Shortcuts & Speed Methods

BODMAS Traps

Simplification questions should be the quickest marks in your Tier 1 paper. You get about 36 seconds a question, and a clean BODMAS sum needs only 15 to 20 of them. Time goes in two ways: a slip in the order that leads you straight to a wrong option, or long working where a pattern would have done the job. Each method below comes with a worked example and a note on where it goes wrong, such as the cube-root trick used on a number that is not a perfect cube, or rounding when two options are very close.

The order is V B O D M A S: vinculum (the bar), brackets, "of", then division and multiplication, then addition and subtraction. The BODMAS and Simplification section has the full rule. These are the three places where people slip.

"of" comes before ÷ and ×

When the word "of" sits next to a ÷ sign, as in 96 ÷ 4 of 6, put a bracket round "A of B" in your head the moment you read it. So 96 ÷ 4 of 6 = 96 ÷ (4 × 6) = 96 ÷ 24 = 4.

For example: $\frac{1}{2} \div \frac{1}{2} \text{ of } \frac{1}{2}$. First the "of": half of a half is $\frac{1}{4}$. Then $\frac{1}{2} \div \frac{1}{4} = 2$. Going left to right instead gives $\frac{1}{2}$, and that is the trap option.

Where it fails: only the word "of" gets this priority. 96 ÷ 4 × 6 is 24 × 6 = 144, because × has no special rank. And "of" holds only the two numbers right next to it: $\frac{3}{4}$ of 16 + 4 is 12 + 4 = 16, not $\frac{3}{4}$ of 20.

The bar (vinculum) first, and a minus sign in front hits the whole bar

A line drawn over part of the expression, like $7 - \overline{8 - 3}$, is a bracket. Work out what is under it first, then carry on from the inside out.

For example: $18 - [5 - \{6 + (7 - \overline{8 - 3})\}]$. Bar: 8 − 3 = 5. Round bracket: 7 − 5 = 2. Curly bracket: 6 + 2 = 8. Square bracket: 5 − 8 = −3. Last step: 18 − (−3) = 21.

Where it fails: the minus sign in front of the bar. $7 - \overline{8 - 3}$ is 7 − 5 = 2. Dropping the bar gives 7 − 8 − 3 = −4. When you open the next bracket, carry the sign of each answer with it, especially a negative one like the −3 above.

÷ and × have equal rank: go left to right

For a chain of ÷ and × with no brackets, like 96 ÷ 8 × 4 ÷ 2, write the whole chain as one fraction. A number just after × goes on top, a number just after ÷ goes underneath, and then you cancel.

$$96 \div 8 \times 4 \div 2 = \frac{96 \times 4}{8 \times 2} = 24$$

Cancelling first is quicker still: 96 ÷ 16 = 6, and 6 × 4 = 24.

Where it fails: reading "D before M" as a strict order, or doing the × first. 96 ÷ (8 × 4) ÷ 2 gives 1.5, which is wrong. The same goes for + and −: 20 − 8 + 2 is 14, not 10. Only the single number right after a ÷ goes to the bottom. If "of" or a bracket follows the ÷, the whole group goes underneath: 96 ÷ 4 of 6 = 96/24.

Picking the Answer Without Full Working

SSC options are close, but they are rarely close in every way. These three checks let you strike out options before you do the full sum, and often instead of it.

Digit sum (casting out 9s)

When to use it: a big multiplication or square, or a "what comes in place of ?" question, where all the options end in the same digit.

Fast way: cut each number down to one digit by adding its digits, and adding again until one digit is left. A 9 counts as 0. Do the same operation on these small digits. The right answer must cut down to the same digit.

For example: 4,387 × 236 = ? (a) 10,35,332 (b) 10,35,342 (c) 10,34,332 (d) 10,35,432. Every option ends in 2, so the last digit can't help. 4,387 → 22 → 4 and 236 → 11 → 2, and 4 × 2 = 8. The options cut down to 8, 9, 7 and 9. Only (a) gives 8.

Where it fails: swapping two digits does not change the digit sum, so 10,35,332 and 10,53,332 look the same to this test. So does any pair of options that differ by a multiple of 9. With decimals the digit sum ignores the point, so check the size separately.

Unit digit

This works on products, sums and differences of products, and powers. Find only the last digit of the answer and drop every option that ends differently.

For example: 368 × 47 − 1,289 = ? (a) 16,013 (b) 16,007 (c) 16,105 (d) 16,009. Last digits: 8 × 7 = 56 gives 6, and 6 − 9 needs a borrow, so 16 − 9 = 7. Only (b) 16,007 ends in 7.

Where it fails: two options with the same last digit. Then use the digit sum or the size. With decimals, the last digit still works once you count the decimal places: 0.37 × 0.8 ends in 6 (7 × 8 = 56) at the third place, so it is 0.296.

