★ KS3-MATH-IMC-0097Front
12
HintAn hour has sixty minutes, not a hundred.
The whyMinutes are sixtieths, not hundredths: 0.1 h = 6 min, 0.25 h = 15 min, 0.75 h = 45 min. The trap is reading 0.15 hours as 15 minutes.
★ KS3-MATH-IMC-0097Back
Professor PiNumber25 cardsFree · no account needed
Answer in your head, then tap to check. Slide or use the buttons to grade.
★ KS3-MATH-IMC-0097Front
12
HintAn hour has sixty minutes, not a hundred.
The whyMinutes are sixtieths, not hundredths: 0.1 h = 6 min, 0.25 h = 15 min, 0.75 h = 45 min. The trap is reading 0.15 hours as 15 minutes.
★ KS3-MATH-IMC-0097Back
★ KS3-MATH-IMC-0098Front
7/11
HintCross-multiply: set 5 × 11 against 7 × 8.
The whyCross-multiplying weighs 5 × 11 = 55 against 7 × 8 = 56, and the larger product sits above the larger fraction, so 7/11 wins. The common-denominator route agrees: 55/88 against 56/88. The quick same-top rule (bigger bottom, smaller fraction) only applies when the numerators already match, which is why it cannot settle this pair.
★ KS3-MATH-IMC-0098Back
★ KS3-MATH-IMC-0099Front
Perpendicular bisector of it
HintTwo conditions: where it crosses, and at what angle.
The whySo to find the mirror line from a point and its image: the midpoint gives a point on the line, and the perpendicular gradient is −1 ÷ the segment's gradient (a gradient of 2 needs −½). Then y = mx + c through the midpoint.
★ KS3-MATH-IMC-0099Back
★ KS3-MATH-IMC-0100Front
−1
HintSteep up against gentle down — the product is a fixed negative number.
The whyA gradient of 3 pairs with −⅓; a gradient of 1 pairs with −1. Parallel lines share a gradient instead. Both facts turn 'reflect this point in the line' into arithmetic.
★ KS3-MATH-IMC-0100Back
★ KS3-MATH-IMC-0101Front
2
HintThink prime.
The why4ⁿ = 2²ⁿ and 8ᵐ = 2³ᵐ; likewise 9ⁿ = 3²ⁿ. Once everything is in one base, halving means subtracting 1 from the exponent (2¹⁴ ÷ 2 = 2¹³), never halving the exponent.
★ KS3-MATH-IMC-0101Back
★ KS3-MATH-IMC-0102Front
One full circle, 2πr
HintThe string only bends where it touches a coin.
The whyWhatever the layout, the string turns through 360° in total, so its arcs together make exactly one circle of radius r; the straight parts equal the gaps between the centres.
★ KS3-MATH-IMC-0102Back
★ KS3-MATH-IMC-0103Front
Percentage profit
HintWhich price is the 'original' that the change is measured against?
The whyPercentage change is always measured against the starting value — cost price for profit, the original for an increase. Selling for £72 at a 20% profit means cost × 1.2 = 72, so the cost was £60.
★ KS3-MATH-IMC-0103Back
★ KS3-MATH-IMC-0104Front
Arc length ÷ circumference
HintThe whole of which the arc is a part.
The whyarc ÷ 2πr = angle ÷ 360 = sector area ÷ πr². Any one of the three fractions gives the other two, so a radius-4 sector with arc 6 has area (6 ÷ 8π) × 16π = 12 with no angle needed.
★ KS3-MATH-IMC-0104Back
★ KS3-MATH-IMC-0105Front
mode; third; spread
HintWhich one is picked straight out of the list?
The whyThe mean need not be in the list, and the median of an even count need not be either; the mode forces a repeat. Setting mode, median and mean against each other is solved by writing the sorted list as a, b, m, ?, ? and translating each rule.
★ KS3-MATH-IMC-0105Back
★ KS3-MATH-IMC-0106Front
It falls by 2k
HintOne shared edge — how many perimeters was it on before?
