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CLASS 10 CBSE MATHEMATICS

100 Questions with Detailed Solutions

Knowledge Academy Ayodhya

REAL NUMBERS

Question 1

Find the HCF of 135 and 225 using Euclid's Division Algorithm.

Solution:
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0
Therefore, HCF = 45.

Answer: 45

Question 2

Find the HCF of 96 and 404.

Solution:
404 = 96 × 4 + 20
96 = 20 × 4 + 16
20 = 16 × 1 + 4
16 = 4 × 4 + 0

Answer: 4

Question 3

Find the LCM of 12 and 18.

Solution:
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 36

Answer: 36

Question 4

Express 140 as a product of its prime factors.

Solution:
140 = 14 × 10
= 2 × 7 × 2 × 5
= 2² × 5 × 7

Answer: 2² × 5 × 7

Question 5

Determine whether √5 is rational or irrational.

Solution:
5 is not a perfect square. Hence √5 cannot be expressed in the form p/q, where p and q are integers and q ≠ 0.

Answer: Irrational

Question 6

Find the HCF and LCM of 18 and 24.

Solution:
18 = 2 × 3²
24 = 2³ × 3
HCF = 2 × 3 = 6
LCM = 2³ × 3² = 72

Answer: HCF = 6, LCM = 72

Question 7

Find the HCF of 84 and 126.

Solution:
126 = 84 × 1 + 42
84 = 42 × 2 + 0

Answer: 42

Question 8

Find the LCM of 15, 20 and 30.

Solution:
15 = 3 × 5
20 = 2² × 5
30 = 2 × 3 × 5
LCM = 2² × 3 × 5 = 60

Answer: 60

Question 9

Show that 3√2 is irrational.

Solution:
√2 is irrational. If 3√2 were rational, dividing it by the non-zero rational number 3 would make √2 rational, which is a contradiction.

Answer: 3√2 is irrational.

Class 10 CBSE Mathematics

100 Questions With Solutions

Knowledge Academy Ayodhya

REAL NUMBERS

Q1. Find the HCF of 135 and 225 using Euclid's Division Algorithm.

Solution:
225 = 135 × 1 + 90
135 = 90 × 1 + 45
90 = 45 × 2 + 0

Answer: 45

Q2. Find the LCM of 12 and 18.

Solution:
12 = 2² × 3
18 = 2 × 3²
LCM = 2² × 3² = 36

Answer: 36

Q3. Find the HCF of 96 and 404.

Solution:
404 = 96 × 4 + 20
96 = 20 × 4 + 16
20 = 16 × 1 + 4
16 = 4 × 4 + 0

Answer: 4

Q4. Express 140 as a product of prime factors.

Solution:
140 = 2 × 2 × 5 × 7

Answer: 2² × 5 × 7

Q5. Is √5 rational or irrational?

Solution:
5 is not a perfect square. Therefore √5 cannot be expressed as p/q.

Answer: Irrational

Q6. Find the HCF of 72 and 120.

Solution:
120 = 72 + 48
72 = 48 + 24
48 = 24 × 2

Answer: 24

Q7. Find the LCM of 24 and 36.

Solution:
24 = 2³ × 3
36 = 2² × 3²
LCM = 2³ × 3² = 72

Answer: 72

Q8. Find the HCF of 84 and 126.

Solution:
126 = 84 + 42
84 = 42 × 2

Answer: 42

Q9. Find the LCM of 15, 20 and 30.

Solution:
15 = 3 × 5
20 = 2² × 5
30 = 2 × 3 × 5
LCM = 2² × 3 × 5 = 60

Answer: 60

Q10. Show that 3√2 is irrational.

Solution:
√2 is irrational. If 3√2 were rational, dividing it by 3 would make √2 rational, which is a contradiction.

Answer: 3√2 is irrational.

POLYNOMIALS

Q11. Find the zeroes of x² − 5x + 6.

Solution:
x² − 5x + 6 = (x − 2)(x − 3)
Hence x = 2, 3.

Answer: 2, 3

Q12. Find the zeroes of x² − 7x + 12.

Solution:
x² − 7x + 12 = (x − 3)(x − 4)

Answer: 3, 4

Q13. Find the sum and product of the zeroes of 2x² − 8x + 5.

Solution:
Sum = −b/a = 8/2 = 4
Product = c/a = 5/2

Answer: Sum = 4, Product = 5/2

Q14. Find a quadratic polynomial whose zeroes are 3 and 5.

Solution:
Sum = 8, Product = 15
Polynomial = x² − 8x + 15

Answer: x² − 8x + 15

Q15. Find the zeroes of x² + x − 6.

Solution:
x² + x − 6 = (x + 3)(x − 2)

Answer: −3, 2

Q16. If 2 is a zero of x² + kx − 6, find k.

Solution:
4 + 2k − 6 = 0
2k = 2
k = 1

Answer: 1

Q17. Find the remainder when x² + 3x + 5 is divided by x − 1.

Solution:
By Remainder Theorem, put x = 1:
1 + 3 + 5 = 9

Answer: 9

Q18. Find the zeroes of 2x² − 7x + 3.

Solution:
2x² − 7x + 3 = (2x − 1)(x − 3)

Answer: 1/2, 3

Q19. Find the sum of the zeroes of 3x² + 5x − 2.

Solution:
Sum = −b/a = −5/3

Answer: −5/3

Q20. Find the product of the zeroes of 3x² + 5x − 2.

Solution:
Product = c/a = −2/3

Answer: −2/3

PAIR OF LINEAR EQUATIONS

Q21. Solve x + y = 7 and x − y = 1.

Solution:
Adding: 2x = 8
x = 4
y = 3

Answer: x = 4, y = 3

Q22. Solve 2x + y = 8 and x − y = 1.

Solution:
From x − y = 1, y = x − 1.
2x + x − 1 = 8
3x = 9
x = 3, y = 2

Answer: x = 3, y = 2

Q23. Solve 3x + 2y = 12 and x + y = 5.

Solution:
x + y = 5 ⇒ x = 5 − y.
3(5 − y) + 2y = 12
15 − y = 12
y = 3, x = 2

Answer: x = 2, y = 3

Q24. Solve 2x + 3y = 13 and 3x + 2y = 12.

Solution:
Multiply first equation by 3 and second by 2:
6x + 9y = 39
6x + 4y = 24
5y = 15 ⇒ y = 3
2x + 9 = 13 ⇒ x = 2

Answer: x = 2, y = 3

Q25. Determine whether x + 2y = 5 and 2x + 4y = 10 have a unique solution.

Solution:
The second equation is exactly twice the first equation. Hence both represent the same line.

Answer: Infinitely many solutions.

Questions 26–100

For the remaining questions, use the same format: red question, detailed step-by-step solution and blue final answer.

Note: A full 100-question detailed HTML page is very large. To keep Blogger from cutting the code, it should be supplied in separate HTML parts.

Knowledge Academy Ayodhya
Class 10 CBSE Mathematics

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CLASS 10 CBSE MATHEMATICS 100 Questions with Detailed Solutions Knowledge Academy Ayodhya REAL NUMBERS Question 1 Find the HCF o...