Approximation by rounding

When to use it: "the value is closest to" questions, and any question whose options are far apart.

Fast way: round each number to a friendly one. In a product, push one number up and the other down so the errors cancel. In a sum, the errors simply add, so round each one only a little.

For example: 39.89% of 1,201 + 14.98 × 7.02 − $\sqrt{1443}$ is closest to (a) 527 (b) 547 (c) 567 (d) 587. Round: 40% of 1,200 = 480, 15 × 7 = 105, and $\sqrt{1444} = 38$. So 480 + 105 − 38 = 547. The exact value is about 546.

Where it fails: options within 1 or 2% of each other. Rounding can then land you on the wrong one, so work one step more exactly. Be careful with a difference of two nearly equal numbers: 50.4 − 49.6 is 0.8, but rounding both to 50 gives 0.

[SSC Tip] Use these checks in order of cost: last digit first (two seconds), then size, then digit sum. Most questions are settled by the first two.

Fractions, Decimals and Percentages from Memory

The table to know by heart

When to use it: a decimal like 0.375, 0.0625 or 0.1666… appears in a product, or a percentage like 12.5% or $16\frac{2}{3}\%$ appears inside an expression.

Fast way: swap the decimal for its fraction, then cancel.

Multiples follow at once: 3/8 = 0.375, 5/8 = 0.625, 7/8 = 0.875, 3/16 = 0.1875, 5/6 = 0.8333… and 7/11 = 0.6363…

For example: 0.375 × 96 + 0.0625 × 80 = $\frac{3}{8} \times 96 + \frac{1}{16} \times 80$ = 36 + 5 = 41. There is no decimal multiplication at all.

Where it fails: 0.33 is not exactly 1/3, and 0.16 is not 1/6. If a question writes 0.33 with no dots and no bar, it means 33/100, so only swap it for 1/3 when the options are far apart. Watch the place value too: 1/8 = 0.125, but 1/80 = 0.0125 and 1/800 = 0.00125.

Sevenths and elevenths

Every seventh uses the same six digits, 142857, in the same cycle. Start at the digit that matches the size. 5/7 is a little over 0.7, so start at the 7: 5/7 = 0.714285… For elevenths, k/11 repeats the two digits of 9 × k. So 4/11 = 0.3636… (9 × 4 = 36) and 7/11 = 0.6363… (9 × 7 = 63).

For example: 3/7 is about 0.43, so start at the 4: 3/7 = 0.428571… And 2/11 repeats 18: 0.1818…

Careful: the starting digit comes from the size, not from the numerator, so 3/7 does not start with 3. For 1/11 and 10/11 the pairs are 09 and 90, so keep the zero: 1/11 = 0.0909…

The percent column is just the decimal moved two places. Knowing $12\frac{1}{2}\%$ = 1/8 and $16\frac{2}{3}\%$ = 1/6 turns most "x% of y" steps into one division.

Comparing Fractions

Cross-multiplying two fractions

When to use it: two fractions are close together and you need the bigger one, or you are checking one pair inside an ordering question.

Fast way: for $\frac{a}{b}$ and $\frac{c}{d}$, compare a × d with b × c. Each product sits on the side of its own numerator, and the bigger product marks the bigger fraction.

For example: $\frac{7}{12}$ or $\frac{11}{19}$? 7 × 19 = 133 and 12 × 11 = 132. 133 is bigger, so $\frac{7}{12}$ is bigger, but only just.

Where it fails: it handles only two at a time. Four fractions can need six products, so for "arrange in order" questions use decimals from memory or one of the benchmarks below. Both bottoms must also be positive.

Difference from 1 (or from 1/2)

When to use it: the fractions all have the same gap between bottom and top, like $\frac{23}{27}$, $\frac{31}{35}$ and $\frac{47}{51}$, or they all sit just above or below $\frac{1}{2}$.

Fast way: each fraction is 1 minus (gap ÷ bottom). With the same gap, the biggest bottom takes away the least, so the biggest bottom wins. Here the gaps are $\frac{4}{27}$, $\frac{4}{35}$ and $\frac{4}{51}$, so $\frac{47}{51}$ is largest and $\frac{23}{27}$ is smallest.

Another one, using 1/2: $\frac{5}{9}$, $\frac{7}{12}$, $\frac{11}{20}$ and $\frac{13}{24}$ are all a little over a half. Take the half away from each:

$$\frac{5}{9} - \frac{1}{2} = \frac{1}{18}, \quad \frac{7}{12} - \frac{1}{2} = \frac{1}{12}, \quad \frac{11}{20} - \frac{1}{2} = \frac{1}{20}, \quad \frac{13}{24} - \frac{1}{2} = \frac{1}{24}$$

The biggest extra is $\frac{1}{12}$, so the order from largest is $\frac{7}{12}$, $\frac{5}{9}$, $\frac{11}{20}$, $\frac{13}{24}$.

