Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
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Key Concepts (Quick Recap)
Linear Equations in One Variable
- Standard form: ax + b = c, where a is not zero
- Rules: Add/subtract the same value on both sides; multiply/divide both sides by the same non-zero value
- Transposition: Moving a term across the equals sign changes its sign (+ becomes -, and vice versa)
- Verification: Substitute the solution back into the original equation to confirm
Algebraic Expressions
- Term: A number, a variable, or a product of numbers and variables (e.g., 5x^2, -3y, 7)
- Like terms: Terms with the same variable part (e.g., 3x and 7x are like terms; 3x and 3x^2 are not)
- Coefficient: The numerical factor of a term (e.g., in 5x^2, the coefficient is 5)
- Addition/Subtraction: Combine like terms by adding/subtracting their coefficients
Algebraic Identities
- (a + b)^2 = a^2 + 2ab + b^2
- (a - b)^2 = a^2 - 2ab + b^2
- a^2 - b^2 = (a + b)(a - b) (difference of squares)
- (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3
- (a - b)^3 = a^3 - 3a^2b + 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)
Factorisation Methods
1. Common factor: Take out the highest common factor (HCF) from all terms 2. Grouping: Group terms to find common factors in pairs 3. Difference of squares: a^2 - b^2 = (a + b)(a - b) 4. Perfect square trinomial: a^2 + 2ab + b^2 = (a + b)^2; a^2 - 2ab + b^2 = (a - b)^2 5. Quadratic trinomial: x^2 + (p+q)x + pq = (x + p)(x + q)
Exercise 3.1 — Linear Equations in One Variable
Q1. Solve: 3x + 7 = 22
3x + 7 = 22
3x = 22 - 7 (transpose +7)
3x = 15
x = 15/3
x = 5
Verification: 3(5) + 7 = 15 + 7 = 22 ✓
Q2. Solve: 2x - 5 = 11
2x - 5 = 11
2x = 11 + 5 (transpose -5)
2x = 16
x = 16/2
x = 8
Verification: 2(8) - 5 = 16 - 5 = 11 ✓
Q3. Solve: 5x - 3 = 3x + 7
5x - 3 = 3x + 7
5x - 3x = 7 + 3 (transpose 3x and -3)
2x = 10
x = 10/2
x = 5
Verification:
LHS: 5(5) - 3 = 25 - 3 = 22
RHS: 3(5) + 7 = 15 + 7 = 22
LHS = RHS ✓
Q4. Solve: 4(x - 2) = 16
4(x - 2) = 16
4x - 8 = 16
4x = 16 + 8 (transpose -8)
4x = 24
x = 24/4
x = 6
Verification: 4(6 - 2) = 4(4) = 16 ✓
Q5. Solve: 3(2x + 1) = 2(x + 4) + 3
3(2x + 1) = 2(x + 4) + 3
6x + 3 = 2x + 8 + 3
6x + 3 = 2x + 11
6x - 2x = 11 - 3
4x = 8
x = 2
Verification:
LHS: 3(2(2) + 1) = 3(5) = 15
RHS: 2(2 + 4) + 3 = 2(6) + 3 = 15
LHS = RHS ✓
Q6. Solve: (2x + 3)/5 = (3x - 1)/4
(2x + 3)/5 = (3x - 1)/4
Cross multiply:
4(2x + 3) = 5(3x - 1)
8x + 12 = 15x - 5
12 + 5 = 15x - 8x
17 = 7x
x = 17/7
Verification:
LHS: (2(17/7) + 3)/5 = (34/7 + 21/7)/5 = (55/7)/5 = 55/35 = 11/7
RHS: (3(17/7) - 1)/4 = (51/7 - 7/7)/4 = (44/7)/4 = 44/28 = 11/7
LHS = RHS ✓
Q7. Solve: 3x - (x - 2)/3 = 4 - (x - 1)/6
3x - (x - 2)/3 = 4 - (x - 1)/6
Multiply both sides by 6 (LCM of 3 and 6):
18x - 2(x - 2) = 24 - (x - 1)
18x - 2x + 4 = 24 - x + 1
16x + 4 = 25 - x
16x + x = 25 - 4
17x = 21
x = 21/17
Verification:
LHS: 3(21/17) - (21/17 - 2)/3 = 63/17 - ((21-34)/17)/3
= 63/17 - (-13/17)/3 = 63/17 + 13/51 = 189/51 + 13/51 = 202/51
RHS: 4 - (21/17 - 1)/6 = 4 - (4/17)/6 = 4 - 4/102 = 4 - 2/51 = 202/51
LHS = RHS ✓
Q8. Solve: (x + 1)/3 + (x + 2)/5 = (x + 3)/2
(x + 1)/3 + (x + 2)/5 = (x + 3)/2
Multiply both sides by 30 (LCM of 3, 5, 2):
10(x + 1) + 6(x + 2) = 15(x + 3)
10x + 10 + 6x + 12 = 15x + 45
16x + 22 = 15x + 45
16x - 15x = 45 - 22
x = 23
Verification:
LHS: (24)/3 + (25)/5 = 8 + 5 = 13
RHS: (26)/2 = 13
LHS = RHS ✓
Q9. If 3x - 4 = 11, find the value of 2x + 3.
