9 → 9, 0 → 0 (note how 2/8 and 3/7 swap). ∛12167: last digit 7 → root ends in 3; strike off the last three digits to leave 12, which lies between 2³ = 8 and 3³ = 27, so the tens digit is 2 → 23, and 23³ = 12167 ✓. ∛175616: last digit 6 → root ends in 6; 175 lies between 5³ = 125 and 6³ = 216 → tens digit 5 → 56, and 56³ = 175616 ✓.

For decimals, count the digits: a square root halves the number of decimal places. √0.0081 has four decimal places, so its root has two: 0.09, and 0.09 × 0.09 = 0.0081 ✓. Do not write 0.9 — that squares to 0.81, a hundred times too big. This single slip is one of the commonest careless losses in the simplification block.

A percentage is a fraction with denominator 100 — nothing more. x% of N = N × x/100. But the moment you convert the percentage into a fraction, arithmetic that looked like long multiplication becomes a single division. 37.5% of 480 is not a multiplication of 480 by 0.375; it is 3/8 of 480 = 180. And 16 2/3% of 4,236 is 1/6 of 4,236 = 706. That is the whole reason the table below matters.

Increase and decrease. A rise of x% multiplies by (100 + x)/100; a fall of x% multiplies by (100 − x)/100. A population of 15,000 rising 8% becomes 15000 × 108/100 = 16,200. And percentage change = (change ÷ original) × 100 — the denominator is always the original, never the new value.

The marks-and-pass-percentage type. A candidate scoring 40% fails by 20 marks; another scoring 45% gets 10 marks more than the pass mark. Find the maximum marks. Let the total be T. Pass mark = 0.40T + 20 and also = 0.45T − 10. Equating: 0.05T = 30 → T = 600. Then the pass mark = 0.40 × 600 + 20 = 260, and the check works: 0.45 × 600 = 270 = 260 + 10 ✓.

The two-candidate election type. In a contest between two candidates the winner gets 56% of the valid votes and wins by 1,344 votes. Total valid votes? The winner has 56% and the loser 44%, so the margin is 12% of the total: 0.12T = 1344 → T = 11,200. Check: 6,272 and 4,928, difference 1,344 ✓. The whole trick is realising the margin, not the winning share, is what you were given.

Two successive percentage changes of a% and b% never simply add. The net change is a + b + ab/100, where a fall is entered as a negative number. Worked: a salary rises 20% and then falls 20%. Net = 20 − 20 + (20 × −20)/100 = 0 − 4 = −4%, a 4% fall, not zero. Verify on ₹5,000: 5000 × 1.20 = 6,000, then 6000 × 0.80 = ₹4,800, which is 4% below 5,000 ✓. Worked upward: +10% then +20% gives 10 + 20 + 200/100 = +32%.

The more/less pair. If A is x% more than B, then B is 100x/(100 + x)% less than A. If A is x% less than B, then B is 100x/(100 − x)% more than A. The two answers are never the same number, and that asymmetry is the whole question. The reason is simply that the base changes: the first percentage sits on B, the second on A.

Price, consumption and expenditure. Expenditure = price × consumption, so if the price rises x% and you want the expenditure unchanged, the consumption must fall by 100x/(100 + x)% — the very same more/less formula. Worked: sugar costs 25% more, so consumption must be cut by 100 × 25/125 = 20%. Check with numbers: 1 kg at ₹40 costs ₹40; the price becomes ₹50; buying 0.8 kg costs 50 × 0.8 = ₹40 ✓. Conversely if the price falls 20%, consumption may rise by 100 × 20/80 = 25% at the same expenditure.

If both price and consumption change, use the successive formula on the expenditure. Price up 20% and consumption up 10% → expenditure up 20 + 10 + (20 × 10)/100 = 32%. The same machinery covers area problems (length up a%, breadth up b%) and population problems (growth rate a% then b%) — one formula, three chapters.

A ratio compares magnitudes and carries no units. To divide a sum in a ratio, add the parts and find one part. Worked: divide ₹1,350 among A, B and C in 2 : 3 : 4. Parts = 9, so one part = 1350 ÷ 9 = ₹150, giving ₹300, ₹450 and ₹600 (sum 1,350 ✓). Never divide by the number of people — divide by the number of parts.

