Shortcut tricksशॉर्टकट ट्रिक्स proven, with their limitsप्रमाणित, अपनी सीमाओं के साथ

Why:क्यों: (1 + a)(1 + b)(1 + c) − 1 expanded gives a + b + c + ab + bc + ca + abc (with a, b, c as fractions); converting to percentages gives the formula.

Limits:सीमाएँ: Exact for any values. The last term matters when the changes are large (e.g. 50% three times: 150 + 75 + 12.5 = 237.5%).

Why:क्यों: If each sells at S, the costs are S/(1 + a) and S/(1 − b). Overall profit = 2S / (S/(1 + a) + S/(1 − b)) − 1, which simplifies to (a − b − 2ab)/(2 + a − b) with a, b as fractions.

Limits:सीमाएँ: Only when both selling prices are equal. If the cost prices are equal instead, the overall % is simply (a − b)/2.

Why:क्यों: MP × (1 − d) = CP × (1 + p) ⇒ MP/CP = (1 + p)/(1 − d); subtracting 1 gives (p + d)/(1 − d).

Limits:सीमाएँ: d must be below 100%. Works with a loss too: use a negative p.

Why:क्यों: Same distance, so (b + s)/(b − s) = k, which rearranges to b/s = (k + 1)/(k − 1).

Limits:सीमाएँ: Same distance both ways; k > 1.

Why:क्यों: From the left, B is at N − q + 1. The people strictly between two seats x and y number |x − y| − 1, which gives both formulas.

Limits:सीमाएँ: Assumes A and B are different people and both positions fit inside the row.

Why:क्यों: All seatings: (n − 1)!. Seatings with the two together: treat them as one unit, (n − 2)! orders around the table, × 2 for their order = 2(n − 2)!. Subtracting, (n − 1)! − 2(n − 2)! = (n − 2)! × (n − 1 − 2) = (n − 3)(n − 2)!.

Limits:सीमाएँ: For n ≥ 3 people in a circle where only relative positions matter (a table, not a garland). In a row the count is n! − 2(n − 1)! = (n − 2)(n − 1)! instead.

Why:क्यों: When A finishes, B has run L − x and C has run (L − x)(L − y)/L. So A wins by L − (L − x)(L − y)/L = x + y − xy/L. It is the successive-discount formula a + b − ab/100 with L in place of 100.

Limits:सीमाएँ: Both margins must be over the same race length L, and runners must keep steady speeds. Convert “by t seconds” margins to metres first.

Why:क्यों: Face value S gives old income Sx₁/100 and sale money SM₁/100. That money buys stock paying SM₁/100 × x₂/M₂. Dividing the two incomes gives M₁x₂/(M₂x₁), independent of S.

Limits:सीमाएँ: No brokerage. With brokerage, use M₁ − b for the sale and M₂ + b for the purchase.

Why:क्यों: Someone in exactly two groups appears once in S₂ and someone in all three appears three times, so S₂ − 3T counts the exactly-two people. In S₁ an exactly-two person appears twice and an all-three person three times; subtracting 2S₂ removes them four and six times, so adding 3T back leaves only the exactly-one people.

Limits:सीमाएँ: Same as inclusion–exclusion: the pair totals must include the people in all three.

Why:क्यों: Choosing the lower floor f fixes the upper floor at f + k + 1, which must be at most n, so there are n − k − 1 choices; swapping who is lower doubles them.

Limits:सीमाएँ: Counts placements of that one pair before any other clue is used.

Why:क्यों: The slice takes start, start + step, … while the index stays below stop, which is exactly ceil((stop − start)/step) indices.

Limits:सीमाएँ: For negative steps the slice runs backwards and the rule is ceil((start − stop)/|step|) after clipping.

Why:क्यों: The main diagonal is the line where across = down, so reflecting in it exchanges the two coordinates. The other diagonal is the line across + down = n, and reflecting in it sends (x, y) to (n − y, n − x).

