SSC GD Simplification

Table of Contents

• The Super Rule: BODMAS
• Working with Brackets (B in BODMAS)
• Fast Math: Squares and Cubes
• Basic Surds and Roots
• Estimation and Approximation

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The Super Rule: BODMAS

If you want to master simplification ssc gd, you must know BODMAS. This rule tells you the exact order to solve any math problem. If you don’t follow the order, your answer will be wrong!

BODMAS is the most important tool for solving any ssc gd simplification questions.

• B stands for Brackets. Solve anything inside brackets first.
• O stands for Of (or Order). This usually means powers, roots, or multiplication linked by the word ‘of’.
• D stands for Division.
• M stands for Multiplication.
• A stands for Addition.
• S stands for Subtraction.

Remember: Division and Multiplication have the same priority. You solve them from left to right. Addition and Subtraction also have the same priority, solved from left to right.

Detailed Theory (BODMAS)

We use BODMAS to make sure everyone gets the same answer for the same problem. Think of it like traffic rules for math.

When you see a complex problem, like $5+3\times 2-10÷5$, you must follow the steps:

1. First, look for Division and Multiplication.
2. Then, look for Addition and Subtraction.

This is the foundation for all simplification ssc gd question types.

4 Solved Examples (BODMAS)

Example 1:

Solve: $10+5\times 2-6÷3$

Solution:

1. First, we look for Division (D) and Multiplication (M).
   • $5\times 2=10$
   • $6÷3=2$
   • The expression becomes: $10+10-2$
2. Then, we solve Addition (A) and Subtraction (S) from left to right.
   • $10+10=20$
   • $20-2=18$
3. So, the answer is $18$.

Example 2:

Solve: $25÷5\times 4+1$

Solution:

1. We have Division (D) and Multiplication (M). We solve from left to right.
   • First, Division: $25÷5=5$
   • The expression becomes: $5\times 4+1$
2. Next, Multiplication: $5\times 4=20$
   • The expression becomes: $20+1$
3. Finally, Addition: $20+1=21$.

Example 3:

Find the value of $18-3\times 5+10$

Solution:

1. First, we do Multiplication (M).
   • $3\times 5=15$
   • The expression becomes: $18-15+10$
2. Now, we solve Addition and Subtraction from left to right.
   • $18-15=3$
   • The expression becomes: $3+10$
3. So, the answer is $13$.

Example 4:

Solve: $40÷8\times 2-5+1$

Solution:

1. We solve Division and Multiplication from left to right.
   • First, Division: $40÷8=5$
   • The expression becomes: $5\times 2-5+1$
2. Next, Multiplication: $5\times 2=10$
   • The expression becomes: $10-5+1$
3. Finally, Addition and Subtraction from left to right.
   • $10-5=5$
   • $5+1=6$
4. The final answer for this ssc gd simplification problem is $6$.

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Working with Brackets (B in BODMAS)

Brackets are the ‘B’ in BODMAS. You must solve everything inside the bracket before you use that number in the rest of the problem.

There are four types of brackets used in simplification ssc gd questions pdf:

1. Vinculum (Bar Bracket): $\overline{a+b}$. This is solved first.
2. Small Bracket (Parentheses): $(a+b)$. Solved second.
3. Curly Bracket (Braces): $a+b$. Solved third.
4. Square Bracket (Box Bracket): $[a+b]$. Solved last.

Detailed Theory (Brackets)

When you have brackets inside other brackets (nested brackets), you always start from the innermost bracket and work your way out.

For example, in $[10+5\times (2+1)]$, you must solve $(2+1)$ first.

Mastering brackets is key to solving complex ssc gd simplification.

4 Solved Examples (Brackets)

Example 1:

Solve: $15-[6+(4-2)]$

Solution:

1. First, we solve the innermost Small Bracket.
   • $(4-2)=2$
   • The expression becomes: $15-[6+2]$
2. Next, we solve the Square Bracket.
   • $[6+2]=8$
   • The expression becomes: $15-8$
3. Finally, we subtract.
   • $15-8=7$
4. The answer is $7$.

Example 2:

Solve: $100÷10\times (5+\overline{6-1})$

Solution:

1. First, we solve the Vinculum (Bar Bracket).
   • $\overline{6-1}=5$
   • The expression becomes: $100÷10\times (5+5)$
2. Next, we solve the Small Bracket.
   • $(5+5)=10$
   • The expression becomes: $100÷10\times 10$
3. Next, we solve the Curly Bracket.
   • $10\times 10=100$
   • The expression becomes: $100÷100$
4. Finally, we divide.
   • $100÷100=1$
5. The answer is $1$.

