Class 10 Test Paper Chapter - 1 and 2 | CBSE Board | Detailed Solutions & Marking Scheme

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 VEDANT SKILL ASSESSMENT SERIES – ACADEMIC YEAR 2026-27

CLASS: X - CBSE
SUBJECT: MATHEMATICS
TOTAL MARKS: 30
TIME: 1 HOUR

SYLLABUS:
Chapter 1 - Real Numbers
Chapter 2 - Polynomials

PATTERN: CBSE 2026-27

GENERAL INSTRUCTIONS

  1. Read all questions carefully before answering.
    (Misreading the question and then blaming the paper will not increase your marks.)

  2. Show proper steps wherever required.
    ("Sir, answer toh yahi aana tha" is not an accepted mathematical method.)

  3. Write neatly and clearly.
    (If your handwriting requires a decoder machine, checking may become an adventure.)

  4. Manage your time wisely.
    (Spending 45 minutes on one question and calling the rest "optional" is not a strategy.)

  5. If you do not know the answer, you may cry silently.
    (Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.)

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SECTION A (7 MARKS)

Q1. If two positive integers a and b are written as

a = x³y²
b = xy³

where x and y are prime numbers, then find HCF(a, b).

A. xy
B. xy²
C. x³y³
D. x²y²

[1 Mark]

Q2. If one zero of the quadratic polynomial

x² + 3x + k

is 2, then the value of k is:

A. 10
B. -10
C. -7
D. -2

[1 Mark]

Q3. The total number of factors of a prime number is:

A. 1
B. 0
C. 2
D. 3

[1 Mark]

Q4. A quadratic polynomial whose zeroes are -3 and 4 is given by:

A. x² - x - 12
B. x² + x + 12
C. x² - x + 12
D. x² + x - 12

[1 Mark]

Q5. The exponent of 2 in the prime factorization of 144 is:

A. 4
B. 5
C. 3
D. 2

[1 Mark]

For Questions 6 and 7, select the correct option:

A. Both A and R are true, and R is the correct explanation of A.
B. Both A and R are true, but R is not the correct explanation of A.
C. A is true, but R is false.
D. A is false, but R is true.

Q6.

Assertion (A):
The product of two numbers is 5780 and their HCF is 17.
Then their LCM is 340.

Reason (R):
For any two positive integers a and b,

HCF(a, b) × LCM(a, b) = a × b

[1 Mark]

Q7.

Assertion (A):
The polynomial

p(x) = x² + 3x + 3

has two real zeroes.

Reason (R):
A quadratic polynomial can have at most two zeroes.

[1 Mark]

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SECTION B (8 MARKS)

Q8. Explain why

7 × 11 × 13 + 13

is a composite number.

[2 Marks]

Q9. Find the zeroes of the quadratic polynomial

4x² - 4x - 3

[2 Marks]

Q10. Find the HCF and LCM of 12, 15 and 21 by applying the prime factorization method.

[2 Marks]

Q11. If α and β are the zeroes of a quadratic polynomial such that

α + β = 6

and

αβ = 4

write the quadratic polynomial.

[2 Marks]

==================================================
SECTION C (6 MARKS)

Q12. Prove that √5 is an irrational number.

[3 Marks]

Q13. If α and β are the zeroes of the quadratic polynomial

p(x) = 3x² - 5x + 2

find the value of

1/α + 1/β

[3 Marks]

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SECTION D (5 MARKS)

Q14. If α and β are the zeroes of the quadratic polynomial

f(x) = x² - p(x + 1) - c

such that

(α + 1)(β + 1) = 0

find the value of c.

Also, if p = 3, find the actual numerical zeroes of the polynomial.

[5 Marks]

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SECTION E (4 MARKS)
CASE STUDY BASED QUESTION

Q15.

An Indian Army contingent of 616 members is to march behind an army band of 32 members in a Republic Day parade.

The two groups are to march in the same number of columns.

To minimize confusion, the parade commander wants to maximize the width of the arrangement by finding the highest common factor of columns.

Based on the above real-life scenario, answer the following:

(i) What is the maximum number of columns in which they can march?

[1 Mark]

(ii) Which mathematical concept/theorem is utilized to find the fundamental alignment of columns here?

[1 Mark]

(iii) If the band size is increased to 48 members and the total contingent becomes 720 members, what will be the new maximum number of columns?

[2 Marks]

VEDANT SKILL ASSESSMENT SERIES – ACADEMIC YEAR 2026-27

CLASS: X
SUBJECT: MATHEMATICS

DETAILED ANSWER KEY WITH MARKING SCHEME

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SECTION A (7 MARKS)

Q1. If

a = x³y²
b = xy³

HCF is obtained by taking the lowest powers of common prime factors.

HCF = x¹y² = xy²

Answer: B. xy²

Marks: 1

Q2. One zero of x² + 3x + k is 2.

Substitute x = 2:

(2)² + 3(2) + k = 0

4 + 6 + k = 0

10 + k = 0

k = -10

Answer: B. -10

Marks: 1

Q3. A prime number has exactly two factors:
1 and itself.

Answer: C. 2

Marks: 1

Q4. Zeroes are -3 and 4.

Polynomial

= (x + 3)(x - 4)

= x² - 4x + 3x - 12

= x² - x - 12

Answer: A. x² - x - 12

Marks: 1

Q5.

144 = 2 × 2 × 2 × 2 × 3 × 3

= 2⁴ × 3²

Exponent of 2 = 4

Answer: A. 4

Marks: 1

Q6.

