Welcome to AssamSchool. This article contains 30 Extra Questions from Real Numbers from Class 10 Mathematics Chapter 1. These Real Numbers Class 10 Extra Questions are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Real Numbers Class 10 Extra Questions
1. Express \(210\) as a product of its prime factors.
2. Find the HCF and LCM of \(144\) and \(198\) using prime factorisation.
3. Prove that \( \sqrt{5} \) is irrational.
4. Explain why \( 5 \times 13 \times 17 + 17 \) is a composite number.
5. Can \( 8^n \) end with the digit 0 for any natural number \( n \)? Justify your answer.
6. Given \( \text{HCF}(306, 1314) = 18 \), find \( \text{LCM}(306, 1314) \).
7. Two buses depart from a station. Bus A takes 24 minutes to complete one round, and Bus B takes 36 minutes. If they depart together at 8 AM, when will they next meet at the starting point?
8. Prove that \( 2 + \sqrt{3} \) is irrational.
9. Use Euclid’s division algorithm to find the HCF of 135 and 225.
10. Show that \( 4\sqrt{7} \) is irrational.
11. Prove that \( \frac{1}{\sqrt{5}} \) is irrational.
12. The HCF of two numbers is 12 and their LCM is 360. If one number is 60, find the other.
13. Determine whether \( \frac{15}{1600} \) has a terminating or non-terminating decimal expansion.
14. Prove that \( \sqrt{7} \) is irrational.
15. Find the HCF and LCM of 24, 36, and 48 using prime factorisation.
16. Verify \( \text{HCF}(24, 18) \times \text{LCM}(24, 18) = 24 \times 18 \)
17. Prove that the sum of a rational number and an irrational number is irrational.
18. Show that \( 5 + 2\sqrt{2} \) is irrational.
19. Find the smallest number divisible by 15, 20, and 25.
20. Explain why \( 3 \times 5 \times 7 + 7 \) is composite.
21. Find the HCF of 867 and 255 using Euclid’s algorithm.
22. Prove that \( \sqrt{6} \) is irrational.
23. Check if 1001 is a prime number.
24. Prove that \( 7 – \sqrt{5} \) is irrational.
25. A rectangular floor is 18m by 12m. Find the largest square tile that can cover the floor without cutting.
26. Show that \( \sqrt{2} + \sqrt{3} \) is irrational.
27. Prove that \( \frac{2}{\sqrt{7}} \) is irrational.
28. Find the HCF and LCM of 96 and 144.
29. Determine the nature of the decimal expansion of \( \frac{77}{210} \)
30. Prove that \( 3\sqrt{5} \) is irrational.
Answers
- Solution:
\( 210 = 2 \times 3 \times 5 \times 7 \) - Solution:
Prime factors:
\( 144 = 2^4 \times 3^2 \), \( 198 = 2 \times 3^2 \times 11 \)
HCF = \( 2 \times 3^2 = 18 \),
LCM = \( 2^4 \times 3^2 \times 11 = 1584 \) - Solution:
Assume \( \sqrt{5} = \frac{a}{b} \) (coprime integers \( a, b\)).
Then \( 5b^2 = a^2 \)
Thus 5 divides \( a^2 \), so 5 divides \( a \)
Let \( a = 5c \).
Then \( 5b^2 = 25c^2 \Rightarrow b^2 = 5c^2 \)
Hence 5 divides \( b \), contradicting coprimality.
Thus \( \sqrt{5} \) is irrational. - Solution:
\( 5 \times 13 \times 17 + 17 = 17(5 \times 13 + 1) = 17 \times 66 \)
It has factors other than 1 and itself, so it is composite. - Solution:
\( 8^n = (2^3)^n = 2^{3n} \)
Prime factorisation has only 2.
Since 5 is not a factor, \( 8^n \) cannot end with 0. - Solution:
\( \text{LCM} = \frac{306 \times 1314}{18} = \frac{402,084}{18} = 22,338 \) - Solution:
LCM of 24 and 36 = 72 minutes.
They meet again at 9:12 AM. - Solution:
Assume \( 2 + \sqrt{3} \) is rational.
Then \( \sqrt{3} = \text{rational} – 2 \), which is rational.
