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Simplification: order of operations, squares, cubes, exponents and the mean

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Key numbers

Key rules and answers

Order of operationsbrackets, indices, ÷ × left to right, + − left to right (class 6, page 17)
24 + 2 × 8 ÷ 2 − 131 (class 6, page 17)
20 + [8 × 2 + {(6 × 3) − (10 ÷ 5)}]52 (class 6, page 17)
√32418 (class 8, page 29-30)
Making 675 a perfect cubemultiply by 5, 3375 = 15^3 (class 8, page 40)
Ramanujan number1729 = 12^3 + 1^3 = 10^3 + 9^3 (class 8, page 38)
81^(5/4) / 64^(−2/3)243 / 1/16 (class 9, page 63-64)
1 + 2 + ... + nn(n + 1)/2 (class 10, page 77)
1 + 3 + ... + (2n − 1)n^2 (class 10, page 78)
1^3 + 2^3 + ... + n^3(n(n + 1)/2)^2 (class 10, page 78-79)
Smallest amicable pair220, 284 (class 10, page 79)
25 students, mean 78.4, 96 → 69correct mean 79.48 (class 9, page 293)

Easy to confuse

8 ÷ 2 × 4 = 16 / 8 ÷ 8 = 1left to right is right (16); doing the multiplication first is wrong
10 − 2 + 3 = 11 / 10 − 5 = 5left to right is right (11); always adding first is wrong
a^m × a^n / (a^m)^nadd the exponents (2^5); multiply the exponents (2^6)
√9 / the solutions of x^2 = 93 only; ±3
√2 ≈ 1.414 / √2an approximation; the exact value
√(1/25) / √21/5, not a surd; a surd
Changing the mean directly / correcting through the sumwrong; right (1960 − 69 + 96)
The full note

Order of operations: BIDMAS

BIDMAS: B brackets, I indices, D division, M multiplication, A addition, S subtraction. The rule is worked from left to right. Do ÷ and × from left to right, then + and − from left to right. 9 + 5 × 2 = 9 + 10 = 19. Example: in the book's buttermilk-stall sum, 5 × (6 + 2) = 40 is right; moving the brackets to write (5 × 6) + 2 = 32 is wrong. Brackets in the right place prevent confusion.

Example: Simplifying with brackets

Simplify 20 + [8 × 2 + {(6 × 3) − (10 ÷ 5)}].

  1. Inner brackets first: 6 × 3 = 18, 10 ÷ 5 = 2; {18 − 2} = 16.
  2. The next bracket: 8 × 2 + 16 = 16 + 16 = 32.
  3. Last, 20 + 32.

Answer: 52.

Example: Division and multiplication in one line

Simplify 24 + 2 × 8 ÷ 2 − 1.

  1. ÷ and × from left to right: 2 × 8 = 16, 16 ÷ 2 = 8 (the book does 8 ÷ 2 = 4 first and then 2 × 4 = 8; the answer is the same).
  2. Then + and − from left to right: 24 + 8 − 1.

Answer: 31.

Square numbers and square roots

If a natural number n can be written as the square of a natural number m (n = m^2), it is a square number: 1, 4, 9, 16, ... Square numbers end only in the digits 0, 1, 4, 5, 6 and 9; a number ending in 2, 3, 7 or 8 is not a perfect square. A perfect square divided by 3 leaves 0 or 1, never 2; divided by 4 it also leaves 0 or 1. A perfect square ending in zero ends in an even number of zeros. For a square root, factorise into primes and pair equal factors: 324 = (2 × 3 × 3)^2, √324 = 18. 250 = 5^2 × 5 × 2; multiplied by 10 it gives 2500 (square root 50), divided by 10 it gives 25 (square root 5). Since 108 = 2^2 × 3^2 × 3, 108 is not a perfect square. Example: the dots of a square kolam grow as 1, 4, 9, 16; they are square numbers.