Assertion:

LCM = (5780)/(17)

= 340

Assertion is True.

Reason:

HCF × LCM = Product of Numbers

This statement is True and explains the assertion.

Answer: A

Marks: 1

Q7.

For p(x) = x² + 3x + 3

Discriminant

D = b² - 4ac

= 3² - 4(1)(3)

= 9 - 12

= -3

Since D < 0, polynomial has no real zeroes.

Assertion is False.

Reason is True because a quadratic polynomial can have at most two zeroes.

Answer: D

Marks: 1

==================================================
SECTION B (8 MARKS)

Q8. Explain why 7 × 11 × 13 + 13 is composite.

7 × 11 × 13 + 13

= 13(7 × 11 + 1)

= 13(77 + 1)

= 13 × 78

= 1014

Since 1014 has factors other than 1 and itself, it is composite.

Answer: Composite Number

Marking Scheme:

Taking 13 common = 1 Mark

Conclusion = 1 Mark

Total = 2 Marks

Q9. Find the zeroes of

4x² - 4x - 3

4x² - 6x + 2x - 3 = 0

2x(2x - 3) + 1(2x - 3) = 0

(2x - 3)(2x + 1) = 0

2x - 3 = 0

x = 3/2

2x + 1 = 0

x = -1/2

Answer:

x = 3/2 and x = -1/2

Marking Scheme:

Factorization = 1 Mark

Both zeroes = 1 Mark

Total = 2 Marks

Q10. Find HCF and LCM of 12, 15 and 21.

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

HCF = 3

LCM = 2² × 3 × 5 × 7

= 420

Answer:

HCF = 3

LCM = 420

Marking Scheme:

Prime Factorization = 1 Mark

HCF and LCM = 1 Mark

Total = 2 Marks

Q11.

α + β = 6

αβ = 4

Quadratic polynomial:

x² - (α + β)x + αβ

= x² - 6x + 4

Answer:

x² - 6x + 4

Marking Scheme:

Formula = 1 Mark

Polynomial = 1 Mark

Total = 2 Marks

==================================================
SECTION C (6 MARKS)

Q12. Prove that √5 is irrational.

Assume √5 is rational.

Then

√5 = a/b

where a and b are coprime integers.

Squaring both sides,

5 = a²/b²

a² = 5b²

Therefore a² is divisible by 5.

Hence a is divisible by 5.

Let a = 5k.

Substituting,

25k² = 5b²

5k² = b²

Thus b is also divisible by 5.

Therefore a and b have common factor 5.

This contradicts that a and b are coprime.

Hence our assumption is wrong.

Therefore √5 is irrational.

Marking Scheme:

Assumption = 1 Mark

Proof and contradiction = 1 Mark

Conclusion = 1 Mark

Total = 3 Marks

Q13.

Given

p(x) = 3x² - 5x + 2

a = 3
b = -5
c = 2

α + β = -b/a

= 5/3

αβ = c/a

= 2/3

Now,

1/α + 1/β

= (α + β)/(αβ)

= (5/3)/(2/3)

= 5/2

Answer:

5/2

Marking Scheme:

Finding α + β = 1 Mark

Finding αβ = 1 Mark

Final value = 1 Mark

Total = 3 Marks

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SECTION D (5 MARKS)

Q14.

Given

f(x) = x² - p(x + 1) - c

f(x) = x² - px - p - c

For polynomial

x² - px - (p + c)

Sum of zeroes

α + β = p

Product of zeroes

αβ = -(p + c)

Given

(α + 1)(β + 1) = 0

αβ + α + β + 1 = 0

Substituting,

-(p + c) + p + 1 = 0

-c + 1 = 0

c = 1

Now p = 3

Polynomial

= x² - 3x - 4

Factorizing,

x² - 4x + x - 4

= x(x + 1) - 4(x + 1)

= (x + 1)(x - 4)

Zeroes:

x = -1

x = 4

Answer:

c = 1

Zeroes = -1 and 4

Marking Scheme:

Finding product of zeroes = 1 Mark

Using condition = 1 Mark

Finding c = 1 Mark

Substituting p = 3 = 1 Mark

Finding zeroes = 1 Mark

Total = 5 Marks

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SECTION E (4 MARKS)
CASE STUDY

Q15.

(i) Maximum number of columns

HCF of 616 and 32

616 = 32 × 19 + 8

32 = 8 × 4 + 0

HCF = 8

Answer: 8 columns

Marks: 1

(ii) Mathematical concept used

Answer:

Highest Common Factor (HCF) using Euclid's Division Algorithm.

Marks: 1

(iii) New arrangement

Find HCF of 720 and 48

720 ÷ 48 = 15

Therefore

HCF = 48

Answer:

48 columns

Marking Scheme:

Method = 1 Mark

Answer = 1 Mark

Total = 2 Marks

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FINAL ANSWER SUMMARY

Q1. B
Q2. B
Q3. C
Q4. A
Q5. A
Q6. A
Q7. D
Q8. Composite Number
Q9. 3/2, -1/2
Q10. HCF = 3, LCM = 420
Q11. x² - 6x + 4
Q12. √5 is irrational
Q13. 5/2
Q14. c = 1, Zeroes = -1 and 4
Q15. (i) 8 columns
(ii) Euclid's Division Algorithm / HCF
(iii) 48 columns

TOTAL MARKS = 30

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