SECTION 1: ALGEBRA
Chapter 1 – Linear Equations in Two Variables
Q1. Solve the following pair of linear equations by substitution method:
i. 2x + 3y = 13 x–y=1
Answer:
From x – y = 1
x=y+1
Substitute in first equation:
2(y + 1) + 3y = 13
2y + 2 + 3y = 13
5y + 2 = 13
5y = 11
y = 11/5
x=y+1
x = 11/5 + 5/5
x = 16/5
Chapter 2 – Quadratic Equations
Q2. Solve using quadratic formula:
x² – 5x + 6 = 0
Answer:
a = 1, b = –5, c = 6
Discriminant D = b² – 4ac
= 25 – 24
=1
x = (–b ± √D) / 2a
= (5 ± 1) / 2
x = 3, 2
Q3. Find nature of roots of:
2x² + 4x + 5 = 0
Answer:
D = b² – 4ac
= 16 – 40
= –24
Since D < 0, roots are imaginary.
Chapter 3 – Arithmetic Progression
Q4. Find the 15th term of AP:
3, 7, 11, 15, …
Answer:
a=3
d=4
n = 15
a■ = a + (n – 1)d
, = 3 + 14 × 4
= 3 + 56
= 59
Chapter 1 – Linear Equations in Two Variables
Q1. Solve the following pair of linear equations by substitution method:
i. 2x + 3y = 13 x–y=1
Answer:
From x – y = 1
x=y+1
Substitute in first equation:
2(y + 1) + 3y = 13
2y + 2 + 3y = 13
5y + 2 = 13
5y = 11
y = 11/5
x=y+1
x = 11/5 + 5/5
x = 16/5
Chapter 2 – Quadratic Equations
Q2. Solve using quadratic formula:
x² – 5x + 6 = 0
Answer:
a = 1, b = –5, c = 6
Discriminant D = b² – 4ac
= 25 – 24
=1
x = (–b ± √D) / 2a
= (5 ± 1) / 2
x = 3, 2
Q3. Find nature of roots of:
2x² + 4x + 5 = 0
Answer:
D = b² – 4ac
= 16 – 40
= –24
Since D < 0, roots are imaginary.
Chapter 3 – Arithmetic Progression
Q4. Find the 15th term of AP:
3, 7, 11, 15, …
Answer:
a=3
d=4
n = 15
a■ = a + (n – 1)d
, = 3 + 14 × 4
= 3 + 56
= 59