The whyJoining hides the shared edge twice over — once from each shape: 20 squares of side 3 in a row have separate perimeters totalling 240, and 19 joins of length 3 take off 19 × 6, leaving 126. Count the joins, not the shapes.
★ KS3-MATH-IMC-0106Back
★ KS3-MATH-IMC-0107Front
120
HintThree of them tile round a point with nothing left over.
The whyIts side equals the distance from centre to vertex, and dropping a perpendicular from one vertex onto a neighbouring side's line gives a 30-60-90 triangle with sides in the ratio 1 : √3 : 2 — the source of every √3 in hexagon answers.
★ KS3-MATH-IMC-0107Back
★ KS3-MATH-IMC-0108Front
List the factor pairs
HintA product of whole numbers has only a few ways to happen.
The whyA product equal to a small whole number has finitely many whole-number factor pairs (including negative ones, which the 'positive' condition then discards). Rearranging xy − 3x − 3y = 3 into (x − 3)(y − 3) = 12 is the standard move for 'how many solutions?'.
★ KS3-MATH-IMC-0108Back
★ KS3-MATH-IMC-0255Front
It scales by that same factor
HintOne multiplier applied to every ingredient carries straight through to the total mix.
The whyA squash recipe's ingredients add up to 240 ml of concentrate; tripling every ingredient triples the total too, giving 720 ml — no need to scale each ingredient separately and re-add. Converted to litres, that's 0.72 l, not 7.2 l or 72 l, which is the easy slip when moving between millilitres and litres.
★ KS3-MATH-IMC-0255Back
★ KS3-MATH-IMC-0256Front
Cancel common trailing zeros
HintIt shrinks both figures to friendlier sizes before you divide.
The why42,000,000 ÷ 6,000: both numbers end in at least three zeros, so cancel three from each first, leaving the much friendlier 42,000 ÷ 6 = 7,000. Dividing the top and bottom of a division by the same power of ten never changes the answer — it just makes the arithmetic easier to see.
★ KS3-MATH-IMC-0256Back
★ KS3-MATH-IMC-0257Front
Divide the total amount by the daily rate
HintFind how many single days it takes to cover the whole job.
The whySaving £15 a week toward a total of £700 needs 700 ÷ 15 ≈ 46.7 weeks — just under a year's 52 weeks, so the honest description is 'just under a year', not 'a year'. The same division — total amount divided by the rate — turns any steady-progress problem into a single number of time units, measured against a sensible benchmark like a week, a month or a year.
★ KS3-MATH-IMC-0257Back
★ KS3-MATH-IMC-0258Front
difference
HintThe gap you get when you subtract one number from another stays fixed.
The why73 − 48 equals 75 − 50, because both numbers went up by 2; it also equals 70 − 45, because both went down by 3. This 'compensation' trick swaps an awkward subtraction for a friendlier one with the same answer, and it's the quickest way to spot which of several expressions secretly matches a given difference without calculating any of them fully.
★ KS3-MATH-IMC-0258Back
★ KS3-MATH-IMC-0259Front
Convert it to an improper fraction first
HintRewrite the mixed number as a single value over one denominator before doing anything else.
The whyTo find ⅔ of 3½, convert 3½ to the improper fraction 7/2 first, then multiply: ⅔ × 7/2 = 14/6 = 2⅓. The usual slip is to apply the fraction to the whole-number part only and carry the ½ through untouched, giving 2½. Splitting the mixed number is only safe if you take ⅔ of BOTH parts and add them: 2 + ⅓ = 2⅓. A percentage behaves the same way — 20% of 3¾ is 1/5 × 15/4 = 3/4.
★ KS3-MATH-IMC-0259Back
★ KS3-MATH-IMC-0260Front
Reverse it with the opposite operation
HintUndo the machine backwards from the answer, one step at a time.
The whyA machine does ×2, then −6, then ×2, then −6, in that order. To work backwards from an output, undo the LAST step first: add 6, halve, add 6, halve. Reversing only the operations but keeping the original order — instead of also reversing the order — sends you to the wrong number.
