Class 10 Maths Test Paper Chapter 1, 2 AND 3 | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026-27
CLASS: X (CBSE) | TIME: 1 Hour | MAX. MARKS: 30 SUBJECT: MATHEMATICS (Ch 1: Real Numbers, Ch 2: Polynomials, Ch 3: Pair of Linear Equations)
GENERAL INSTRUCTIONS:
Read all questions carefully before answering. — Misreading the question and then blaming the paper will not increase your marks.
Show proper steps wherever required. — “Sir, answer toh yahi aana tha” is not an accepted mathematical method.
Write neatly and clearly. — If your handwriting requires a decoder machine, checking may become an adventure.
Manage your time wisely. — Spending 45 minutes on one question and calling the rest “optional” is not a strategy.
If you do not know the answer, you may cry silently. — Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.
SECTION A (Objective Type Questions)
[7 × 1 = 7 Marks]
Q1. If the LCM of two positive integers 26 and k is 338, and their HCF is 13, then the exact numerical value of k is: A. 13 B. 26 C. 169 D. 338
Q2. The graphical trajectory of a quadratic polynomial y = ax² + bx + c is shown to be a parabola opening entirely upwards, which touches the horizontal x-axis at exactly one single distinct point. Which structural deduction must be true? A. a > 0 and b² - 4ac > 0 B. a < 0 and b² - 4ac = 0 C. a > 0 and b² - 4ac = 0 D. a < 0 and b² - 4ac < 0
Q3. For which real value of the parameter 'p' does the pair of linear equations 2x + py = 8 and px + 8y = p² possess an infinite number of matching solutions? A. p = 4 B. p = -4 C. p = 2 D. p = 0
Q4. If α and β are the real zeros of the quadratic polynomial f(x) = x² - p(x + 1) - c, then the simplified value of the mathematical term (α + 1)(β + 1) is equal to: A. c - 1 B. 1 - c C. p - c D. p + c + 1
Q5. A fraction becomes equal to 1/3 when 1 is subtracted from its numerator, and it becomes equal to 1/4 when 8 is added to its denominator. The sum of the original numerator and denominator of this fraction is: A. 12 B. 17 C. 7 D. 23
For Question 6 and Question 7, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option from the following:
A. Both A and R are true, R is correct explanation of A
B. Both A and R are true, R is not correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q6. Assertion (A): The number (7 + 3√2) is an irrational number. Reason (R): The sum or difference of a rational number and an irrational number is always guaranteed to be an irrational number.
Q7. Assertion (A): The pair of linear mathematical equations 3x - 4y = 7 and 6x - 8y = 15 represents a system of parallel line tracks with zero solutions. Reason (R): A pair of simultaneous linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 is consistent if the coefficient ratios satisfy the structural condition (a1 / a2) = (b1 / b2) ≠ (c1 / c2).
SECTION B (Very Short Answer Questions)
[4 × 2 = 8 Marks]
Q8. Prove using prime factor structures that the number 6ⁿ can never end with the digit 0 for any natural number power n belonging to the set of natural numbers. [2]
Q9. If α and β are the zeros of the quadratic polynomial g(x) = 2x² + 5x + k, calculate the value of the parameter k such that the given relation holds perfectly true: α² + β² + αβ = 21/4. [2]
Q10. Determine the values of 'm' and 'n' for which the following system of linear equations has infinitely many solutions: [2]
2x + 3y = 7 and (m + n)x + (2m - n)y = 21
Q11. Solve the given pair of simultaneous linear equations using the elimination method: [2]
x/2 + 2y/3 = -1 and x - y/3 = 3
SECTION C (Short Answer Questions)
[2 × 3 = 6 Marks]
Q12. Prove that √3 is an irrational number using algebraic proof by contradiction. [3]
Q13. If α and β are the zeros of the quadratic polynomial f(x) = 3x² - 4x + 1, construct a brand new quadratic polynomial whose zeros are given by the expressions (α² / β) and (β² / α). [3]
SECTION D (Long Answer Question)
[1 × 5 = 5 Marks]
Q14. Solve the following pair of linear equations graphically:
2x + y = 6
2x - y + 2 = 0
Find the exact coordinates of the points where these plotted lines intersect the vertical y-axis. Hence, calculate the area of the triangular region enclosed between these two lines and the vertical y-axis line. [5]
SECTION E (Case Study Based Question)
[1 × 4 = 4 Marks]
Q15. Read the following real-world scenario carefully and answer the structural questions that follow:
Logistics Optimization and Fleet Operations Management: A regional transport logistics agency based out of Rajkot is optimizing its delivery operations. The agency manages a mixed delivery fleet containing small trucks and large carrier containers to transport industrial parts. On Monday, a shipment consisting of 3 small trucks and 2 large containers moved a total cargo load of 22 tonnes. On Tuesday, under matching load configurations, a fleet consisting of 4 small trucks and 3 large containers moved a total cargo load of 31 tonnes. The baseline operational cost can be broken down into a fixed maintenance tariff per truck and a variable transit premium dependent on the capacity parameters. Engineers model these daily workflows using simultaneous systems of linear functions to track resource efficiency metrics and maintain cost-effective consistency across transport links.
