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Class 10 Chapter 5 Full Unit Test 2027 | Mathematics Algebraic Fractions | Punjab Boards
Class 10 Math Chapter 5 full unit Test
Class 10 Chapter 5 Full Unit Test 2027 for Mathematics is designed according to the new Board Pattern 2027 for Punjab Boards. This Algebraic Fractions test provides objective and subjective questions for complete chapter preparation, including rational expressions, algebraic fractions, simplification, quadratic equations, exponential equations, reciprocal equations, and word problems. Students can use this Class 10 Mathematics Chapter 5 Test for board exam preparation, revision, practice, and self-assessment.
Full Unit Test
Unit 5: Algebraic Fractions
Board Paper Pattern
Objective Type
Time Allowed: 20 Min.
Max. Marks: 15
Note: Four possible answer A, B, C and D to each question are given.
The choice which you think is correct, fill that circle in front of that
question with Marker or ink pen in the answer book. Cutting filling two
or more circles will result in zero mark in that question.
Answer Key: (i) D (ii) D (iii) C (iv) C (v) A
(vi) B (vii) C (viii) C (ix) D (x) C
(xi) B (xii) D (xiii) D (xiv) B (xv) C
(i) Which of the following is a rational fraction?
Solution:
A rational expression is p(x)/q(x), where p(x) and q(x) are polynomials and q(x) ≠ 0.
(A) x = x/1: numerator x and denominator 1 are both polynomials → rational ✓
(B) 1/√x: the denominator √x = x1/2 is not a polynomial (power is not a whole number) → not rational ✗
(C) 1/x: numerator 1 and denominator x are polynomials → rational ✓
(A) and (C) are both rational ⟹ Answer: (D)
(ii) The rational expression
x + y
x2 + y2
in lowest form is:
Solution:
An expression is in lowest form when numerator and denominator have no common factor (other than 1).
Numerator: x + y — cannot be factorised further.
Denominator: x² + y² — a sum of squares, which cannot be factorised (only x² − y² = (x + y)(x − y) can).
So there is no common factor to cancel. Note: x² + y² ≠ (x + y)², so we cannot cancel (x + y).
The expression is already in lowest form ⟹ Answer: (D)
(iii) Simplifying
1
x
+
1
y
=
Solution:
Step 1: LCM of denominators x and y = xy
Step 2: Convert each fraction to denominator xy:
1·yx·y +
1·xy·x
Step 3: Add the numerators:
y + xxy
=
x + yxy
⟹ Answer: (C)
(iv)
3
x − y
−
1
y − x
=
Solution:
Step 1: Make the denominators the same. Since y − x = −(x − y):
1y − x =
1−(x − y) =
−1x − y
Step 2: Substitute:
3x − y −
( −1x − y ) =
3x − y +
1x − y
Step 3: Add the numerators (same denominator):
=
3 + 1x − y =
4x − y
⟹ Answer: (C)
(v) Reducing equation
x4 + 2x2 + 1 = 0
into a quadratic equation by letting
x2 = t to get:
Solution:
Step 1: Write x⁴ = (x²)². The equation becomes (x²)² + 2(x²) + 1 = 0.
Step 2: Put x² = t:
t² + 2t + 1 = 0
(Extra: (t + 1)² = 0 ⟹ t = −1 ⟹ x² = −1, which has no real solution.)
Answer: (A)
(vi) What we need to substitute to convert the equation
(x2 +
1
x2
)
− 3(x +
1
x
)
+ 4 = 0
into a quadratic equation.
Solution:
Step 1: Let t = x + 1/x.
Step 2: Square both sides: t² = x² + 2 + 1/x² ⟹ x² + 1/x² = t² − 2
Step 3: Substitute in the equation: (t² − 2) − 3t + 4 = 0
t² − 3t + 2 = 0, which is a quadratic equation in t. ✓
(If we took t = x² + 1/x² (option A), the other bracket (x + 1/x) would not be expressed in t, so it would not reduce.)
(Extra: (t − 1)(t − 2) = 0 ⟹ t = 1 or t = 2. For t = 2: x + 1/x = 2 ⟹ x = 1. For t = 1: x² − x + 1 = 0 has no real roots.)
Answer: (B) t = x + 1/x
(vii) To convert
(x + 2)(x − 5)(x + 3)(x − 6) = 12
we need to suppose:
Solution:
Step 1: Pair the factors so that the sum of the constants in each pair is the same: (+2, −5) gives −3 and (+3, −6) gives −3.
