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AFCAT Numerical Ability · Formula sheet

Squares, Cubes and Roots formulas

20 formulas, 2 reference tables and 43 common traps for AFCAT Numerical Ability Squares, Cubes and Roots, grouped by subtopic.

Full notes with worked examples

Squares, Their Last Digits, and Perfect Squares

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Squaring a number, and the square identities

Square of a sum or a difference

(a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2

Testing a perfect square by prime factors

Perfect square test

n=paqb⋯ is a square when a,b,… are all evenn = p^{a} q^{b} \cdots \text{ is a square when } a, b, \ldots \text{ are all even}

The last digit of a square

Last digit of nLast digit of n²Examples
00202=40020^2 = 400
1 or 91112=12111^2 = 121, 192=36119^2 = 361
2 or 84122=14412^2 = 144, 182=32418^2 = 324
3 or 79232=52923^2 = 529, 172=28917^2 = 289
4 or 66142=19614^2 = 196, 162=25616^2 = 256
55252=62525^2 = 625
The last two digits are always 25.

Common traps

The middle term

(4+5)2=81(4 + 5)^2 = 81, not 16+25=4116 + 25 = 41. The square of a sum keeps the cross term 2ab2ab.

Brackets decide the sign

(−7)2=49(-7)^2 = 49, but −72=−49-7^2 = -49: without brackets only the 7 is squared.

A good ending is not proof

2,231 ends in 1, a possible ending, but it lies between 472=220947^2 = 2209 and 482=230448^2 = 2304. The last digit can rule a number out, never in.

Squares ending in 5

2,675 ends in 5, but in 75, not 25, so it is not a perfect square.

One odd power is enough to fail

300=22×3×52300 = 2^2 \times 3 \times 5^2 has two pairs, but the single 3 means it is not a perfect square.

Factor all the way down

72=8×972 = 8 \times 9 looks like two squares' worth, but 8=238 = 2^3. So 72=23×3272 = 2^3 \times 3^2 has an odd power and is not a square.

Square Roots of Whole Numbers, Decimals and Fractions

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Square root of a large perfect square

Root by bracketing

(10a)2<N<(10a+10)2  ⇒  10a<N<10a+10(10a)^2 < N < (10a + 10)^2 \;\Rightarrow\; 10a < \sqrt N < 10a + 10

Square roots in an expression

Roots of products and quotients

ab=a b,ab=ab\sqrt{ab} = \sqrt a\,\sqrt b, \quad \sqrt{\tfrac ab} = \dfrac{\sqrt a}{\sqrt b}

Square roots of decimals and fractions

Root of a decimal

a102k=a10k\sqrt{\dfrac{a}{10^{2k}}} = \dfrac{\sqrt a}{10^{k}}

Equations with a square root

Clearing a root

ax+b=k  ⇒  ax+b=k2\sqrt{ax + b} = k \;\Rightarrow\; ax + b = k^2

Common traps

Two candidates for the units digit

2,116 ends in 6, so its root ends in 4 or 6. 452=202545^2 = 2025 is below it, so the root is 46, not 44.

Convert rupees to paise

₹20.25 collected as 'as many paise each as members' is 2,025 paise, so there are 45 members. Taking the root of 20.25 gives 4.5, which cannot be a count.

No splitting over a sum

25+144=169=13\sqrt{25 + 144} = \sqrt{169} = 13, not 5+12=175 + 12 = 17. Add first, then take the root.

The root sign is positive

49=7\sqrt{49} = 7, never −7-7, even though (−7)2=49(-7)^2 = 49. So (a−b)2=∣a−b∣\sqrt{(a - b)^2} = |a - b|, which is b−ab - a when bb is the larger.

An odd number of decimal places

0.9\sqrt{0.9} is not 0.3, because 0.32=0.090.3^2 = 0.09. With an odd number of places, write 0.9=0.900.9 = 0.90: its root is about 0.949.

Halve the places, do not keep them

0.000004=0.002\sqrt{0.000004} = 0.002: six places become three. Writing 0.000002 keeps all six.

Square the whole side

x+7=6\sqrt{x + 7} = 6 gives x+7=36x + 7 = 36, so x=29x = 29. Squaring only the xx inside gives a wrong equation.

A root is never negative

x=−3\sqrt x = -3 has no solution: x=9x = 9 gives 9=3\sqrt 9 = 3, not −3-3.