First, find x:
3x - 4 = 11
3x = 15
x = 5
Now find 2x + 3:
2(5) + 3 = 10 + 3 = 13
Q10. Solve: 0.5x + 1.5 = 3.5
0.5x + 1.5 = 3.5
0.5x = 3.5 - 1.5
0.5x = 2.0
x = 2.0/0.5
x = 4
Verification: 0.5(4) + 1.5 = 2.0 + 1.5 = 3.5 ✓
Q11. Solve: (2x - 1)/3 = (x + 2)/2 + 1
(2x - 1)/3 = (x + 2)/2 + 1
Multiply both sides by 6 (LCM of 3 and 2):
2(2x - 1) = 3(x + 2) + 6
4x - 2 = 3x + 6 + 6
4x - 2 = 3x + 12
4x - 3x = 12 + 2
x = 14
Verification:
LHS: (2(14) - 1)/3 = 27/3 = 9
RHS: (14 + 2)/2 + 1 = 16/2 + 1 = 8 + 1 = 9
LHS = RHS ✓
Q12. Solve: 5(x - 3) = 3(x + 1)
5(x - 3) = 3(x + 1)
5x - 15 = 3x + 3
5x - 3x = 3 + 15
2x = 18
x = 9
Verification:
LHS: 5(9 - 3) = 5(6) = 30
RHS: 3(9 + 1) = 3(10) = 30
LHS = RHS ✓
Q13. Solve: (3x + 2)/7 + (4x + 1)/5 = (2x + 3)/2
(3x + 2)/7 + (4x + 1)/5 = (2x + 3)/2
Multiply both sides by 70 (LCM of 7, 5, 2):
10(3x + 2) + 14(4x + 1) = 35(2x + 3)
30x + 20 + 56x + 14 = 70x + 105
86x + 34 = 70x + 105
86x - 70x = 105 - 34
16x = 71
x = 71/16
Verification:
LHS: (3(71/16) + 2)/7 + (4(71/16) + 1)/5
= (213/16 + 32/16)/7 + (284/16 + 16/16)/5
= (245/16)/7 + (300/16)/5
= 245/112 + 300/80
= 35/16 + 15/4 = 35/16 + 60/16 = 95/16
RHS: (2(71/16) + 3)/2 = (142/16 + 48/16)/2 = (190/16)/2 = 190/32 = 95/16
LHS = RHS ✓
Q14. Solve: 7x - 2 = 4x + 7
7x - 2 = 4x + 7
7x - 4x = 7 + 2
3x = 9
x = 3
Verification:
LHS: 7(3) - 2 = 21 - 2 = 19
RHS: 4(3) + 7 = 12 + 7 = 19
LHS = RHS ✓
Q15. Solve: (x - 3)/4 + (x + 5)/3 = (2x + 1)/6
(x - 3)/4 + (x + 5)/3 = (2x + 1)/6
Multiply both sides by 12 (LCM of 4, 3, 6):
3(x - 3) + 4(x + 5) = 2(2x + 1)
3x - 9 + 4x + 20 = 4x + 2
7x + 11 = 4x + 2
7x - 4x = 2 - 11
3x = -9
x = -3
Verification:
LHS: (-3-3)/4 + (-3+5)/3 = (-6)/4 + 2/3 = -3/2 + 2/3 = -9/6 + 4/6 = -5/6
RHS: (2(-3)+1)/6 = (-5)/6 = -5/6
LHS = RHS ✓
Q16. Solve: 2(3x - 1) - 3(2x + 1) = 5x - 7
2(3x - 1) - 3(2x + 1) = 5x - 7
6x - 2 - 6x - 3 = 5x - 7
-5 = 5x - 7
-5 + 7 = 5x
2 = 5x
x = 2/5
Verification:
LHS: 2(3(2/5) - 1) - 3(2(2/5) + 1) = 2(6/5 - 1) - 3(4/5 + 1)
= 2(1/5) - 3(9/5) = 2/5 - 27/5 = -25/5 = -5
RHS: 5(2/5) - 7 = 2 - 7 = -5
LHS = RHS ✓
Q17. If x/3 + x/4 = 14, find x.
x/3 + x/4 = 14
Multiply both sides by 12 (LCM of 3 and 4):
4x + 3x = 168
7x = 168
x = 24
Verification: 24/3 + 24/4 = 8 + 6 = 14 ✓
Q18. Solve: 3(x + 2) - 2(x - 3) = x + 12
3(x + 2) - 2(x - 3) = x + 12
3x + 6 - 2x + 6 = x + 12
x + 12 = x + 12
This is always true — the equation is an identity.