[(x + 2)(x − 5)] · [(x + 3)(x − 6)] = 12
Step 2: Multiply each pair:
(x + 2)(x − 5) = x² − 5x + 2x − 10 = x² − 3x − 10
(x + 3)(x − 6) = x² − 6x + 3x − 18 = x² − 3x − 18
Step 3: Both contain x² − 3x, so put x² − 3x = t:
(t − 10)(t − 18) = 12 ⟹ t² − 28t + 180 = 12 ⟹ t² − 28t + 168 = 0 (quadratic in t)
Answer: (C) x² − 3x = t
(viii) The equation
5x4 − 19x2 + 12 = 0
in reduced quadratic form is:
Solution:
Step 1: Write x⁴ = (x²)²: 5(x²)² − 19(x²) + 12 = 0
Step 2: Put x² = t:
5t² − 19t + 12 = 0
(Extra: t = (19 ± √(361 − 240))/10 = (19 ± 11)/10 ⟹ t = 3 or t = 4/5, so x = ±√3 or x = ±2/√5.)
Answer: (C)
(ix) To reduce
(x + 1)(x − 4)(x + 2)(x − 5) = −8
into quadratic equation, we need to suppose:
Solution:
Step 1: Pair the factors so that the sums of the constants are equal: (+1, −4) gives −3 and (+2, −5) gives −3.
[(x + 1)(x − 4)] · [(x + 2)(x − 5)] = −8
Step 2: Multiply each pair:
(x + 1)(x − 4) = x² − 3x − 4
(x + 2)(x − 5) = x² − 3x − 10
Step 3: Put x² − 3x = t:
(t − 4)(t − 10) = −8 ⟹ t² − 14t + 40 + 8 = 0 ⟹ t² − 14t + 48 = 0 (quadratic in t)
Answer: (D) t = x² − 3x
(x) Solution of the equation
x +
1
x
= 2
is x =
Solution:
x + 1/x = 2 (x ≠ 0)
Step 1: Multiply both sides by x: x² + 1 = 2x
Step 2: Bring all terms to one side: x² − 2x + 1 = 0
Step 3: Factorise: (x − 1)² = 0 ⟹ x = 1
Check: 1 + 1/1 = 2 ✓. (x = −1 gives −1 − 1 = −2 ✗, and x = 0 is not allowed.)
Answer: (C) x = 1
(xi) To convert
9x − 2(3x) + 1 = 0
into quadratic equation, we need to suppose:
Solution:
Step 1: Write 9x = (3²)x = (3x)².
(3x)² − 2(3x) + 1 = 0
Step 2: Put 3x = t:
t² − 2t + 1 = 0 (quadratic in t)
(Extra: (t − 1)² = 0 ⟹ t = 1 ⟹ 3x = 30 ⟹ x = 0.)
Answer: (B) t = 3x
(xii) Solution of the equation
22x − 3 × 2x + 2 = 0
is x =
Solution:
Step 1: Write 22x = (2x)². Put 2x = t:
t² − 3t + 2 = 0
Step 2: Factorise: t² − t − 2t + 2 = 0 ⟹ t(t − 1) − 2(t − 1) = 0 ⟹ (t − 1)(t − 2) = 0
t = 1 or t = 2
Step 3: Go back to x:
If t = 1: 2x = 1 = 20 ⟹ x = 0
If t = 2: 2x = 21 ⟹ x = 1
Check: x = 0 → 1 − 3 + 2 = 0 ✓ x = 1 → 4 − 6 + 2 = 0 ✓
x = 0 and x = 1 ⟹ Answer: (D) Both (B) and (C)
(xiii) To reduce the equation
x4 − 16x2 + 63 = 0
into a quadratic equation, we need to suppose that:
Solution:
Step 1: Write x⁴ = (x²)²: (x²)² − 16(x²) + 63 = 0
Step 2: Put x² = t:
t² − 16t + 63 = 0 (quadratic in t)
(Extra: (t − 7)(t − 9) = 0 ⟹ t = 7 or 9 ⟹ x = ±√7 or x = ±3.)
Answer: (D) t = x²
(xiv) 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:
Solution:
Job rate = part of the job done in 1 hour.
