Mathematical Reasoning & Aptitude

A number-series question hides a rule and shows you four or five of its outputs. The disciplined method has three steps, always in this order. Step 1 — write the differences between consecutive terms below the row. Step 2 — if the differences are not constant, take the differences of the differences. Step 3 — if differences lead nowhere, check ratios (divide each term by the previous one) and then check the terms against the squares, cubes and prime lists you know by heart. Ninety per cent of series break at Step 1 or Step 2.

Worked example 1 — constant second difference. Series: 2, 6, 12, 20, 30, ? The first differences are 6−2 = 4, 12−6 = 6, 20−12 = 8, 30−20 = 10. These are not constant, but their own differences are 2, 2, 2 — perfectly constant. So the next first difference must be 10 + 2 = 12, and the missing term is 30 + 12 = 42. A second reading of the same series: 2 = 1×2, 6 = 2×3, 12 = 3×4, 20 = 4×5, 30 = 5×6, so the next is 6×7 = 42. Two independent routes agreeing is the cheapest possible check.

Worked example 2 — a multiply-and-add rule. Series: 5, 11, 23, 47, ? Differences are 6, 12, 24 — they are doubling, which is the signature of a "×2 then +something" rule. Test it: 5×2 + 1 = 11 ✓, 11×2 + 1 = 23 ✓, 23×2 + 1 = 47 ✓. So the next term is 47×2 + 1 = 95. Worked example 3 — squares in disguise. Series: 3, 8, 15, 24, 35, ? Add 1 to every term and you get 4, 9, 16, 25, 36 — the squares of 2, 3, 4, 5, 6. So the rule is n² − 1 and the next term is 7² − 1 = 49 − 1 = 48.

Odd-one-out and wrong-term questions are the same skill worn differently. Series: 8, 27, 64, 100, 125, 216 — find the term that does not belong. Read them as cubes: 2³ = 8, 3³ = 27, 4³ = 64, 5³ = 125, 6³ = 216. Every term is a perfect cube except 100, which is 10² but not a cube (4³ = 64 and 5³ = 125 sit on either side of it). So 100 is the intruder; the term that should be there is 125 — and 125 already appears, confirming 100 is the extra one.

Every letter-series question becomes a number-series question the moment you replace each letter by its position in the alphabet. So the one thing you must be able to do instantly is convert both ways. The forward positions are A = 1 up to Z = 26; the reverse position of any letter is 27 minus its forward position, because 1 + 26 = 27. So L is 12 forward, and 27 − 12 = 15, which is O — L and O are mirror partners.

Worked example 1. Complete: B, D, G, K, P, ? Convert: B = 2, D = 4, G = 7, K = 11, P = 16. Differences are 2, 3, 4, 5 — so the next difference is 6, giving 16 + 6 = 22. And 22 is V (since T = 20, U = 21, V = 22). Worked example 2. Complete: Z, X, V, T, ? Convert: 26, 24, 22, 20 — falling by 2 each time, so the next is 18, which is R. Worked example 3. Complete: A, C, F, J, O, ? Positions 1, 3, 6, 10, 15 are the triangular numbers, rising by 2, 3, 4, 5; the next rise is 6, giving 21 = U.

An alphanumeric series is a single line mixing digits, letters and sometimes symbols, followed by counting questions. There is no pattern to find — only careful reading. Take the line: 5 B 9 K 3 T P 7 M 2 J 8 D 4. Question: how many letters are immediately preceded by a digit and immediately followed by a digit? Walk left to right and test each letter. B sits between 5 and 9 ✓. K sits between 9 and 3 ✓. T is preceded by 3 but followed by the letter P ✗. P is preceded by the letter T ✗. M sits between 7 and 2 ✓. J sits between 2 and 8 ✓. D sits between 8 and 4 ✓. Five letters qualify — B, K, M, J and D — so the correct count is 5.

A coding question gives you one word with its code and asks you to encode or decode a second word. The whole job is to discover the transformation, and there are only a handful of possibilities. Test them in this order:
(1) is each letter shifted forward or backward by a fixed number?
(2) is each letter replaced by its opposite letter, A↔Z?
(3) has the word been reversed before or after shifting?
(4) are letters replaced by their position numbers or a sum of them?
(5) is the code a substitution where whole words stand for other words?

