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Class 10 Mathematics Chapter 5 Test 1 – Algebraic Fractions | New Paper Pattern 2027 | Punjab Boards
Class 10 Math Chapter 5 Test 1 Board Paper Pattern 2027
Class 10 Mathematics Chapter 5 Test 1 – Algebraic Fractions is designed according to the new paper pattern 2027 for Punjab Boards. This chapter-wise test provides practice for Class 10 Mathematics Algebraic Fractions, including objective and subjective questions based on the expected Punjab Board examination format. Students can use this Chapter 5 Test 1 for revision, self-assessment, and preparation for the Class 10 Mathematics Board Exam 2027.
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Unit 05:
Total Time: 35 min
CHAPTER WISE CLASS TEST
(Test # 1)
Mathematics
Total Marks: 25
1
A
B
C
D
2
A
B
C
D
3
A
B
C
D
4
A
B
C
D
5
A
B
C
D
6
A
B
C
D
7
A
B
C
D
8
A
B
C
D
NOTE:
Four possible answers 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 Pen ink. Cutting or filling two or more circles
will result in zero mark in that question.
(8)
Q.1:
1.
What type of expression is:
2x − 1
x² + 4
is:
Solution:
A polynomial has no variable in a denominator, so (A) is wrong.
An equation needs an “=” sign, so (D) is wrong.
A numerical fraction has only numbers in the numerator and denominator, but here x is present, so (C) is wrong.
The given expression is a quotient of two polynomials, 2x − 1 and x² + 4, where the denominator is never zero (x² + 4 ≥ 4).
Answer: (B) an algebraic fraction
2.
Simplified form of
x² − y²
x + y
(x − y) is:
Solution:
Step 1: Factorise the numerator using x² − y² = (x + y)(x − y).
(x + y)(x − y)x + y
(x − y)
Step 2: Cancel the common factor (x + y):
= (x − y)(x − y)
= (x − y)² ⟹ Answer: (D)
3.
Roots of y² − 24y + 128 = 0 are:
Solution:
y² − 24y + 128 = 0
Find two numbers whose product is +128 and whose sum is −24: they are −8 and −16.
y² − 8y − 16y + 128 = 0
y(y − 8) − 16(y − 8) = 0
(y − 8)(y − 16) = 0
y − 8 = 0 or y − 16 = 0 ⟹ y = 8 or y = 16
Check: 8² − 24(8) + 128 = 64 − 192 + 128 = 0 ✓ 16² − 24(16) + 128 = 256 − 384 + 128 = 0 ✓
Roots: 8, 16 ⟹ Answer: (B)
4.
Simplify
5x + 10
x + 2
:
Solution:
Step 1: Take 5 common from the numerator: 5x + 10 = 5(x + 2)
5(x + 2)x + 2
Step 2: Cancel the common factor (x + 2) (where x ≠ −2).
= 5 ⟹ Answer: (A)
5.
What is the LCM for
1
x
+
1
x + 2
?
Solution:
The LCM of the fractions means the LCM of their denominators: x and (x + 2).
x and (x + 2) have no common factor, so their LCM is their product.
LCM = x · (x + 2) = x(x + 2) = x² + 2x
Answer: (A) x(x + 2)
6.
An equation which remains unchanged when variable is replaced by its
reciprocal is called:
Solution:
By definition, an equation that remains unchanged when x is replaced by 1/x is called a reciprocal equation.
(Exponential equation: variable in the exponent; radical equation: variable under a root sign.)
Answer: (A)
7.
To reduce ax2n + bxn + c = 0 into quadratic
equation we put
Solution:
Write x2n = (xn)². The equation becomes:
a(xn)² + b(xn) + c = 0
Put xn = y:
ay² + by + c = 0, which is a quadratic equation in y.
Answer: (B) xn = y
8.
If 3x = 1 then x equals
Solution:
Write 1 as a power of 3: 1 = 30
3x = 30
The bases are equal, so the exponents must be equal: x = 0
Check: 30 = 1 ✓
Answer: (C) 0
(PART – I)
Q.2.
Write short answers to following questions.
(6×2=12)
(i) Define a 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:
x + 1x² − 4
(x ≠ ±2).
(ii)
Is
19
x
an algebraic fraction.
Answer: Yes.
An algebraic fraction is a fraction whose numerator and denominator are algebraic expressions, with denominator ≠ 0.
Here the numerator 19 is a constant (a polynomial of degree 0) and the denominator x is a polynomial, with x ≠ 0.
So 19/x is an algebraic fraction (x ≠ 0).
(iii)
Reduce to the lowest form
4x + 8
4
Solution:
Step 1: Take 4 common from the numerator: 4x + 8 = 4(x + 2)
4(x + 2)4
Step 2: Cancel the common factor 4.
= x + 2
(iv)
Simplify
x
x − y
+
2x
x + y
Solution:
Step 1: LCM of denominators = (x − y)(x + y) = x² − y²
Step 2: Make the denominators equal:
x(x + y) + 2x(x − y)(x − y)(x + y)
Step 3: Expand the numerator: x² + xy + 2x² − 2xy = 3x² − xy
Step 4: Take x common: 3x² − xy = x(3x − y)
=
x(3x − y)x² − y²
(v)
Simplify
1
6x
+
1
4x²
Solution:
Step 1: LCM of 6x and 4x² = 12x²
Step 2: Divide LCM by each denominator and multiply the numerator:
12x² ÷ 6x = 2x and 12x² ÷ 4x² = 3
2x12x² +
312x²
Step 3: Add the numerators over the common denominator:
=
2x + 312x²
(vi) Define exponential equation:
Answer:
An equation in which the variable occurs in the exponent (power) is called an exponential equation.
Examples: 3x = 81, 2x+1 = 16.
(PART – II)
Note: Attempt the following question.
(5)
Q.3:
Reduce to the lowest form:
6x² − 8xy
9xy − 12y²
Solution:
6x² − 8xy9xy − 12y²
Step 1: Factorise the numerator. The common factor of 6x² and 8xy is 2x:
6x² − 8xy = 2x(3x − 4y)
Step 2: Factorise the denominator. The common factor of 9xy and 12y² is 3y:
9xy − 12y² = 3y(3x − 4y)
Step 3: Write the fraction with the factors:
2x(3x − 4y)3y(3x − 4y)
Step 4: Cancel the common factor (3x − 4y) (where 3x ≠ 4y and y ≠ 0):
=
2x3y
(lowest form)