Example 3:

Find the value of $50-2\times [10+5-(3-1)]$

Solution:

1. Start with the innermost Small Bracket:
   • $(3-1)=2$
   • Expression: $50-2\times [10+5-2]$
2. Solve the Curly Bracket:
   • $5-2=3$
   • Expression: $50-2\times [10+3]$
3. Solve the Square Bracket:
   • $[10+3]=13$
   • Expression: $50-2\times 13$
4. Use BODMAS (Multiplication before Subtraction):
   • $2\times 13=26$
   • Expression: $50-26$
5. Subtract: $50-26=24$.

Example 4:

Solve: $12\times 3÷(9-3)$

Solution:

1. First, we solve the Bracket.
   • $(9-3)=6$
   • The expression becomes: $12\times 3÷6$
2. We solve Multiplication and Division from left to right.
   • $12\times 3=36$
   • The expression becomes: $36÷6$
3. Finally, we divide.
   • $36÷6=6$
4. The answer is $6$. This is a common type of ssc gd simplification question.

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Fast Math: Squares and Cubes

In ssc gd simplification, you need to calculate squares and cubes quickly. These are part of the ‘O’ (Order) in BODMAS.

• Square: Multiplying a number by itself. $x^2=x\times x$.
• Cube: Multiplying a number by itself three times. $x^3=x\times x\times x$.

You should memorize squares up to 30 and cubes up to 15 for the exam.

Detailed Theory (Squares and Cubes)

When you see a number raised to a power, you must calculate that value before doing any multiplication or division (unless it is inside a bracket).

Example: $5^2=25$. $4^3=64$.

If the problem is $10+3^2$, you must calculate $3^2=9$ first. Then $10+9=19$.

These operations are often hidden inside complex simplification ssc gd questions pdf.

4 Solved Examples (Squares and Cubes)

Example 1:

Solve: $4^2+5^2-1^3$

Solution:

1. First, we calculate all the powers (Squares and Cubes).
   • $4^2=4\times 4=16$
   • $5^2=5\times 5=25$
   • $1^3=1\times 1\times 1=1$
   • The expression becomes: $16+25-1$
2. Next, we solve Addition and Subtraction.
   • $16+25=41$
   • $41-1=40$
3. The answer is $40$.

Example 2:

Find the value of $2^3\times 3^2÷6$

Solution:

1. First, calculate the powers.
   • $2^3=2\times 2\times 2=8$
   • $3^2=3\times 3=9$
   • The expression becomes: $8\times 9÷6$
2. Solve Multiplication and Division from left to right.
   • $8\times 9=72$
   • The expression becomes: $72÷6$
3. Finally, divide.
   • $72÷6=12$
4. The answer is $12$.

Example 3:

Solve: $(10-7{)}^3+5^2$

Solution:

1. First, we must solve the Bracket (B).
   • $(10-7)=3$
   • The expression becomes: $3^3+5^2$
2. Next, we calculate the powers (O).
   • $3^3=27$
   • $5^2=25$
   • The expression becomes: $27+25$
3. Finally, we add.
   • $27+25=52$
4. The answer is $52$.

Example 4:

Calculate: $100-4^3+2\times 5$

Solution:

1. First, calculate the power (O).
   • $4^3=64$
   • The expression becomes: $100-64+2\times 5$
2. Next, calculate Multiplication (M).
   • $2\times 5=10$
   • The expression becomes: $100-64+10$
3. Solve Addition and Subtraction from left to right.
   • $100-64=36$
   • $36+10=46$
4. The final answer for this ssc gd simplification is $46$.

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Basic Surds and Roots

Surds are just numbers written under a root sign, like $\sqrt{4}$ or $\sqrt[3]{8}$. Finding the root is the opposite of squaring or cubing. This is also part of the ‘O’ (Order) in BODMAS.

• Square Root: $\sqrt{x}$. What number multiplied by itself gives $x$? Example: $\sqrt{9}=3$.
• Cube Root: $\sqrt[3]{x}$. What number multiplied by itself three times gives $x$? Example: $\sqrt[3]{27}=3$.

Detailed Theory (Surds)

For simplification ssc gd, you need to know the perfect squares and cubes well.

Sometimes, the number under the root is not a perfect square (like $\sqrt{12}$). You simplify it by finding a perfect square factor inside.

Example: $\sqrt{12}=\sqrt{4\times 3}=\sqrt{4}\times \sqrt{3}=2\sqrt{3}$.