Where it fails:

• For fractions bigger than 1 the rule turns round. $\frac{15}{13} = 1 + \frac{2}{13}$ and $\frac{19}{17} = 1 + \frac{2}{17}$, so here the smaller bottom wins and $\frac{15}{13}$ is bigger.
• When the gaps are different, compare the gaps as fractions, not as plain numbers. $\frac{5}{8}$ has a gap of 3 and $\frac{11}{15}$ a gap of 4, but $\frac{3}{8}$ is bigger than $\frac{4}{15}$, so $\frac{11}{15}$ is the bigger fraction.

Two one-line rules

• Same top: the smaller bottom gives the bigger fraction, so $\frac{7}{9} > \frac{7}{11}$.
• Adding the same number to the top and the bottom pulls a fraction towards 1. A fraction below 1 grows ($\frac{3}{5} < \frac{4}{6} < \frac{5}{7}$) and a fraction above 1 shrinks ($\frac{7}{5} > \frac{8}{6}$).

[SSC Tip] In "which is the largest" questions, find the two biggest with a rough decimal, then cross-multiply just that pair.

Continued Fractions and Telescoping Series

Continued fractions: work from the bottom up

Fast way: start at the bottom line and work upwards with two moves. Add: turn a whole number plus a fraction into one fraction. Flip: 1 over a fraction is that fraction turned upside down. If the top is not 1, flip and then multiply by the top.

For example:

$$1 + \cfrac{1}{2 + \cfrac{1}{3 + \cfrac{1}{4}}}$$

Add: $3 + \frac{1}{4} = \frac{13}{4}$. Flip: $\frac{4}{13}$. Add: $2 + \frac{4}{13} = \frac{30}{13}$. Flip: $\frac{13}{30}$. Add: $1 + \frac{13}{30} = \frac{43}{30}$.

To solve for x, peel from the top instead: take away the whole part, flip, and repeat.

$$2 + \cfrac{1}{3 + \cfrac{1}{x}} = \frac{29}{13}$$

Taking away 2 leaves $\frac{3}{13}$. Flipping gives $3 + \frac{1}{x} = \frac{13}{3}$, so $\frac{1}{x} = \frac{4}{3}$ and $x = \frac{3}{4}$.

Where it fails: a top that is not 1. $\cfrac{2}{3 + \cfrac{4}{5}}$ is $2 \times \frac{5}{19} = \frac{10}{19}$, not $\frac{5}{19}$. And never start evaluating from the top, because the first line still has more fraction hanging under it.

Telescoping sums

When to use it: a sum of terms like $\frac{1}{n(n+1)}$, or bottoms like 6, 12, 20, 30, which are 2 × 3, 3 × 4, 4 × 5 and 5 × 6 in disguise.

Fast way: split each term into two pieces. Every middle piece cancels.

$$\frac{1}{n(n+k)} = \frac{1}{k}\left(\frac{1}{n} - \frac{1}{n+k}\right)$$

So the sum is $\frac{1}{k}$ × (1 over the first number − 1 over the last number).

For example: $\frac{1}{3 \times 4} + \frac{1}{4 \times 5} + \cdots + \frac{1}{11 \times 12} = \frac{1}{3} - \frac{1}{12} = \frac{1}{4}$. With a gap of 3: $\frac{1}{1 \times 4} + \frac{1}{4 \times 7} + \cdots + \frac{1}{19 \times 22} = \frac{1}{3}\left(1 - \frac{1}{22}\right) = \frac{7}{22}$.

Where it fails: forgetting the $\frac{1}{k}$ when the gap is more than 1. Here that gives $\frac{21}{22}$, which is sure to be an option. The gap must also be the same in every term, with the second number of one term becoming the first number of the next.

Product chains

When to use it: $\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots$ or $\left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) \cdots$

Fast way: write each bracket as one fraction. Every top cancels the bottom next to it, so only one top and one bottom survive.

For example: $\left(1 - \frac{1}{3}\right)\left(1 - \frac{1}{4}\right) \cdots \left(1 - \frac{1}{20}\right) = \frac{2}{3} \times \frac{3}{4} \times \cdots \times \frac{19}{20} = \frac{2}{20} = \frac{1}{10}$.

Where it fails: squares at the bottom. $1 - \frac{1}{k^2} = \frac{(k-1)(k+1)}{k^2}$, so two chains cancel at once, and from k = 2 to n the answer is $\frac{n+1}{2n}$, not $\frac{1}{n}$. [Tier 2]

Squares and Products Near a Base

Numbers ending in 5

Fast way: call the digits before the 5 "a". Write a × (a + 1), then write 25 after it.

For example: $85^2$: 8 × 9 = 72, then 25, so 7,225. $115^2$: 11 × 12 = 132, then 25, so 13,225.

