Model Problems-Quadratic Equations-10th Class-Mathematics-English Medium

                                                                     Model Problems

Chapter Name: Quadratic Equations

1. Product of two consecutive even nubers is 48. Express the given data in the form of quadratic equation.

2. Check whether (x+1)(x-2)=2(x+2) is a quadratic equation or not?

3. Sum of the squares of two consecutive whole numbers is 61. Express the given data in the form of quadratic equation.

4. Rani said that "(x+2)² + 15=(x-1)² is a quadratic equation." Do you agree with her? Justify.

5. Find the roots of x² + 7x + 12= 0 by factorisation.

6. Find the roots of 3x² - 2x - 1= 0 by completing square method.

7. Find the roots of 2x² + 7x + 3 = 0 by factorisation.

8. Find the roots of x² + 3x - 28 = 0 by completing square method.

9. Find the roots of x² - 2x - 1= 0 by using formula.

10. Find the roots of 4x² = 3x + 1.

11. Sum of the squares of three consecutive multiples of 3 is 261. Find them.

12. In a Right angled triangle, Hieght is 3m more than 3 times of its Base and Hypotenuse is 25m. Find the hieght of triangle. 

13. What is the discriminant of 4x² - 5x - 6= 0.

14. What is Nature of roots of 2x² + 3x + 4= 0.

15. Varun said that "4x² - 4x + 1= 0 has equal and real roots." Do you agree with him? Justify.

16. For what value of 'm', mx² - 3mx + 9= 0 has equal roots?

17. Find the value of 'k', if 9x² + 6kx + 2k= 0 has equal roots.

18. If x² - px + 4= 0 has equal roots then find the value of 'p'.

19.

20.

Model Problems-Pair of Linear Equations-10th Class-Mathematics

Model Problems

Chapter Name: Pair of Linear Equations in two Variables

1. Cost of 5 pens and 6 notebooks is Rs. 200. Express the given data in the form of linear equation in two variables.

2. Sum of two numbers is 23. Express the given data in the form of linear equation in two variables.

3. In a mathematics test, Ramu got 2 marks more than 3 times of marks of Raju. Express the given data in the form of linear equation in two variables.

4. Cost of a pen is 1 rupee more than twice of cost of pencil. Express the given data in the form of linear equation in two variables.

5. Solve 2x + 3y - 6 = 20 and x + 2y = 2 by Elimination method.

6. Solve 3x + y = 2 and 2x + y = 3 by Elimination method.

7. Solve x + y = 4 and x + 3y = 10 by Substitution method.

8. Solve 2x - y = 10 and 3x - 2y = 14 by Substitution method.

9. Check whether 5x + 2y = 8 and 2x - y = 4 are consistent or inconsistent?

10. Lakshmi said that "x - 2y - 6 = 0 and 3x - 6y - 18 = 0 have infinity many solutions". Do you agree with her? Justify.

11. Surya said that "3x + y = 10 and 6x + 2 y - 5 = 0 are represent parallel lines." Do you agree? Justify your answer.

12. Check whether x - y = 3 and 3x - 3y = 4 are intersecting lines or parallel lines?

13. If "2x - 3py = 7 and 3x - 9y = 10" are represent a railway track, then find the value of 'p'.

14. For what value of 'k', "3x + 2y = 1 and 2kx + 4y = 2 having No solution".

15. If "x - 2y = 4 and 3x - 6y = 5 are inconsistent", then find the value of 'm'.

16. For which value of 'n', "x + 2y = 4 and 3x + ny = 10 have only one solution".

17. Solve: 3x + 4y = 13 and 2x - 3y = 3 graphically.

18. Find the solution of  2x + y = 7 and x + 3y = 6 by graphical representation.

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Worksheet Solutions-Real numbers-10th class-Mathematics-English Medium-2026-2027

WORKSHEET - 1 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Division Algorithm & Euclid's Lemma

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Q1. In a = bq + r (Division algorithm):
a is _______, q is _______, r is _______, b is _______ and condition for "r" is _______.

a = Dividend,    b = Divisor

q = Quotient,    r = Remainder

Condition for r: 0 ≤ r < b

Q2. Find the HCF of 56 and 98, using Euclid's Division algorithm.