Together they complete the whole job (1 job) in 4 hours, so their combined rate is:
Combined rate =
1 job4 hours =
14
job per hour
Combined rate = rate of Arshia + rate of Ibrahim = x + y
x + y = 1/4 ⟹ Answer: (B)
(xv) Least common multiples of
x − y, x + y, x2 − y2
is:
Solution:
Step 1: Factorise each expression:
x − y = (x − y)
x + y = (x + y)
x² − y² = (x − y)(x + y)
Step 2: LCM = product of every distinct factor, taken the greatest number of times it appears in any one expression. The factors (x − y) and (x + y) each appear at most once.
LCM = (x − y)(x + y)
= x² − y² ⟹ Answer: (C)
Here is Subjective Part of Model Paper 2027.
Subjective Type
Time Allowed: 2:10 hours
(Part-I)
Max. Marks: 60
2. Write short answers to any six (06) questions:
(2 × 6 = 12)
(i) Reduce to lowest form:
6x2 − 8xy 9xy − 12y2
6x2 − 8xy 9xy − 12y2
Solution:
Step 1: Factorise the numerator (common factor 2x): 6x² − 8xy = 2x(3x − 4y)
Step 2: Factorise the denominator (common factor 3y): 9xy − 12y² = 3y(3x − 4y)
Step 3: Write the fraction: 2x(3x − 4y)3y(3x − 4y)
Step 4: Cancel the common factor (3x − 4y) (where 3x ≠ 4y).
Lowest form = 2x3y
(ii) Reduce the following rational expressions to lowest forms:
ac a2x2 − ax
ac a2x2 − ax
Solution:
Step 1: Factorise the denominator (common factor ax): a²x² − ax = ax(ax − 1)
Step 2: acax(ax − 1)
Step 3: Cancel the common factor a (a ≠ 0).
Lowest form = cx(ax − 1)
(iii) Reduce the following rational expressions to lowest forms:
x(2a2 − 3ax) a(4a2x − 9x3)
x(2a2 − 3ax) a(4a2x − 9x3)
Solution:
Step 1: Numerator: x(2a² − 3ax) = x · a(2a − 3x) = ax(2a − 3x)
Step 2: Denominator: a(4a²x − 9x³) = a · x(4a² − 9x²) = ax[(2a)² − (3x)²] = ax(2a − 3x)(2a + 3x)
Step 3: ax(2a − 3x)ax(2a − 3x)(2a + 3x)
Step 4: Cancel the common factor ax(2a − 3x).
Lowest form = 12a + 3x
(iv) Simplify:
x − 2y
4
+
x + y
6
−
2x − y
15
Solution:
Step 1: LCM of 4, 6 and 15 = 60
Step 2: Write each fraction with denominator 60 (60 ÷ 4 = 15, 60 ÷ 6 = 10, 60 ÷ 15 = 4):
15(x − 2y) + 10(x + y) − 4(2x − y)60
Step 3: Expand the numerator: 15x − 30y + 10x + 10y − 8x + 4y
Step 4: Collect like terms: (15 + 10 − 8)x + (−30 + 10 + 4)y = 17x − 16y
Answer = 17x − 16y60
(v) Simplify:
4a
a2 − 1
−
a + 1
a − 1
Solution:
Step 1: Factorise: a² − 1 = (a − 1)(a + 1). LCM = (a − 1)(a + 1)
Step 2: Make the second denominator the same (multiply top and bottom by a + 1):
4a(a − 1)(a + 1) − (a + 1)(a + 1)(a − 1)(a + 1) = 4a − (a + 1)²(a − 1)(a + 1)
Step 3: Expand the numerator: 4a − (a² + 2a + 1) = −a² + 2a − 1 = −(a² − 2a + 1) = −(a − 1)²
Step 4: −(a − 1)²(a − 1)(a + 1) — cancel one factor (a − 1):
Answer = −(a − 1)a + 1 = 1 − aa + 1
(vi) Simplify:
x + y
12x − 6y
+
x − y
18x − 9y
Solution:
