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Class 10 Mathematics Chapter 5 Test 3 | Algebraic Fractions | Punjab Boards 2027
Class 10 Math Chapter 5 Test 3 New paper pattern 2027
Class 10 Mathematics Chapter 5 Test 3 – Algebraic Fractions is a useful practice paper for students preparing for the Punjab Boards 2027 examinations. This test covers important concepts and questions from Chapter 5: Algebraic Fractions, including rational expressions, simplification, algebraic fractions, factorization, and related problems. Students can use this Class 10 Math Test 3 for revision, self-assessment, and preparation according to the new board paper pattern.
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Unit Wise Class Test
T-3
Unit 5: Algebraic Fractions
Time: 5 minutes
(Objective Type)
Total Marks: 5
(1 × 5 = 5)
(1 × 5 = 5)
1. Tick the correct answer.
(i) The rational expression x − yx3 − y3 in lowest form is:
(ii) Least common multiple of x + 1 and x2 − 1 is:
(iii) a3a − b − b3a − b =
(iv) 3x − y − 1y − x =
(v) What we need to substitute to convert the equation (x2 + 1x2) − 3(x + 1x) + 4 = 0 into a quadratic equation.
(i) x³ − y³ = (x − y)(x² + xy + y²) [difference of cubes]
x − y(x − y)(x² + xy + y²) → cancel (x − y) → 1x² + xy + y² Ans: (b)
(ii) x² − 1 = (x − 1)(x + 1), so x + 1 is already a factor of x² − 1.
LCM = (x − 1)(x + 1) = x² − 1 Ans: (c)
(iii) Same denominator, so subtract numerators: a³ − b³a − b
a³ − b³ = (a − b)(a² + ab + b²) → cancel (a − b) → a² + ab + b² Ans: (a)
(iv) y − x = −(x − y), so 1y − x = −1x − y
3x − y + 1x − y = 4x − y Ans: (c)
(v) Let t = x + 1x. Square both sides: t² = x² + 2 + 1x²
So x² + 1x² = t² − 2. Put in equation: (t² − 2) − 3t + 4 = 0 → t² − 3t + 2 = 0 (quadratic)
Substitute t = x + 1x Ans: (b)
x − y(x − y)(x² + xy + y²) → cancel (x − y) → 1x² + xy + y² Ans: (b)
(ii) x² − 1 = (x − 1)(x + 1), so x + 1 is already a factor of x² − 1.
LCM = (x − 1)(x + 1) = x² − 1 Ans: (c)
(iii) Same denominator, so subtract numerators: a³ − b³a − b
a³ − b³ = (a − b)(a² + ab + b²) → cancel (a − b) → a² + ab + b² Ans: (a)
(iv) y − x = −(x − y), so 1y − x = −1x − y
3x − y + 1x − y = 4x − y Ans: (c)
(v) Let t = x + 1x. Square both sides: t² = x² + 2 + 1x²
So x² + 1x² = t² − 2. Put in equation: (t² − 2) − 3t + 4 = 0 → t² − 3t + 2 = 0 (quadratic)
Substitute t = x + 1x Ans: (b)
Unit Wise Class Test
T-3
Unit 5: Algebraic Fractions
Time: 35 minutes
(Subjective Type)
Total Marks: 20
(6 × 2 = 12)
2. Write short answers to the following questions:
(i) Reduce the following rational expressions to lowest forms: 2x2 + 17x + 213x2 + 26x + 35
Step 1: Factorise numerator: 2x² + 17x + 21 = 2x² + 14x + 3x + 21 = 2x(x + 7) + 3(x + 7) = (x + 7)(2x + 3)
Step 2: Factorise denominator: 3x² + 26x + 35 = 3x² + 21x + 5x + 35 = 3x(x + 7) + 5(x + 7) = (x + 7)(3x + 5)
Step 3: Cancel common factor (x + 7): (x + 7)(2x + 3)(x + 7)(3x + 5)
Ans: 2x + 33x + 5
Step 2: Factorise denominator: 3x² + 26x + 35 = 3x² + 21x + 5x + 35 = 3x(x + 7) + 5(x + 7) = (x + 7)(3x + 5)
Step 3: Cancel common factor (x + 7): (x + 7)(2x + 3)(x + 7)(3x + 5)
Ans: 2x + 33x + 5
(ii) Simplify: x + y12x − 6y + x − y18x − 9y
Step 1: Factorise denominators: 12x − 6y = 6(2x − y) and 18x − 9y = 9(2x − y)
Step 2: LCM of denominators = 18(2x − y)
Step 3: Make equal denominators: 3(x + y)18(2x − y) + 2(x − y)18(2x − y)
Step 4: Add numerators: 3x + 3y + 2x − 2y18(2x − y)
Ans: 5x + y18(2x − y)
Step 2: LCM of denominators = 18(2x − y)
Step 3: Make equal denominators: 3(x + y)18(2x − y) + 2(x − y)18(2x − y)
Step 4: Add numerators: 3x + 3y + 2x − 2y18(2x − y)
Ans: 5x + y18(2x − y)
(iii) Subtract 1x2 + 2 from 2x3 + x2 + 3(x2 + 2)2.
