Mathematical Operations

In this type the question changes the meaning of the signs and then asks for the value of an expression. A legend line comes first, such as 'if + means ×, − means ÷, × means − and ÷ means +', and then an expression with four to six signs, often with a bracket.

Why it works the way it does. The signs printed on the page are only labels. The real operation is whatever the legend says. So the first job is translation, and calculation comes only after the translation is complete. After translation the printed order is forgotten: the new signs follow the normal VBODMAS order, because rank belongs to the operation, not to the printed symbol.

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The method, in three steps.
1. Write the legend as two columns: the printed sign and its new meaning. Read the arrow carefully: '+ means ×' says that wherever + is printed, you write ×.
2. Rewrite the whole expression with the real signs. Brackets stay where they are, but the signs inside them change too. Do not calculate while rewriting.
3. Apply VBODMAS to the rewritten expression and calculate once.

Two kinds of legend. In an interchange legend the signs swap in pairs (+ with −, × with ÷). In a cyclic legend all four signs move in a circle, for example + → ×, × → −, − → ÷, ÷ → +. In both, every printed sign is replaced exactly once. A sign that the question does not mention keeps its usual meaning.

Example: if ‘÷’ means ‘+’, ‘+’ means ‘×’, ‘×’ means ‘−’ and ‘−’ means ‘÷’, then 18 ÷ 6 + 4 × 6 − 3 becomes 18 + 6 × 4 − 6 ÷ 3 = 18 + 24 − 2 = 40.

Here letters or odd symbols such as P, Q, $ or # stand for the four operations. The legend may use symbols ('A means ÷, B means ×, C means +, D means −') or words ('A denotes add, B denotes multiply …'). Then the question asks one of three things: the value of an expression; what comes in place of ? when the ? stands on the right of '='; or which of four equations is correct.

The logic. This is symbol swap without the confusion of old meanings: a letter has no meaning of its own, so you cannot slip back to the printed sign. The new danger is reading the legend backwards, or mixing up neighbouring entries. Remember that the letters change from question to question: A is not always ÷. The letters carry no rank of their own; the operation they stand for decides the rank.

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The method.
1. Write the legend in one row, letter beside its sign: A ÷, B ×, C +, D −.
2. Translate the whole line into real signs without calculating.
3. Apply VBODMAS.

Three shapes of the question.

• Evaluate: translate, then calculate.
• *? on the right:* work out the left side as one number L. If the right side reads '? C 20', that is ? + 20, so ? = L − 20. You undo the operation that stands next to ?.
• Which equation is correct: translate the options one by one and stop at the first that balances. Unless the question says otherwise, one legend serves all the options.

Example: P is +, Q is −, R is ×, S is ÷. Then 48 S 6 P 5 R 4 Q 7 becomes 48 ÷ 6 + 5 × 4 − 7 = 8 + 20 − 7 = 21.

Here the equation is false as printed. If two signs change places it becomes true. The options are pairs of signs such as '+ and ×', and your job is to find the pair that works. A cousin of this form asks, for two equations given in the question, whether they are correct after a stated interchange.

The logic. Only the signs move; the numbers stay where they are. Interchanging '+ and ×' means every + becomes × and every × becomes +. Because the signs sit between fixed numbers, the swap changes which numbers get multiplied, divided, added or taken away. So think about where the big numbers go. A × or + next to a big number makes the result large; a ÷ or − next to it makes the result small.

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The method.
1. Calculate the printed left side with VBODMAS and compare it with the right side. The gap tells you whether the left side must grow or shrink.
2. Find the biggest number and the signs on either side of it. The correct swap usually changes one of these.
3. Test the options one at a time: rewrite with the two signs interchanged, calculate, compare. The first that balances is the answer. With four options you never need more than four tests.

Example: 8 − 6 × 4 ÷ 2 + 3 = 17. As printed the left side is 8 − 12 + 3 = −1, far below 17. Interchange − and +: 8 + 6 × 4 ÷ 2 − 3 = 8 + 12 − 3 = 17. It balances.

In this type two numbers swap places so that the equation holds. The signs stay and the whole numbers move. A number means the full number as printed: in 58 + 36 × 4 the numbers are 58, 36 and 4, never 5, 8, 3 or 6. The exam asks it in two forms: which two numbers should be interchanged to make the equation correct, or what is the value after interchanging two given numbers (or digits).

The logic. The printed left side is wrong by some amount, so some number is in the wrong place. A big number inside a product makes the value big; a big number used as a divisor makes it small. So find the culprit: the biggest product, or the number that is being divided. If the left side is too big, a large number must leave a product or become a divisor. If it is too small, the opposite.

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The method.
1. Calculate the printed left side and note the gap.
2. Spot the culprit term. Look at divisions as well: a ÷ asks for a number that divides cleanly, and a number that is a multiple of the divisor (like 798 for ÷ 6) is a likely mover.
3. Test the options: swap the two numbers at every place they stand, rewrite, and calculate with VBODMAS.
4. For the 'find the value' form there is nothing to hunt: swap, then calculate.

Example: 3 + 7 × 5 − 8 ÷ 4 = 20. As printed the left side is 3 + 35 − 2 = 36, which is 16 too big. Interchange 3 and 7: 7 + 3 × 5 − 8 ÷ 4 = 7 + 15 − 2 = 20. It balances.

Here the numbers are printed with blanks (or stars, or triangles) between them, and the options are sets of signs to be put in order into the blanks. In many recent papers one of the signs is '=' itself: you choose the sequence that makes the equation balanced. In older questions the '=' is printed and you fill only + − × ÷. A sibling form gives a few numbers and asks which expression, made from them once each, equals a target.

