Algebra – Important Formulas and Concepts for Placement Aptitude

Algebra is a vast and honestly little difficult section. Also Placement aptitude doesn’t test you much on algebra except for few companies. So if you are very well covered with rest of the portion and aiming for exams like CAT, then may be it makes sense to go deeper into this topic. Sometimes you will be given word problem which will make use of basics of equation like Problem of ages. You will not necessarily need much of a deeper understanding of Algebra for such problems and you can pick that up simply by solving some easy examples.

1. Basics Definitions

  • Constants – fixed values [1, 102, -8, 7/2]
  • Variables – does not have fixed value [x, y, m, p]
  • Linear Equation in 1 Variable: (x-5 = 0)
  • Linear Equation in 2 Variable: (2x+5y = 10)
  • Quadratic Equation: (x2+5x+9 = 0)
  • Cubic Equation: (x-1 = x3)

2. Algebraic Identities

  • Identities with two symbols (degree 2) –
    • \( (a+b)^2= a^2+2ab+b^2 \)
    • \( (a-b)^{2}=a^{2}-2ab+b^{2} \)
    • \( (a^2-b^2)= (a+b)(a-b) \)
  • Identities with two symbols (degree 3) –
    • \( (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} \)
    • \( (a-b)^3= a^{3}-3a^{2}b+ 3ab^{2}-b^3 \)
    • \( (a^{3}+b^{3}) = (a+b)(a^{2}-ab+b^{2})\)
    • \( ( a^{3}-b^{3})=(a-b)(a^{2}+ab+b^{2})\)
  • Identities with three symbols (degree 2) –
    • \( (a+b+c)^2= a^{2} + b^{2}+c^{2} + 2ab + 2ac+2bc \)
    • \( (x+a)(x+b)=x^{2}+x(a+b)+ab \)
  • Identities with three symbols (degree 3) –
    • \( (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = (a^3 + b^3+c^3-3abc) \)

Q: Simplify the following expression: $(2x+3)^2−(2x−3)^2$

Sol: $(2x+3)^2−(2x−3)^2 = [(2x)^2+2⋅2x⋅3+3^2]−[(2x)^2−2⋅2x⋅3+3^2] = (4x^2+12x+9)−(4x^2−12x+9) = 24x$

3. Solutions of a Linear Equation

Given there are two Linear Equations in 2 variables $a_1x+b_1y+c_1 = 0 \ , \ a_2x+b_2y+c_2 = 0$

Case 1: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, then equation has Infinite solutions

Case 2: If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ then equation has No Solutions

Case 3: If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, then equation has Unique Solution

4. Digit Based Problems

Numbers with different number of digits like 2-digit, 3-digit and more can be expressed in the form of equations. Further, Digit based questions can be solved using Equations.

  • 2-digit –> ab = (10a+b)
  • 3-digit –> abc = (100a+10b+c)
  • 4-digit –> abcd = (1000a+100b+10c+d)

Q: A 2-digit number is such that the sum of the number and its reverse is 121. Find the number.

Sol: Let the 2-digit number be 10a+b, where a is the tens digit and b is the units digit. The reverse of the number is 10b+a.

According to the problem:(10a+b)+(10b+a)=121

Simplifying the equation:11(a+b)=121

a+b=121/11 = 11

The sum of the digits a+b=11, the possible pairs for (a,b) are (2,9), (3,8), (4,7), (5,6) and of-course the reverse order of it.

Thus, the possible 2-digit numbers are 29, 92, 38, 83, 47, 74, 56 and 65

Q: A 2-digit number is such that the difference between the number and its reverse is 36. Find the number

Sol: Let the 2-digit number be 10a+b.

The reverse of the number is 10b+a

According to the problem:(10a+b)−(10b+a)=36

Simplifying the equation:9(a−b)=36

a−b=36/9=4

The difference of the digits a−b=4. The possible pairs for (a,b) are (5,1), (6,2), (7,3), (8,4) and (9,5)

Thus, the possible 2-digit numbers are 51, 62, 73, 84, 95, 15, 26, 37, 48 and 59.

5. problem on Ages

Q: The sum of the ages of a father and his son is 50 years. Five years ago, the father’s age was four times his son’s age. Find their present ages.

Sol: Let the present age of the son be x, and the present age of the father be 50−x (since the sum of their ages is 50).

Five years ago:

  • The son’s age was x−5
  • The father’s age was (50−x)−5=45−x

According to the problem:45−x=4(x−5)

Simplifying the equation: 45 − x = 4x − 20

x ​=13

So, the son’s present age is 13 years, and the father’s present age is 50−13 = 37 years.

Q: A man is three times as old as his son. Five years ago, the man was four times as old as his son. Find their present ages.

Sol: Let the present age of the son be x, and the present age of the man be 3x (since the man is three times as old as his son).

Five years ago:

  • The son’s age was x−5
  • The man’s age was 3x−5

According to the problem:3x−5=4(x−5)

Simplifying the equation: x = 15

So, the son’s present age is 15 years, and the man’s present age is 3×15 =45 years.

Q: The present ages of a father and his daughter are in the ratio of 7:3. After 6 years, the ratio of their ages will be 5:3. Find their present ages.

Sol: Let the present ages of the father and daughter be 7x and 3x, respectively.

After 6 years:

  • The father’s age will be 7x+6
  • The daughter’s age will be 3x+6

According to the problem: (7x+6)/(3x+6) = 5/3

Cross-multiplying: 3(7x+6) = 5(3x+6)

Simplifying the equation: 21x+18 = 15x+30

So, daughter’s present age is = 6 years and father’s present age is = 7×6/3 = 14 years.

6. Quadratic Equation

  • An equation of the form –> (ax2+bx+c = 0) where a, b, c are all Real and a not equals 0 is called a quadratic equation.
  • Highest degree of “n” = 2
  • Number of Roots / Solution = 2 (real or imaginary)

Methods of finding Roots

1. Factorizing Method
  • A quadratic equation ax2+bx+c = 0 can be written in the form (x-α)(x-β)=0 where α and β are the roots. The method of factorizing the expression on the left hand side of the quadratic equation is explained with an example — x2+7x+12 = 0.
    • 1st Step – Write down b (the coefficient of x) as the sum of two quantities whose product is equal to ac. In this case “7x” can be written as the sum of (-3x) + (-4x)so that the product is equal to +12.
    • 2nd Step – Rewrite the equation with the “bx” term split in the above manner. In this case, the above equation can be written as x2-3x-4x+12 = 0.
    • 3rd Step – Take the first two term and rewrite them together after taking out the common factor between them and then do the same with the third and the fourth term. You should ensure that the term left after taking out the common factor from the first and second term is the same as that of the third and fourth term. In this case, the equation can be rewritten as x(x-3)-4(x-3) = 0.
    • 4th Step – Re write the entire left hand side to get the form (x-α)(x-β)=0. In this case, we can rewrite the given equation as (x-3)(x-4) = 0.
    • 5th Step – Now, α and β are the roots of the quadratic equation. For x2+7x+12 = 0, the roots of the equation are 3 and 4.
2. Discriminant Method
  • For a quadratic equation of the form ax2+bx+c = 0 ; the standard formula for finding the root is given by – $$x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \\ where, \ Discriminant \ (D) \ = \sqrt{b^2-4ac}$$
  • Certain rules regarding roots of a quadratic equation equation –
    • Sum of Roots = -b/a
    • Product of Roots = c/a

You can also refer following videos to grasp basics of percentage in more details

Practice

Refer Topic: Statistics: https://www.learntheta.com/placement-aptitude-statistics/

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