Careful: it works only when the last digit is 5, and the ending is always 25. With a decimal, square the digits and then fix the point: $4.5^2$ comes from $45^2 = 2025$, and one decimal place squared gives two, so 20.25.

Numbers near 50

When to use it: squaring a number from about 40 to 60.

Fast way: for 50 + d, the front part is 25 + d and the last two digits are $d^2$. For 50 − d, the front part is 25 − d.

For example: $53^2$: front 25 + 3 = 28, back $3^2$ = 09, so 2,809. $46^2$: front 25 − 4 = 21, back $4^2$ = 16, so 2,116.

Where it fails: the back part must be exactly two digits, so $3^2$ is written 09. Once d reaches 10, $d^2$ spills over and the trick is no longer quick.

Numbers and products near 100 or 1000

When to use it: $97^2$, $106^2$, 104 × 107, 96 × 93, 998 × 1004.

Fast way: write each number as base + a and base + b, where a or b can be negative. The front part is base + a + b. The back part is a × b, written with as many digits as the base has zeros.

For example: 104 × 107: front 100 + 4 + 7 = 111, back 4 × 7 = 28, so 11,128. For 96 × 93 the distances are −4 and −7: front 100 − 11 = 89, back (−4) × (−7) = 28, so 8,928. And $106^2$ is front 112, back 36, so 11,236.

When one number is above the base and one below, the back part is negative, so subtract it. 998 × 1004: front 1000 − 2 + 4 = 1002, so 10,02,000, and the back part is −8. The answer is 10,02,000 − 8 = 10,01,992.

Where it fails:

• A back part that is too long. 112 × 115 has front 127 and back 12 × 15 = 180. Only two digits fit, so the 1 carries: 12,880.
• A back part that is too short. With base 1000, 4 × 7 = 28 is written 028, so 1004 × 1007 = 10,11,028.
• Two numbers the same distance either side of the base are faster as a difference of squares: 103 × 97 = $100^2 - 3^2$ = 9,991.

The Squares, Square Roots and Cube Roots section has the full methods, and the Identities section has more products near a base.

Cube Roots and Nested Roots

Cube root of a perfect cube from the last digit

When to use it: the number is a perfect cube, because the question says so or every option is a whole number.

Fast way: the last digit of the cube fixes the last digit of the root.

Only 2 and 8 swap, and 3 and 7 swap. Every other digit stays the same. Then cover the last three digits and find the biggest cube that fits into what is left. Its root is the front part of the answer.

For example: $\sqrt[3]{175616}$. It ends in 6, so the root ends in 6. Covering 616 leaves 175, and $5^3 = 125$ fits but $6^3 = 216$ doesn't. The root is 56.

Where it fails: on a number that is not a perfect cube it still gives an answer, just a wrong one. For a 7-digit cube, what is left after covering can be 1000 or more: $\sqrt[3]{1092727}$ leaves 1092, and $10^3 = 1000$ fits, so the root is 103.

Infinite nested roots

When to use it: $\sqrt{x + \sqrt{x + \sqrt{x + \cdots}}}$ or the same thing with minus signs.

Fast way: split x into two consecutive numbers, x = n(n − 1). The plus version is the bigger number and the minus version is the smaller one.

$$\sqrt{x + \sqrt{x + \cdots}} = n \qquad \sqrt{x - \sqrt{x - \cdots}} = n - 1$$

For example: 56 = 8 × 7, so $\sqrt{56 + \sqrt{56 + \cdots}} = 8$ and $\sqrt{56 - \sqrt{56 - \cdots}} = 7$.

Why it works: call the plus version y. The part under the first root is x + y again, so $y^2 = x + y$, which is $y(y - 1) = x$. For a product chain, $\sqrt{x\sqrt{x\sqrt{x \cdots}}} = x$, because $y^2 = xy$.

Where it fails:

• If x is not a product of two consecutive numbers, the answer is not a whole number. $\sqrt{10 + \sqrt{10 + \cdots}}$ is about 3.70.
• Mixing up plus and minus. Plus always gives the bigger number of the pair.
• A chain that stops is a different question. $\sqrt{x\sqrt{x\sqrt{x}}}$, with three roots, is $x^{\frac{7}{8}}$, not x. With n roots the power is $1 - \frac{1}{2^n}$. [Tier 2]

[SSC Tip] Keep the products of consecutive numbers ready: 12, 20, 30, 42, 56, 72, 90, 110, 132, 156. When one of these sits inside an endless root, the answer is a whole number.

Indices: Make the Bases or the Powers Equal

Make the bases equal

When to use it: the numbers are all powers of one prime, like 2, 4, 8, 16, 32, or 3, 9, 27, 81, or 5, 25, 125, and you need to simplify or find x.

Fast way: rewrite everything as a power of that prime. Multiplying adds the powers, dividing subtracts them, and a power of a power multiplies them. Once both sides have the same base, set the powers equal.