Worked example 1 — a backward shift of one. If GARDEN is written as FZQCDM, how is FLOWER written? Line the words up letter by letter: G→F, A→Z, R→Q, D→C, E→D, N→M. Every letter has moved one place back, and A wrapped around to Z. Apply the same to FLOWER: F→E, L→K, O→N, W→V, E→D, R→Q. The code is EKNVDQ. Notice how the A→Z wrap in the given pair was the confirmation — without it you might have suspected something more complicated.

Worked example 2 — the opposite-letter code. In a certain code LIGHT is written as ORTSG. Check with the 27-rule: L = 12, and 27 − 12 = 15 = O ✓. I = 9, 27 − 9 = 18 = R ✓. G = 7, 27 − 7 = 20 = T ✓. H = 8, 27 − 8 = 19 = S ✓. T = 20, 27 − 20 = 7 = G ✓. So every letter is replaced by its mirror. Now encode DARK: D = 4 → 23 = W; A = 1 → 26 = Z; R = 18 → 9 = I; K = 11 → 16 = P. DARK becomes WZIP.

Substitution coding is the type where whole words are swapped, and it is answered by pure elimination. Suppose: "pit dar sam" means "you are good", "dar tir jo" means "they are honest", and "sam kal jo" means "you and honest". Compare the first two lines: the only shared English word is are and the only shared code word is dar, so dar = are. Compare lines one and three: the shared English word is you and the shared code word is sam, so sam = you. From line one, what is left is pit = good. From line three the shared word with line two is jo, and the shared English word is honest, so jo = honest and therefore kal = and.

Never solve a relationship question in your head. Draw a family tree in horizontal layers: everyone of the same generation on one line, parents on the line above, children on the line below. Mark males with a plus sign and females with a minus sign; a horizontal double line joins husband and wife; a vertical line drops from parent to child. Read the statements one at a time, adding exactly one mark per statement, and never guess a gender the question has not given.

Worked example 1. A is the brother of B. B is the daughter of C. C is the son of D. How is A related to D? Take it one statement at a time. "B is the daughter of C" puts C one layer above B. "A is the brother of B" puts A on the same layer as B, and tells us A is male — so A is also a child of C. "C is the son of D" puts D one layer above C and tells us C is male, so C is the father of A and B. Now count layers from D: C is D's child, and A is C's child, so A is two layers below D and male. A is therefore the grandson of D. Note carefully that the question never reveals whether D is male or female, so "A is D's grandson" is the most we can say.

Worked example 2 — the photograph type. Pointing to a photograph, a man says, "He is the son of the only son of my grandfather." How is the man in the photograph related to the speaker? Unwrap the phrase from the inside out, which is the only reliable order. The innermost piece is "my grandfather". Next, "the only son of my grandfather" — since the grandfather has exactly one son and the speaker descends from him, that son must be the speaker's father. Finally, "the son of my father" is either the speaker himself or a brother; because the sentence says "he", pointing to another person, the photograph shows the speaker's brother.

Coded relations replace words with symbols. Suppose: A + B means A is the father of B; A − B means A is the wife of B; A × B means A is the brother of B; A ÷ B means A is the daughter of B. What does P + Q − R tell us about P and R? Read left to right in pairs. "P + Q" makes P the father of Q. "Q − R" makes Q the wife of R, which also tells us Q is female — consistent with being someone's daughter. So R is Q's husband and P is Q's father; a man is his daughter's husband's father-in-law. Hence P is the father-in-law of R.

Almost every arithmetic question in this unit is really a percentage question in disguise, and percentage work is fastest when you convert to a fraction. Finding 25% of a number is dividing by 4; finding 12.5% is dividing by 8. Learn the standard conversions once and a whole class of questions becomes mental arithmetic. To compare fractions without decimals, cross-multiply: to test 3/7 against 5/11, compute 3 × 11 = 33 and 5 × 7 = 35; since 33 < 35, the fraction 3/7 is the smaller one. Adding 7/15 to the comparison: 5 × 15 = 75 against 7 × 11 = 77, so 5/11 < 7/15. Hence the ascending order is 3/7 < 5/11 < 7/15 (as decimals, 0.4286, 0.4545 and 0.4667).

Ratio: making a chain. If A : B = 3 : 4 and B : C = 6 : 7, we cannot simply write 3 : 4 : 7, because B stands for 4 in one ratio and 6 in the other. Make B equal in both by taking the LCM of 4 and 6, which is 12. Multiply the first ratio by 3 to get A : B = 9 : 12; multiply the second by 2 to get B : C = 12 : 14. Now the chain reads A : B : C = 9 : 12 : 14. Divide Rs 1750 in this ratio: the parts total 9 + 12 + 14 = 35, so one part is 1750 ÷ 35 = 50, giving A = 9 × 50 = 450, B = 12 × 50 = 600, C = 14 × 50 = 700. Check: 450 + 600 + 700 = 1750 ✓.

