Class 10 Maths Test Paper Chapter 1, 2 AND 3 | GSEB Board | Detailed Solutions & Marking Scheme
VEDANT CLASSES – IGNITE TEST SERIES
Class: 10
Subject: Mathematics
Board: Gujarat Secondary and Higher Secondary Education Board (GSHSEB)
Chapters: Real Numbers, Polynomials, Pair of Linear Equations in Two Variables
Time: 60 Minutes
Maximum Marks: 30
GENERAL INSTRUCTIONS
All questions are based on NCERT textbook exercises and examples.
All questions are numerical-based only.
Use the Euclid Division Algorithm wherever applicable.
Show all necessary calculations.
Internal choice is provided as per sections.
Answer only the required number of questions.
SECTION A – OBJECTIVE QUESTIONS
(7 Marks)
Q.1 Choose the correct option. (3 × 1 = 3)
(i) The HCF of 96 and 404 is:
(A) 2
(B) 4
(C) 8
(D) 12
(ii) If the zeroes of the polynomial x² − 5x + 6 are α and β, then α + β is:
(A) 5
(B) 6
(C) −5
(D) −6
(iii) The pair of linear equations 2x + 3y = 12 and 4x + 6y = 24 has:
(A) One solution
(B) No solution
(C) Infinitely many solutions
(D) Exactly two solutions
Q.2 Fill in the blanks using the correct option given in brackets. (2 × 1 = 2)
(i) The HCF of 18 and 30 is ________. (3, 6, 9)
(ii) If one zero of the polynomial x² − 9 is 3, then the other zero is ________. (−3, 3, 9)
Q.3 State whether True or False. (2 × 1 = 2)
(i) The HCF of two prime numbers is always 1. ________
(ii) The pair of equations x + y = 5 and 2x + 2y = 10 has a unique solution. ________
SECTION B – ANSWER ANY 3 QUESTIONS
(3 × 2 = 6)
Q.4 Real Numbers
(a) Find the HCF of 135 and 225 using the Euclid Division Algorithm.
(b) Verify:
HCF × LCM = Product of the two numbers.
Q.5 Irrational Numbers
(a) Prove that √3 is irrational.
(b) Prove that 2√5 is irrational.
Q.6 Polynomials
(a) Find p(2) for p(x) = x² − 4x + 7.
(b) Verify whether 2 is a zero of the polynomial.
Q.7 Zeroes of Polynomials
For the polynomial x² − 7x + 12:
(a) Find its zeroes by factorisation.
(b) Verify the relation between zeroes and coefficients.
Q.8 Pair of Linear Equations
Solve the following pair of equations by substitution method:
x + y = 7
x − y = 3
SECTION C – ANSWER ANY 3 QUESTIONS
(3 × 3 = 9)
Q.9 Real Numbers
(a) Find the HCF of 306 and 657 using the Euclid Division Algorithm.
(b) Find the LCM using:
HCF × LCM = Product of the numbers.
Q.10 Irrational Numbers
(a) Prove that √5 is irrational.
(b) Hence, prove that 2 + √5 is irrational.
Q.11 Polynomials
(a) Factorise x² − 11x + 30.
(b) Find its zeroes.
(c) Verify:
Sum of zeroes = −b/a
Product of zeroes = c/a
Q.12 Pair of Linear Equations – Problem Sum
The sum of two numbers is 27 and their difference is 9.
(a) Form the pair of linear equations.
(b) Find the two numbers.
(c) Verify your answer.
Q.13 Pair of Linear Equations – Problem Sum
A notebook costs ₹20 and a pen costs ₹10. The total cost of 4 notebooks and 5 pens is ₹130.
(a) Form a pair of linear equations.
(b) Determine whether the given information is sufficient to find the cost of one notebook and one pen.
(c) If the cost of one notebook is ₹20, find the cost of one pen.
SECTION D – ANSWER ANY 2 QUESTIONS
(2 × 4 = 8)
Q.14 Real Numbers
(a) Find the HCF of 455 and 42 using the Euclid Division Algorithm.
(b) Find the LCM of the numbers.
(c) Verify:
HCF × LCM = Product of the numbers.
Q.15 Polynomials
For the polynomial 2x² − 9x + 10:
(a) Find the zeroes by factorisation.
(b) Find the sum and product of the zeroes.
(c) Verify the relation between zeroes and coefficients.
Q.16 Pair of Linear Equations – Problem Sum
The sum of the digits of a two-digit number is 9. If 27 is added to the number, its digits get reversed.
(a) Form the pair of linear equations.
(b) Find the original number.
(c) Verify your answer.
(d) State the digits of the new number obtained after adding 27.
------------------- END OF QUESTION PAPER -------------------
ANSWER KEY WITH MARKING SCHEME
SECTION A (7 Marks)
Q.1
(i) HCF of 96 and 404:
404 = 96 × 4 + 20
96 = 20 × 4 + 16
20 = 16 × 1 + 4
16 = 4 × 4 + 0
HCF = 4
Answer: (B) 4 [1 Mark]
(ii) For x² − 5x + 6:
α + β = −(−5)/1 = 5
Answer: (A) 5 [1 Mark]
(iii)
2x + 3y = 12
4x + 6y = 24
a₁/a₂ = 2/4 = 1/2
b₁/b₂ = 3/6 = 1/2
c₁/c₂ = 12/24 = 1/2
Therefore, infinitely many solutions.