Every real number x is a solution.
Exercise 3.2 — Algebraic Identities
Q1. Expand using identity: (x + 5)^2
Using (a + b)^2 = a^2 + 2ab + b^2
Here a = x, b = 5
(x + 5)^2 = x^2 + 2(x)(5) + 5^2
= x^2 + 10x + 25
Q2. Expand using identity: (3x - 4)^2
Using (a - b)^2 = a^2 - 2ab + b^2
Here a = 3x, b = 4
(3x - 4)^2 = (3x)^2 - 2(3x)(4) + 4^2
= 9x^2 - 24x + 16
Q3. Expand using identity: (2a + 3b)^2
Using (a + b)^2 = a^2 + 2ab + b^2
Here a = 2a, b = 3b
(2a + 3b)^2 = (2a)^2 + 2(2a)(3b) + (3b)^2
= 4a^2 + 12ab + 9b^2
Q4. Find the value of 103^2 using identity.
103^2 = (100 + 3)^2
Using (a + b)^2 = a^2 + 2ab + b^2
= 100^2 + 2(100)(3) + 3^2
= 10000 + 600 + 9
= 10609
Q5. Find the value of 98^2 using identity.
98^2 = (100 - 2)^2
Using (a - b)^2 = a^2 - 2ab + b^2
= 100^2 - 2(100)(2) + 2^2
= 10000 - 400 + 4
= 9604
Q6. Find the value of 997^2 using identity.
997^2 = (1000 - 3)^2
Using (a - b)^2 = a^2 - 2ab + b^2
= 1000^2 - 2(1000)(3) + 3^2
= 1000000 - 6000 + 9
= 994009
Q7. Evaluate 105 x 95 using a suitable identity.
105 x 95 = (100 + 5)(100 - 5)
Using a^2 - b^2 = (a + b)(a - b)
= 100^2 - 5^2
= 10000 - 25
= 9975
Q8. Evaluate 103 x 97 using a suitable identity.
103 x 97 = (100 + 3)(100 - 3)
Using a^2 - b^2 = (a + b)(a - b)
= 100^2 - 3^2
= 10000 - 9
= 9991
Q9. If x + 1/x = 5, find x^2 + 1/x^2.
Given: x + 1/x = 5
Square both sides:
(x + 1/x)^2 = 5^2
x^2 + 2(x)(1/x) + 1/x^2 = 25
x^2 + 2 + 1/x^2 = 25
x^2 + 1/x^2 = 25 - 2
x^2 + 1/x^2 = 23
Q10. If x + 1/x = 6, find x^2 + 1/x^2 and x^4 + 1/x^4.
Given: x + 1/x = 6
Finding x^2 + 1/x^2:
(x + 1/x)^2 = 6^2
x^2 + 2 + 1/x^2 = 36
x^2 + 1/x^2 = 34
Finding x^4 + 1/x^4:
(x^2 + 1/x^2)^2 = 34^2
x^4 + 2 + 1/x^4 = 1156
x^4 + 1/x^4 = 1154
Q11. If x - 1/x = 3, find x^2 + 1/x^2.
Given: x - 1/x = 3
Square both sides:
(x - 1/x)^2 = 3^2
x^2 - 2(x)(1/x) + 1/x^2 = 9
x^2 - 2 + 1/x^2 = 9
x^2 + 1/x^2 = 11
Q12. Expand: (2x + 3y + 4z)^2
Using (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca
Here a = 2x, b = 3y, c = 4z
(2x + 3y + 4z)^2 = (2x)^2 + (3y)^2 + (4z)^2
+ 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)
= 4x^2 + 9y^2 + 16z^2 + 12xy + 24yz + 16zx
Q13. If a + b + c = 12 and ab + bc + ca = 15, find a^2 + b^2 + c^2.
Using (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
12^2 = a^2 + b^2 + c^2 + 2(15)
144 = a^2 + b^2 + c^2 + 30
a^2 + b^2 + c^2 = 144 - 30
a^2 + b^2 + c^2 = 114
Q14. Simplify: (3x + 7)^2 - (3x - 7)^2
Using (a + b)^2 - (a - b)^2 = 4ab
Here a = 3x, b = 7
(3x + 7)^2 - (3x - 7)^2 = 4(3x)(7)
= 84x
Alternative (direct expansion):
(9x^2 + 42x + 49) - (9x^2 - 42x + 49)
= 9x^2 + 42x + 49 - 9x^2 + 42x - 49
= 84x
Q15. Simplify: (a + b)^2 + (a - b)^2 and express in terms of a^2 + b^2.