Limits:सीमाएँ: Only for a square sheet; a rectangle’s diagonal is not a line of symmetry, so folding along it does not lay the halves on each other.

Why:क्यों: Counted by size, the full figure has 16 smallest triangles, 16 made of two pieces, 8 of four pieces and 4 halves of the square: 16 + 16 + 8 + 4 = 44. The smaller figures are the same count with fewer lines.

Limits:सीमाएँ: Only for these exact figures. A single extra line changes the total, so count by size whenever the figure differs.

Why:क्यों: (1 + 1/n) × (1 − 1/(n + 1)) = ((n + 1)/n) × (n/(n + 1)) = 1, so the two changes cancel exactly.

Limits:सीमाएँ: Works directly for unit fractions. For other rises use r/(100 + r) × 100 (a 30% rise needs a 300/13 ≈ 23.08% cut).

Why:क्यों: Each percentage change multiplies the value by (1 ± r/100); a fraction is just that multiplier written exactly, and products of fractions cancel.

Limits:सीमाएँ: Best when the percentages are “fraction friendly” (multiples of 5, 12.5, 16.67, 33.33). For 17% or 23% use the decimal 1.17 / 0.77.

Why:क्यों: It is the LCM method with the total work fixed at 100 units, so the fraction table you already know does the work.

Limits:सीमाएँ: Messy for days like 7 or 13 (14.28%, 7.69%): there the LCM method is cleaner.

Why:क्यों: (x + y)/2 − 2xy/(x + y) = ((x + y)² − 4xy) / (2(x + y)) = (x − y)² / (2(x + y)). So the harmonic mean is always below the plain average by exactly that gap.

Limits:सीमाएँ: Only for equal distances. For equal times the average speed is the plain average.

Why:क्यों: (10t + 1)^n = 1 + n·10t + (terms with 100 or more). Mod 100 only 1 + 10·(n·t) survives.

Limits:सीमाएँ: Only for bases ending in 1. For other bases, first raise to a power that ends in 1 (for example 3⁴ = 81, 7⁴ = 2401, 9² = 81).

Why:क्यों: 1000 ≡ −1 (mod 1001), and 1001 = 7 × 11 × 13, so each group of three digits flips sign.

Limits:सीमाएँ: Gives the remainder up to sign: if the alternating sum is negative, add 1001 first.

Why:क्यों: For a sum s ≤ 7 the first die can show 1 to s − 1 (each fixing the second die), giving s − 1 ways; for s ≥ 7 it can show s − 6 to 6, giving 13 − s ways. Both are 6 − |s − 7|.

Limits:सीमाएँ: Only for two ordinary six-sided dice. For three dice the counts follow a different pattern (1, 3, 6, 10, 15, 21, 25, 27, 27, 25, …).

Why:क्यों: Going round once from A back to A passes every other person exactly once. Everyone except A and B is between them on exactly one of the two sides, so the two counts add up to n − 2.

Limits:सीमाएँ: Each seat holds one person and every seat is filled. With empty chairs, count chairs instead of people.

Why:क्यों: Each step lifts one item out of the unarranged part and drops it next to the arranged block. Removing an item from a line never changes the order of the items left behind, so the unarranged part is always the input with the arranged items deleted.

Limits:सीमाएँ: Only for “one item moves to one end” machines. Machines that swap pairs, shift by positions or change the items themselves (for example add 1 to each number) need step-by-step working.

Why:क्यों: Turning by 45° eight times is a full turn, so directions behave like the hours on an 8-hour clock. Adding notches and reducing modulo 8 is exactly repeated rotation.

Limits:सीमाएँ: Works for turns that are whole multiples of 45°. For 30° or 60° steps use 12 notches instead of 8.

Why:क्यों: A number and its digit sum leave the same remainder when divided by 9, and remainders survive addition, subtraction and multiplication (“casting out nines”).