(a) Formulate a pair of simultaneous linear equations in two variables 'x' and 'y' to represent the operational delivery configurations outlined above. [1]
(b) Check whether the formulated system of equations is consistent or inconsistent by comparing their respective tracking coefficient metrics. [1]
(c) Calculate the individual cargo carrying capacities (in tonnes) of a single small truck and a single large container using an algebraic method of your choice. Show step-by-step verification. [2]
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026–27
CLASS: X (CBSE) | SUBJECT: MATHEMATICS
Chapter 1: Real Numbers, Chapter 2: Polynomials, Chapter 3: Pair of Linear Equations
SOLVED ANSWER KEY WITH MARKING SCHEME
Max. Marks: 30
SECTION A – OBJECTIVE TYPE QUESTIONS
[7 × 1 = 7 Marks]
Q1.
Using:
HCF × LCM = Product of numbers
13 × 338 = 26 × k
4394 = 26k
k = 4394/26
k = 169
Correct Answer: C. 169
Marking Scheme: 1 Mark
Q2.
Parabola opens upward:
a > 0
Touches x-axis at exactly one point:
Discriminant = 0
b² − 4ac = 0
Correct Answer: C
Marking Scheme: 1 Mark
Q3.
For infinitely many solutions:
a1/a2 = b1/b2 = c1/c2
Given:
2x + py = 8
px + 8y = p²
2/p = p/8 = 8/p²
First:
2/p = p/8
p² = 16
p = ±4
Now check:
For p = 4
2/4 = 4/8 = 8/16
1/2 = 1/2 = 1/2 ✓
For p = -4
2/-4 ≠ 8/16
Not possible
Correct Answer: A. p = 4
Marking Scheme: 1 Mark
Q4.
Given:
f(x) = x² − p(x + 1) − c
= x² − px − p − c
Sum of zeros:
α + β = p
Product:
αβ = −(p + c)
Now:
(α + 1)(β + 1)
= αβ + α + β + 1
= −(p + c) + p + 1
= 1 − c
Correct Answer: B. 1 − c
Marking Scheme: 1 Mark
Q5.
Let numerator = x
Denominator = y
Given:
(x − 1)/y = 1/3
3x − y = 3 ……(1)
Also:
x/(y + 8) = 1/4
4x − y = 8 ……(2)
Subtract:
x = 5
Substitute:
15 − y = 3
y = 12
Sum = 5 + 12
= 17
Correct Answer: B. 17
Marking Scheme: 1 Mark
Q6.
7 + 3√2
Rational + Irrational
= Irrational
Assertion = True
Reason = True
Reason correctly explains assertion.
Correct Answer: A
Marking Scheme: 1 Mark
Q7.
Given:
3x − 4y = 7
6x − 8y = 15
a1/a2 = 3/6 = 1/2
b1/b2 = -4/-8 = 1/2
c1/c2 = 7/15
Since:
a1/a2 = b1/b2 ≠ c1/c2
Lines are parallel.
No solution.
Assertion = True
Reason = False
(This condition represents inconsistent system, not consistent)
Correct Answer: C
Marking Scheme: 1 Mark
SECTION B – VERY SHORT ANSWER QUESTIONS
[4 × 2 = 8 Marks]
Q8. Why 6ⁿ can never end with 0? [2]
6ⁿ = (2 × 3)ⁿ
= 2ⁿ × 3ⁿ
A number ending with 0 must contain factor 10 = 2 × 5
Prime factorization of 6ⁿ contains only 2 and 3
No factor 5 exists.
Hence 6ⁿ can never end with digit 0.
