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Class 10 Mathematics Unit 5 Algebraic Fractions Test 2 | Punjab Boards Paper Pattern 2027
Class 10 chapter 5 Test 2 New Paper pattern 2027
Prepare for Class 10 Mathematics Unit 5: Algebraic Fractions with this complete Unit Wise Class Test 2 based on the Punjab Boards paper pattern. This test includes objective-type MCQs and subjective questions covering exponential equations, reciprocal equations, reduced quadratic equations, algebraic equations and word problems. It is useful for Class 10 Mathematics preparation, Unit 5 test practice, Punjab Board 2027 paper pattern and school-level examination preparation.
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Unit Wise Class Test
T-2
Unit 5: Algebraic Fractions
Time: 5 minutes
(Objective Type)
Total Marks: 5
1. Tick the correct answer. (1 × 5 = 5)
(i) The equation 5x4 − 19x2 + 12 = 0 in reduced quadratic form is:
Let t = x² so x⁴ = t².
Substitute: 5t² − 19t + 12 = 0 Ans: (c)
Substitute: 5t² − 19t + 12 = 0 Ans: (c)
(ii) Lowest form of 15xyzz2 + z is:
Factor denominator: z² + z = z(z + 1)
15xyzz(z + 1) → cancel z → 15xyz + 1 Ans: (c)
15xyzz(z + 1) → cancel z → 15xyz + 1 Ans: (c)
(iii) Arshia and Ibrahim complete a job together in 4 hours. If job rate of Arshia is x and of Ibrahim is y, then mathematical equation in this scenario is:
Rate = work done per hour. Combined rate = x + y.
Whole job (1) in 4 hours: (x + y) × 4 = 1 → x + y = 14 Ans: (b)
Whole job (1) in 4 hours: (x + y) × 4 = 1 → x + y = 14 Ans: (b)
(iv) To reduce the equation x4 − 16x2 + 63 = 0 into a quadratic equation, we need to suppose that:
Since x⁴ = (x²)², put t = x²: t² − 16t + 63 = 0 Ans: (d)
(v) Solution of the equation x + 1x = 2 is x =
Multiply by x: x² + 1 = 2x → x² − 2x + 1 = 0
(x − 1)² = 0 → x = 1 Ans: (c)
Check: 1 + 1 = 2 ✓
(x − 1)² = 0 → x = 1 Ans: (c)
Check: 1 + 1 = 2 ✓
Unit Wise Class Test
T-2
Unit 5: Algebraic Fractions
Time: 35 minutes
(Subjective Type)
Total Marks: 20
2. Write short answers to the following questions: (6 × 2 = 12)
(i) Define Exponential Equation with example.
Definition: An equation in which the variable appears in the exponent is called an exponential equation.
Example: 2x = 8 → 2x = 2³ → x = 3
Example: 2x = 8 → 2x = 2³ → x = 3
(ii) Solve:
4x − 3 · 2x + 3 + 128 = 0
Step 1: Write 4x = (2x)² and 2x+3 = 2x · 2³ = 8 · 2x
Step 2: Let t = 2x: t² − 3(8t) + 128 = 0 → t² − 24t + 128 = 0
Step 3: Factorise (8 × 16 = 128, 8 + 16 = 24): (t − 8)(t − 16) = 0 → t = 8 or t = 16
Step 4: 2x = 8 = 2³ → x = 3; 2x = 16 = 2⁴ → x = 4
Ans: x = 3, 4
Step 2: Let t = 2x: t² − 3(8t) + 128 = 0 → t² − 24t + 128 = 0
Step 3: Factorise (8 × 16 = 128, 8 + 16 = 24): (t − 8)(t − 16) = 0 → t = 8 or t = 16
Step 4: 2x = 8 = 2³ → x = 3; 2x = 16 = 2⁴ → x = 4
Ans: x = 3, 4
(iii) Solve the following equations:
3x + 33 − x − 12 = 0
Step 1: 33−x = 3³ ÷ 3x = 27 ÷ 3x
Step 2: Let t = 3x: t + 27t − 12 = 0
Step 3: Multiply by t: t² + 27 − 12t = 0 → t² − 12t + 27 = 0
Step 4: Factorise (3 × 9 = 27, 3 + 9 = 12): (t − 3)(t − 9) = 0 → t = 3 or 9
Step 5: 3x = 3 → x = 1; 3x = 9 = 3² → x = 2
Ans: x = 1, 2
Step 2: Let t = 3x: t + 27t − 12 = 0
Step 3: Multiply by t: t² + 27 − 12t = 0 → t² − 12t + 27 = 0
Step 4: Factorise (3 × 9 = 27, 3 + 9 = 12): (t − 3)(t − 9) = 0 → t = 3 or 9
Step 5: 3x = 3 → x = 1; 3x = 9 = 3² → x = 2
Ans: x = 1, 2
(iv) Solve:
(x + 1)(x + 2)(x − 5)(x − 6) = 144
Step 1: Pair factors so the x-terms match: (x + 1)(x − 5) = x² − 4x − 5 and (x + 2)(x − 6) = x² − 4x − 12
Step 2: Let t = x² − 4x: (t − 5)(t − 12) = 144
Step 3: t² − 17t + 60 = 144 → t² − 17t − 84 = 0
Step 4: Factorise: (t − 21)(t + 4) = 0 → t = 21 or t = −4
Step 5: t = 21: x² − 4x − 21 = 0 → (x − 7)(x + 3) = 0 → x = 7, −3
Step 6: t = −4: x² − 4x + 4 = 0 → (x − 2)² = 0 → x = 2
Step 7: Check x = 2: (3)(4)(−3)(−4) = 144 ✓
Ans: x = −3, 2, 7
Step 2: Let t = x² − 4x: (t − 5)(t − 12) = 144
Step 3: t² − 17t + 60 = 144 → t² − 17t − 84 = 0
Step 4: Factorise: (t − 21)(t + 4) = 0 → t = 21 or t = −4
Step 5: t = 21: x² − 4x − 21 = 0 → (x − 7)(x + 3) = 0 → x = 7, −3
Step 6: t = −4: x² − 4x + 4 = 0 → (x − 2)² = 0 → x = 2
Step 7: Check x = 2: (3)(4)(−3)(−4) = 144 ✓
Ans: x = −3, 2, 7
(v) Define Reciprocal Equation with example.