Step 1: Subtract = first fraction − second: 2x3 + x2 + 3(x2 + 2)2 − 1x2 + 2
Step 2: LCD = (x2 + 2)2. Multiply second fraction by x2 + 2x2 + 2: 1x2 + 2 = x2 + 2(x2 + 2)2
Step 3: 2x3 + x2 + 3 − (x2 + 2)(x2 + 2)2 = 2x3 + x2 + 3 − x2 − 2(x2 + 2)2
Ans: 2x3 + 1(x2 + 2)2
Step 2: LCD = (x2 + 2)2. Multiply second fraction by x2 + 2x2 + 2: 1x2 + 2 = x2 + 2(x2 + 2)2
Step 3: 2x3 + x2 + 3 − (x2 + 2)(x2 + 2)2 = 2x3 + x2 + 3 − x2 − 2(x2 + 2)2
Ans: 2x3 + 1(x2 + 2)2
(iv) Simplify: 3a2 − 6ab6ab × 10b24ab − 8b2
Step 1: Factorise: 3a² − 6ab = 3a(a − 2b) and 4ab − 8b² = 4b(a − 2b)
Step 2: Rewrite: 3a(a − 2b)6ab × 10b24b(a − 2b)
Step 3: Simplify first fraction: 3a(a − 2b)6ab = a − 2b2b
Step 4: Simplify second: 10b24b(a − 2b) = 5b2(a − 2b)
Step 5: Multiply and cancel (a − 2b) and b: (a − 2b) × 5b2b × 2(a − 2b) = 54
Ans: 54
Step 2: Rewrite: 3a(a − 2b)6ab × 10b24b(a − 2b)
Step 3: Simplify first fraction: 3a(a − 2b)6ab = a − 2b2b
Step 4: Simplify second: 10b24b(a − 2b) = 5b2(a − 2b)
Step 5: Multiply and cancel (a − 2b) and b: (a − 2b) × 5b2b × 2(a − 2b) = 54
Ans: 54
(v) Simplify: x2 − 3x + 2x2 + 3x − 4 × 2x2 + 8x3x + 6
Step 1: Factorise: x² − 3x + 2 = (x − 1)(x − 2) and x² + 3x − 4 = (x + 4)(x − 1)
Step 2: Factorise: 2x² + 8x = 2x(x + 4) and 3x + 6 = 3(x + 2)
Step 3: Rewrite: (x − 1)(x − 2)(x + 4)(x − 1) × 2x(x + 4)3(x + 2)
Step 4: Cancel (x − 1) and (x + 4) from numerator and denominator
Ans: 2x(x − 2)3(x + 2)
Step 2: Factorise: 2x² + 8x = 2x(x + 4) and 3x + 6 = 3(x + 2)
Step 3: Rewrite: (x − 1)(x − 2)(x + 4)(x − 1) × 2x(x + 4)3(x + 2)
Step 4: Cancel (x − 1) and (x + 4) from numerator and denominator
Ans: 2x(x − 2)3(x + 2)
(vi) Simplify: a3a − b − b3b − a
Step 1: Since b − a = −(a − b): b3b − a = −b3a − b
Step 2: So expression = a3a − b + b3a − b (minus × minus = plus)
Step 3: Same denominator, add numerators: a3 + b3a − b
Step 4: Sum of cubes: a³ + b³ = (a + b)(a² − ab + b²)
Ans: (a + b)(a2 − ab + b2)a − b (no common factor, so this is the lowest form)
Step 2: So expression = a3a − b + b3a − b (minus × minus = plus)
Step 3: Same denominator, add numerators: a3 + b3a − b
Step 4: Sum of cubes: a³ + b³ = (a + b)(a² − ab + b²)
Ans: (a + b)(a2 − ab + b2)a − b (no common factor, so this is the lowest form)
Note: Answer the following Questions.