Marking Scheme:
- Prime factorization = 1 Mark
- Correct explanation = 1 Mark
Q9. Find value of k [2]
Given:
g(x) = 2x² + 5x + k
α + β = -5/2
αβ = k/2
Now:
α² + β² + αβ = 21/4
α² + β²
= (α + β)² − 2αβ
Required:
= (α + β)² − αβ
21/4 = (25/4) − k/2
Multiply by 4:
21 = 25 − 2k
2k = 4
k = 2
Final Answer:
k = 2
Marking Scheme:
- Formula use = 1 Mark
- Correct answer = 1 Mark
Q10. Find m and n [2]
Given:
2x + 3y = 7
(m + n)x + (2m − n)y = 21
For infinite solutions:
2/(m+n) = 3/(2m−n) = 7/21
= 1/3
So:
m + n = 6 ……(1)
2m − n = 9 ……(2)
Add:
3m = 15
m = 5
Substitute:
5 + n = 6
n = 1
Final Answer:
m = 5, n = 1
Marking Scheme:
- Equation formation = 1 Mark
- Correct values = 1 Mark
Q11. Solve using elimination [2]
Given:
x/2 + 2y/3 = -1
x − y/3 = 3
Multiply by 6:
3x + 4y = -6 ……(1)
6x − 2y = 18 ……(2)
Multiply (1) by 2:
6x + 8y = -12
Subtract (2):
10y = -30
y = -3
Substitute:
x − (-3/3) = 3
x + 1 = 3
x = 2
Final Answer:
x = 2, y = -3
Marking Scheme:
- Elimination process = 1 Mark
- Correct answer = 1 Mark
SECTION C – SHORT ANSWER QUESTIONS
[2 × 3 = 6 Marks]
Q12. Prove √3 irrational [3]
Assume:
√3 = p/q
where p and q are coprime.
Squaring:
3 = p²/q²
p² = 3q²
Thus p divisible by 3.
Let:
p = 3k
Substitute:
9k² = 3q²
q² = 3k²
Thus q divisible by 3.
Hence p and q have common factor 3.
Contradiction.
Therefore:
√3 is irrational.
Marking Scheme:
- Assumption = 1 Mark
- Contradiction = 1 Mark
- Conclusion = 1 Mark
Q13. Form New Polynomial [3]
Given:
f(x) = 3x² − 4x + 1
α + β = 4/3
αβ = 1/3
New roots:
α²/β and β²/α
Sum:
(α³ + β³)/αβ
Now:
α³ + β³
= (α + β)³ − 3αβ(α + β)
= (4/3)³ − 3(1/3)(4/3)
= 64/27 − 4/3
= 28/27
Sum:
= (28/27)/(1/3)
= 28/9
Product:
(α²/β)(β²/α)
= αβ
= 1/3
Polynomial:
x² − (28/9)x + 1/3
Multiply by 9:
9x² − 28x + 3
Marking Scheme:
- Sum = 1 Mark
- Product = 1 Mark
- Polynomial = 1 Mark
SECTION D – LONG ANSWER QUESTION
[1 × 5 = 5 Marks]
Q14. Graphical Solution [5]
Given:
2x + y = 6
y = 6 − 2x
2x − y + 2 = 0
y = 2x + 2
Coordinates on y-axis
Put x = 0
For first line:
y = 6
Point = (0, 6)
For second line:
y = 2
Point = (0, 2)
Point of intersection
2x + y = 6
2x − y = -2
Add:
4x = 4
x = 1
Substitute:
y = 4
Intersection = (1, 4)
Area of triangle
Base = 6 − 2 = 4
Height = distance from (1,4) to y-axis = 1
Area
= 1/2 × Base × Height
= 1/2 × 4 × 1
= 2 sq units
Final Answer:
- Y-axis intercepts = (0,6) and (0,2)
- Intersection point = (1,4)
- Area of triangle = 2 sq units
Marking Scheme:
- Graph/intercepts = 2 Marks
- Intersection point = 1 Mark
- Area calculation = 2 Marks
SECTION E – CASE STUDY QUESTION
[1 × 4 = 4 Marks]
Q15.
(a) Form equations [1]
Let:
x = capacity of small truck
y = capacity of large container
Monday:
3x + 2y = 22
Tuesday:
4x + 3y = 31
Marking Scheme:
Correct equations = 1 Mark
(b) Consistency check [1]
3/4 ≠ 2/3
Since:
a1/a2 ≠ b1/b2
System is consistent and has unique solution.
Marking Scheme:
Correct conclusion = 1 Mark
(c) Find capacities [2]
Equations:
3x + 2y = 22 ……(1)
4x + 3y = 31 ……(2)
Multiply (1) by 3:
9x + 6y = 66
Multiply (2) by 2:
8x + 6y = 62
Subtract:
x = 4
Substitute:
3(4) + 2y = 22
12 + 2y = 22
2y = 10
y = 5
Final Answer:
Small truck = 4 tonnes
Large container = 5 tonnes
Verification:
3(4) + 2(5) = 22 ✓
4(4) + 3(5) = 31 ✓
Marking Scheme:
- Solving = 1 Mark
- Correct answer + verification = 1 Mark
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