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Knowing Our Numbers - Using Brackets

Grade 6CBSE

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

🔑Concepts

Brackets are used to group numbers and operations together, indicating that the operations inside the brackets must be performed first. For example, in the expression (10+5)×2(10 + 5) \times 2, we first add 1010 and 55 to get 1515, then multiply by 22.

The use of brackets helps in simplifying calculations by breaking down large numbers. For instance, multiplying 102×7102 \times 7 can be visualized as splitting a large area into two smaller, manageable rectangles: one of 100×7100 \times 7 and another of 2×72 \times 7.

The Distributive Property is the core concept behind using brackets. It allows us to distribute a multiplier to each term inside the bracket, such as a×(b+c)=(a×b)+(a×c)a \times (b + c) = (a \times b) + (a \times c).

Brackets help maintain the correct order of operations. When you see an expression like 10×(2+3)10 \times (2 + 3), the brackets act as a 'priority wall,' ensuring the addition happens before the multiplication, which would otherwise be performed in a different order under standard BODMAS rules.

Expansion of brackets is a process where we remove brackets by multiplying the term outside with every term inside. If we have (10+2)×(10+3)(10 + 2) \times (10 + 3), we can think of it as distributing the first group over the second, resulting in a sum of four separate products.

Using brackets makes mental math easier. To calculate 9×1089 \times 108, we can visualize it as 9×(100+8)9 \times (100 + 8). This transforms one difficult multiplication into two simple ones: 900900 and 7272, which are then summed to find the total.

📐Formulae

a×(b+c)=a×b+a×ca \times (b + c) = a \times b + a \times c

a×(bc)=a×ba×ca \times (b - c) = a \times b - a \times c

(a+b)×(c+d)=a×(c+d)+b×(c+d)(a + b) \times (c + d) = a \times (c + d) + b \times (c + d)

(a+b)×(c+d)=ac+ad+bc+bd(a + b) \times (c + d) = ac + ad + bc + bd

💡Examples

Problem 1:

Evaluate 6×1076 \times 107 using the expansion of brackets.

Solution:

Step 1: Write 107107 as (100+7)(100 + 7). So, the expression becomes 6×(100+7)6 \times (100 + 7). Step 2: Use the distributive property to multiply 66 with both terms inside the bracket: (6×100)+(6×7)(6 \times 100) + (6 \times 7). Step 3: Calculate the individual products: 600+42600 + 42. Step 4: Add the results: 642642.

Explanation:

We simplified the multiplication by breaking 107107 into a sum of a hundred and a single-digit number, then distributed the 66 over the addition.

Problem 2:

Find the value of 102×103102 \times 103 using brackets.

Solution:

Step 1: Expand both numbers: (100+2)×(100+3)(100 + 2) \times (100 + 3). Step 2: Multiply the first bracket by each term of the second: (100+2)×100+(100+2)×3(100 + 2) \times 100 + (100 + 2) \times 3. Step 3: Expand again: (100×100)+(2×100)+(100×3)+(2×3)(100 \times 100) + (2 \times 100) + (100 \times 3) + (2 \times 3). Step 4: Calculate each part: 10000+200+300+610000 + 200 + 300 + 6. Step 5: Sum all values: 1050610506.

Explanation:

By expanding both factors into (100+n)(100 + n) forms, we converted a multi-digit multiplication problem into a series of simple multiplications and additions.