Applying Euclid's Division Lemma (a = bq + r):

98 = 56 × 1 + 42    (r = 42 ≠ 0)

56 = 42 × 1 + 14    (r = 14 ≠ 0)

42 = 14 × 3 + 0     (r = 0)

Since remainder is 0, HCF(56, 98) = 14.

Q3. Use Euclid's Division Algorithm to find the HCF of 135 and 225.

Here 225 > 135. Using a = bq + r:

225 = 135 × 1 + 90    (r = 90 ≠ 0)

135 = 90 × 1 + 45      (r = 45 ≠ 0)

90 = 45 × 2 + 0        (r = 0)

Therefore, HCF(135, 225) = 45.

Q4. If the HCF of 12 and 84 is expressed as 7m - 2, then find the value of "m".

First, find HCF of 12 and 84:

84 = 12 × 7 + 0  ⇒  HCF(12, 84) = 12

Given HCF = 7m - 2

⇒ 7m - 2 = 12 ⇒ 7m = 14 ⇒ m = 2

Q5. Find the greatest number that divides 245 and 1029 leaving the remainder 5 in each case.

Subtract remainder 5 from given numbers:

245 - 5 = 240  and  1029 - 5 = 1024

Now, find HCF(240, 1024) using Euclid's Algorithm:

1024 = 240 × 4 + 64  |  240 = 64 × 3 + 48

64 = 48 × 1 + 16       |  48 = 16 × 3 + 0

The required greatest number is 16.

WORKSHEET - 2 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Prime Factorisation

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Q1. Find the prime factorisation of 60.

Dividing 60 by prime factors step-by-step:

60 = 2 × 30 = 2 × 2 × 15 = 2 × 2 × 3 × 5

Prime factorisation of 60 = 2² × 3 × 5

Q2. Find the prime factorisation of the following:

(a) 84
84 = 2 × 42
= 2 × 2 × 21
= 2² × 3 × 7

(b) 180
180 = 2 × 90
= 2 × 2 × 45 = 2 × 2 × 3 × 15
= 2² × 3² × 5

(c) 360
360 = 2 × 180
= 2 × 2 × 2 × 3² × 5
= 2³ × 3² × 5

(d) 945
945 = 3 × 315
= 3 × 3 × 105 = 3 × 3 × 3 × 35
= 3³ × 5 × 7

Q3. Find the smallest number that must be multiplied to 540 to make it a perfect square.

Prime factorisation of 540:

540 = 2 × 2 × 3 × 3 × 3 × 5 = (2²) × (3²) × 3 × 5

Here, prime factors 3 and 5 do not have pairs.

To make it a perfect square, we must multiply by 3 × 5 = 15.

Required smallest number = 15.

Q4. Check whether 8ⁿ can end with digit '0' for any natural number n. Justify your answer.

If a number ends with digit 0, its prime factorisation must contain both 2 and 5.

Now, 8ⁿ = (2³)ⁿ = 2³ⁿ

The only prime factor of 8ⁿ is 2. It does not contain 5 as a prime factor.

By Fundamental Theorem of Arithmetic, 8ⁿ can NEVER end with digit 0 for any n ∈ N.

WORKSHEET - 3 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Fundamental Theorem of Arithmetic - HCF & LCM

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Q1. Two numbers have prime factorizations: A = 2² × 3 × 5 and B = 2 × 3² × 7.
(a) HCF = _______      (b) LCM = _______

(a) HCF = Product of smallest powers of common prime factors

         = 2¹ × 3¹ = 6

(b) LCM = Product of greatest powers of all prime factors

         = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260

Q2. Find HCF of 24 and 36 by prime factorisation.

24 = 2 × 2 × 2 × 3 = 2³ × 3¹

36 = 2 × 2 × 3 × 3 = 2² × 3²

HCF(24, 36) = 2² × 3¹ = 4 × 3 = 12

Q3. Using prime factorisation, find HCF and LCM of 84 and 180. Also verify HCF × LCM = product of numbers.