Step 1: Factorise the denominators: 12x − 6y = 6(2x − y) and 18x − 9y = 9(2x − y)
Step 2: LCM = 18(2x − y) (LCM of 6 and 9 is 18)
Step 3: 3(x + y)18(2x − y) + 2(x − y)18(2x − y) = 3(x + y) + 2(x − y)18(2x − y)
Step 4: Expand the numerator: 3x + 3y + 2x − 2y = 5x + y
Answer = 5x + y18(2x − y)
(vii) Simplify:
3a2 − 6ab
6ab
×
10b2
4ab − 8b2
Solution:
Step 1: Factorise: 3a² − 6ab = 3a(a − 2b) and 4ab − 8b² = 4b(a − 2b)
Step 2: 3a(a − 2b)6ab × 10b²4b(a − 2b)
Step 3: Multiply numerators and denominators: 30ab²(a − 2b)24ab²(a − 2b)
Step 4: Cancel the common factor 6ab²(a − 2b):
Answer = 54
(viii) Simplify:
5a2
4a − 8
·
6a − 12
10a
Solution:
Step 1: Factorise: 4a − 8 = 4(a − 2) and 6a − 12 = 6(a − 2)
Step 2: 5a²4(a − 2) × 6(a − 2)10a = 30a²(a − 2)40a(a − 2)
Step 3: Cancel the common factor 10a(a − 2):
Answer = 3a4
(ix) Simplify:
24/m
5l + 10m
×
5l2 − 20m2
16mn
Solution (solved exactly as printed, with 24/m):
Step 1: Factorise: 5l + 10m = 5(l + 2m) and 5l² − 20m² = 5(l² − 4m²) = 5(l − 2m)(l + 2m)
Step 2: Remove the small fraction by multiplying top and bottom of the first fraction by m: 24/m5(l + 2m) = 245m(l + 2m)
Step 3: 245m(l + 2m) × 5(l − 2m)(l + 2m)16mn = 120(l − 2m)(l + 2m)80m²n(l + 2m)
Step 4: Cancel the common factor 40(l + 2m):
Answer = 3(l − 2m)2m²n
Note: if the question was meant to have 24m (not 24/m) in the numerator, the answer would be 3(l − 2m)2n.
3. Write short answers to any six (06) questions:
(2 × 6 = 12)
(i) Solve:
2x4 − 11x2 + 12 = 0
Solution:
Step 1: Write x⁴ = (x²)² and put x² = t: 2t² − 11t + 12 = 0
Step 2: Factorise (two numbers with product 2 × 12 = 24 and sum −11: −8 and −3):
2t² − 8t − 3t + 12 = 0 ⟹ 2t(t − 4) − 3(t − 4) = 0 ⟹ (t − 4)(2t − 3) = 0
t = 4 or t = 32
Step 3: Replace t by x²:
x² = 4 ⟹ x = ±2
x² = 32 ⟹ x = ±√(32) = ±√62
Check: x = 2 → 2(16) − 11(4) + 12 = 32 − 44 + 12 = 0 ✓
Solution set = { 2, −2, √6/2, −√6/2 }
(ii) Define Exponential Equation with example.
Answer:
An equation in which the variable occurs in the exponent (power) is called an exponential equation.
Example: 3x = 81 ⟹ 3x = 34 ⟹ x = 4
(iii) Define algebraic fraction.
Answer:
A fraction whose numerator and denominator are algebraic expressions (with denominator ≠ 0) is called an algebraic fraction.
Example: x + 1x − 2 (x ≠ 2), 3y (y ≠ 0)
(iv) Reduce the following rational expressions to lowest forms:
15a2b2c 100(a2 − a2b)
15a2b2c 100(a2 − a2b)
Solution:
Step 1: Take a² common in the denominator: 100(a² − a²b) = 100a²(1 − b)
Step 2: 15a²b²c100a²(1 − b)
Step 3: Cancel the common factor 5a² (15 ÷ 5 = 3 and 100 ÷ 5 = 20):
Lowest form = 3b²c20(1 − b)
(v) Define rational expression.
Answer:
An expression of the form p(x)q(x), where p(x) and q(x) are polynomials and q(x) ≠ 0, is called a rational expression.