The logic. An equation has two sides, and '=' splits the numbers into a left group and a right group. Each side obeys VBODMAS on its own. Very often one side is a single number standing alone, at the very start or the very end. That is why n numbers need n − 1 signs with exactly one of them '=', and why the search is short.

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The method.
1. Count the blanks. Exactly one sign is '='. An option with no '=' or with two '=' is out at once.
2. Pick the lone number: try '=' at the last blank first, then at the first blank, and only then in the middle.
3. Make that number from the rest. A target much bigger than the numbers needs ×; a target smaller than the first number needs ÷ or −. A clean division like 18 ÷ 9 is a good sign; a division that leaves a fraction is rejected when everything else is whole.
4. Test the whole equation with VBODMAS, not the last blank alone.

Expression-equals-target form. Evaluate each option with VBODMAS and throw out far-off values by size. Brackets decide a lot: 2 + 5 × 9 = 47 but (2 + 5) × 9 = 63. Check that each option uses the given numbers exactly once.

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Example: for 12 _ 4 _ 3 _ 6 = 15, try ÷, ×, +: 12 ÷ 4 = 3, then 3 × 3 = 9, then 9 + 6 = 15. The set ×, ÷, + gives 12 × 4 ÷ 3 + 6 = 22 and fails.

Here the question invents a symbol and defines it, so the symbol is not +, −, × or ÷: it is a machine. You put numbers in, and the definition tells you what comes out. The recent papers set it in three shapes:

• a two-number formula: 'A @ B means A is added to the square of B', or 'a ⊕ b = 2a + 3b − 4';
• two symbols defined together and combined: x # y = (x + y)(x − y) and x @ y = (x² + y²) ÷ 2, then find (5 # 3) @ (8 # 6);
• a one-number definition on digits: 'n ≈ means the sum of the cubes of the digits of n'.

The logic. The first number goes where a (or A) stands, and the second number where b (or B) stands. In 6 ⊕ 2, a = 6 and b = 2, so 2 × 6 + 3 × 2 − 4 = 14. Write a = 2 and b = 6 and you get 18, a different answer. Most mistakes in this type are this one mix-up.

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The method.
1. Write the definition once and mark the slots: first = a, second = b.
2. Substitute, then calculate with VBODMAS.
3. In a chain such as (6 ⊕ 2) ⊕ 3, go from the inner bracket outwards; the result of step one becomes the first number of step two: 14 ⊕ 3 = 33. Brackets matter, because 6 ⊕ (2 ⊕ 3) gives 35.
4. With two symbols, keep each formula on its own line and use it only for its own symbol.
5. For a digit definition, split the number into digits first and apply the definition to the digits. A cube table helps: 1, 8, 27, 64, 125, 216, 343, 512, 729.

When the formula is symmetric in a and b (like a² + b²), the order does not matter; for 3a − 2b the order is everything.

In this type no formula is given. The question shows a few solved cases, and you must discover the relation and use it on a new case. It comes in three shapes.

• Sample lines: 2 + 2 = 6, 3 + 3 = 12, then 4 + 4 = ? The symbol is only a label.
• Pairs (number analogy): 6 : 40 :: 9 : 61 and then 12 : ?, or pairs of two numbers such as 31 : 52 = 4836.
• Triads: sets of three numbers, (5, 3, 16) and (6, 3, 19), and you pick the set that follows the same relation, or fill the blank in a row of a table.

The logic. One case can be explained by many relations, so a relation is accepted only when it fits every given case. The size of the output guides you. A little more or less than the product: the leftover is usually a, b or 1. Far larger than the inputs: squares. Between the inputs or smaller: a sum or a difference.

Search order, simple first: a ± b; ab ± c; 2a + b and a + 2b; a(b ± 1); a² ± b and a² − b²; a fixed multiple of ab (like 3ab). Test each guess on every case and drop it at the first failure.

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Pairs with one number on each side. The simplest machine is linear: multiply, then add or subtract a constant. Multiplier = change in output ÷ change in input; constant = output − multiplier × input. If the multiplier changes from pair to pair, the relation is not linear: try n² ± c.

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Triads. Find the relation from the complete rows, then test every option set against it. When the blank is in the middle, solve the relation backwards. Whole numbers are used as they are; never break a number into digits unless the question allows it.

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'a mod b' means the remainder when a is divided by b. For positive numbers it lies from 0 to b − 1: 47 mod 6 = 5, because 47 = 6 × 7 + 5. It is the leftover, not the quotient; a student who writes 7 has answered a different question.

The exam asks mod in three forms: an expression of mods, such as (125 mod 9) + (47 mod 6) × 2; a formula on given values, such as ((A + B) × C) mod 97; and a plain 'remainder when 11 × 22 × 33 × 44 × 55 × 66 is divided by 7'.

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The logic. Each mod is a small bracket: find its value first, then finish with VBODMAS. And the remainder of a sum or a product depends only on the remainders of its parts: (a × b) mod n = ((a mod n) × (b mod n)) mod n. That is why you may cut every factor down first and keep the numbers tiny. It holds for + − ×, never for ÷.

Shortcuts for the divisor.

For 9, a digit sum of 9 means remainder 0, not 9. For any other divisor, divide directly with your tables.

The method. (1) Solve the mods inside the brackets first. (2) Reduce each factor of a product to its own remainder. (3) Multiply step by step and reduce again after every step. (4) Finish the rest of the expression with VBODMAS. Every mod in this chapter works on positive numbers, so a remainder is never negative.