Algebra: Notes और MCQ Practice

आसान भाषा में:\nAlgebra वो tool है जिससे हम unknown numbers को symbols (x, y) में बदल कर equations बनाते हैं और solve करते हैं. ये SSC में speed और accuracy दोनों माँगता है, इसलिए basics clear रखें.\n\nImportant Formulas / Rules\n- Linear Equation: ax + b = 0 → x = -b ÷ a\n- Quadratic Formula: ax² + bx + c = 0 → x = (-b ± √(b²-4ac)) ÷ (2a)\n- Sum of AP: Sn = n/2 × (2a + (n-1)d)\n- Sum of GP: Sn = a × (1‑rⁿ) ÷ (1‑r) (r ≠ 1)\n- Difference of Squares: a² - b² = (a+b) × (a‑b)\n- Common Factor: a×b + a×c = a × (b + c)\n\nQuick Tricks\n1. Cross‑multiplication for fractions – If a/b = c/d, then a×d = b×c. Use it to compare or solve quickly.\n2. Factor‑by‑grouping shortcut – For ax² + bx + c, look for two numbers whose product = a×c and sum = b; split middle term and factor.\n\nCommon Mistakes\n- Sign error: हमेशा “‑b” को ठीक से लिखें, खासकर quadratic formula में.\n- Extraneous root: Square both sides → check solution back in original equation, नहीं तो extra root रह जाता है.

Concept

Algebra is the branch of mathematics that uses letters (variables) to represent numbers and studies the rules for manipulating these symbols. In SSC exams you mainly deal with linear equations, quadratic equations, expressions, and simple word problems. The goal is to translate a word problem into an equation, solve for the unknown, and interpret the answer.

Important formulas / rules / facts

• Linear equation in one variable: ax + b = 0 → x = -b ÷ a
• Sum of two numbers: if a + b = S, then b = S - a
• Difference of two numbers: if a - b = D, then a = D + b
• Product of two numbers: if a × b = P, then b = P ÷ a
• Quadratic equation: ax² + bx + c = 0 → x = [-b ± √(b² - 4ac)] ÷ (2a)
• Difference of squares: a² - b² = (a - b) × (a + b)
• Sum of squares: a² + b² = (a + b)² - 2ab
• Simple interest (used in some algebra word‑problems): SI = P × R × T ÷ 100
• Ratio to fraction: a : b = a ÷ (a + b)

Solved Examples

Example 1

Question: The sum of two numbers is 48 and their difference is 8. Find the numbers.

Solution:

1. Let the numbers be x and y with x > y.
2. x + y = 48 ……(1)
3. x - y = 8 ……(2)
4. Add (1) and (2): 2x = 56 → x = 28.
5. Substitute x in (1): 28 + y = 48 → y = 20.

Answer: 28 and 20.

Example 2

Question: Solve for x: 3(x - 4) = 2x + 5.

Solution:

1. Expand left side: 3x - 12 = 2x + 5.
2. Bring variables to one side: 3x - 2x = 5 + 12.
3. x = 17.

Answer: x = 17.

Example 3

Question: A number when increased by 20% becomes 72. What was the original number?

Solution:

1. Let original number be N.
2. After 20% increase: N + 20% of N = N × (1 + 20/100) = N × 1.2 = 72.
3. N = 72 ÷ 1.2 = 60.

Answer: 60.

Example 4

Question: Find the two consecutive integers whose product is 156.

Solution:

1. Let the integers be n and n+1.
2. n(n + 1) = 156 → n² + n - 156 = 0.
3. Solve quadratic: Discriminant = 1² + 4×156 = 1 + 624 = 625 → √625 = 25.
4. n = [-1 ± 25] ÷ 2.

• Positive root: (-1 + 25) ÷ 2 = 24 ÷ 2 = 12.
• Negative root gives -13 (reject for positive consecutive integers).

1. Numbers are 12 and 13.

Answer: 12 and 13.

Example 5

Question: The ratio of ages of A and B is 3:5. After 4 years, the ratio becomes 4:7. Find their present ages.

Solution:

1. Let ages be 3k and 5k.
2. After 4 years: (3k + 4) : (5k + 4) = 4 : 7.
3. Cross‑multiply: 7(3k + 4) = 4(5k + 4).
4. 21k + 28 = 20k + 16 → 21k - 20k = 16 - 28 → k = -12.

(Negative k indicates we set the ratio opposite; try swapping.)

1. Assume ages are 5k and 3k (i.e., B older). Then (5k + 4) : (3k + 4) = 7 : 4.

Cross‑multiply: 4(5k + 4) = 7(3k + 4) → 20k + 16 = 21k + 28 → k = -12 again.

1. Mistake: The ratio after 4 years must be (3k+4)/(5k+4)=4/7.

So 7(3k+4)=4(5k+4) → 21k+28 = 20k+16 → k = -12 → impossible.

1. Re‑check: The ratio after 4 years is 4:7, meaning (3k+4)/(5k+4)=4/7.

Solve: 7(3k+4)=4(5k+4) → 21k+28=20k+16 → k= -12.

Since k cannot be negative, the given data is inconsistent. (This shows importance of verifying data.)

Answer: No valid solution; the problem data is contradictory.

Shortcuts and Tricks

• Cross‑multiplication for ratio problems: Directly set up (a + extra) / (b + extra) = new ratio and solve without expanding fully.
• Use the “product‑sum” method for quadratic with integer roots: For ax² + bx + c = 0, find two numbers whose product = a×c and sum = b.
• Percentage shortcut: To find original value when increased by p%, divide the final value by (1 + p/100).
• Difference of squares: When you see a² - b², factor it quickly as (a - b)(a + b) to simplify calculations.
• Quick discriminant check: For ax² + bx + c, if b² - 4ac is a perfect square, the roots are rational – useful for speed.

Common Mistakes to Avoid

• Forgetting to distribute negative signs while expanding brackets.
• Mixing up “increase by” and “decrease by” percentages – always use (1 ± p/100).
• Ignoring the condition that variables represent whole numbers when the problem states “integers” or “natural numbers”.
• Misreading ratio statements; write the ratio in the same order as the quantities.
• Skipping the step of checking whether the discriminant is non‑negative for real roots.