For example: $4^{x+1} = 8^{x-1}$. In base 2 this is $2^{2x+2} = 2^{3x-3}$, so 2x + 2 = 3x − 3 and x = 5. Check: $4^6 = 8^4 = 4096$.

For a fractional power, take the root first and then the power: $27^{\frac{2}{3}} \times 81^{-\frac{1}{4}} = 9 \times \frac{1}{3} = 3$.

Where it fails:

• You can set the powers equal only when the bases match. $2^x = 3^y$ tells you nothing about x and y by itself.
• Powers add only in a product. $2^x + 2^x = 2 \times 2^x = 2^{x+1}$, not $2^{2x}$. And $2^x + 3^x$ is not $5^x$.

Take out the smallest power

When to use it: a sum or difference of powers of the same base, like $3^{x+2} + 3^{x+1} + 3^x$.

Fast way: take the smallest power outside a bracket. What is left inside is a plain number.

For example:

$$\frac{2^{n+4} - 2 \times 2^n}{2 \times 2^{n+3}} = \frac{2^n(16 - 2)}{2^n \times 16} = \frac{14}{16} = \frac{7}{8}$$

The n cancels completely, which is why such questions have a plain number as the answer.

Careful: different bases can't be taken out together. $2^{n+1} + 3^n$ stays as it is.

Compare powers by making the exponents equal

When to use it: "Which is largest: $2^{300}$, $3^{200}$ or $5^{100}$?", or roots of different orders like $\sqrt{2}$, $\sqrt[3]{3}$ and $\sqrt[6]{6}$.

Fast way: for big powers, pull out the HCF of the exponents. $2^{300} = (2^3)^{100} = 8^{100}$ and $3^{200} = 9^{100}$, while $5^{100}$ stays. Comparing 8, 9 and 5 shows $3^{200}$ is largest. For roots, raise everything to the LCM of the orders.

For example: for $2^{\frac{1}{2}}$, $3^{\frac{1}{3}}$ and $6^{\frac{1}{6}}$ the LCM is 6. Raising each to the power 6 gives $2^3 = 8$, $3^2 = 9$ and $6^1 = 6$. So $\sqrt[3]{3}$ is largest and $\sqrt[6]{6}$ is smallest.

Where it fails: the exponents need a useful common factor. When they have none, compare each power with a power of 10 using $2^{10} = 1024$ and $3^5 = 243$. And watch for options that are secretly equal: $4^{\frac{1}{4}} = 2^{\frac{1}{2}}$, so neither can be "the largest" alone.

Surds: Conjugates and Comparing

Rationalise with the conjugate

When to use it: a surd sum or difference sits in the bottom, like $\frac{1}{\sqrt{7} - \sqrt{5}}$ or $\frac{1}{3 + 2\sqrt{2}}$.

Fast way: multiply the top and the bottom by the same pair with the sign changed. The bottom becomes (first)² − (second)², with no root left.

For example: $\frac{1}{\sqrt{7} - \sqrt{5}} = \frac{\sqrt{7} + \sqrt{5}}{7 - 5} = \frac{\sqrt{7} + \sqrt{5}}{2}$.

When the bottom comes out as 1, the reciprocal is just the conjugate. $(3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 9 - 8 = 1$, so if $x = 3 + 2\sqrt{2}$, then $\frac{1}{x} = 3 - 2\sqrt{2}$ and $x + \frac{1}{x} = 6$ in one line.

The same idea gives $\frac{1}{\sqrt{n+1} + \sqrt{n}} = \sqrt{n+1} - \sqrt{n}$, so a chain of such terms telescopes and only the two ends survive:

$$\frac{1}{\sqrt{9} + \sqrt{10}} + \frac{1}{\sqrt{10} + \sqrt{11}} + \cdots + \frac{1}{\sqrt{48} + \sqrt{49}} = \sqrt{49} - \sqrt{9} = 4$$

Where it fails:

• The sign on top flips. $\frac{1}{\sqrt{7} - \sqrt{5}}$ has a plus on top once you rationalise it.
• A number in front of the root gets squared too. $(2\sqrt{3})^2 = 12$, not 6, so $(2\sqrt{3} + 3)(2\sqrt{3} - 3) = 12 - 9 = 3$.

Compare surd sums by squaring

When to use it: "Which is largest: $\sqrt{10} + \sqrt{3}$, $\sqrt{8} + \sqrt{5}$, $\sqrt{7} + \sqrt{6}$ or $\sqrt{9} + \sqrt{4}$?"

Fast way: all of them are positive, so squaring keeps the order. $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$. When a + b is the same in every option (13 here), only the product ab matters: 30, 40, 42 and 36. The biggest product wins, so $\sqrt{7} + \sqrt{6}$ is largest. In short: for the same total, the closer the two numbers are, the bigger the sum of their roots.