Limits:सीमाएँ: Cannot tell apart options that differ by a multiple of 9, and does not work for division or decimals. Pair it with the last-digit check.

Why:क्यों: The last digit of a sum or product depends only on the last digits of the inputs (arithmetic mod 10).

Limits:सीमाएँ: Useless when all options share a last digit. Then use the digit-sum check, which catches different errors.

Why:क्यों: Any such probability is k/N for a whole number k. Reducing k/N divides N by gcd(k, N), so the reduced denominator is a factor of N.

Limits:सीमाएँ: Only for a single equally likely sample space. With two stages without replacement the space is ordered pairs (for two cards, 52 × 51), so use that product as N.

Why:क्यों: “I alone” and “either” are exactly the two options in which statement I works by itself; the other three all say that I alone is not enough. So the result of the first test always splits the five options into two and three.

Limits:सीमाएँ: Works only with the standard five-option format. Some papers add a sixth option for “questions that need both but give contradictory data”; check the option list first.

Why:क्यों: In “A ⊕ B means A is the … of B”, the relation word describes A, so it fixes A’s gender and says nothing about B’s. The only other source of gender is a spouse, because the couple must be one man and one woman.

Limits:सीमाएँ: Neutral answers (for example “cousin”) do not need a gender, and some papers do not offer “cannot be determined”; then pick the option that holds for both genders, if any.

Why:क्यों: The half turn equals a mirror image only when the figure looks the same in water (it has a horizontal line of symmetry), and equals a water image only when it looks the same in a mirror. For a figure with no symmetry, the half turn is never the answer.

Limits:सीमाएँ: If the figure is symmetric about a horizontal or vertical line, the half turn can coincide with the right answer; check the symmetry first.

Why:क्यों: Setters build the trap from the word itself: they put its synonym and its opposite together so that a rushed reader picks the wrong one. That leaves the pair as the only real contenders. The check generates hundreds of questions from this topic and confirms the answer is always one of an opposite pair among the options.

Limits:सीमाएँ: In papers where every option is a near-synonym, there is no pair to find; fall back on the sentence test.

Why:क्यों: Exam setters build series from a handful of rules. Checking them from cheapest to costliest finds the rule with the least arithmetic, and step 4 turns the guess “× something + something” into a calculation.

Limits:सीमाएँ: Rare series mix two operations alternately (×2, +3, ×2, +3). If all six checks fail, try alternating operations on consecutive terms.

Why:क्यों: Relation names in exams depend on two things only: generation gap and gender (plus paternal/maternal side for uncles and aunts). The score finds the gap without drawing the whole tree.

Limits:सीमाएँ: For in-laws and the paternal/maternal question you still need a quick tree; the score only fixes the generation.

Why:क्यों: Checking a candidate is a few multiplications; with sorted options each check halves the choices, so two checks cover four options.

Limits:सीमाएँ: Not for “cannot be determined” options or questions with several unknowns. Write the equation then.

Why:क्यों: For products and quotients, relative errors add: (1 + e₁)(1 + e₂) ≈ 1 + e₁ + e₂. Opposite roundings have opposite signs and cancel.

Limits:सीमाएँ: When options are within 2–3% of each other, calculate exactly (and use the digit-sum check).

Why:क्यों: A conclusion is definite only when no allowed diagram contradicts it, so one counter-diagram is enough to reject it. This is exactly how the questions are checked.

Limits:सीमाएँ: It takes longer than the rules, so use the rules first and keep this for doubtful cases.

Why:क्यों: Cost per item is 1/x and selling price per item is 1/y; (1/y − 1/x)/(1/x) = (x − y)/y.

Limits:सीमाएँ: Assumes the same money unit on both sides; for “x for ₹p, y for ₹q” compare p/x with q/y.

Why:क्यों: Year 1 is identical. In year 2 CI also earns interest on year 1’s interest: P(r)(r). In year 3 it earns that twice more plus interest on interest-on-interest, giving P r² (3 + r).