Example: x² + 1x − 3 (x ≠ 3)
(vi) Solve the following equations:
4x4 − 16x2 + 15 = 0
Solution:
Step 1: Put x² = t: 4t² − 16t + 15 = 0
Step 2: Factorise (product 4 × 15 = 60, sum −16: −10 and −6):
4t² − 10t − 6t + 15 = 0 ⟹ 2t(2t − 5) − 3(2t − 5) = 0 ⟹ (2t − 5)(2t − 3) = 0
t = 52 or t = 32
Step 3: Replace t by x²:
x² = 52 ⟹ x = ±√(52) = ±√102
x² = 32 ⟹ x = ±√(32) = ±√62
Solution set = { ±√10/2, ±√6/2 }
(vii) Solve the following equations:
4 · 22x+1 − 9 · 2x + 1 = 0
Solution:
Step 1: 22x+1 = 22x · 2 = 2(2x)². The equation becomes 4 · 2(2x)² − 9 · 2x + 1 = 0
Step 2: Put 2x = t: 8t² − 9t + 1 = 0
Step 3: Factorise: 8t² − 8t − t + 1 = 0 ⟹ 8t(t − 1) − 1(t − 1) = 0 ⟹ (t − 1)(8t − 1) = 0
t = 1 or t = 18
Step 4: If 2x = 1 = 20 ⟹ x = 0
If 2x = 18 = 2−3 ⟹ x = −3
Check: x = 0 → 4(2) − 9 + 1 = 0 ✓ x = −3 → 4 · 2−5 − 9 · 2−3 + 1 = 18 − 98 + 1 = 0 ✓
x = 0, x = −3
(viii) Define Irrational expression with example.
Answer:
An expression in which the variable appears under a radical sign (root), i.e. with a fractional power, is called an irrational expression.
Example: √x + 5 or 1√x + 1
(ix) Solve the following equations:
3x + 33−x − 12 = 0
Solution:
Step 1: 33−x = 3³ · 3−x = 273x
Step 2: Put 3x = t: t + 27t − 12 = 0
Step 3: Multiply by t: t² + 27 − 12t = 0 ⟹ t² − 12t + 27 = 0
Step 4: Factorise: t² − 3t − 9t + 27 = 0 ⟹ t(t − 3) − 9(t − 3) = 0 ⟹ (t − 3)(t − 9) = 0
t = 3 or t = 9
Step 5: 3x = 31 ⟹ x = 1 ; 3x = 32 ⟹ x = 2
Check: x = 1 → 3 + 9 − 12 = 0 ✓ x = 2 → 9 + 3 − 12 = 0 ✓
x = 1, x = 2
4. Write short answers to any six (06) questions:
(2 × 6 = 12)
(i) A tradesman buys a number of articles for Rs. 300.
Four of them are broken in transit but by selling the remaining
at a profit of Rs. 1.25 each, he gains Rs.350 altogether.
How many articles did he buy?
Solution:
⚠ If the gain were literally Rs. 350, the equation below gives n ≈ 290.7, which is not a whole number. The standard version of this question gives n = 60 when the total sale is Rs. 350 (a gain of Rs. 50), so it is solved that way here. Please check the printed figure.
Step 1: Let he buy n articles. Cost of each = Rs. 300n. Remaining articles = n − 4.
Step 2: Selling price of each = 300n + 1.25 = 300n + 54
Step 3: Total sale = Rs. 350 (cost 300 + gain 50):
(n − 4)(300n + 54) = 350
Step 4: Multiply both sides by 4n: (n − 4)(1200 + 5n) = 1400n
1200n + 5n² − 4800 − 20n = 1400n ⟹ 5n² − 220n − 4800 = 0 ⟹ n² − 44n − 960 = 0
Step 5: Factorise (60 × 16 = 960, 60 − 16 = 44): (n − 60)(n + 16) = 0 ⟹ n = 60 (n = −16 is rejected)
Check: cost each = 300 ÷ 60 = Rs. 5; selling price each = 5 + 1.25 = Rs. 6.25; 56 × 6.25 = Rs. 350 ✓
He bought 60 articles.
(ii) Arshia and Ibraheem complete a job together in 4 hours.
If Arshia takes 6 hours, then find how much time will Ibraheem take?
Solution:
Step 1: Work rate = part of job done in 1 hour.
Arshia’s rate = 16; together rate = 14
Step 2: Ibraheem’s rate = 14 − 16 = 312 − 212 = 112
Step 3: Time for Ibraheem alone = 1 ÷ 112 = 12 hours
Ibraheem will take 12 hours.