Contradiction.
Hence, \( 2 + \sqrt{3} \) is irrational. - Solution:
\( 225 = 135 \times 1 + 90 \)
\( 135 = 90 \times 1 + 45 \)
\( 90 = 45 \times 2 + 0 \)
HCF = 45 - Solution:
Assume \( 4\sqrt{7} = \frac{a}{b} \)
Then \( \sqrt{7} = \frac{a}{4b} \), implying \( \sqrt{7} \) is rational.
Contradiction. - Solution:
If \( \frac{1}{\sqrt{5}} = \frac{a}{b} \), then \( \sqrt{5} = \frac{b}{a} \), implying \( \sqrt{5} \) is rational. Contradiction. - Solution:
Other number = \( \frac{12 \times 360}{60} = 72 \) - Solution:
Denominator \( 1600 = 2^6 \times 5^2 \)
Only primes 2 and 5.
Hence, decimal terminates. - Solution:
Similar to \( \sqrt{3} \).
Assume \( \sqrt{7} = \frac{a}{b} \)
Then \( 7b^2 = a^2 \)
7 divides \( a\), then 7 divides \( b \). Contradiction. - Solution:
Prime factors:
24 = \( 2^3 \times 3 \),
36 = \( 2^2 \times 3^2 \),
48 = \( 2^4 \times 3 \)
HCF = \( 2^2 \times 3 = 12 \)
LCM = \( 2^4 \times 3^2 = 144 \) - Solution:
HCF\((24, 18\)) = 6,
LCM\((24, 18\)) = 72
\( 6 \times 72 = 432 = 24 \times 18 \)
Verified. - Solution:
Let \( r \) be rational and \( i \) irrational.
Assume \( r + i = s \) (rational).
Then \( i = s – r \), which is rational. Contradiction. - Solution:
Assume \( 5 + 2\sqrt{2} \) is rational.
Then \( \sqrt{2} = \frac{\text{rational} – 5}{2} \),
implying \( \sqrt{2} \) is rational. Contradiction. - Solution:
LCM of 15, 20, 25 = LCM(15, 20) = 60; LCM(60, 25) = 300. - Solution:
\( 3 \times 5 \times 7 + 7 = 7(15 + 1) = 7 \times 16 \). Composite. - Solution:
\( 867 = 255 \times 3 + 102 \)
\( 255 = 102 \times 2 + 51 \)
\( 102 = 51 \times 2 + 0 \).
HCF = 51 - Solution:
Assume \( \sqrt{6} = \frac{a}{b} \)
Then \( 6b^2 = a^2 \)
Primes 2 and 3 divide \( a \), hence divide \( b \). Contradiction. - Solution:
1001 = 7 × 11 × 13. Not prime. - Solution:
Assume \( 7 – \sqrt{5} \) is rational.
Then \( \sqrt{5} = 7 – \text{rational} \), implying \( \sqrt{5} \) is rational. Contradiction. - Solution:
HCF of 18 and 12 = 6. Largest tile is 6m × 6m. - Solution:
Assume \( \sqrt{2} + \sqrt{3} = r \)
Squaring: \( 5 + 2\sqrt{6} = r^2 \)
Hence \( \sqrt{6} = \frac{r^2 – 5}{2} \), rational. Contradiction. - Solution:
If \( \frac{2}{\sqrt{7}} = \frac{a}{b} \), then \( \sqrt{7} = \frac{2b}{a} \), implying \( \sqrt{7} \) is rational. Contradiction. - Solution:
Prime factors:
96 = \( 2^5 \times 3 \)
144 = \( 2^4 \times 3^2 \)
HCF = \( 2^4 \times 3 = 48 \)
LCM = \( 2^5 \times 3^2 = 288 \) - Solution:
Simplify \( \frac{77}{210} = \frac{11}{30} \).
Denominator 30 = 2 × 3 × 5. Non-terminating (prime 3 present). - Solution:
Assume \( 3\sqrt{5} = \frac{a}{b} \)
Then \( \sqrt{5} = \frac{a}{3b} \), implying \( \sqrt{5} \) is rational. Contradiction.
The Bottom Line
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You can also check the related MCQs for this chapter here: MCQ on Real Numbers
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