(a + b)^2 + (a - b)^2
= (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2)
= 2a^2 + 2b^2
= 2(a^2 + b^2)
Q16. Expand using identity: (2a - 5b)^2
Using (a - b)^2 = a^2 - 2ab + b^2
Here a = 2a, b = 5b
(2a - 5b)^2 = (2a)^2 - 2(2a)(5b) + (5b)^2
= 4a^2 - 20ab + 25b^2
Q17. Find the product: (3x + 4)(3x - 4)
Using (a + b)(a - b) = a^2 - b^2
Here a = 3x, b = 4
(3x + 4)(3x - 4) = (3x)^2 - 4^2
= 9x^2 - 16
Q18. If a + b = 7 and a - b = 3, find a^2 + b^2.
Given: a + b = 7, a - b = 3
Step 1: Find ab using (a + b)^2 - (a - b)^2 = 4ab
7^2 - 3^2 = 4ab
49 - 9 = 4ab
40 = 4ab
ab = 10
Step 2: Find a^2 + b^2
(a + b)^2 = a^2 + 2ab + b^2
49 = a^2 + 2(10) + b^2
49 = a^2 + 20 + b^2
a^2 + b^2 = 29
Q19. Evaluate: 53^2 - 47^2 using identity.
Using a^2 - b^2 = (a + b)(a - b)
Here a = 53, b = 47
53^2 - 47^2 = (53 + 47)(53 - 47)
= 100 x 6
= 600
Q20. If 2x + 3y = 8 and xy = 2, find 4x^2 + 9y^2.
Given: 2x + 3y = 8, xy = 2
Square both sides of 2x + 3y = 8:
(2x + 3y)^2 = 64
4x^2 + 2(2x)(3y) + 9y^2 = 64
4x^2 + 12xy + 9y^2 = 64
4x^2 + 12(2) + 9y^2 = 64
4x^2 + 24 + 9y^2 = 64
4x^2 + 9y^2 = 64 - 24
4x^2 + 9y^2 = 40
Exercise 3.3 — Factorisation
Q1. Factorise: 6x^2 + 12x
Common factor method:
6x^2 + 12x = 6x(x + 2)
Check: 6x(x + 2) = 6x^2 + 12x ✓
Q2. Factorise: 15xy + 25x
HCF of 15xy and 25x = 5x
15xy + 25x = 5x(3y + 5)
Check: 5x(3y + 5) = 15xy + 25x ✓
Q3. Factorise: x^2 - 9
This is a difference of squares: a^2 - b^2 = (a + b)(a - b)
Here a = x, b = 3
x^2 - 9 = x^2 - 3^2
= (x + 3)(x - 3)
Check: (x + 3)(x - 3) = x^2 - 9 ✓
Q4. Factorise: 4x^2 - 25
Difference of squares:
4x^2 - 25 = (2x)^2 - 5^2
= (2x + 5)(2x - 5)
Check: (2x + 5)(2x - 5) = 4x^2 - 25 ✓
Q5. Factorise: x^2 + 5x + 6
Quadratic trinomial: find two numbers with sum = 5 and product = 6.
The numbers are 2 and 3.
x^2 + 5x + 6 = x^2 + 2x + 3x + 6
= x(x + 2) + 3(x + 2)
= (x + 2)(x + 3)
Check: (x + 2)(x + 3) = x^2 + 5x + 6 ✓
Q6. Factorise: x^2 + 7x + 12
Find two numbers with sum = 7 and product = 12.
The numbers are 3 and 4.
x^2 + 7x + 12 = x^2 + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Check: (x + 3)(x + 4) = x^2 + 7x + 12 ✓
Q7. Factorise: x^2 - 7x + 12
Find two numbers with sum = -7 and product = 12.
The numbers are -3 and -4.
x^2 - 7x + 12 = x^2 - 3x - 4x + 12
= x(x - 3) - 4(x - 3)
= (x - 3)(x - 4)
Check: (x - 3)(x - 4) = x^2 - 7x + 12 ✓
Q8. Factorise: x^2 + x - 12
Find two numbers with sum = 1 and product = -12.
The numbers are 4 and -3.
x^2 + x - 12 = x^2 + 4x - 3x - 12
= x(x + 4) - 3(x + 4)
= (x + 4)(x - 3)
Check: (x + 4)(x - 3) = x^2 + x - 12 ✓
Q9. Factorise: x^2 - 5x + 6
Find two numbers with sum = -5 and product = 6.
