CBSE Class 10 Mathematics Real Numbers Worksheet Set A - Free Printable
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Step-by-step solution for: CBSE Class 10 Mathematics Real Numbers Worksheet Set A
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Step-by-step solution for: CBSE Class 10 Mathematics Real Numbers Worksheet Set A
17. Complete the missing entries in the following factor tree.
- Top left box: 324
- Box below 324: 108
- Box to the right of 108: 36
- Box below 36: 12
- Box to the right of 12: 4
- Box below 4: 2
18. Which of the following has a terminating decimal expansion?
(a) 11/125
19. Which of the following are irrational?
(b) ∛8 is rational (equals 2), so not irrational.
(c) 5.27414141... is repeating, so rational.
(d) π is irrational.
The correct answer is (d) π.
20. Determine the values of p and q so that the prime factorization of 2520 is expressible as 2^p × 3^q × n.
- Prime factorization of 2520: 2³ × 3² × 5 × 7
- Therefore, p = 3, q = 2
21. Using Euclid’s division algorithm, whether the pair of numbers 847, 2160 are co-prime or not.
- GCD(847, 2160):
2160 = 2×847 + 466
847 = 1×466 + 381
466 = 1×381 + 85
381 = 4×85 + 41
85 = 2×41 + 3
41 = 13×3 + 2
3 = 1×2 + 1
2 = 2×1 + 0
- GCD = 1, so they are co-prime.
22. Using Euclid’s division lemma, show that the square of any positive integer is of the form 2m or 2m+1 for some integer m.
- Let a be any positive integer. By Euclid’s lemma, a = 2q or a = 2q+1 for some integer q.
- If a = 2q, then a² = 4q² = 2(2q²) = 2m, where m = 2q².
- If a = 2q+1, then a² = 4q² + 4q + 1 = 2(2q² + 2q) + 1 = 2m + 1, where m = 2q² + 2q.
- Thus, a² is either 2m or 2m+1.
23. Find the largest number of four digits exactly divisible by 12, 15, 18 and 27.
- LCM of 12, 15, 18, 27:
12 = 2²×3, 15=3×5, 18=2×3², 27=3³ → LCM = 2²×3³×5 = 540
- Largest 4-digit number: 9999
- 9999 ÷ 540 = 18 with remainder 279
- So, 9999 - 279 = 9720
- Answer: 9720
24. Using fundamental theorem of arithmetic, find the LCM and HCF of 816 and 170.
- Prime factorizations:
816 = 2⁴ × 3 × 17
170 = 2 × 5 × 17
- HCF = product of lowest powers of common primes = 2¹ × 17¹ = 34
- LCM = product of highest powers of all primes = 2⁴ × 3 × 5 × 17 = 16 × 3 × 5 × 17 = 2040
25. Let d be the HCF of 24 and 36. Find two numbers a and b, such that d = 24a + 36b.
- HCF of 24 and 36 is 12.
- Use extended Euclidean algorithm:
36 = 1×24 + 12 → 12 = 36 - 1×24
So, 12 = (-1)×24 + (1)×36
- Thus, a = -1, b = 1
26. Find two numbers which on multiplication with √180 gives the smallest rational number. Are these numbers rational or irrational?
- √180 = √(36×5) = 6√5
- To make it rational, multiply by √5: 6√5 × √5 = 6×5 = 30
- So, one number is √5. Another could be 1/√5, but the smallest rational number (positive) is 30 when multiplying by √5.
- Actually, the question asks for two numbers whose product with √180 is rational and smallest. The smallest positive rational is achieved by multiplying by √5, giving 30.
- But if we consider two numbers, say x and y, such that x * y * √180 is rational and minimal, we can take x = √5, y = 1, then product is 30.
- Alternatively, if we interpret as two numbers to multiply with √180 separately, then each should be √5 to get 30, but that's not minimal if we allow fractions.
- Actually, the smallest rational number obtainable is 0, by multiplying by 0, but that’s trivial.
- Probably intended: multiply by √5 to get 30. So one number is √5 (irrational).
- To get the smallest non-zero rational, multiply by √5. So the number is √5, which is irrational.
27. The product of three consecutive positive integers is divisible by 6. This statement true or false? Justify your answer.
- True.
- Among any three consecutive integers, at least one is divisible by 2 (even), and at least one is divisible by 3 (since every third number is divisible by 3).
- So the product is divisible by 2×3=6.
28. If n is an odd integer, then show that n² - 1 is divisible by 8.
- Let n = 2k+1 for some integer k.
- n² - 1 = (2k+1)² - 1 = 4k² + 4k + 1 - 1 = 4k(k+1)
- k(k+1) is always even because it's the product of two consecutive integers.
- So 4k(k+1) is divisible by 8.
29. Write whether 2√45 + 3√20 / 2√5 on simplification gives a rational or an irrational number.
- Simplify numerator: 2√45 = 2×3√5 = 6√5; 3√20 = 3×2√5 = 6√5
- Numerator: 6√5 + 6√5 = 12√5
- Denominator: 2√5
- So, 12√5 / 2√5 = 6
- 6 is rational.