(iii) Solve the following equations:
x4 − 16x2 + 63 = 0
Solution:
Step 1: Put x² = t: t² − 16t + 63 = 0
Step 2: Factorise (product 63, sum −16: −7 and −9): t² − 7t − 9t + 63 = 0 ⟹ (t − 7)(t − 9) = 0
t = 7 or t = 9
Step 3: x² = 7 ⟹ x = ±√7 ; x² = 9 ⟹ x = ±3
Check: x = 3 → 81 − 144 + 63 = 0 ✓
Solution set = { ±3, ±√7 }
(iv) Simplify:
x3 − 27
x2 − 9
÷
x2 + 3x + 9
x2 + 6x + 9
Solution:
Step 1: Factorise: x³ − 27 = (x − 3)(x² + 3x + 9); x² − 9 = (x − 3)(x + 3); x² + 6x + 9 = (x + 3)²
Step 2: Change ÷ into × by inverting the second fraction:
(x − 3)(x² + 3x + 9)(x − 3)(x + 3) × (x + 3)²x² + 3x + 9
Step 3: Cancel (x − 3), (x² + 3x + 9) and one factor (x + 3):
Answer = x + 3
(v) Reduce the following to the lowest form:
x2 − 7x + 12
x2 − 6x + 9
Solution:
Step 1: x² − 7x + 12 = x² − 3x − 4x + 12 = x(x − 3) − 4(x − 3) = (x − 3)(x − 4)
Step 2: x² − 6x + 9 = (x − 3)²
Step 3: (x − 3)(x − 4)(x − 3)(x − 3) — cancel (x − 3) (x ≠ 3):
Lowest form = x − 4x − 3
(vi) One pipe fills water in 5 hours, another in 8 hours.
Second pipe closed after 2 hours. Find the total time.
Solution (both pipes are opened together at the start):
Step 1: Rate of first pipe = 15 tank per hour; rate of second pipe = 18 tank per hour
Step 2: Work done by both in 2 hours = 2(15 + 18) = 2 × 1340 = 1320 of the tank
Step 3: Remaining = 1 − 1320 = 720
Step 4: Time for the first pipe alone to fill the rest = 720 ÷ 15 = 720 × 5 = 74 hours
Step 5: Total time = 2 + 74 = 154 hours
Total time = 3¾ hours = 3 hours 45 minutes.
(vii) A tap can fill a tank in 6 hours. Another tap can fill it
in 9 hours. If both taps are opened together, how long will it
take to fill the empty tank?
Solution:
Step 1: Rates: 16 and 19 tank per hour
Step 2: Together = 16 + 19 = 318 + 218 = 518 tank per hour
Step 3: Time = 1 ÷ 518 = 185 hours
Time = 3.6 hours = 3 hours 36 minutes.
(viii) Define Reciprocal Equation with example.
Answer:
An equation which remains unchanged when the variable x is replaced by its reciprocal 1x is called a reciprocal equation.
Example: 2x² − 5x + 2 = 0. Replace x by 1x: 2x² − 5x + 2 = 0. Multiply by x²: 2 − 5x + 2x² = 0, which is the same equation. ✓
(ix) Ahmad takes 2 hours to paint 50 glasses. Faiza takes 2 hours
to paint 45 glasses. Working together, how long should it take
them to paint 150 glasses?
Solution:
Step 1: Ahmad’s rate = 50 ÷ 2 = 25 glasses per hour
Step 2: Faiza’s rate = 45 ÷ 2 = 22.5 glasses per hour
Step 3: Together = 25 + 22.5 = 47.5 glasses per hour
Step 4: Time = 15047.5 = 30095 = 6019 hours
Time = 6019 hours ≈ 3.16 hours ≈ 3 hours 9½ minutes.
(Part-II) Note: Attempt any two (02) questions.
(2 × 8 = 16)
5:(a) Simplify:
1
x2 − 5x + 6
+
1
x2 − 3x + 2
−
1
x2 − 8x + 15
Solution:
Step 1: Factorise each denominator:
x² − 5x + 6 = (x − 2)(x − 3)
x² − 3x + 2 = (x − 1)(x − 2)
x² − 8x + 15 = (x − 3)(x − 5)
Step 2: LCM = (x − 1)(x − 2)(x − 3)(x − 5)
Step 3: Multiply each numerator by the missing factors:
(x − 1)(x − 5) + (x − 3)(x − 5) − (x − 1)(x − 2)(x − 1)(x − 2)(x − 3)(x − 5)
Step 4: Expand the numerator:
(x − 1)(x − 5) = x² − 6x + 5
(x − 3)(x − 5) = x² − 8x + 15
(x − 1)(x − 2) = x² − 3x + 2
Numerator = (x² − 6x + 5) + (x² − 8x + 15) − (x² − 3x + 2) = x² − 11x + 18
Step 5: Factorise: x² − 11x + 18 = (x − 2)(x − 9)
Step 6: (x − 2)(x − 9)(x − 1)(x − 2)(x − 3)(x − 5) — cancel (x − 2):
Answer = x − 9(x − 1)(x − 3)(x − 5)
Check at x = 0: left side = 16 + 12 − 115 = 35; answer = −9−15 = 35 ✓
(b) Simplify:
x + 2
x2 − x − 12
−
x
x2 + 6x + 9
Solution:
Step 1: Factorise the denominators:
x² − x − 12 = x² − 4x + 3x − 12 = (x − 4)(x + 3)
x² + 6x + 9 = (x + 3)²
Step 2: LCM = (x − 4)(x + 3)²
Step 3: (x + 2)(x + 3) − x(x − 4)(x − 4)(x + 3)²
Step 4: Expand the numerator: (x² + 5x + 6) − (x² − 4x) = 9x + 6
Step 5: 9x + 6 = 3(3x + 2), which has no common factor with the denominator.