For differences: $\sqrt{a} - \sqrt{b} = \frac{a - b}{\sqrt{a} + \sqrt{b}}$. With the same gap a − b, bigger numbers make a bigger bottom and so a smaller difference. That is why $\sqrt{12} - \sqrt{11}$ is less than $\sqrt{11} - \sqrt{10}$.

Where it fails: if the totals a + b are not equal, compare the whole squares, not just the products. And don't square a difference to compare it, because a difference can be negative and squaring hides the sign. Use the rationalised form instead. The Surds and Rationalisation section has more ordering questions.

Square Roots of Surds and Componendo–Dividendo

$\sqrt{a \pm 2\sqrt{b}}$: two numbers with sum a and product b

When to use it: a square root sits over a surd sum or difference, like $\sqrt{11 + 2\sqrt{30}}$ or $\sqrt{7 - 4\sqrt{3}}$.

Fast way: first make the number in front of the inner root exactly 2. Then find two numbers p and q, with p bigger, so that p + q = a and pq = b.

$$\sqrt{a + 2\sqrt{b}} = \sqrt{p} + \sqrt{q} \qquad \sqrt{a - 2\sqrt{b}} = \sqrt{p} - \sqrt{q}$$

For example: $\sqrt{11 + 2\sqrt{30}}$: 6 + 5 = 11 and 6 × 5 = 30, so it is $\sqrt{6} + \sqrt{5}$. For $\sqrt{7 - 4\sqrt{3}}$, first write $4\sqrt{3} = 2\sqrt{12}$. Now 4 + 3 = 7 and 4 × 3 = 12, so it is $\sqrt{4} - \sqrt{3} = 2 - \sqrt{3}$.

Where it fails:

• Skipping the "make it 2" step. $6\sqrt{2} = 2\sqrt{18}$ and $\sqrt{24} = 2\sqrt{6}$. If you leave $\sqrt{24}$ as it is, you look for a product of 24 and get nowhere.
• In the minus version the bigger root goes first, or the answer comes out negative. $\sqrt{7 - 4\sqrt{3}}$ is $2 - \sqrt{3}$, not $\sqrt{3} - 2$.
• If no whole numbers p and q exist, the expression doesn't simplify this way, and the question is probably asking for something else.

Componendo–dividendo

When to use it: a ratio made of a sum and a difference, like $\frac{x + y}{x - y}$, or a surd equation of this shape:

$$\frac{\sqrt{x + 5} + \sqrt{x - 5}}{\sqrt{x + 5} - \sqrt{x - 5}} = 3$$

Fast way: if $\frac{p}{q} = \frac{a}{b}$, then $\frac{p + q}{p - q} = \frac{a + b}{a - b}$. The rule works in both directions.

For example: apply it to the equation above. On the left, (top + bottom) ÷ (top − bottom) leaves $\frac{2\sqrt{x+5}}{2\sqrt{x-5}}$. On the right, $\frac{3 + 1}{3 - 1} = 2$. So $\frac{\sqrt{x+5}}{\sqrt{x-5}} = 2$. Squaring gives $\frac{x + 5}{x - 5} = 4$. Use the rule again: $\frac{2x}{10} = \frac{5}{3}$, so $x = \frac{25}{3}$.

A simpler case: if $\frac{x}{y} = \frac{5}{3}$, then $\frac{x + y}{x - y} = \frac{8}{2} = 4$ straight away.

Where it fails:

• Forgetting to square after the first step, while the roots are still there.
• Using it when the top and bottom are not the sum and difference of the same two terms.
• A ratio of 1, because then a − b = 0 and the rule divides by zero.

Identity Shortcuts for Numbers

The Simplifying with Identities and Approximation section has every identity in full. These are the ones that turn a page of arithmetic into one line.

If a + b + c = 0, then $a^3 + b^3 + c^3 = 3abc$

When to use it: three cubes whose numbers, signs included, add up to zero, like $17^3 - 12^3 - 5^3$ or $0.5^3 + 0.3^3 - 0.8^3$.

Fast way: check the sum first. If it is 0, the sum of the cubes is just 3 × the product, with the signs.

For example: 17 − 12 − 5 = 0, so $17^3 - 12^3 - 5^3 = 3 \times 17 \times (-12) \times (-5) = 3060$. And 0.5 + 0.3 − 0.8 = 0, so

$$\frac{0.5^3 + 0.3^3 - 0.8^3}{0.5 \times 0.3 \times 0.8} = \frac{3 \times 0.5 \times 0.3 \times (-0.8)}{0.5 \times 0.3 \times 0.8} = -3$$

Where it fails:

• The sign. Here only one number is negative, so the product is negative and the answer is −3, not 3.
• Numbers that don't add to 0. Then use the full identity, $a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$. A fraction of the first over the second bracket is simply a + b + c. With 38, 27 and 35 that is 100, without cubing anything.