Limits:सीमाएँ: Annual compounding only. For half-yearly, use r/2 and double the number of periods first.

Why:क्यों: 1/B = 1/t − 1/a = (a − t)/(a t).

Limits:सीमाएँ: a must be greater than t (one person alone is always slower than the pair).

Why:क्यों: Before meeting both run the same time t. A covers what B still has to run, so u·t = v·b and v·t = u·a; dividing gives (u/v)² = b/a.

Limits:सीमाएँ: Both must start at the same moment.

Why:क्यों: If a and b leave the same remainder when divided by d, then d divides b − a; the largest such d is the HCF of all the differences.

Limits:सीमाएँ: Only for “the same remainder”. For different given remainders, subtract each remainder from its number first.

Why:क्यों: The minute hand gains 5.5° a minute and must make up 30H degrees: 30H / 5.5 = 60H/11.

Limits:सीमाएँ: Between 11 and 12 the overlap is at 12:00 itself, so there are 11 overlaps in 12 hours, not 12.

Why:क्यों: Order relations are transitive in one direction only; a single opposite sign lets the far ends land in either order.

Limits:सीमाएँ: For “either A > D or A ≤ D” answers, check whether the two conclusions together cover every case.

Why:क्यों: Every arrangement with the vowels together is obtained exactly once by choosing an order for the units and then an order for the letters inside the block, so the two counts multiply (the multiplication principle). Repeated letters are divided out in whichever part they sit.

Limits:सीमाएँ: Only for one block of letters that must be adjacent. Two separate blocks each need their own inner count, and “no two vowels together” needs the gap method instead.

Why:क्यों: Every selection either has no woman or has at least one, and never both, so the two counts add up to the total. Choosing “one woman, then anyone” instead counts a team with two women twice, once for each woman picked first.

Limits:सीमाएँ: Works whenever “none” is easy to count. If the condition is “at most” or “exactly”, count the allowed cases directly.

Why:क्यों: A number with d digits lies between 10ᵈ⁻¹ and 10ᵈ, so its log lies in [d − 1, d). The whole part of log N is therefore d − 1.

Limits:सीमाएँ: Needs enough decimal places in the given log: if the product lands very close to a whole number, four-figure logs may not be accurate enough.

Why:क्यों: Write each factor as a ratio of common logs: (log 7/log 2) × (log 11/log 7) × (log 64/log 11). Every middle term cancels, leaving log 64/log 2 = log₂ 64.

Limits:सीमाएँ: The links must match exactly (argument of one = base of the next). A product like log₂ 3 × log₅ 7 does not simplify this way.

Why:क्यों: (10a + b) − (10b + a) = 9(a − b) and (10a + b) + (10b + a) = 11(a + b).

Limits:सीमाएँ: Two-digit numbers only. For three digits, reversing changes the number by 99 × (first digit − last digit).

Why:क्यों: B’s speed is x/t, so B needs L/(x/t) = Lt/x seconds for the race. A finishes t seconds earlier: Lt/x − t = t(L − x)/x.

Limits:सीमाएँ: Needs steady speeds. The x metres and t seconds must describe the same margin, measured when the winner finishes.

Why:क्यों: PW grows to A at simple interest, so A = PW(1 + RT/100). Then A − PW = PW × RT/100, which is the interest on PW, and dividing A by (100 + RT)/100 gives PW.

Limits:सीमाएँ: Assumes simple interest, the usual convention for true discount. It does not apply to compound-interest present value.

Why:क्यों: BD is the interest on A and BG = BD − TD is the interest on TD, at the same rate and time. Interest is proportional to the principal, so A/TD = BD/BG, giving A = BD × TD/(BD − TD).

Limits:सीमाएँ: BD and TD must be for the same bill, rate and time, under simple interest.

Why:क्यों: Distances from the foot are h√3 (at 30°) and h/√3 (at 60°). Their difference h(√3 − 1/√3) = 2h/√3 equals d, so h = d√3/2 and h/√3 = d/2. The 60°-30° triangle at the top is isosceles, which is why the walk equals the line of sight from the nearer point.