Answer = 9x + 6(x − 4)(x + 3)² = 3(3x + 2)(x − 4)(x + 3)²
Check at x = 0: left side = 2−12 − 0 = −16; answer = 6(−4)(9) = −16 ✓
6:(a) Simplify:
x3 − a3
x2 − 2bx + b2
×
x2 − bx − cx + bc
ax − x2
÷
x2 − cx
x − b
Solution:
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) = −x(x − a)
x² − cx = x(x − c)
Step 2: Substitute, and change ÷ into × by inverting the last fraction:
(x − a)(x² + ax + a²)(x − b)² × (x − b)(x − c)−x(x − a) × x − bx(x − c)
Step 3: Multiply everything into one fraction:
(x − a)(x² + ax + a²)(x − b)(x − c)(x − b)(x − b)² · (−x)(x − a) · x(x − c)
Step 4: Cancel (x − a), (x − b)² and (x − c). What is left:
x² + ax + a²−x · x
Answer = −x² + ax + a²x²
Check with a = 1, b = 2, c = 3, x = 4: 634 × (−16) ÷ 2 = −2116; answer = −2116 ✓
(b) Solve the equation:
(
x2 +
1
x2
)
+ 2
(
x +
1
x
)
− 33 = 0
Solution:
Step 1: Let t = x + 1x. Square both sides: t² = x² + 2 + 1x² ⟹ x² + 1x² = t² − 2
Step 2: Substitute: (t² − 2) + 2t − 33 = 0 ⟹ t² + 2t − 35 = 0
Step 3: Factorise (product −35, sum 2: 7 and −5): t² + 7t − 5t − 35 = 0 ⟹ (t + 7)(t − 5) = 0
t = −7 or t = 5
Step 4: Case t = 5: x + 1x = 5 ⟹ x² − 5x + 1 = 0
x = 5 ± √(25 − 4)2 = 5 ± √212
Case t = −7: x + 1x = −7 ⟹ x² + 7x + 1 = 0
x = −7 ± √(49 − 4)2 = −7 ± √452 = −7 ± 3√52
Solution set = { (5 ± √21)/2 , (−7 ± 3√5)/2 }
7:(a)
A train is scheduled to cover a distance of 120 km at a certain
average speed, owing to a service checkup this average speed is
reduced by 5 km/h, and in consequence the journey takes 20 minutes
more than the scheduled time. What was the scheduled speed?
Solution:
Step 1: Let the scheduled speed = v km/h. Reduced speed = (v − 5) km/h.
Scheduled time = 120v h; new time = 120v − 5 h; extra time = 20 min = 13 h
Step 2: Form the equation: 120v − 5 − 120v = 13
Step 3: Take LCM v(v − 5): 120v − 120(v − 5)v(v − 5) = 13 ⟹ 600v(v − 5) = 13
Step 4: Cross-multiply: v(v − 5) = 1800 ⟹ v² − 5v − 1800 = 0
Step 5: Factorise (45 × 40 = 1800, 45 − 40 = 5): (v − 45)(v + 40) = 0 ⟹ v = 45 or v = −40 (rejected, speed cannot be negative)
Check: 120 ÷ 45 = 2 h 40 min; 120 ÷ 40 = 3 h; difference = 20 min ✓
Scheduled speed = 45 km/h.
(b)
Fahad runs 600m at a certain pace, and then doubling his pace,
does another 600m. If he took
5
2
to cover the distance 1200m, find the pace he started at,
in meters per seconds.