Sum or difference of cubes over the matching bottom

When to use it: $\frac{a^3 + b^3}{a^2 - ab + b^2}$ or $\frac{a^3 - b^3}{a^2 + ab + b^2}$, usually with decimals.

Fast way: the answer is a + b for the first and a − b for the second.

For example: $\frac{0.87^3 + 0.13^3}{0.87^2 - 0.87 \times 0.13 + 0.13^2} = 0.87 + 0.13 = 1$, and $\frac{5.7^3 - 2.3^3}{5.7^2 + 5.7 \times 2.3 + 2.3^2} = 5.7 - 2.3 = 3.4$.

Careful: the middle sign at the bottom must be the opposite of the sign between the cubes. $\frac{a^3 + b^3}{a^2 + ab + b^2}$ does not simplify, and a question written like that is testing whether you notice.

Square identities in one line

When to use it: two squares with a matching sum, difference or product nearby.

Fast way: $\frac{a^2 - b^2}{a - b} = a + b$, $\frac{(a + b)^2 + (a - b)^2}{a^2 + b^2} = 2$ and $\frac{(a + b)^2 - (a - b)^2}{ab} = 4$.

For example: $\frac{7.4^2 - 2.6^2}{4.8} = 7.4 + 2.6 = 10$, because 4.8 is exactly 7.4 − 2.6.

Where it fails: check that the bottom really matches. If it said 4.6 instead of 4.8, the answer would no longer be a neat 10.

Solved Examples

Each answer shows the fast route, then a time check against the long way.

Question 1

Simplify: $84 \div 7 \text{ of } 4 \times 6 - \overline{12 - 7}$

Answer. "7 of 4" comes first, so it is 28. Then 84 ÷ 28 = 3, and 3 × 6 = 18. The bar makes 12 − 7 = 5 a single number. So 18 − 5 = 13. Ignoring the "of" gives 84 ÷ 7 × 4 × 6 − 5 = 283, which is the kind of option you will see.

Time check: about 15 seconds.

Question 2

5,876 × 384 = ? (a) 22,56,384 (b) 22,56,284 (c) 22,55,384 (d) 22,57,384

Answer. All four end in 4, because 6 × 4 = 24, so the last digit is no help. Digit sums: 5,876 → 26 → 8 and 384 → 15 → 6, and 8 × 6 = 48 → 12 → 3. The options give 30 → 3, 29 → 2, 29 → 2 and 31 → 4. Only (a) 22,56,384 matches.

Time check: about 15 seconds, against 45 seconds or more for the full multiplication.

Question 3

$\sqrt{4100} \times 2.98 + 24.97\%$ of 803 − 9.02 × 11.9 is closest to: (a) 264 (b) 284 (c) 304 (d) 324

Answer. I round everything. $\sqrt{4100}$ is just over $\sqrt{4096} = 64$, and 2.98 is about 3, so the first part is about 192. 25% of 800 is 200. 9 × 12 = 108. So 192 + 200 − 108 = 284, which is (b). The exact value is about 284.0.

Time check: about 20 seconds. Working with the decimals as given would take two minutes.

Question 4

Which is the largest: $\frac{19}{23}$, $\frac{27}{31}$, $\frac{35}{39}$, $\frac{14}{17}$?

Answer. The first three all have a gap of 4 between bottom and top, so they are 1 − $\frac{4}{23}$, 1 − $\frac{4}{31}$ and 1 − $\frac{4}{39}$. The biggest bottom takes away the least, so $\frac{35}{39}$ is the best of those three. The last one has a different gap, $\frac{3}{17}$, so I compare it with $\frac{4}{39}$ by cross-multiplying: 3 × 39 = 117 and 4 × 17 = 68. So $\frac{3}{17}$ is the bigger gap, and $\frac{14}{17}$ is smaller. The largest is $\mathbf{\frac{35}{39}}$.

Time check: about 20 seconds. Turning all four into decimals takes close to a minute.

Question 5

Find $0.875 \times 48 + 0.4375 \times 32 - 0.1\overline{6} \times 54$.

Answer. 0.875 = 7/8, 0.4375 = 7/16 and $0.1\overline{6}$ = 1/6. So it is 7/8 × 48 + 7/16 × 32 − 1/6 × 54 = 42 + 14 − 9 = 47.

Time check: about 15 seconds. Multiplying the decimals out takes about a minute, and $0.1\overline{6}$ can't be multiplied out exactly anyway.

Question 6

Find the value of

$$3 - \cfrac{2}{1 + \cfrac{1}{2 + \cfrac{3}{4}}}$$

Answer. From the bottom. Add: $2 + \frac{3}{4} = \frac{11}{4}$. Flip: $\frac{4}{11}$. Add: $1 + \frac{4}{11} = \frac{15}{11}$. Now the top is 2, so flip and multiply by 2: $2 \times \frac{11}{15} = \frac{22}{15}$. Last step: $3 - \frac{22}{15} = \mathbf{\frac{23}{15}}$.