Proportion and the age question. In a proportion a : b = c : d, the product of the extremes equals the product of the means, so ad = bc. The mean proportional between 9 and 25 is the number x with 9 : x = x : 25, so x² = 225 and x = 15. Now a classic ratio-with-time problem. The present ages of A and B are in the ratio 4 : 5, and after 6 years the ratio becomes 5 : 6. Let the ages be 4x and 5x. Then (4x + 6) : (5x + 6) = 5 : 6, so cross-multiplying, 6(4x + 6) = 5(5x + 6), that is 24x + 36 = 25x + 30, giving x = 6. So the ages are 4 × 6 = 24 and 5 × 6 = 30. Check: after 6 years they are 30 and 36, and 30 : 36 = 5 : 6 ✓.

Percentage: two formulas that answer most questions. First, successive change: if a quantity changes by a% and then by b%, the net change is a + b + ab/100 per cent, where a decrease is entered as a negative number. A 20% rise followed by a 20% fall gives 20 − 20 − 400/100 = −4%, a net loss of 4 per cent — never zero. Verify on Rs 100: it becomes 120, then 120 × 0.8 = 96. Second, the reciprocal rule: if the price of a commodity rises by R%, consumption must fall by 100R/(100 + R) per cent to keep expenditure unchanged. For a 25% rise that is 2500/125 = 20%. Verify: 100 units at Rs 100 costs Rs 10,000; at the new price Rs 125, spending Rs 10,000 buys 80 units — exactly 20% less.

Worked example — pass marks. A candidate who scores 40% of the total fails by 20 marks; another who scores 50% gets 15 marks more than the pass mark. Find the total marks and the pass mark. Let the total be T. From the first candidate, the pass mark is 0.40T + 20. From the second, the pass mark is 0.50T − 15. These describe the same pass mark, so 0.40T + 20 = 0.50T − 15, giving 35 = 0.10T and T = 350. The pass mark is 0.40 × 350 + 20 = 140 + 20 = 160. Check with the second candidate: 0.50 × 350 = 175, and 175 − 160 = 15 ✓.

This whole family of questions rests on knowing which number the percentage is taken on. Profit and loss percentages are always on the cost price. Discount is always on the marked price (also called list price). Interest is always on the principal — for simple interest, on the original principal every year; for compound interest, on the principal plus the interest already added. Get the base right and the rest is one multiplication.

Worked example 1 — marked price, discount and profit together. A shopkeeper marks an article at Rs 800, allows a discount of 25%, and still makes a profit of 20%. Find the cost price. First the selling price, because discount acts on the marked price: SP = 800 × (1 − 0.25) = 800 × 0.75 = Rs 600. Now the profit acts on the cost price, so SP = CP × 1.20, giving CP = 600 ÷ 1.20 = Rs 500. Check both ways: a 25% discount on 800 is a reduction of 200, leaving 600 ✓; and 20% of 500 is 100, so 500 + 100 = 600 ✓.

Worked example 2 — the same selling price twice. A trader sells two articles at the same price, gaining 20% on one and losing 20% on the other. What is the overall result? Take the common selling price as Rs 100 each. The first article cost 100 ÷ 1.20 = Rs 83.33; the second cost 100 ÷ 0.80 = Rs 125. Total cost = 83.33 + 125 = Rs 208.33, total selling price = Rs 200. So there is a loss of Rs 8.33 on Rs 208.33, which is 8.33 ÷ 208.33 × 100 = 4% loss. The shortcut: when the same price is used with a gain of x% and a loss of x%, the result is always a loss of x²/100 per cent — here 400/100 = 4%.

Worked example 3 — compound interest step by step. Find the compound interest on Rs 8000 at 5% per annum for 2 years. Year one: interest = 5% of 8000 = 400, so the amount becomes 8400. Year two: interest = 5% of 8400 (not 8000) = 420, so the amount becomes 8820. Compound interest = 8820 − 8000 = Rs 820. Simple interest for the same period would be 8000 × 5 × 2 ÷ 100 = Rs 800, so the gap is Rs 20 — exactly what the formula P(R/100)² predicts: 8000 × 0.05 × 0.05 = 20 ✓. If instead the interest were compounded half-yearly, we would halve the rate and double the periods: 8000 × (1.025)⁴ = Rs 8830.50, giving CI of about Rs 830.50 — more frequent compounding always earns more.