(4 + 4 = 8)
3. Simplify: x3 − a3x2 − 2bx + b2 × x2 − bx − cx + bcax − x2 ÷ x2 − cxx − b
Step 1: Factorise every part:
x³ − a³ = (x − a)(x² + ax + a²)
x² − 2bx + b² = (x − b)²
x² − bx − cx + bc = x(x − b) − c(x − b) = (x − b)(x − c)
ax − x² = x(a − x)
x² − cx = x(x − c)
Step 2: Change ÷ into × by taking reciprocal of the last fraction: x − bx(x − c)
Step 3: (x − a)(x2 + ax + a2)(x − b)² × (x − b)(x − c)x(a − x) × x − bx(x − c)
Step 4: Cancel (x − b)(x − b) with (x − b)², and cancel (x − c)
Step 5: Left: (x − a)(x2 + ax + a2)x(a − x) · x. Since (x − a) = −(a − x), x − aa − x = −1
Ans: − x2 + ax + a2x2
x³ − a³ = (x − a)(x² + ax + a²)
x² − 2bx + b² = (x − b)²
x² − bx − cx + bc = x(x − b) − c(x − b) = (x − b)(x − c)
ax − x² = x(a − x)
x² − cx = x(x − c)
Step 2: Change ÷ into × by taking reciprocal of the last fraction: x − bx(x − c)
Step 3: (x − a)(x2 + ax + a2)(x − b)² × (x − b)(x − c)x(a − x) × x − bx(x − c)
Step 4: Cancel (x − b)(x − b) with (x − b)², and cancel (x − c)
Step 5: Left: (x − a)(x2 + ax + a2)x(a − x) · x. Since (x − a) = −(a − x), x − aa − x = −1
Ans: − x2 + ax + a2x2
4. Simplify: 4xx2 − 3x + 2 − 41 − x − 5x − 2
Step 1: Factorise: x² − 3x + 2 = (x − 1)(x − 2)
Step 2: Make denominators match: 1 − x = −(x − 1), so −41 − x = +4x − 1
Step 3: Expression = 4x(x − 1)(x − 2) + 4x − 1 − 5x − 2
Step 4: LCD = (x − 1)(x − 2). Multiply 4/(x − 1) by (x − 2)/(x − 2) and 5/(x − 2) by (x − 1)/(x − 1)
Step 5: 4x + 4(x − 2) − 5(x − 1)(x − 1)(x − 2) = 4x + 4x − 8 − 5x + 5(x − 1)(x − 2) = 3x − 3(x − 1)(x − 2)
Step 6: 3x − 3 = 3(x − 1); cancel (x − 1): 3(x − 1)(x − 1)(x − 2)
Ans: 3x − 2
Step 2: Make denominators match: 1 − x = −(x − 1), so −41 − x = +4x − 1
Step 3: Expression = 4x(x − 1)(x − 2) + 4x − 1 − 5x − 2
Step 4: LCD = (x − 1)(x − 2). Multiply 4/(x − 1) by (x − 2)/(x − 2) and 5/(x − 2) by (x − 1)/(x − 1)
Step 5: 4x + 4(x − 2) − 5(x − 1)(x − 1)(x − 2) = 4x + 4x − 8 − 5x + 5(x − 1)(x − 2) = 3x − 3(x − 1)(x − 2)
Step 6: 3x − 3 = 3(x − 1); cancel (x − 1): 3(x − 1)(x − 1)(x − 2)
Ans: 3x − 2