Cubes, Cube Roots, and Square and Cube Roots Together

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Cubing a number, and the cube identities

Sum and difference of cubes

a3±b3=(a±b)(a2∓ab+b2)a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)

Cube roots by prime factors

Cube root by prime factors

p3aq3b3=paqb\sqrt[3]{p^{3a} q^{3b}} = p^{a} q^{b}

Cube roots of negative numbers, and cube root equations

Clearing a cube root

ax+b3=k  ⇒  ax+b=k3\sqrt[3]{ax + b} = k \;\Rightarrow\; ax + b = k^3

Cube roots of decimals and fractions

Cube root of a decimal

a103k3=a310k\sqrt[3]{\dfrac{a}{10^{3k}}} = \dfrac{\sqrt[3]{a}}{10^{k}}

Square roots and cube roots in one expression

One root at a time

a2=a,b33=b\sqrt{a^2} = a, \quad \sqrt[3]{b^3} = b

The cubes from 1 to 10, and spotting a perfect cube

nn³Last digit of n³
111
288
2 and 8 swap: a cube ending in 8 has a root ending in 2.
3277
3 and 7 swap: a cube ending in 7 has a root ending in 3.
4644
51255
62166
73433
85122
97299
101,0000

Common traps

An odd power keeps the sign

(−5)3=−125(-5)^3 = -125, but (−5)2=25(-5)^2 = 25. Cubes of negative numbers are negative.

A cube of a sum is not a sum of cubes

(1+4)3=125(1 + 4)^3 = 125, not 1+64=651 + 64 = 65. The expansion has two middle terms.

Cubes can end in any digit

A cube can end in 2, 3, 7 or 8: 83=5128^3 = 512 and 73=3437^3 = 343. The last digit never rules a number out as a cube, unlike a square.

Bracket with cubes, not squares

For 9,261, strike off three digits, not two, and compare 9 with the cubes 8 and 27. Comparing 92 with squares sends you to the wrong tens digit.

Groups of three, not pairs

263\sqrt[3]{2^6} is 22=42^2 = 4, not 23=82^3 = 8. Divide each exponent by 3 for a cube root, by 2 only for a square root.

Cube, do not square

x3=4\sqrt[3]{x} = 4 gives x=64x = 64, not 16. Clearing a cube root needs the cube of both sides.

A negative cube root exists

−2163=−6\sqrt[3]{-216} = -6. Answering 'undefined' mixes up cube roots with square roots.

Decimal places in threes

0.00273\sqrt[3]{0.0027} is not 0.03, because 0.033=0.0000270.03^3 = 0.000027. Four decimal places do not give a neat cube root: 0.00273\sqrt[3]{0.0027} is about 0.139.

A third of the places

0.0000643=0.04\sqrt[3]{0.000064} = 0.04: six places become two. Halving them, as for a square root, gives 0.004.

Read the small index

643=4\sqrt[3]{64} = 4 but 64=8\sqrt{64} = 8. Missing the 3 on the root sign doubles the answer here.

The Least Number That Makes a Perfect Square or Cube

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Least multiplier or divisor, by prime factors

What a perfect power needs

square: every exponent even,cube: every exponent a multiple of 3\text{square: every exponent even}, \quad \text{cube: every exponent a multiple of } 3

Least number to add or subtract for a perfect square

Nearest squares

k2≤n<(k+1)2:add (k+1)2−n,subtract n−k2k^2 \le n < (k + 1)^2: \quad \text{add } (k + 1)^2 - n, \quad \text{subtract } n - k^2

Common traps

Least means only the missing primes

72=23×3272 = 2^3 \times 3^2. Multiplying by 8 also gives a square (576), but the least multiplier is 2, giving 144.

A cube needs threes

40=23×540 = 2^3 \times 5. A cube needs two more 5s, so multiply by 25 to reach 1,000. Multiplying by 5 gives 23×522^3 \times 5^2, still not a cube.

Read the verb

Dividing removes the leftover primes; multiplying adds the missing ones. For a cube the two answers usually differ, so answer the one asked.

Which neighbour

Adding uses the square above; subtracting uses the square below. For 850 the answers are 50 (up to 900) and 9 (down to 841); swapping the neighbours mixes them up.

Step from a round square

Guessing the root by eye often lands one off. Start from a square you know, such as 302=90030^2 = 900, and step up or down by 2k+12k + 1.