Limits:सीमाएँ: Only for 30° and 60° with both points on the same side, in line with the foot, on level ground and with the eye height ignored.

Why:क्यों: tan θ = h/a and tan(90° − θ) = cot θ = h/b. Multiplying, tan θ × cot θ = 1 = h²/(ab), so h = √(ab).

Limits:सीमाएँ: The two angles must add to exactly 90° and both distances must be measured from the foot on level ground. The points can be on the same side or on opposite sides.

Why:क्यों: Each rule says what is true in every Venn diagram the two statements allow. The “nothing definite” cases are exactly those where one fitting diagram makes the A–C conclusion true and another makes it false.

Limits:सीमाएँ: The middle term (B) must link the two statements in the order shown. If the second statement runs C → B, only “No” and “Some” statements can be turned round; “All” cannot.

Why:क्यों: The seats strictly between p and q are p + 1, …, q − 1, which is q − p − 1 seats.

Limits:सीमाएँ: For a circle there are two ways round; the question must say which direction.

Why:क्यों: A person facing the centre has the centre straight ahead; turning that direction a quarter-turn to their left gives the clockwise tangent of the circle at their seat (seen from above). Facing outward reverses the forward direction, so the left-hand side reverses too.

Limits:सीमाएँ: Only for drawings seen from above. If a book draws the table from below or mirrors it, the words “clockwise” and “anticlockwise” swap.

Why:क्यों: Each fact is one equation in two unknowns. One equation leaves a whole line of solutions (many whole-number ages), while two different equations cross at a single point.

Limits:सीमाएँ: The two facts must be independent: “Ravi is 8 years older” and “Sita is 8 years younger” say the same thing. Very small ratios with a tiny age range can occasionally leave one whole-number answer by accident.

Why:क्यों: A person in exactly two groups is counted twice in the circle totals and removed once by the pair total. A person in all three is counted three times, removed three times, then added once. Everyone in at least one group ends up counted exactly once.

Limits:सीमाएँ: The pair figures must be “both A and B” (including those in all three). If a question gives “only A and B” instead, add those regions directly without subtracting the triple.

Why:क्यों: The floors strictly between p and q are p + 1, …, q − 1, which is q − p − 1 floors, each with one person.

Limits:सीमाएँ: One person per floor. If a floor can hold two flats or people, count people, not floors.

Why:क्यों: Exam family trees assume full siblings share both parents and that a married couple are the parents of each other’s children. Under those rules each pair of links above describes the same position in the tree as the single link it collapses to.

Limits:सीमाएँ: Half-siblings and step-relations are never intended in these questions. Spouse-of-sibling links do not collapse to a blood relation; they give in-laws.

Why:क्यों: Each host bit doubles the addresses; the all-zeros and all-ones host parts are the network and broadcast addresses.

Limits:सीमाएँ: /31 point-to-point links and /32 single hosts are special cases.

Why:क्यों: A vertical mirror sends (x, y) to (w − x, y) and water sends (x, y) to (x, h − y). Applying both sends (x, y) to (w − x, h − y), which is exactly a half turn about the centre.

Limits:सीमाएँ: For mirrors placed on the side or below. A mirror at an angle (a diagonal mirror) swaps the across and down positions instead.

Why:क्यों: The punch goes through every layer, and f folds stack 2^f layers. Each layer gets its own copy of every hole, and no two copies land in the same place unless a hole sits on a crease.

Limits:सीमाएँ: A hole punched exactly on a crease opens into one hole, not two, so the count is lower.

Why:क्यों: Every triangle in such a figure has the top vertex as one corner, two of the n lines as sides, and one horizontal level (a cross line or the base) as its third side. Choosing 2 lines from n and 1 level from m + 1 gives the count.