84 = 2² × 3 × 7  |  180 = 2² × 3² × 5

HCF = 2² × 3 = 12  |  LCM = 2² × 3² × 5 × 7 = 1260

Verification: HCF × LCM = 12 × 1260 = 15,120

Product of numbers = 84 × 180 = 15,120. (Verified!)

Q4. Determine the least number that is divisible by 15, 20 and 24.

The required least number is the LCM(15, 20, 24):

15 = 3 × 5  |  20 = 2² × 5  |  24 = 2³ × 3

LCM = 2³ × 3 × 5 = 8 × 3 × 5 = 120

The least divisible number is 120.

Q5. Find the greatest number dividing 306 and 657 exactly.

The required greatest number is HCF(306, 657):

306 = 2 × 3² × 17  |  657 = 3² × 73

HCF(306, 657) = 3² = 9.

WORKSHEET - 4 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Rational Numbers & Decimal Expansions

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Q1. (a) When does a rational number have a terminating decimal expansion?
(b) What type of decimal expansion does a rational number have if its denominator (in lowest form) contains primes other than 2 or 5?

(a) A rational number p/q has a terminating decimal expansion when prime factorisation of denominator q is of the form 2ⁿ × 5^m (n, m ≥ 0).

(b) It will have a Non-terminating recurring (repeating) decimal expansion.

Q2. Without actual division, state whether the decimal expansion of the following is terminating or non-terminating recurring?
(a) 7/8          (b) 9/15

(a) 7/8: Denominator = 8 = 2³ = 2³ × 5⁰. Form 2ⁿ × 5^m ⇒ Terminating.

(b) 9/15: In simplest form = 3/5. Denominator = 5 = 2⁰ × 5¹. Form 2ⁿ × 5^m ⇒ Terminating.

Q3. Write the decimal expansion of the following without actual division:
(a) 1/8          (b) 13/30

(a) 1/8 = 1 / 2³ = (1 × 5³) / (2³ × 5³) = 125 / 1000 = 0.125

(b) 13/30 = 13 / (3 × 10) = 0.4333... = 0.43̄ (Non-terminating recurring)

Q4. Express 0.375 as a rational form (p/q).

0.375 = 375 / 1000

Dividing numerator and denominator by 125:

0.375 = 3 / 8

Q5. Express 0.27 (0.2727...) as a rational number.

Let x = 0.2727... — (1)

Multiplying by 100: 100x = 27.2727... — (2)

Subtract (1) from (2): 99x = 27

x = 27 / 99 = 3 / 11

WORKSHEET - 5 Class X — Mathematics

Chapter: 1. Real Numbers Topic: Proofs of Irrational Numbers

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Q1. Prove that √3 is irrational.

Assume to the contrary that √3 is rational. Then √3 = a/b (where a, b are co-prime integers, b ≠ 0).

⇒ a = b√3 ⇒ Squaring both sides: a² = 3b² — (1)

Thus, 3 divides a², which means 3 divides a. Let a = 3c.

Substitute in (1): (3c)² = 3b² ⇒ 9c² = 3b² ⇒ b² = 3c²

Thus, 3 divides b², so 3 divides b.

Hence, 3 is a common factor of both a and b, contradicting that a, b are co-prime. Therefore, √3 is irrational.

Q2. Prove that 2 + √3 is irrational.

Let us assume 2 + √3 is rational.

Then 2 + √3 = a/b (where a, b are integers, b ≠ 0).

⇒ √3 = (a/b) - 2 ⇒ √3 = (a - 2b) / b

Since a, b are integers, (a - 2b)/b is rational.

This implies √3 is rational, which contradicts the known fact that √3 is irrational.

Hence, 2 + √3 is irrational.

Q3. Prove that √3 + 5√2 is irrational.

Assume x = √3 + 5√2 is rational.

⇒ x - 5√2 = √3 ⇒ Squaring both sides: (x - 5√2)² = (√3)²

⇒ x² - 10√2x + 50 = 3 ⇒ x² + 47 = 10√2x

⇒ √2 = (x² + 47) / (10x)

Since x is rational, (x² + 47)/(10x) is rational, implying √2 is rational.