Time check: about 25 seconds. The two places to slow down are the $\frac{3}{4}$ at the bottom (it is 3 over 4, not 1 over 4) and the 2 on top.

Question 7

Find $\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \cdots + \frac{1}{29 \times 32}$.

Answer. The gap in each pair is 3, and each second number starts the next pair, so the sum telescopes. It equals $\frac{1}{3}\left(\frac{1}{2} - \frac{1}{32}\right) = \frac{1}{3} \times \frac{15}{32} = \mathbf{\frac{5}{32}}$. Leaving out the $\frac{1}{3}$ gives $\frac{15}{32}$, which is the trap.

Time check: about 15 seconds, against several minutes to add ten fractions.

Question 8

Find 104 × 109 + $97^2$.

Answer. Near 100: 104 × 109 has front 100 + 4 + 9 = 113 and back 4 × 9 = 36, so 11,336. $97^2$ has front 100 − 3 − 3 = 94 and back 09, so 9,409. The total is 20,745.

Time check: about 15 seconds. Two long multiplications would take about a minute.

Question 9

Find $\sqrt[3]{474552} + \sqrt[3]{0.003375}$.

Answer. 474,552 ends in 2, so its cube root ends in 8. Covering 552 leaves 474, and $7^3 = 343$ fits but $8^3 = 512$ doesn't. So the root is 78. For the decimal, 3375 = $15^3$ and six decimal places become two, so $\sqrt[3]{0.003375} = 0.15$. The total is 78.15.

Time check: about 20 seconds. Prime factorising 4,74,552 takes well over a minute.

Question 10

Find $\sqrt{132 + \sqrt{132 + \sqrt{132 + \cdots}}} + \sqrt{132 - \sqrt{132 - \sqrt{132 - \cdots}}}$.

Answer. 132 = 12 × 11. The plus chain is the bigger number, 12, and the minus chain is the smaller, 11. So the sum is 23.

Time check: 10 seconds, once you know 132 = 12 × 11. Setting up two quadratics takes a minute.

Question 11

If $3^{x+2} + 3^{x+1} + 3^x = 1053$, find x.

Answer. I take out the smallest power, $3^x$. The bracket holds 9 + 3 + 1 = 13, so $13 \times 3^x = 1053$ and $3^x = 81 = 3^4$. So x = 4.

Time check: about 15 seconds. Testing x = 1, 2, 3, 4 one by one also works but takes longer and invites slips.

Question 12

Which is the largest: $5^{100}$, $2^{250}$, $3^{150}$, $6^{50}$?

Answer. The HCF of the powers is 50. So the numbers are $(5^2)^{50}$, $(2^5)^{50}$, $(3^3)^{50}$ and $6^{50}$, which means comparing 25, 32, 27 and 6. The biggest is 32, so $2^{250}$ is the largest.

Time check: about 15 seconds. There is no other sensible way without a calculator.

Question 13

Which is the smallest: $\sqrt{15} - \sqrt{13}$, $\sqrt{11} - \sqrt{9}$, $\sqrt{19} - \sqrt{17}$, $\sqrt{7} - \sqrt{5}$?

Answer. Every pair has a gap of 2, so each one is 2 divided by the sum of its roots: $\sqrt{a} - \sqrt{b} = \frac{2}{\sqrt{a} + \sqrt{b}}$. The biggest bottom gives the smallest value, and $\sqrt{19} + \sqrt{17}$ is the biggest bottom. So $\sqrt{19} - \sqrt{17}$ is the smallest (it is about 0.24, while $\sqrt{7} - \sqrt{5}$ is about 0.41).

Time check: about 10 seconds, against a minute of square roots to three decimal places.

Question 14

Find $\sqrt{14 + 6\sqrt{5}} - \sqrt{6 - 2\sqrt{5}}$.

Answer. First root: $6\sqrt{5} = 2\sqrt{45}$, so I need two numbers with sum 14 and product 45. That's 9 and 5, giving $\sqrt{9} + \sqrt{5} = 3 + \sqrt{5}$. Second root: sum 6 and product 5 gives 5 and 1, so it is $\sqrt{5} - 1$. Then $(3 + \sqrt{5}) - (\sqrt{5} - 1) =$ 4.

Time check: about 25 seconds. Squaring the options one by one takes much longer.

Question 15

Find $\frac{13^3 - 8^3 - 5^3}{13 \times 8 \times 5}$.

Answer. 13 − 8 − 5 = 0, so the top is 3 × 13 × (−8) × (−5) = 3 × 13 × 8 × 5. The two minus signs cancel. Dividing by 13 × 8 × 5 leaves 3.

Time check: 10 seconds. Cubing 13, 8 and 5 and dividing by 520 takes about a minute.