Answer: (C) Infinitely many solutions [1 Mark]
Q.2
(i) HCF(18, 30) = 6 [1 Mark]
(ii) x² − 9 = (x − 3)(x + 3)
Other zero = −3 [1 Mark]
Q.3
(i) True [1 Mark]
(ii) False [1 Mark]
SECTION B (Answer Any 3)
(3 × 2 = 6)
Q.4
135 = 225 × 0 + 135
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
HCF = 45 [1 Mark]
LCM = (135 × 225)/45
= 675 [0.5 Mark]
Verification:
HCF × LCM = 45 × 675 = 30375
135 × 225 = 30375
Verified. [0.5 Mark]
Q.5
(a) Assume √3 is rational.
√3 = p/q, where p and q are coprime integers.
Squaring:
3q² = p²
Thus p² is divisible by 3, so p is divisible by 3.
Let p = 3k.
Substituting:
3q² = 9k²
q² = 3k²
Thus q is also divisible by 3.
This contradicts that p and q are coprime.
Therefore, √3 is irrational. [1 Mark]
(b) Assume 2√5 is rational.
Then:
2√5 = p/q
√5 = p/(2q)
This implies √5 is rational, which is impossible.
Therefore, 2√5 is irrational. [1 Mark]
Q.6
p(x) = x² − 4x + 7
p(2) = 2² − 4(2) + 7
= 4 − 8 + 7
= 3 [1 Mark]
Since p(2) ≠ 0,
2 is not a zero of the polynomial. [1 Mark]
Q.7
x² − 7x + 12
= x² − 3x − 4x + 12
= x(x − 3) − 4(x − 3)
= (x − 3)(x − 4)
Zeroes are 3 and 4. [1 Mark]
α + β = 3 + 4 = 7 = −(−7)/1
αβ = 3 × 4 = 12 = 12/1
Verified. [1 Mark]
Q.8
x + y = 7
x − y = 3
From first equation:
x = 7 − y
Substitute in second equation:
7 − y − y = 3
7 − 2y = 3
y = 2
x = 5
Solution: (5, 2) [2 Marks]
SECTION C (Answer Any 3)
(3 × 3 = 9)
Q.9
657 = 306 × 2 + 45
306 = 45 × 6 + 36
45 = 36 × 1 + 9
36 = 9 × 4 + 0
HCF = 9 [2 Marks]
LCM = (306 × 657)/9
= 22338 [1 Mark]
Q.10
(a) Proof of irrationality of √5 follows the contradiction method similar to √3.
Hence, √5 is irrational. [2 Marks]
(b) Assume 2 + √5 is rational.
Then:
√5 = (2 + √5) − 2
Difference of two rational numbers is rational.
This implies √5 is rational, which is a contradiction.
Therefore, 2 + √5 is irrational. [1 Mark]
Q.11
(a)
x² − 11x + 30
= x² − 5x − 6x + 30
= x(x − 5) − 6(x − 5)
= (x − 5)(x − 6) [1 Mark]
(b)
Zeroes are 5 and 6. [1 Mark]
(c)
α + β = 5 + 6 = 11
−b/a = −(−11)/1 = 11
αβ = 5 × 6 = 30
c/a = 30/1 = 30
Verified. [1 Mark]
Q.12
(a)
Let the numbers be x and y.
x + y = 27
x − y = 9 [1 Mark]
(b)
Adding:
2x = 36
x = 18
y = 9 [1.5 Marks]
(c)
18 + 9 = 27
18 − 9 = 9
Verified. [0.5 Mark]
Q.13
(a)
Let notebook cost = x
Pen cost = y
4x + 5y = 130 [1 Mark]
(b)
Only one equation with two variables is insufficient.
Hence, information is insufficient. [1 Mark]
(c)
Substitute x = 20:
4(20) + 5y = 130
80 + 5y = 130
5y = 50
y = 10
Pen cost = ₹10 [1 Mark]
SECTION D (Answer Any 2)
(2 × 4 = 8)
Q.14
455 = 42 × 10 + 35
42 = 35 × 1 + 7
35 = 7 × 5 + 0
HCF = 7 [2 Marks]
LCM = (455 × 42)/7
= 2730 [1 Mark]
Verification:
7 × 2730 = 19110
455 × 42 = 19110
Verified. [1 Mark]
Q.15
2x² − 9x + 10
= 2x² − 5x − 4x + 10
= x(2x − 5) − 2(2x − 5)
= (x − 2)(2x − 5)
Zeroes are 2 and 5/2. [2 Marks]
Sum of zeroes:
2 + 5/2 = 9/2
Product:
2 × 5/2 = 5 [1 Mark]
−b/a = −(−9)/2 = 9/2
c/a = 10/2 = 5
Verified. [1 Mark]
Q.16
Let the tens digit be x and units digit be y.
(a)
x + y = 9 [1 Mark]
Original number = 10x + y
After adding 27:
10x + y + 27 = 10y + x
9x − 9y = −27
x − y = −3 [1 Mark]
(b)
Solving:
x + y = 9
x − y = −3
Adding:
2x = 6
x = 3
y = 6
Original number = 36 [1 Mark]
(c)
36 + 27 = 63
Digits are reversed.
Verified. [0.5 Mark]
(d)
New number = 63
Digits are 6 and 3. [0.5 Mark]
------------------- TOTAL MARKS: 30 -------------------
Comments
Post a Comment