This is a contradiction. Thus, √3 + 5√2 is irrational.

WORKSHEET - 6 (Part 1) Class X — Mathematics

Chapter: 1. Real Numbers Topic: Logarithms — Forms, Evaluations & Laws

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Q1. Convert into logarithmic form:
(i) 10³ = 1000      (ii) 2⁴ = 16      (iii) 3² = 9

(i) 10³ = 1000 ⇒ log₁₀ 1000 = 3

(ii) 2⁴ = 16 ⇒ log₂ 16 = 4

(iii) 3² = 9 ⇒ log₃ 9 = 2

Q2. Convert into exponential form:
(i) log₂ 8 = 3      (ii) log₅ 25 = 2      (iii) log₁₀ 100 = 2

(i) log₂ 8 = 3 ⇒ 2³ = 8

(ii) log₅ 25 = 2 ⇒ 5² = 25

(iii) log₁₀ 100 = 2 ⇒ 10² = 100

Q3. Evaluate:
(i) log₃ 3    (ii) log₃ 81    (iii) log₁₀ 1    (iv) log₅ 125    (v) log₁₀ 1000    (vi) log₁₀ 0.01

(i) log₃ 3 = 1

(ii) log₃ 3⁴ = 4

(iii) log₁₀ 1 = 0

(iv) log₅ 5³ = 3

(v) log₁₀ 10³ = 3

(vi) log₁₀ 10^-2 = -2

Q4. Write the laws of logarithms:

a) Product Law: log_a (mn) = log_a m + log_a n

b) Quotient Law: log_a (m/n) = log_a m - log_a n

c) Power Law: log_a (m^k) = k · log_a m

Q5. Expand using laws of logarithms:

a) log(5x) = log 5 + log x

b) log(3y/2) = log 3 + log y - log 2

c) log(7a³) = log 7 + 3 log a

d) log(4x²/y) = log 4 + 2 log x - log y

WORKSHEET - 6 (Part 2) Class X — Mathematics

Chapter: 1. Real Numbers Topic: Logarithms — Simplification & Proofs

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Q6. Simplify:

a) log 8 + log 5 = log(40)

b) log 12 - log 3 = log(4)

c) 3 log 2 = log(2³) = log 8

d) 2 log x - log y = log(x²/y)

e) log₁₀ 8 - log₁₀ 2 = log₁₀ 4

f) log₁₀ 50 + log₁₀ 2 = log₁₀ 100 = 2

Q7. Evaluate (Given: log 2 = 0.3010, log 3 = 0.4771):
(a) log 6     (b) log 9     (c) log(8/3)     (d) log₁₀ 20

a) log 6 = log(2 × 3) = log 2 + log 3 = 0.3010 + 0.4771 = 0.7781

b) log 9 = log(3²) = 2 log 3 = 2 × 0.4771 = 0.9542

c) log(8/3) = 3 log 2 - log 3 = 3(0.3010) - 0.4771 = 0.9030 - 0.4771 = 0.4259

d) log₁₀ 20 = log₁₀ (2 × 10) = log₁₀ 2 + 1 = 0.3010 + 1 = 1.3010

Q8. If log a = 2 and log b = 3, find the values of:
(a) log(ab)          (b) log(a²/b)

a) log(ab) = log a + log b = 2 + 3 = 5

b) log(a²/b) = 2 log a - log b = 2(2) - 3 = 4 - 3 = 1

Q9. If x² + y² = 18xy, then prove that 2 log(x - y) = 4 log 2 + log x + log y.

Given: x² + y² = 18xy

Subtract 2xy from both sides:

x² + y² - 2xy = 18xy - 2xy  ⇒  (x - y)² = 16xy

Taking log on both sides:

log(x - y)² = log(16xy)

⇒ 2 log(x - y) = log 16 + log x + log y

⇒ 2 log(x - y) = log(2⁴) + log x + log y

⇒ 2 log(x - y) = 4 log 2 + log x + log y. (Hence Proved!)