Standard vocabulary, asked directly: the compound ratio of a : b and c : d is ac : bd — so 2 : 3 compounded with 4 : 5 is 8 : 15. The duplicate ratio of 2 : 3 is 4 : 9 (squares); the sub-duplicate is √2 : √3; the triplicate is 8 : 27 (cubes); the sub-triplicate is ∛2 : ∛3. An inverse ratio simply flips it: the inverse of 2 : 3 is 3 : 2.

Ratio changes when a number is added. The ratio of two numbers is 3 : 5. When 9 is added to each, the ratio becomes 2 : 3. Find the numbers. Write them as 3x and 5x: (3x + 9)/(5x + 9) = 2/3 → 3(3x + 9) = 2(5x + 9) → 9x + 27 = 10x + 18 → x = 9, so the numbers are 27 and 45. Check: 36/54 = 2/3 ✓. The single most important habit here is to introduce x, not to guess from the options.

Income–expenditure–saving. The incomes of A and B are in the ratio 5 : 4 and their expenditures in the ratio 3 : 2. If each saves ₹6,000, find their incomes. Let incomes be 5x and 4x, expenditures 3y and 2y. Then 5x − 3y = 6000 and 4x − 2y = 6000. Subtracting, x − y = 0, so x = y. Substituting, 5x − 3x = 2x = 6000 → x = 3000. Incomes are ₹15,000 and ₹12,000 (expenditures ₹9,000 and ₹6,000, savings ₹6,000 each ✓).

Proportion words. If a : b = c : d then ad = bc, and d is the fourth proportional; the fourth proportional to 3, 5 and 21 is (5 × 21)/3 = 35. The mean proportional between a and b is √(ab): between 9 and 16 it is √144 = 12. The third proportional to a and b is b²/a: to 4 and 12 it is 144/4 = 36.

Coins in a ratio — mind the values. A bag has one-rupee, 50-paise and 25-paise coins in the ratio 5 : 6 : 8, totalling ₹150. Find the number of each. Take 5x, 6x, 8x coins. Value = 5x(1) + 6x(0.5) + 8x(0.25) = 5x + 3x + 2x = 10x = 150 → x = 15. So 75, 90 and 120 coins, with value 75 + 45 + 30 = ₹150 ✓. The ratio of numbers is not the ratio of values — that is the whole question.

Partnership is ratio with a time dimension. In simple partnership the capitals stay in for the same period, so profit divides in the capital ratio: A puts in ₹8,000 and B ₹12,000 for a year, profit ₹15,000 → ratio 2 : 3 → ₹6,000 and ₹9,000. In compound partnership the periods differ, so profit divides in the ratio of capital × time: A invests ₹20,000 for 12 months, B joins later with ₹30,000 for 6 months → (20000 × 12) : (30000 × 6) = 240000 : 180000 = 4 : 3. On a profit of ₹35,000 that is ₹20,000 and ₹15,000 — not the ₹14,000/₹21,000 you would get from the bare capital ratio.

Working versus sleeping partner. A working partner manages the business and is paid a fee — usually a percentage of the profit — before the rest is split by capital. Worked: A and B invest ₹50,000 and ₹70,000. A, the working partner, gets 10% of the profit for managing; the remainder is divided in the capital ratio. Total profit ₹24,000. Fee = ₹2,400, leaving ₹21,600 to split as 5 : 7 → one part = 21600 ÷ 12 = ₹1,800 → A gets ₹9,000 and B gets ₹12,600. So A receives 2400 + 9000 = ₹11,400 and B ₹12,600 (total ₹24,000 ✓).

Three partners joining at different times. A invests ₹12,000 at the start; B joins after 3 months with ₹16,000; C joins 6 months after B with ₹18,000. Split a year-end profit of ₹57,000. Capital × months: A = 12000 × 12 = 1,44,000; B = 16000 × 9 = 1,44,000; C = 18000 × 3 = 54,000. Divide all by 18,000 → 8 : 8 : 3, that is 19 parts, one part = 57000 ÷ 19 = ₹3,000. So A ₹24,000, B ₹24,000, C ₹9,000 ✓.

Average = total ÷ number of items, so total = average × number — and that second form is what you actually use. Almost every average question is solved by converting averages into totals, doing the arithmetic on totals, and converting back at the end.