An average is only a disguised total: average × count = total. That single rearrangement solves nearly every average question. The average of the first n natural numbers is (n + 1)/2, so the average of the first 10 natural numbers is 11/2 = 5.5. If the average of 5 numbers is 27 and the number 35 is removed, the total falls from 5 × 27 = 135 to 135 − 35 = 100, and the average of the remaining 4 becomes 100 ÷ 4 = 25. If the average age of 30 students is 14 years and rises to 15 when the teacher is included, the teacher's age is 31 × 15 − 30 × 14 = 465 − 420 = 45 years.

Worked example — the overlapping-average question. The average of 11 numbers is 50. The average of the first 6 is 49 and the average of the last 6 is 52. Find the sixth number. Convert every average to a total: all eleven sum to 11 × 50 = 550; the first six sum to 6 × 49 = 294; the last six sum to 6 × 52 = 312. Notice that 294 + 312 = 606 counts the sixth number twice, because it belongs to both groups. So the sixth number is 606 − 550 = 56.

Time, speed and distance. The one conversion you need is 1 km/h = 5/18 m/s, so multiply by 5/18 going into metres per second and by 18/5 coming out. Thus 54 km/h = 54 × 5/18 = 15 m/s. A train 180 m long moving at 54 km/h crosses a pole in 180 ÷ 15 = 12 seconds, because for a pole the train covers only its own length. The same train crossing a platform 270 m long must cover 180 + 270 = 450 m, taking 450 ÷ 15 = 30 seconds. For boats, downstream speed = boat + stream and upstream speed = boat − stream: a boat of 12 km/h in a 3 km/h stream does 15 km/h down and 9 km/h up, so 45 km downstream takes 3 hours and the return takes 5 hours.

Worked example — finding the third worker. A, B and C together finish a job in 10 days, while A and B together take 12 days. How long would C take alone? Take total work = LCM(10, 12) = 60 units. All three together do 60 ÷ 10 = 6 units/day; A and B together do 60 ÷ 12 = 5 units/day. Subtracting, C alone does 6 − 5 = 1 unit/day, so C alone needs 60 ÷ 1 = 60 days. The related man-days idea is just as simple: if 12 men build a wall in 18 days, the job is 12 × 18 = 216 man-days, so 8 men would need 216 ÷ 8 = 27 days.

Factorise into primes and both the HCF and the LCM fall out at once. Take 72 and 120: 72 = 2³ × 3² and 120 = 2³ × 3 × 5. The HCF takes each shared prime to its lowest power, giving 2³ × 3 = 24. The LCM takes every prime to its highest power, giving 2³ × 3² × 5 = 360. Verify with the identity HCF × LCM = product of the numbers: 24 × 360 = 8640, and 72 × 120 = 8640 ✓. Remember which is which by size: the HCF can never exceed the smaller number, and the LCM can never be less than the larger one.

Worked example — the remainder question. Find the smallest number which, when divided by 12, 15 and 20, leaves a remainder of 5 in each case. A number divisible by all three must be a multiple of their LCM. Factorise: 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5, so the LCM is 2² × 3 × 5 = 60. The smallest number leaving remainder 5 with each divisor is therefore 60 + 5 = 65. Check all three: 65 ÷ 12 = 5 remainder 5 ✓; 65 ÷ 15 = 4 remainder 5 ✓; 65 ÷ 20 = 3 remainder 5 ✓. Note the companion pattern: if instead the number were short by the same amount each time — say it left remainders 7, 10 and 15 — you would compute LCM minus the common shortfall of 5, giving 55.

Basic mensuration. A handful of formulas covers everything NET asks. A rectangle of 12 cm by 5 cm has area 12 × 5 = 60 cm², perimeter 2(12 + 5) = 34 cm, and diagonal √(144 + 25) = √169 = 13 cm. A circle of radius 7 cm has area (22/7) × 49 = 154 cm² and circumference 2 × (22/7) × 7 = 44 cm — this is why setters always choose radius 7 or 14. A cube of side 6 cm has volume 216 cm³ and total surface area 6 × 36 = 216 cm², a coincidence worth noticing but not trusting. A cylinder of radius 7 cm and height 10 cm has volume (22/7) × 49 × 10 = 1540 cm³ and curved surface 2 × (22/7) × 7 × 10 = 440 cm².