The numbers are -2 and -3.
x^2 - 5x + 6 = x^2 - 2x - 3x + 6
= x(x - 2) - 3(x - 2)
= (x - 2)(x - 3)
Check: (x - 2)(x - 3) = x^2 - 5x + 6 ✓
Q10. Factorise: x^2 - 16
Difference of squares:
x^2 - 16 = x^2 - 4^2
= (x + 4)(x - 4)
Check: (x + 4)(x - 4) = x^2 - 16 ✓
Q11. Factorise: a^2 + 10a + 25
This is a perfect square trinomial.
a^2 + 10a + 25 = a^2 + 2(a)(5) + 5^2
Using (a + b)^2 = a^2 + 2ab + b^2:
= (a + 5)^2
= (a + 5)(a + 5)
Check: (a + 5)^2 = a^2 + 10a + 25 ✓
Q12. Factorise: 4x^2 - 20x + 25
This is a perfect square trinomial.
4x^2 - 20x + 25 = (2x)^2 - 2(2x)(5) + 5^2
Using (a - b)^2 = a^2 - 2ab + b^2:
= (2x - 5)^2
= (2x - 5)(2x - 5)
Check: (2x - 5)^2 = 4x^2 - 20x + 25 ✓
Q13. Factorise: ax + bx + ay + by
Grouping method:
ax + bx + ay + by
= (ax + bx) + (ay + by)
= x(a + b) + y(a + b)
= (a + b)(x + y)
Check: (a + b)(x + y) = ax + ay + bx + by ✓
Q14. Factorise: 2x^2 + 7x + 6
Product of coefficient of x^2 and constant = 2 x 6 = 12
Find two numbers with sum = 7 and product = 12.
The numbers are 3 and 4.
2x^2 + 7x + 6 = 2x^2 + 4x + 3x + 6
= 2x(x + 2) + 3(x + 2)
= (2x + 3)(x + 2)
Check: (2x + 3)(x + 2) = 2x^2 + 4x + 3x + 6 = 2x^2 + 7x + 6 ✓
Q15. Factorise: 3x^2 + 11x + 10
Product of coefficient of x^2 and constant = 3 x 10 = 30
Find two numbers with sum = 11 and product = 30.
The numbers are 5 and 6.
3x^2 + 11x + 10 = 3x^2 + 6x + 5x + 10
= 3x(x + 2) + 5(x + 2)
= (3x + 5)(x + 2)
Check: (3x + 5)(x + 2) = 3x^2 + 6x + 5x + 10 = 3x^2 + 11x + 10 ✓
Q16. Factorise: 9x^2 - 3x - 2
Product of coefficient of x^2 and constant = 9 x (-2) = -18
Find two numbers with sum = -3 and product = -18.
The numbers are -6 and 3.
9x^2 - 3x - 2 = 9x^2 - 6x + 3x - 2
= 3x(3x - 2) + 1(3x - 2)
= (3x + 1)(3x - 2)
Check: (3x + 1)(3x - 2) = 9x^2 - 6x + 3x - 2 = 9x^2 - 3x - 2 ✓
Q17. Factorise: 8a^3 - 27b^3
Using a^3 - b^3 = (a - b)(a^2 + ab + b^2)
Here a = 2a, b = 3b
8a^3 - 27b^3 = (2a)^3 - (3b)^3
= (2a - 3b)((2a)^2 + (2a)(3b) + (3b)^2)
= (2a - 3b)(4a^2 + 6ab + 9b^2)
Q18. Factorise: x^3 + 64
Using a^3 + b^3 = (a + b)(a^2 - ab + b^2)
x^3 + 64 = x^3 + 4^3
= (x + 4)(x^2 - 4x + 16)
Check: (x + 4)(x^2 - 4x + 16)
= x^3 - 4x^2 + 16x + 4x^2 - 16x + 64
= x^3 + 64 ✓
Q19. Factorise: 27 - 8a^3
Using a^3 - b^3 = (a - b)(a^2 + ab + b^2)
27 - 8a^3 = 3^3 - (2a)^3
= (3 - 2a)(3^2 + 3(2a) + (2a)^2)
= (3 - 2a)(9 + 6a + 4a^2)
Q20. Factorise: a^2 - b^2 - a + b
Group and factorise:
a^2 - b^2 - a + b
= (a^2 - b^2) - (a - b)
= (a + b)(a - b) - (a - b)
= (a - b)[(a + b) - 1]
= (a - b)(a + b - 1)
Check: (a - b)(a + b - 1)
= a^2 + ab - a - ab - b^2 + b
= a^2 - b^2 - a + b ✓
Exercise 3.4 — Word Problems
Q1. The sum of three consecutive even numbers is 78. Find the numbers.
Let the three consecutive even numbers be x, x + 2, x + 4.
x + (x + 2) + (x + 4) = 78
3x + 6 = 78
3x = 72
x = 24
The three numbers are 24, 26, 28.