Limits:सीमाएँ: Every cross line must run from one side to the other and cut all the lines from the vertex. Lines from other vertices break the formula; count by size instead.

Why:क्यों: The Constitution is organised in Parts, so related subjects have nearby numbers. Knowing the block narrows four options to one or two.

Limits:सीमाएँ: Blocks help you place an article; exact numbers inside a block still need practice.

Why:क्यों: English took these words whole from Greek and Latin, so the root still carries its meaning. Every rooted word in this topic’s table is checked to contain the idea its root promises.

Limits:सीमाएँ: Words like “Veteran” or “Fatal” have no helpful root; learn those as they come.

Why:क्यों: x/100 × y = y/100 × x, because multiplication is commutative.

Limits:सीमाएँ: Always true; it only helps when one of the two numbers is a friendly percentage (10, 20, 25, 50…).

Why:क्यों: Doubling time is ln 2 / ln(1 + r/100) ≈ 69.3/r for small r; 72 corrects for the curvature at typical exam rates of 6–12% and divides evenly by many numbers.

Limits:सीमाएँ: Doubling and tripling are within half a year, quadrupling within 0.7 years, for rates from 4% to 20%. For simple interest use 100 ÷ r for doubling instead.

Why:क्यों: 1 km/h = 1000 m / 3600 s = 5/18 m/s, so 18k km/h = 5k m/s.

Limits:सीमाएँ: Other speeds still need × 5/18.

Why:क्यों: n² − (n − 1)² = 2n − 1 = (n − 1) + n, and (a + d)(a − d) = a² − d².

Limits:सीमाएँ: Always true; it is fastest when the neighbour is a round number.

Why:क्यों: The day counts between these dates are all multiples of 7, which John Conway noticed and turned into his Doomsday rule.

Limits:सीमाएँ: Within one year. To move between years, add 1 weekday per ordinary year and 2 per leap year.

Why:क्यों: A vertical mirror reflects every angle θ (measured from 12) to 360° − θ; for both hands that is the same as the time 12:00 − t.

Limits:सीमाएँ: For a mirror placed on the side only. Water (upside-down) images need a different rule, which is not covered here.

Why:क्यों: These are the smallest whole-number solutions of a² + b² = c², and examiners pick them so the answer is a whole number.

Limits:सीमाएँ: Other cases need √(a² + b²); check whether the options are given in surd form.

Why:क्यों: Reversing the alphabet maps position p to 27 − p.

Limits:सीमाएँ: Only for the full 26-letter alphabet.

Why:क्यों: Multiplying both sides of a/b > c/d by the positive number b·d keeps the inequality and removes the division.

Limits:सीमाएँ: Positive numbers only; with negative values the direction can flip.

Why:क्यों: For positive numbers, x < y exactly when xᴸ < yᴸ for any positive power L, because raising to a positive power keeps the order. Choosing L as the LCM turns every root into a whole number.

Limits:सीमाएँ: Positive numbers only. With many surds the powers grow quickly; compare two at a time if the numbers get large.

Why:क्यों: A square’s last digit depends only on the root’s last digit, and (10t)² ≤ N < (10t + 10)² pins down the leading part t. Between 10t + d and 10t + 10 − d, the square of 10t + 5 lies in the middle, so the comparison picks the right one.

Limits:सीमाएँ: Only for exact perfect squares. A non-square ending in 1, 4, 5, 6, 9 or 0 still gives a confident but wrong answer, so check by squaring when in doubt.

Why:क्यों: The cubes of 0 to 9 end in 0, 1, 8, 7, 4, 5, 6, 3, 2, 9: all different, so the last digit of a cube fixes the last digit of its root. Since (10t)³ = 1000t³, striking off three digits and bracketing gives t.

Limits:सीमाएँ: Perfect cubes only, and the two-step version handles roots up to 99 (cubes up to six digits). Larger roots need the bracket step done on a longer leading part.