Surds: Multiplying, Simplifying and Rationalising

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Multiplying and dividing roots

Product and quotient rules

an bn=abn,anbn=abn\sqrt[n]{a}\,\sqrt[n]{b} = \sqrt[n]{ab}, \quad \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}} = \sqrt[n]{\dfrac ab}

Simplest form of a surd, and like surds

Taking out a square or cube factor

a2b=ab,a3b3=ab3\sqrt{a^2 b} = a\sqrt b, \quad \sqrt[3]{a^3 b} = a\sqrt[3]{b}

Squaring a two-term surd, and its square root

Square of a two-term surd

(a±b)2=a+b±2ab(\sqrt a \pm \sqrt b)^2 = a + b \pm 2\sqrt{ab}

Rationalising a denominator

Rationalising with the conjugate

1a+b=a−ba−b\dfrac{1}{\sqrt a + \sqrt b} = \dfrac{\sqrt a - \sqrt b}{a - b}

Common traps

Products combine, sums do not

3+12\sqrt 3 + \sqrt{12} is not 15\sqrt{15}. Only a product or a quotient goes under one root sign.

One kind of root at a time

5×53\sqrt 5 \times \sqrt[3]{5} is neither 25\sqrt{25} nor 253\sqrt[3]{25}. A square root and a cube root combine only as fractional powers, which come on the last page.

Take out the largest square

200=250\sqrt{200} = 2\sqrt{50} is true but not finished. The largest square factor is 100, so 200=102\sqrt{200} = 10\sqrt 2.

Unlike surds stay apart

25+322\sqrt 5 + 3\sqrt 2 cannot be written as 575\sqrt 7 or 5105\sqrt{10}. It is already as simple as it gets.

Keep the cross term

(11+5)2(\sqrt{11} + \sqrt 5)^2 is 16+25516 + 2\sqrt{55}, not 16. The cross term 2ab2\sqrt{ab} is the part most often dropped.

Larger first with a minus

A square root of 13−24013 - 2\sqrt{40} is 8−5\sqrt 8 - \sqrt 5, that is 22−52\sqrt 2 - \sqrt 5. Writing 5−8\sqrt 5 - \sqrt 8 gives a negative number.

Multiply the top too

Multiplying only the bottom changes the value. 16−2\dfrac{1}{\sqrt 6 - 2} becomes 6+22\dfrac{\sqrt 6 + 2}{2} only when top and bottom are both multiplied by 6+2\sqrt 6 + 2.

A cube root needs two more

To clear 33\sqrt[3]{3}, multiply by 93\sqrt[3]{9}, not by 33\sqrt[3]{3}: 33×33=93\sqrt[3]{3} \times \sqrt[3]{3} = \sqrt[3]{9} is still a root.

Roots as Fractional Powers, and the Laws of Indices

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A root as a fractional power

Fractional power

am/n=amn=(an)ma^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m

The laws of indices

Laws of indices

aman=am+n,aman=am−n,(am)n=amna^m a^n = a^{m + n}, \quad \dfrac{a^m}{a^n} = a^{m - n}, \quad (a^m)^n = a^{mn}

Equations with equal powers

Equal powers

ax=by=k  ⇒  a=k1/x, b=k1/ya^x = b^y = k \;\Rightarrow\; a = k^{1/x},\ b = k^{1/y}

Common traps

The bottom is the root

x2/3x^{2/3} is the cube root of x2x^2, not the square root of x3x^3, which is x3/2x^{3/2}.

A half power is not half

361/2=636^{1/2} = 6, not 18. The power one half takes a square root.

Add, do not multiply

x1/2×x1/4=x3/4x^{1/2} \times x^{1/4} = x^{3/4}, not x1/8x^{1/8}. Multiplying powers of one base adds the indices.

A negative index is not a negative number

5−2=1255^{-2} = \tfrac{1}{25}, a positive number, not −25-25.

Same base first

4x=264^x = 2^6 does not give x=6x = 6. Write 4 as 222^2: 22x=262^{2x} = 2^6, so x=3x = 3.

A product becomes a sum

If a=k1/xa = k^{1/x} and b=k1/yb = k^{1/y}, then ab=k1/x+1/yab = k^{1/x + 1/y}, not k1/(xy)k^{1/(xy)}.

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