When adding or subtracting surds, they must have the same root part (like $2\sqrt{3}+5\sqrt{3}$).

4 Solved Examples (Basic Surds)

Example 1:

Solve: $\sqrt{16}+\sqrt[3]{8}-5$

Solution:

1. First, we calculate the roots (O).
   • $\sqrt{16}=4$ (because $4\times 4=16$)
   • $\sqrt[3]{8}=2$ (because $2\times 2\times 2=8$)
   • The expression becomes: $4+2-5$
2. Next, we solve Addition and Subtraction.
   • $4+2=6$
   • $6-5=1$
3. The answer is $1$.

Example 2:

Simplify: $3\sqrt{5}+2\sqrt{5}-\sqrt{5}$

Solution:

1. Since all the terms have the same root ($\sqrt{5}$), we can treat them like regular numbers (like $3x+2x-x$).
2. We add and subtract the numbers outside the root.
   • $3+2=5$
   • $5-1=4$
3. So, the answer is $4\sqrt{5}$.

Example 3:

Solve: $\sqrt{25}\times \sqrt{4}+10$

Solution:

1. First, calculate the roots (O).
   • $\sqrt{25}=5$
   • $\sqrt{4}=2$
   • The expression becomes: $5\times 2+10$
2. Next, calculate Multiplication (M).
   • $5\times 2=10$
   • The expression becomes: $10+10$
3. Finally, add.
   • $10+10=20$
4. The answer is $20$. This is a common ssc gd simplification question.

Example 4:

Simplify: $\sqrt{50}+\sqrt{8}$

Solution:

1. We must simplify each surd first to see if they can be added.
   • $\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$
   • $\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$
2. The expression becomes: $5\sqrt{2}+2\sqrt{2}$
3. Now they have the same root, so we add the coefficients.
   • $5+2=7$
4. The final answer is $7\sqrt{2}$. This skill is vital for simplification ssc gd question.

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Estimation and Approximation

Sometimes in ssc gd simplification questions, the numbers are very large or have many decimals. The question might ask for the “approximate value.” This means you don’t need the exact answer.

Approximation means rounding the numbers first to make the calculation much easier and faster. This saves time in the exam.

Detailed Theory (Approximation)

We round numbers to the nearest whole number, 10, or 100 before we start calculating.

Rule of Rounding:

• If the digit after the rounding place is 5 or more, round up.
• If the digit is 4 or less, keep the number as it is.

Example: $49.8$ rounds to $50$. $10.3$ rounds to $10$.

Approximation is a key technique for quickly solving ssc gd simplification questions pdf when exact answers are not required.

4 Solved Examples (Approximation)

Example 1:

Approximate the value of: $49.9\times 10.1+20.03$

Solution:

1. First, we round the numbers to the nearest whole number.
   • $49.9\approx 50$
   • $10.1\approx 10$
   • $20.03\approx 20$
2. The approximate expression is: $50\times 10+20$
3. Use BODMAS (Multiplication first).
   • $50\times 10=500$
   • $500+20=520$
4. The approximate answer is $520$.

Example 2:

Approximate the value of: $199÷9.8+50.1$

Solution:

1. Round the numbers.
   • $199\approx 200$
   • $9.8\approx 10$
   • $50.1\approx 50$
2. The approximate expression is: $200÷10+50$
3. Use BODMAS (Division first).
   • $200÷10=20$
   • $20+50=70$
4. The approximate answer is $70$. This is a fast way to handle ssc gd simplification.

Example 3:

Approximate the value of: $\sqrt{63.9}+{3.01}^2$

Solution:

1. Round the numbers to make them easy to work with (look for perfect squares/cubes).
   • $63.9\approx 64$ (because 64 is a perfect square)
   • $3.01\approx 3$
2. The approximate expression is: $\sqrt{64}+3^2$
3. Calculate the root and the power (O).
   • $\sqrt{64}=8$
   • $3^2=9$
   • The expression becomes: $8+9$
4. Add: $8+9=17$.

Example 4:

Approximate the value of: $79.7\times 4.9÷2.01$

Solution:

1. Round the numbers.
   • $79.7\approx 80$
   • $4.9\approx 5$
   • $2.01\approx 2$
2. The approximate expression is: $80\times 5÷2$
3. Solve Multiplication and Division from left to right.
   • $80\times 5=400$
   • $400÷2=200$
4. The approximate answer is $200$. This technique is essential for quick simplification ssc gd.