Why:क्यों: BD = A × RT/100 and TD = A × RT/(100 + RT). Their ratio is (100 + RT)/100 = 1 + RT/100, whatever the face value.

Limits:सीमाएँ: Simple interest only, with BD and TD for the same bill and time.

Why:क्यों: Each ₹M invested buys ₹100 of stock that pays ₹x, so every rupee invested earns x/M. Income = investment × x/M = investment × (100x/M)/100.

Limits:सीमाएँ: Add brokerage to the price first when buying (M + b). Yield is on the money paid, not on the face value.

Why:क्यों: The roots multiply to c and add to −b. A positive product means equal signs, and the sum then has the sign of −b.

Limits:सीमाएँ: Needs real roots (check b² ≥ 4c). When both equations have roots of the same sign, you still have to solve them.

Why:क्यों: Fixed clues shrink the possibilities the most, so placing them first keeps the number of rows you must track small.

Limits:सीमाएँ: Some puzzles have no fixed clue; then start from the person mentioned in the most clues.

Why:क्यों: Reading both statements together first makes the combined picture stick, and it becomes very hard to notice that one statement was enough on its own. Physically hiding one statement forces the fair test.

Limits:सीमाएँ: It costs a few seconds per question; with very easy statements you can do it mentally.

Why:क्यों: Most machines repeat one simple move, so a single changed item usually reveals it. Checking Step 2 guards against coincidences, such as an item that is both the largest number and the last in the line.

Limits:सीमाएँ: Mixed machines (words and numbers handled alternately) need Steps 1 and 2 to find two rules; check Step 3 if it is shown.

Why:क्यों: Most output traps come from aliasing and in-place mutation, which are invisible if you track values but not objects. Marking in-place operations makes them visible.

Limits:सीमाएँ: For recursion, use a call stack (one row per call) instead of one row per variable.

Why:क्यों: Each normal form forbids one more kind of dependency, so testing them in order gives the highest form reached.

Limits:सीमाएँ: Multivalued dependencies (4NF) need a different test.

Why:क्यों: The four conditions are jointly necessary (Coffman conditions), so removing any one rules out deadlock.

Limits:सीमाएँ: Avoidance (banker’s algorithm) works differently: it keeps the system in a safe state instead of removing a condition.

Why:क्यों: Exam series are built from independent simple rules. Each rule removes the options that break it, so two quick checks beat one hard look at the whole figure.

Limits:सीमाएँ: A few series alternate between two rules (odd and even figures). If no constant step appears, compare figures 1 and 3, and 2 and 4.

Why:क्यों: 6 of the 28 dates in this set are birth anniversaries, so one fact (the person) unlocks the date.

Limits:सीमाएँ: International days set by the UN have no such link; learn them by season.

Why:क्यों: Of the 25 States in this set, most capitals are the obvious city, so exam setters get their wrong answers from the few that are not. Knowing those few covers most of the hard questions.

Limits:सीमाएँ: It does not help with States whose capital is also the largest city; those are simply recall.

Why:क्यों: Four of the eight classical dances in this 20-row table carry their region in the name, so half the classical questions need no memory at all.

Limits:सीमाएँ: Folk dances rarely carry the State in their name; learn those in State pairs.

Why:क्यों: Current-affairs questions test a fact together with its context: a number without its scheme or unit is the commonest wrong answer. A four-part card stores exactly what the question needs.

Limits:सीमाएँ: It costs a minute per card; save it for the questions you get wrong, not every one.

Why:क्यों: An idiom is by definition figurative, so a literal reading of its words cannot be its meaning. Setters add one literal option because it catches people who rush.

Limits:सीमाएँ: A few phrases are half-literal (“in the nick of time”); then rely on the sentence test.

Why:क्यों: Each wrong option is the correct word with one slip, each at a different spot. So at each spot where the options differ, only one option is wrong there and the other three agree: the majority at every spot spells the word.

Limits:सीमाएँ: It needs the options to be single slips of one word, which is how most papers set them, but not all.