Cube numbers and cube roots

A number multiplied by itself twice more gives a cube number: 1, 8, 27, 64, 125, 216, ... A number ending in 0, 1, 4, 5, 6 or 9 has a cube ending in the same digit; ending in 2 its cube ends in 8, ending in 8 in 2; ending in 3 its cube ends in 7, ending in 7 in 3. A perfect cube never ends in exactly two zeros (1000 ends in three). 675 = 3 × 3 × 3 × 5 × 5; to make it a perfect cube, multiply by 5 (3375 = 15^3). 27000 = 2^3 × 3^3 × 5^3, and its cube root is 30. 1729 = 12^3 + 1^3 = 10^3 + 9^3; it is the smallest number that is the sum of two cubes in two ways (the Ramanujan number).

Laws of exponents

Any non-zero number to the power zero is 1: a^0 = 1 (a ≠ 0). A negative exponent: a^(−m) = 1/a^m. With the same base: a^m × a^n = a^(m+n) (product law); a^m ÷ a^n = a^(m−n) (quotient law); (a^m)^n = a^(mn) (power law). Also a^m × b^m = (ab)^m and a^m / b^m = (a/b)^m. 2^3 × 2^2 = 2^5 = 32; it is not 2^6. In scientific notation a number is written S × 10^a with 1 ≤ S < 10 and a an integer; 0.00123 = 1.23 × 10^(−3), 1230000 = 1.23 × 10^6; a positive exponent shows a large number and a negative exponent a small one.

Example: The zero exponent is a shortcut

What is (4^(−1) + 4^(−2) + 4^(−3))^0 × (4^(−1) + 4^(−2) + 4^(−3))^1?

  1. Any number (except zero) to the power zero is 1; there is no need to work inside the first bracket.
  2. The second part: 1/4 + 1/16 + 1/64.
  3. 1 × (1/4 + 1/16 + 1/64) = 21/64.

Answer: 21/64 (2025 question 194).

Roots and surds

If r^n = x, then r is the nth root of x; the parts are the radical sign, the index n and the radicand x. For even n the radical means the positive root only: √4 = 2, −√4 = −2; writing √4 = ±2 is wrong. Fractional index: x^(m/n) = (the nth root of x)^m; 81^(5/4) = 3^5 = 243; 64^(−2/3) = 1/16. A surd is an irrational root of a rational number: √2 and ∛3 are surds; √(1/25) = 1/5 is not; π is irrational but is not a surd. The order of a surd is the index of its root: the fifth root of 99 has order 5. √2 ≈ 1.414 and √3 ≈ 1.732 are approximations: (1.414)^2 = 1.999396, (1.732)^2 = 2.999824. The Std 9 book misprints both squares; the values above are right.

Special series

The sum of the first n natural numbers: 1 + 2 + ... + n = n(n + 1)/2. The sum of the first n odd natural numbers: 1 + 3 + 5 + ... + (2n − 1) = n^2. The sum of squares: 1^2 + 2^2 + ... + n^2 = n(n + 1)(2n + 1)/6. The sum of cubes: 1^3 + 2^3 + ... + n^3 = (n(n + 1)/2)^2; that is, the sum of the cubes is the square of the sum of the numbers. So if the sum of the cubes is 72900, the sum of the numbers is √72900 = 270. 220 and 284 are amicable numbers: the sum of the divisors of one (other than itself) is the other; they are the smallest amicable pair. Example: if rows of 1, 3, 5, ... people join to pull a temple car rope, the first n rows hold n^2 people in all.

The arithmetic mean

Arithmetic mean = sum of all the values ÷ number of values: the mean of 25, 32, 36, 38, 45, 41, 35, 36 is 288/8 = 36. By the assumed mean method, correct mean = assumed mean + mean of the differences: with 38 as the assumed mean the mean difference is −2, so the mean is 36. To correct a misread value: find the sum, subtract the wrong value, add the right one; do not change the mean directly. Example: this is how a class teacher corrects the class mean after writing one student's mark wrongly.

Example: Correcting a misread mark

The mean mark of 25 students is 78.4. A score of 96 was taken as 69 by mistake. What is the correct mean?

  1. Wrong sum = 25 × 78.4 = 1960.
  2. Correct sum = 1960 − 69 + 96 = 1987.
  3. Correct mean = 1987/25.

Answer: 79.48 (2025 question 195 uses these numbers).