★ KS3-MATH-IMC-0260Back
★ KS3-MATH-IMC-0261Front
Multiply them
HintTwo chained percentages collapse into one product of decimals.
The why25% of 60% of a number is the same as working out 0.25 × 0.6 = 0.15, i.e. 15% of the number directly. Chaining doesn't mean adding the percentages (25 + 60 = 85% would be wrong) or finding each 'of' separately and subtracting — it means multiplying the two decimal forms together.
★ KS3-MATH-IMC-0261Back
★ KS3-MATH-IMC-0262Front
Write x/5 = x + 12 as an equation
HintTurn the sentence into symbols first, then rearrange to solve for the number.
The whySetting up x/5 = x + 12 and multiplying both sides by 5 gives x = 5x + 60, so −4x = 60 and x = −15. A negative answer isn't a mistake here — some word-to-algebra puzzles are only satisfied by a negative number, and trusting the algebra over an instinct that the answer 'should' be positive is exactly what solves them.
★ KS3-MATH-IMC-0262Back
★ KS3-MATH-IMC-0263Front
Area ratio equals the length ratio squared
HintIf every side scales ×3, the covered surface scales by 3×3, not by 3.
The whyTwo similar triangles have corresponding sides in the ratio 2 : 5. Their areas come out in the ratio 4 : 25 (that's 2² : 5²), not 2 : 5. This squared relationship is why doubling a shape's linear size roughly quadruples the material needed to cover it, and why similar 3D solids scale volume by the cube of the length ratio instead.
★ KS3-MATH-IMC-0263Back
★ KS3-MATH-IMC-0264Front
zero
HintNothing else can make a multiplication vanish to nothing.
The whySolving (x − 3)(x + 5)x = 0 gives three answers without expanding a thing: x = 3, x = −5 or x = 0, because a product can only come out as nothing if one of its pieces does. This is the fast way to finish an equation once you've forced one side down to a product of factors — for example, y(y − 4)(y + 4) = 0 hands you all three solutions of y³ = 16y immediately, without multiplying anything out.
★ KS3-MATH-IMC-0264Back
★ KS3-MATH-IMC-0265Front
Represent the unknown with a letter and reason generally
HintGive the starting value a symbol and follow it through algebraically, instead of picking a single case.
The whyTesting 'the sum of three consecutive integers is always a multiple of 3' with 4, 5, 6 gives 15 — a multiple of 3, which looks convincing but only proves that one case. Writing the three numbers as n − 1, n and n + 1 and adding gives 3n, obviously a multiple of 3 whatever whole number n is — now it holds for every case at once, not just the one you tried.
★ KS3-MATH-IMC-0265Back
★ KS3-MATH-IMC-0266Front
Pythagoras' theorem
HintThe right angle between the two shorter lengths is the clue — square and compare them to the joining one.
The whyTwo touching semicircles can hide a right angle between two of their radii; if the larger semicircle has radius 5 and two equal legs of length r meet it at that right angle, then r² + r² = 5², so r² = 12.5. The smaller semicircle's area, ½πr², is then ½π(12.5) = 6.25π — found without ever solving for r itself, since only r² was needed.
★ KS3-MATH-IMC-0266Back
★ KS3-MATH-IMC-0267Front
Together they equal the whole rectangle's perimeter
HintPair one corner with its diagonal partner and add their two side-totals — the cuts cancel out.
The whySplitting a 10 cm by 7 cm rectangle this way, however the two cuts fall, the top-left piece's perimeter plus the bottom-right piece's perimeter always comes to 2 × (10 + 7) = 34 cm. The cut lengths appear once as an addition and once as a subtraction across the diagonal pair, so they always cancel, leaving exactly the original rectangle's perimeter.
★ KS3-MATH-IMC-0267Back
24 points on the KS3 Mathematics map, across Y7 · Y8 · Y9 · Y10 · Y11. The faint stars are the rest of the subject — this deck is the lit part.
Positional, never a mastery claim — the map shows where these cards live, not what your child has learned.
Here, nothing is saved. In your child’s own sky every card is scheduled — it comes back just before they’d forget it — and the professor who wrote it is one tap away.