Consecutive numbers. For any evenly spaced list, average = (first + last)/2, which is also the middle term. So the average of 1 to 50 is (1 + 50)/2 = 25.5, and the average of the first n natural numbers is (n + 1)/2. Two more worth memorising: the average of the first n odd numbers is n (1, 3, 5, 7, 9 average 5 ✓) and the average of the first n even numbers is n + 1 (2, 4, 6, 8, 10 average 6 ✓).

Weighted average. When two groups of different sizes are merged, the combined average is (n₁a₁ + n₂a₂)/(n₁ + n₂). Worked: 30 students average 42 marks and 20 students average 52. Combined = (30 × 42 + 20 × 52)/50 = (1260 + 1040)/50 = 2300/50 = 46. Note it is not the plain average 47 — the larger group pulls the answer towards itself.

Replacing a member. The average weight of 8 persons increases by 2.5 kg when a new person replaces one weighing 65 kg. The new person weighs 65 + 8 × 2.5 = 85 kg. The logic: the total rose by 8 × 2.5 = 20 kg, and that entire rise is the difference between the two individuals. Adding a member works the same way — if adding one number to 10 numbers averaging 23 raises the average to 25, the new number is 11 × 25 − 10 × 23 = 275 − 230 = 45.

The cricket-average classic. A batsman in his 17th innings scores 85 and thereby raises his average by 3. Find the new average. The 85 has to cover the new average once and pay 3 extra runs to each of the 16 earlier innings, so new average = 85 − 3 × 16 = 37. Verify fully: earlier average 34 over 16 innings = 544 runs; 544 + 85 = 629; 629 ÷ 17 = 37 ✓.

Correcting a wrong entry. The average of 25 observations is 36, but one value 48 was wrongly read as 23. Correct average? The total was short by 48 − 23 = 25, so the average rises by 25 ÷ 25 = 1 → 37. Never redo the whole sum; only the change in total matters.

Alligation is the weighted average read backwards: you know the mean and both ingredients, and you want the ratio. The rule of alligation says cheaper quantity : dearer quantity = (dearer price − mean) : (mean − cheaper price). Draw a cross, put the two prices at the top corners and the mean in the middle; each diagonal difference lands under the opposite price.

Worked example 1 — prices. In what ratio must rice at ₹40/kg be mixed with rice at ₹60/kg so that the mixture is worth ₹45/kg? Ratio = (60 − 45) : (45 − 40) = 15 : 5 = 3 : 1. Check: three kg at ₹40 plus one kg at ₹60 costs 120 + 60 = ₹180 for 4 kg, that is ₹45/kg ✓. Notice the answer leans towards the cheaper rice, exactly as it must, since ₹45 is nearer ₹40.

Worked example 2 — free water. In what ratio must water be mixed with milk costing ₹36 a litre so that the mixture can be sold at ₹30 a litre without loss? Water costs ₹0, so ratio = (36 − 30) : (30 − 0) = 6 : 30 = 1 : 5, i.e. one litre of water to five of milk. Check: 5 litres of milk cost ₹180, and 6 litres of mixture at ₹30 fetches ₹180 ✓.

The two-groups row deserves its own working. A class of 45 has an average mark of 60; the boys average 70 and the girls 55. How many boys? Cross rule on 70 and 55 about 60: boys : girls = (60 − 55) : (70 − 60) = 5 : 10 = 1 : 2, so boys = 45 × 1/3 = 15 and girls = 30. Verify: (15 × 70 + 30 × 55)/45 = (1050 + 1650)/45 = 2700/45 = 60 ✓. And the interest row: ₹5,000 lent partly at 5% and partly at 8% yields ₹340 in a year, so the overall rate is 340/5000 × 100 = 6.8%; ratio = (8 − 6.8) : (6.8 − 5) = 1.2 : 1.8 = 2 : 3 → ₹2,000 and ₹3,000. Check: 100 + 240 = ₹340 ✓.

Repeated replacement. If a vessel holds x units of pure liquid and y units are drawn out and replaced with water, and this is done n times, the pure liquid left = x(1 − y/x)ⁿ. Worked: from a 40-litre cask of pure milk, 4 litres are drawn out and replaced by water, twice. Milk left = 40 × (1 − 4/40)² = 40 × (0.9)² = 40 × 0.81 = 32.4 litres, so water = 7.6 litres and milk : water = 81 : 19. Do the second step slowly the first time: after the first replacement the cask holds 36 L milk and 4 L water, and when you now draw 4 L of mixture you remove 3.6 L milk, leaving 32.4 L ✓.