Verification: 24 + 26 + 28 = 78 ✓
Q2. A number is 5 more than three times another number. Their sum is 29. Find the numbers.
Let the other number be x.
Then the first number = 3x + 5.
x + (3x + 5) = 29
4x + 5 = 29
4x = 24
x = 6
The other number = 6.
The first number = 3(6) + 5 = 23.
Verification: 23 + 6 = 29 ✓
The first number is 5 more than 3 times 6: 3(6) + 5 = 23 ✓
Q3. The length of a rectangle is 5 cm more than its breadth. If the perimeter is 54 cm, find the length and breadth.
Let breadth = x cm
Then length = (x + 5) cm
Perimeter = 2(length + breadth) = 54
2(x + 5 + x) = 54
2(2x + 5) = 54
4x + 10 = 54
4x = 44
x = 11
Breadth = 11 cm
Length = 11 + 5 = 16 cm
Verification: Perimeter = 2(16 + 11) = 2(27) = 54 cm ✓
Q4. The ages of Rahul and his father are in the ratio 2:7. After 5 years, the sum of their ages will be 55 years. Find their present ages.
Let the present ages be 2x and 7x.
After 5 years: (2x + 5) + (7x + 5) = 55
9x + 10 = 55
9x = 45
x = 5
Rahul's present age = 2(5) = 10 years
Father's present age = 7(5) = 35 years
Verification: After 5 years: (10+5) + (35+5) = 15 + 40 = 55 ✓
Ratio: 10:35 = 2:7 ✓
Q5. A purse contains Rs 250 in notes of Rs 10 and Rs 50 denominations. If the number of Rs 10 notes is one more than twice the number of Rs 50 notes, find the number of each type of note.
Let the number of Rs 50 notes = x
Then the number of Rs 10 notes = 2x + 1
Total amount = 50x + 10(2x + 1) = 250
50x + 20x + 10 = 250
70x = 240
x = 240/70 = 24/7
Since x must be a whole number, let us recheck.
Actually, 70x = 240 gives x = 24/7 which is not an integer.
This suggests there may be a typo in the textbook. If the total
is Rs 260 instead:
50x + 10(2x + 1) = 260
70x + 10 = 260
70x = 250
x = 250/70 (still not integer)
If total = Rs 280:
70x + 10 = 280
70x = 270 (not integer)
Working with the given data as-is, the method is correct.
For exam purposes, set up the equation and solve using the
procedure shown. The key skill is translating the word problem
into an equation.
Q6. The sum of two numbers is 45 and their difference is 13. Find the numbers.
Let the larger number = x
Let the smaller number = y
x + y = 45 ... (1)
x - y = 13 ... (2)
Adding (1) and (2):
2x = 58
x = 29
Substituting in (1):
29 + y = 45
y = 16
The two numbers are 29 and 16.
Verification: 29 + 16 = 45 ✓ and 29 - 16 = 13 ✓
Q7. Divide 180 into two parts such that one part is twice the other.
Let one part = x
Then the other part = 2x
x + 2x = 180
3x = 180
x = 60
The two parts are 60 and 120.
Verification: 60 + 120 = 180 ✓ and 120 = 2(60) ✓
Q8. The present age of a man is 3 times the age of his son. After 10 years, the man will be twice the age of his son. Find their present ages.
Let son's present age = x years
Then man's present age = 3x years
After 10 years:
Man's age = 3x + 10
Son's age = x + 10
Condition: 3x + 10 = 2(x + 10)
3x + 10 = 2x + 20
3x - 2x = 20 - 10
x = 10
Son's present age = 10 years
Man's present age = 3(10) = 30 years
Verification: After 10 years: Son = 20, Man = 40. Is 40 = 2(20)? Yes ✓
Q9. A train travels 300 km at a uniform speed. If the speed had been 10 km/h more, the journey would have taken 1 hour less. Find the speed of the train.
Let the speed of the train = x km/h
Time taken at speed x = 300/x hours
Time taken at speed (x + 10) = 300/(x + 10) hours
Condition: 300/x - 300/(x + 10) = 1
Multiply by x(x + 10):
300(x + 10) - 300x = x(x + 10)
300x + 3000 - 300x = x^2 + 10x
3000 = x^2 + 10x
x^2 + 10x - 3000 = 0
x^2 + 60x - 50x - 3000 = 0
x(x + 60) - 50(x + 60) = 0
(x + 60)(x - 50) = 0
x = 50 (rejecting x = -60 since speed cannot be negative)
Speed of the train = 50 km/h
Verification: Time at 50 km/h = 300/50 = 6 h
Time at 60 km/h = 300/60 = 5 h
Difference = 6 - 5 = 1 hour ✓
Q10. The numerator of a fraction is 3 less than the denominator. If the numerator and denominator are both increased by 2, the fraction becomes 3/4. Find the fraction.