Definition: An equation that remains unchanged when x is replaced by 1/x is called a reciprocal equation. Its coefficients read the same from both ends (e.g. a, b, c, b, a).
Example: 2x⁴ + 3x³ − 4x² + 3x + 2 = 0 (coefficients 2, 3, −4, 3, 2)
Example: 2x⁴ + 3x³ − 4x² + 3x + 2 = 0 (coefficients 2, 3, −4, 3, 2)
(vi) Solve the following equations:
5x4 − 19x2 + 12 = 0
Step 1: Let t = x²: 5t² − 19t + 12 = 0
Step 2: Split middle term: 5t² − 15t − 4t + 12 = 5t(t − 3) − 4(t − 3) = (t − 3)(5t − 4) = 0
Step 3: t = 3 or t = 45
Step 4: x² = 3 → x = ±√3; x² = 45 → x = ±2√5 = ±2√55
Ans: x = ±√3, ±2√5
Step 2: Split middle term: 5t² − 15t − 4t + 12 = 5t(t − 3) − 4(t − 3) = (t − 3)(5t − 4) = 0
Step 3: t = 3 or t = 45
Step 4: x² = 3 → x = ±√3; x² = 45 → x = ±2√5 = ±2√55
Ans: x = ±√3, ±2√5
Note: Answer the following Questions.(4 + 4 = 8)
3. Solve:
3(x2 + 1x2) + 4(x + 1x) − 14 = 0
Step 1: Let t = x + 1x. Square: t² = x² + 2 + 1x2 → x² + 1x2 = t² − 2
Step 2: Substitute: 3(t² − 2) + 4t − 14 = 0 → 3t² + 4t − 20 = 0
Step 3: Split middle term: 3t² + 10t − 6t − 20 = t(3t + 10) − 2(3t + 10) = (3t + 10)(t − 2) = 0
Step 4: t = 2 or t = −103
Step 5: t = 2: x + 1x = 2 → x² − 2x + 1 = 0 → (x − 1)² = 0 → x = 1
Step 6: t = −103: 3x² + 10x + 3 = 0 → (3x + 1)(x + 3) = 0 → x = −13 or x = −3
Ans: x = 1, −3, −13
Step 2: Substitute: 3(t² − 2) + 4t − 14 = 0 → 3t² + 4t − 20 = 0
Step 3: Split middle term: 3t² + 10t − 6t − 20 = t(3t + 10) − 2(3t + 10) = (3t + 10)(t − 2) = 0
Step 4: t = 2 or t = −103
Step 5: t = 2: x + 1x = 2 → x² − 2x + 1 = 0 → (x − 1)² = 0 → x = 1
Step 6: t = −103: 3x² + 10x + 3 = 0 → (3x + 1)(x + 3) = 0 → x = −13 or x = −3
Ans: x = 1, −3, −13
4. A cyclist travels 30 km at a certain speed. If the speed had been 5 km/h faster, the journey would have taken 1 hour less. Find the speed.
Step 1: Let original speed = v km/h. Then speed after increase = v + 5.
Step 2: Time = distance ÷ speed: 30v − 30v + 5 = 1
Step 3: Multiply by v(v + 5): 30(v + 5) − 30v = v(v + 5) → 150 = v² + 5v
Step 4: v² + 5v − 150 = 0 → (v + 15)(v − 10) = 0 → v = −15 or v = 10
Step 5: Speed cannot be negative, so v = 10.
Step 6: Check: 30 ÷ 10 = 3 h and 30 ÷ 15 = 2 h, difference 1 h ✓
Ans: Speed = 10 km/h
Step 2: Time = distance ÷ speed: 30v − 30v + 5 = 1
Step 3: Multiply by v(v + 5): 30(v + 5) − 30v = v(v + 5) → 150 = v² + 5v
Step 4: v² + 5v − 150 = 0 → (v + 15)(v − 10) = 0 → v = −15 or v = 10
Step 5: Speed cannot be negative, so v = 10.
Step 6: Check: 30 ÷ 10 = 3 h and 30 ÷ 15 = 2 h, difference 1 h ✓
Ans: Speed = 10 km/h
