TERMINOLOGIES
The unknown quantities in an algebraic expression are known as variables. An equation is a statement of equality of 2 algebraic expressions.
Linear equation in 1 variable: ax + b = c; Here a,b,c are constants while x is Unknown. An equation satisfied by all values of x is called identity.
Linear equation in 2 variables : a₁x+by+c₁ = 0 and a2x+ b₂y+c₂ = 0; a,b is not equal to 0

The number of variables = the Number of equations required to solve it.
Consider the linear function, 𝑦 = 𝑎𝑥 + 𝑏 The variable on the left-hand side, 𝑦, is called the dependent variable as its value depends on the value of the independent variable. The variable on the right-hand side, 𝑥, is called the independent variable.
For example, given the two variables “number of days of rain per month” and “number of umbrella sales”, the number of days of rain would be the independent variable, whereas the number of umbrella sales would be the dependent variable. This is because the number of umbrella sales will likely depend on how much it has rained, rather than the other way around.
We can also assume these linear equations to be equations of a line, then –
1. If the two lines intersect, there is exactly one solution.
2. If the lines are parallel then there is no solution.
3. If both the lines coincide exactly or both of them are the same then there will be infinite solutions.
System of Linear Equations:
Consistent - If at least 1 solution exists.
Inconsistent - It has no solutions.
Conditions for solutions to existing for a system of equations – a1x + b1y = c1 and a2x + b2y = c2

Case 1. Unique solutions exist when a1/a2 is not equal to b1/b2
Case 2. If (a1/a2) = (b1/b2) = (c1/c2), then there will be infinitely many solutions. This type of equation is called a dependent pair of linear equations in two variables. If we plot the graph of this equation, the lines will coincide.
Case 3. If (a1/a2) = (b1/b2) ≠ (c1/c2), then there will be no solution. This type of equation is called an inconsistent pair of linear equations. If we plot the graph, the lines will be parallel.

Question 1
A man's age is p% of what it was 20 years ago but q% of what he will be after 10 years. What is his present age?
a) q/p
b) p/q
c) {10 (q+2p)}/p-q
d) {5 (q+3p)}/p-q
Solution
Let the present age be x
x=p% of (x - 20)
x= q% of (x + 10)
p% of (x-20) =q% of (x+10)
px - 20p = qx + 10q
x(p - q) = 10q + 20p = 10(q + 2p)
(p - q) * x = 10q + 20p = 10(q + 2p)
x = 10(q + 2p) / p – q
Answer Option c

Question 2
Mayank, Mirza, Little and Jaspal bought a motorbike for $60. Mayank paid one-half of the sum of the amounts paid by the other boys. Mirza paid one-third of the sum of the amounts paid by the other boys. Little paid one-fourth of the sum of the amounts paid by the other boys. How much did Jaspal have to pay?
a) $15.
b) $13
c) $17
d) None of these
Solution
Let the amount paid by Mayank be y.
So, the amount paid by the other three =2y.
This gives us the total bill => y+2y= 60 => 3y = 60 => y = 20
Similarly, if the amount paid by mirza is z. Paid by others = 3z
z + 3z = 60 => 4z = 60 => z = 15
Similarly, if the amount paid by little be k. Paid by others = 4k
k + 4k = 60 => 5k=60 => k=12
Amount paid by Jaspal = 60-(20+15+12)= $13
Answer Option b

QUADRATIC EQUATION
-An equation is any expression in the form f(x) = 0
-An equation of the form: ax2+ bx + c = 0 can be described as a quadratic equation where a,b and c are all real and a is not equal to 0.
-D = (b2 - 4ac) is the discriminant of the quadratic equation
-If D < 0 then the equation has no real roots
-If D > 0, then the equation has two distinct roots
-For equation ax2+ bx + c = 0 ,to find the value of x we use the formula x = (-b ± √(b2 - 4ac)) / 2a.
-Quadratic Equations have the highest degree of the variable as 2.
-The values of 'x' that will satisfy the quadratic equation are called its roots.
-The roots of the quadratic equation can be rational or irrational.
-There can be 3 cases of different types of roots:
1)Both the roots are real and equal.
2)Both the roots are real and unequal.
3)Both the roots are imaginary
-Sum and Product of the roots of the equation
1. Suppose we have an equation ax2+ bx + c = 0 and its roots are x1 and x2.
The sum of the roots can be given by : x1 + x2 = -b/a
The product of the roots can be given by : x1 * x2= c/a
2. A quadratic equation can also be expressed as :
x2 - (Sum of the roots) x + (Product of the roots) = 0
-Maxima and Minima of Quadratic Equations:
If a > 0 , then we can find the minimum value of the Quadratic equation as
y= -D / 4a D=>Discriminant
The value of x at the minima is x = (- b)/ (2a)
-If a < 0 then we can find the maximum value of the Quadratic equation as
y = - D/4a D => Discriminant.
The value of x at the maxima is x = (- b) / (2a)

Question 3
Sum of the roots of a quadratic equation is 5 less than the product of the roots. If one root is 1 more than the other root, find the product of the roots?
a) 6 or 3
b) 12 or 2
c) 8 or 4
d) 12 or 4
Solution
Sum of the roots = (product of the roots)-5
Let r1 and r2 be the two roots
r1=r2+1
Therefore,
r1 + 1 + r2 = (r1r2) * r2 - 5
r2 + 1 + r2 = (r2+1) * r2 - 5
2r2 + 6 = (r2)^2 +r2
(r2)^ 2 - r2 - 6=0
Solving, we get r2 = 3 or - 2
Therefore, r2 = 4 or -1
Hence the product of the two roots is either 12 or 2
Answer: (b)

Question 4
The number of distinct real roots of the equation(x + {1 / x})2- 3(x + {1 / x}) + 2 = 0 equals
a) 1
b) 2
c) 0
d) None of these
Solution
x + 1 / x = y
y2- 3y + 2 = 0
y>=2 & y<=-2
(y - 1)(y - 2) = 0
y=1(Not possible) or y = 2
Now, x + 1 / x = 2
This has only one option x = 1
Answer : (a)

Question 5
If f(y) = x2+ (2p + 1) x + p2 - 1 and x is a real number, for what values of ‘p' the function becomes 0?
The function f(y) is a quadratic equation.
It is given that x is real.
So the discriminant of f(y) ≥ 0
i.e. D = b2
- 4ac ≥ 0 or
(2p + 1)2 – 4(p2
- 1) ≥ 0
4p2 + 4p + 1 – 4 (p2
- 1) ≥ 0
4p + 5 ≥ 0
Or p ≥ −5/4
Hence the answer is p ≥ −5/4

Question 5
The quadratic equations x² - 6x + a = 0 and x² - cx + 6 = 0 have one root in common. The other roots
of the first and second equations are integers in the ratio 4:3 . Then the common root is
a) 1
b) 2
c) 3
d) 4
Solution
Since the roots are in the ratio 4: 3, let them be 4k and 3k
Let the common root be t
Now, x²-6x+a=0 =>Sum of roots = 6 => 6=4k + t -(1)
Product of roots=> a => a= 4kt - (2)
From :x² - cx + 6 = 0 => Sum of roots = c => c = 3k + t -(3)
Product of roots = 6 => 6= 3kt -(4)
From (2) and (4)
a/6= 4/3
a=8
So 8= 4kt
kt = 2
Now x²-6x+a=0
x²-6x+8 = 0
Roots are (4, 2)
So, 4k and t are 4 and 2, but they are integers, common root t = 2
Answer : (b)