30. Find the HCF and LCM of 288, 360 and 384 by prime factorization method.
- Prime factorizations:
288 = 2⁵ × 3²
360 = 2³ × 3² × 5
384 = 2⁷ × 3
- HCF = product of lowest powers of common primes = 2³ × 3¹ = 8×3 = 24
- LCM = product of highest powers of all primes = 2⁷ × 3² × 5 = 128 × 9 × 5 = 5760
- Top left box: 324
- Box below 324: 108
- Box to the right of 108: 36
- Box below 36: 12
- Box to the right of 12: 4
- Box below 4: 2
18. Which of the following has a terminating decimal expansion?
(a) 11/125
19. Which of the following are irrational?
(b) ∛8 is rational (equals 2), so not irrational.
(c) 5.27414141... is repeating, so rational.
(d) π is irrational.
The correct answer is (d) π.
20. Determine the values of p and q so that the prime factorization of 2520 is expressible as 2^p × 3^q × n.
- Prime factorization of 2520: 2³ × 3² × 5 × 7
- Therefore, p = 3, q = 2
21. Using Euclid’s division algorithm, whether the pair of numbers 847, 2160 are co-prime or not.
- GCD(847, 2160):
2160 = 2×847 + 466
847 = 1×466 + 381
466 = 1×381 + 85
381 = 4×85 + 41
85 = 2×41 + 3
41 = 13×3 + 2
3 = 1×2 + 1
2 = 2×1 + 0
- GCD = 1, so they are co-prime.
22. Using Euclid’s division lemma, show that the square of any positive integer is of the form 2m or 2m+1 for some integer m.
- Let a be any positive integer. By Euclid’s lemma, a = 2q or a = 2q+1 for some integer q.
- If a = 2q, then a² = 4q² = 2(2q²) = 2m, where m = 2q².
- If a = 2q+1, then a² = 4q² + 4q + 1 = 2(2q² + 2q) + 1 = 2m + 1, where m = 2q² + 2q.
- Thus, a² is either 2m or 2m+1.
23. Find the largest number of four digits exactly divisible by 12, 15, 18 and 27.
- LCM of 12, 15, 18, 27:
12 = 2²×3, 15=3×5, 18=2×3², 27=3³ → LCM = 2²×3³×5 = 540
- Largest 4-digit number: 9999
- 9999 ÷ 540 = 18 with remainder 279
- So, 9999 - 279 = 9720
- Answer: 9720
24. Using fundamental theorem of arithmetic, find the LCM and HCF of 816 and 170.
- Prime factorizations:
816 = 2⁴ × 3 × 17
170 = 2 × 5 × 17
- HCF = product of lowest powers of common primes = 2¹ × 17¹ = 34
- LCM = product of highest powers of all primes = 2⁴ × 3 × 5 × 17 = 16 × 3 × 5 × 17 = 2040
25. Let d be the HCF of 24 and 36. Find two numbers a and b, such that d = 24a + 36b.
- HCF of 24 and 36 is 12.
- Use extended Euclidean algorithm:
36 = 1×24 + 12 → 12 = 36 - 1×24
So, 12 = (-1)×24 + (1)×36
- Thus, a = -1, b = 1
26. Find two numbers which on multiplication with √180 gives the smallest rational number. Are these numbers rational or irrational?
- √180 = √(36×5) = 6√5
- To make it rational, multiply by √5: 6√5 × √5 = 6×5 = 30
- So, one number is √5. Another could be 1/√5, but the smallest rational number (positive) is 30 when multiplying by √5.
- Actually, the question asks for two numbers whose product with √180 is rational and smallest. The smallest positive rational is achieved by multiplying by √5, giving 30.
- But if we consider two numbers, say x and y, such that x * y * √180 is rational and minimal, we can take x = √5, y = 1, then product is 30.
- Alternatively, if we interpret as two numbers to multiply with √180 separately, then each should be √5 to get 30, but that's not minimal if we allow fractions.
- Actually, the smallest rational number obtainable is 0, by multiplying by 0, but that’s trivial.
- Probably intended: multiply by √5 to get 30. So one number is √5 (irrational).
- To get the smallest non-zero rational, multiply by √5. So the number is √5, which is irrational.
27. The product of three consecutive positive integers is divisible by 6. This statement true or false? Justify your answer.
- True.
- Among any three consecutive integers, at least one is divisible by 2 (even), and at least one is divisible by 3 (since every third number is divisible by 3).
- So the product is divisible by 2×3=6.
28. If n is an odd integer, then show that n² - 1 is divisible by 8.
- Let n = 2k+1 for some integer k.
- n² - 1 = (2k+1)² - 1 = 4k² + 4k + 1 - 1 = 4k(k+1)
- k(k+1) is always even because it's the product of two consecutive integers.
- So 4k(k+1) is divisible by 8.
29. Write whether 2√45 + 3√20 / 2√5 on simplification gives a rational or an irrational number.
- Simplify numerator: 2√45 = 2×3√5 = 6√5; 3√20 = 3×2√5 = 6√5
- Numerator: 6√5 + 6√5 = 12√5
- Denominator: 2√5
- So, 12√5 / 2√5 = 6
- 6 is rational.
30. Find the HCF and LCM of 288, 360 and 384 by prime factorization method.
- Prime factorizations:
288 = 2⁵ × 3²
360 = 2³ × 3² × 5
384 = 2⁷ × 3
- HCF = product of lowest powers of common primes = 2³ × 3¹ = 8×3 = 24
- LCM = product of highest powers of all primes = 2⁷ × 3² × 5 = 128 × 9 × 5 = 5760
Parent Tip: Review the logic above to help your child master the concept of sets of real numbers worksheet.