Let the denominator = x
Then the numerator = x - 3
The fraction = (x - 3)/x
Condition: (x - 3 + 2)/(x + 2) = 3/4
(x - 1)/(x + 2) = 3/4
Cross multiply:
4(x - 1) = 3(x + 2)
4x - 4 = 3x + 6
4x - 3x = 6 + 4
x = 10
Denominator = 10, Numerator = 10 - 3 = 7
The fraction = 7/10
Verification: (7+2)/(10+2) = 9/12 = 3/4 ✓
Q11. The perimeter of a triangle is 36 cm. The sides are in the ratio 3:4:5. Find the length of each side.
Let the sides be 3x, 4x, and 5x.
Perimeter: 3x + 4x + 5x = 36
12x = 36
x = 3
The sides are:
3(3) = 9 cm
4(3) = 12 cm
5(3) = 15 cm
Verification: 9 + 12 + 15 = 36 cm ✓
Ratio: 9:12:15 = 3:4:5 ✓
Q12. Three-fifth of a number is 4 more than half of the number. Find the number.
Let the number = x
(3/5)x = (1/2)x + 4
(3/5)x - (1/2)x = 4
(6/10)x - (5/10)x = 4
(1/10)x = 4
x = 40
The number is 40.
Verification: (3/5)(40) = 24 and (1/2)(40) + 4 = 24 ✓
Q13. A number consists of two digits. The digit in the tens place is twice the digit in the units place. If 18 is added to the number, the digits interchange their places. Find the number.
Let the units digit = x
Then the tens digit = 2x
The number = 10(2x) + x = 21x
After adding 18: 21x + 18 = 10x + 2x (wait, reversed digits)
The new number after interchange = 10(x) + 2x = 12x
Condition: 21x + 18 = 12x
21x - 12x = -18
This gives 9x = -18, x = -2, which is not valid.
Let me reconsider. The new number with interchanged digits:
If original number = 10(2x) + x, digits are 2x (tens) and x (units).
After interchanging: tens = x, units = 2x.
New number = 10x + 2x = 12x.
So: 21x + 18 = 12x does not work. Let me re-read the problem.
Actually, x must be a single digit (1-4, since 2x must also be a digit).
If tens digit = 2x and units digit = x:
Original number = 10(2x) + x = 20x + x = 21x
After interchange: number = 10(x) + 2x = 10x + 2x = 12x
21x + 18 = 12x gives negative, so the condition must be:
Wait — let me try x = 4:
Tens = 8, Units = 4. Number = 84
After adding 18: 84 + 18 = 102. Not a digit swap.
Let me try: the number with tens digit 2x and units digit x.
Possible: x = 1, number = 21. 21 + 18 = 39. Not a swap.
x = 2, number = 42. 42 + 18 = 60. Not a swap.
x = 3, number = 63. 63 + 18 = 81. Not a swap.
x = 4, number = 84. 84 + 18 = 102. Not valid.
Hmm, x = 3 gives 63 → 81. The digits swap from 6,3 to 8,1 — no.
Actually if tens = 2x and units = x:
x=1: 21, swap = 12. 21 + 18 = 39. No.
x=2: 42, swap = 24. 42 + 18 = 60. No.
x=3: 63, swap = 36. 63 + 18 = 81. No.
x=4: 84, swap = 48. 84 + 18 = 102. No.
The standard Samacheer version states: tens digit is twice the
units digit. If 18 is subtracted, digits interchange.
Let: number = 10(2x) + x = 21x
After subtracting 18: 21x - 18 = 10(x) + 2x = 12x
21x - 12x = 18
9x = 18
x = 2
Original number = 10(4) + 2 = 42
After subtracting 18: 42 - 18 = 24
Digits interchanged: 42 → 24 ✓
Answer: The number is 42.
Q14. Raju is 4 years older than Ravi. Six years ago, Raju's age was double Ravi's age. Find their present ages.
Let Ravi's present age = x years
Then Raju's present age = x + 4 years
Six years ago:
Ravi's age = x - 6
Raju's age = (x + 4) - 6 = x - 2
Condition: x - 2 = 2(x - 6)
x - 2 = 2x - 12
-2 + 12 = 2x - x
10 = x
Ravi's present age = 10 years
Raju's present age = 14 years
Verification: 6 years ago, Ravi = 4 and Raju = 8. Is 8 = 2(4)? Yes ✓
Q15. The total cost of 5 pens and 3 pencils is Rs 55, while the total cost of 3 pens and 5 pencils is Rs 45. Find the cost of each.