Source: கணக்கு, ஆறாம் வகுப்பு, முதல் பருவம், தொகுதி 2, தமிழ் வழி (Mathematics, Standard VI, Term I, Volume 2, Tamil medium; Samacheer Kalvi) (6ஆம் வகுப்புக் கணக்கு, பருவம் 1, இயல் 1 (1.7 செயலிகளின் வரிசை), பக்கம் 16-17; 8ஆம் வகுப்புக் கணக்கு, இயல் 1 எண்கள், பக்கம் 27-46; 9ஆம் வகுப்புக் கணக்கு, இயல் 2 மெய்யெண்கள், பக்கம் 62-66, இயல் 8 (8.4 கூட்டுச் சராசரி), பக்கம் 285-293; 10ஆம் வகுப்புக் கணிதம், இயல் 2 (2.11 சிறப்புத் தொடர்கள்), பக்கம் 77-79)

PYQ (2012-2025 papers)

Provisional key: until TNPSC publishes the final key; the answer may change.

Free questions

  1. Question 1

    Consider the following statements on perfect squares. 1. A perfect square that ends in zero ends in an odd number of zeros. 2. A perfect square divided by 3 never leaves remainder 2. 3. A perfect square divided by 4 leaves remainder 0 or 1. Which of these are correct?

    1. A. 2 only
    2. B. 1 and 2 only
    3. C. 2 and 3 only
    4. D. 1, 2 and 3
    Answer and explanation

    Correct answer: C. 2 and 3 only

    A square ending in zero ends in an even number of zeros (2500, 10000); a perfect square leaves 0 or 1 on division by 3 and 0 or 1 on division by 4; so 2 and 3 are correct and 1 is wrong.

    Source: கணக்கு, எட்டாம் வகுப்பு, தமிழ் வழி (Mathematics, Standard VIII, Tamil medium; Samacheer Kalvi) (இயல் 1 எண்கள், p. 28 (PDF p. 34))

  2. Question 2

    Assertion [A]: 10 − 2 + 3 = 11. Reason [R]: Although A (addition) is written before S (subtraction) in BIDMAS, addition and subtraction are done from left to right.

    1. A. Both [A] and [R] are true and [R] is the correct explanation of [A]
    2. B. Both [A] and [R] are true but [R] is not the correct explanation of [A]
    3. C. [A] is true but [R] is false
    4. D. [A] is false but [R] is true
    Answer and explanation

    Correct answer: A. Both [A] and [R] are true and [R] is the correct explanation of [A]

    BIDMAS is worked from left to right: 10 − 2 + 3 = 8 + 3 = 11; doing 10 − 5 = 5 is wrong; so both are true and [R] explains [A].

    Source: கணக்கு, ஆறாம் வகுப்பு, முதல் பருவம், தொகுதி 2, தமிழ் வழி (Mathematics, Standard VI, Term I, Volume 2, Tamil medium; Samacheer Kalvi) (இயல் 1 எண்கள் (1.7 செயலிகளின் வரிசை), p. 17 (PDF p. 21))

  3. Question 3

    What is the difference between the values of 5 × (6 + 2) and (5 × 6) + 2?

    1. A. 0
    2. B. 8
    3. C. 2
    4. D. 10
    Answer and explanation

    Correct answer: B. 8

    5 × (6 + 2) = 5 × 8 = 40 and (5 × 6) + 2 = 30 + 2 = 32; the difference is 40 − 32 = 8; putting brackets in the right place avoids confusion.

    Source: கணக்கு, ஆறாம் வகுப்பு, முதல் பருவம், தொகுதி 2, தமிழ் வழி (Mathematics, Standard VI, Term I, Volume 2, Tamil medium; Samacheer Kalvi) (இயல் 1 எண்கள் (1.7 செயலிகளின் வரிசை), pp. 16-17 (PDF pp. 20-21))

The unit's 20 free questions

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Frequently asked questions

Simplify 20 + [8 × 2 + {(6 × 3) − (10 ÷ 5)}].

52.

Simplify 24 + 2 × 8 ÷ 2 − 1.

31.

What is (4^(−1) + 4^(−2) + 4^(−3))^0 × (4^(−1) + 4^(−2) + 4^(−3))^1?

21/64 (2025 question 194).

Full guide: TNPSC Group 4 Part C Aptitude: 25 questions, 2 units and how to practise them

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