krit.club logo

Number - The Four Operations

Grade 9IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The four basic operations are addition (++), subtraction (−-), multiplication (×\times), and division (÷\div).

•

For addition and subtraction of large numbers, align the digits according to their place value (units, tens, hundreds, etc.).

•

When multiplying decimals, multiply the numbers as if they were whole numbers, then place the decimal point so that the answer has the same number of decimal places as the sum of the decimal places in the numbers being multiplied.

•

For division by a decimal, multiply both the divisor and the dividend by a power of 1010 to make the divisor a whole number.

•

The Order of Operations follows the BIDMAS/BODMAS rule: Brackets, Indices (or Orders), Division and Multiplication (from left to right), and Addition and Subtraction (from left to right).

•

Rules for signs: (+)×(+)=(+)(+) \times (+) = (+), (−)×(−)=(+)(-) \times (-) = (+), (+)×(−)=(−)(+) \times (-) = (-), and (−)×(+)=(−)(-) \times (+) = (-). The same rules apply for division.

📐Formulae

a+b=b+a (Commutative Law)a + b = b + a \text{ (Commutative Law)}

a×b=b×a (Commutative Law)a \times b = b \times a \text{ (Commutative Law)}

a×(b+c)=(a×b)+(a×c) (Distributive Law)a \times (b + c) = (a \times b) + (a \times c) \text{ (Distributive Law)}

Dividend=(Divisor×Quotient)+Remainder\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}

💡Examples

Problem 1:

Evaluate the expression: 15+2×(8−3)2÷515 + 2 \times (8 - 3)^2 \div 5.

Solution:

Step 1: Brackets: 8−3=58 - 3 = 5. Step 2: Indices: 52=255^2 = 25. Step 3: Multiplication and Division (left to right): 2×25=502 \times 25 = 50, then 50÷5=1050 \div 5 = 10. Step 4: Addition: 15+10=2515 + 10 = 25.

Explanation:

Applying BIDMAS: Brackets first, then Indices, then Multiplication/Division, and finally Addition.

Problem 2:

Calculate 4005−28674005 - 2867 using the vertical method.

Solution:

4005−28671138\begin{array}{r} 4005 \\ -2867 \\ \hline 1138 \end{array}

Explanation:

Subtract the units: 15−7=815 - 7 = 8 (borrowing from the thousands place through the hundreds and tens). Subtract the tens: 9−6=39 - 6 = 3. Subtract the hundreds: 9−8=19 - 8 = 1. Subtract the thousands: 3−2=13 - 2 = 1.

Problem 3:

Find the product of 1.25×0.041.25 \times 0.04.

Solution:

Step 1: Multiply as whole numbers: 125×4=500125 \times 4 = 500. Step 2: Count decimal places: 1.251.25 has 22 places and 0.040.04 has 22 places, totaling 44 places. Step 3: Place the decimal: 500500 becomes 0.05000.0500 or 0.050.05.

Explanation:

To multiply decimals, ignore the decimals first, then re-insert the decimal point based on the total number of decimal places in the original factors.

Problem 4:

Solve −12÷(−3)+(−5)×2-12 \div (-3) + (-5) \times 2.

Solution:

Step 1: Division: −12÷(−3)=4-12 \div (-3) = 4 (negative divided by negative is positive). Step 2: Multiplication: (−5)×2=−10(-5) \times 2 = -10 (negative times positive is negative). Step 3: Addition: 4+(−10)=4−10=−64 + (-10) = 4 - 10 = -6.

Explanation:

The order of operations and sign rules are strictly followed.