Let the cost of one pen = Rs x
Let the cost of one pencil = Rs y
Equation 1: 5x + 3y = 55 ... (1)
Equation 2: 3x + 5y = 45 ... (2)
Multiply (1) by 3: 15x + 9y = 165 ... (3)
Multiply (2) by 5: 15x + 25y = 225 ... (4)
Subtract (3) from (4):
16y = 60
y = 60/16 = 15/4
Wait — the cost should be a whole number. Let me recheck.
Multiply (1) by 5: 25x + 15y = 275 ... (3)
Multiply (2) by 3: 9x + 15y = 135 ... (4)
Subtract (4) from (3):
16x = 140
x = 140/16 = 35/4
This gives non-integer values. Using the typical Samacheer
textbook values of 5 pens + 3 pencils = Rs 55 and
3 pens + 5 pencils = Rs 45:
From (1): 5x + 3y = 55
From (2): 3x + 5y = 45
Multiply (1) by 5 and (2) by 3:
25x + 15y = 275 ... (5)
9x + 15y = 135 ... (6)
(5) - (6): 16x = 140, x = 8.75
Multiply (1) by 3 and (2) by 5:
15x + 9y = 165 ... (7)
15x + 25y = 225 ... (8)
(8) - (7): 16y = 60, y = 3.75
Cost of one pen = Rs 8.75 = Rs 35/4
Cost of one pencil = Rs 3.75 = Rs 15/4
Verification: 5(35/4) + 3(15/4) = 175/4 + 45/4 = 220/4 = 55 ✓
And: 3(35/4) + 5(15/4) = 105/4 + 75/4 = 180/4 = 45 ✓
Common Mistakes
1. Forgetting to flip the sign when transposing: When you move a term across the equals sign, its sign changes. Moving +3x to the other side makes it -3x. Moving -7 to the other side makes it +7. Always check your transpositions.
2. Not distributing the minus sign in brackets: When expanding -(x - 3), students often write -x - 3 instead of -x + 3. The minus sign distributes to every term inside the brackets. Write it as -1(x - 3) = -x + 3.
3. Confusing (a + b)^2 with a^2 + b^2: (a + b)^2 = a^2 + 2ab + b^2, NOT just a^2 + b^2. The cross-term 2ab is always present. Similarly, (a - b)^2 = a^2 - 2ab + b^2, not a^2 - b^2.
4. Wrong factorisation sign in sum/difference of cubes:
Many students swap these signs. Remember: the quadratic factor always has the opposite sign of the linear factor's middle term.
- a^3 + b^3 = (a + b)(a^2 - ab + b^2) — note the MINUS sign in the middle
- a^3 - b^3 = (a - b)(a^2 + ab + b^2) — note the PLUS sign in the middle
5. Dividing only one side of the equation: If you divide both sides by 3, you must divide EVERY term on both sides by 3. Dividing only the left side is a common error. Example: 3x + 6 = 15 gives x + 2 = 5, NOT x + 6 = 5.
6. Setting up word problems incorrectly: Always define the variable clearly and write what it represents. "Let x = age" is ambiguous — write "Let x = son's age in years." Then translate the word condition into an equation before solving.
7. Not verifying the answer: Always substitute your answer back into the original equation (not your simplified version) to check for calculation errors. A wrong sign or arithmetic slip can give an incorrect answer that looks plausible.
8. Mixing up factor pairs: For factorising x^2 + bx + c, you need two numbers whose SUM is b and PRODUCT is c. Students sometimes find numbers with the right product but wrong sum. Always check: p + q = b AND p x q = c.
Exam Checklist
- [ ] I can solve linear equations in one variable (including equations with fractions and brackets)
- [ ] I can transpose terms correctly (changing signs when crossing the equals sign)
- [ ] I can expand expressions using all seven algebraic identities from memory
- [ ] I can apply identities to evaluate expressions like 103^2, 997^2, 105 x 95
- [ ] I can find x^2 + 1/x^2 given x + 1/x or x - 1/x
- [ ] I can factorise using common factor, grouping, difference of squares, and perfect square trinomials
- [ ] I can factorise quadratic trinomials of the form x^2 + bx + c and ax^2 + bx + c
- [ ] I can factorise sums and differences of cubes (a^3 + b^3 and a^3 - b^3)
- [ ] I can set up word problems by defining variables and forming equations
- [ ] I can verify my solution by substituting back into the original equation
- [ ] I remember that a^2 + b^2 is NOT the same as (a + b)^2
- [ ] I can simplify algebraic fractions by factorising numerators and denominators
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Frequently Asked Questions
The key identities are: (a+b)^2 = a^2+2ab+b^2, (a-b)^2 = a^2-2ab+b^2, a^2-b^2 = (a+b)(a-b), (a+b)^3 = a^3+3a^2b+3ab^2+b^3, (a-b)^3 = a^3-3a^2b+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).