Solution (the unit after 5/2 is missing in the paper; it must be minutes):
Step 1: Let the starting pace = p m/s. Doubled pace = 2p m/s.
Step 2: Total time = 52 minutes = 52 × 60 = 150 seconds
Step 3: Time for first 600 m = 600p s; time for next 600 m = 6002p = 300p s
Step 4: 600p + 300p = 150 ⟹ 900p = 150
Step 5: p = 900150 = 6
Check: 600 ÷ 6 = 100 s; 600 ÷ 12 = 50 s; total 150 s = 2.5 min ✓
He started at 6 meters per second.
(Part-III) Note: Attempt any one (01) question.
(1 × 8 = 8)
8:(a) Simplify:
a2 − (b − c)2
(a + b)2 − c2
×
a2 − (b + c)2
(a − b)2 − c2
Solution:
Use the identity p² − q² = (p − q)(p + q) on each part:
a² − (b − c)² = (a − b + c)(a + b − c)
(a + b)² − c² = (a + b − c)(a + b + c)
a² − (b + c)² = (a − b − c)(a + b + c)
(a − b)² − c² = (a − b − c)(a − b + c)
Substitute:
(a − b + c)(a + b − c)(a + b − c)(a + b + c) × (a − b − c)(a + b + c)(a − b − c)(a − b + c)
Cancel (a + b − c), (a + b + c), (a − b − c) and (a − b + c) — every factor cancels.
Answer = 1
Check with a = 3, b = 1, c = 1: 915 × 53 = 4545 = 1 ✓
(b) Solve the equation:
4x4 − 16x2 + 15 = 0
Solution:
Step 1: Put x² = t: 4t² − 16t + 15 = 0
Step 2: Factorise (product 60, sum −16: −10 and −6):
4t² − 10t − 6t + 15 = 0 ⟹ 2t(2t − 5) − 3(2t − 5) = 0 ⟹ (2t − 5)(2t − 3) = 0
t = 52 or t = 32
Step 3: x² = 52 ⟹ x = ±√102 ; x² = 32 ⟹ x = ±√62
Solution set = { ±√10/2, ±√6/2 }
9:(a)
Solve the equation:
(x + 9)(x − 3)(x − 7)(x + 5) = 385
Solution:
Step 1: Pair the factors so that the constants add up to the same number: 9 + (−7) = 2 and (−3) + 5 = 2.
[(x + 9)(x − 7)] · [(x − 3)(x + 5)] = 385
Step 2: Multiply each pair:
(x + 9)(x − 7) = x² + 2x − 63
(x − 3)(x + 5) = x² + 2x − 15
Step 3: Put x² + 2x = t: (t − 63)(t − 15) = 385
t² − 78t + 945 = 385 ⟹ t² − 78t + 560 = 0
Step 4: Factorise (8 × 70 = 560, 8 + 70 = 78): (t − 8)(t − 70) = 0 ⟹ t = 8 or t = 70
Step 5: Case t = 8: x² + 2x − 8 = 0 ⟹ (x + 4)(x − 2) = 0 ⟹ x = −4 or x = 2
Case t = 70: x² + 2x − 70 = 0 ⟹ x = −2 ± √(4 + 280)2 = −2 ± √2842 = −2 ± 2√712 = −1 ± √71
Check: x = 2 → (11)(−1)(−5)(7) = 385 ✓ x = −4 → (5)(−7)(−11)(1) = 385 ✓
Solution set = { 2, −4, −1 + √71, −1 − √71 }
(b)
Solve the equation:
(
x2 +
1
x2
)
+
(
x +
1
x
)
= 4
Solution:
Step 1: Let t = x + 1x. Then x² + 1x² = t² − 2
Step 2: Substitute: (t² − 2) + t = 4 ⟹ t² + t − 6 = 0
Step 3: Factorise: t² + 3t − 2t − 6 = 0 ⟹ (t + 3)(t − 2) = 0 ⟹ t = −3 or t = 2
Step 4: Case t = 2: x + 1x = 2 ⟹ x² − 2x + 1 = 0 ⟹ (x − 1)² = 0 ⟹ x = 1
Case t = −3: x + 1x = −3 ⟹ x² + 3x + 1 = 0 ⟹ x = −3 ± √(9 − 4)2 = −3 ± √52
Check: x = 1 → (1 + 1) + (1 + 1) = 4 ✓
Solution set = { 1, (−3 + √5)/2, (−3 − √5)/2 }
