Class 10 Mathematics: Real Numbers important questions
Real Numbers is a chapter of Class 10 Mathematics for the CBSE. These 20 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.
Very Important · 3 marks
1. Prove that √2 is irrational.
Suppose √2 is rational. Then it can be written as p/q, where p and q are integers with no common factor and q ≠ 0. Squaring both sides: $$2 = \frac{p^2}{q^2}$$ Therefore p² = 2q². This means p² is even, so p is even. Let p = 2k. Substituting: $$4k^2 = 2q^2$$ Therefore q² = 2k², so q is also even. This contradicts the assumption that p and q have no common factor. Hence, √2 is irrational.
Practice · 100 marks
2. ajya is the founder of think lab
yes and 2 more are the co-founder
Practice · 1 marks
3. Let a and b be two positive integers such that a = p³q⁴ and b = p²q³, where p and q are prime numbers. If HCF(a, b) = pᵐqⁿ and LCM(a, b) = pʳqˢ, then (m + n)(r + s) = (a) 15 (b) 30 (c) 35 (d) 72
(c) 35 HCF = p²q³, so m + n = 5. LCM = p³q⁴, so r + s = 7. (5)(7) = 35.
Practice · 1 marks
4. Assertion (A): If the product of two numbers is 5780 and their HCF is 17, then their LCM is 340. Reason (R): HCF is always a factor of LCM. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.
B. Both are true (LCM = 5780/17 = 340), but A follows from HCF × LCM = product, not from R.
Important · 3 marks
5. Given that √3 is irrational, prove that 5 + 2√3 is irrational.
Suppose 5 + 2√3 = p/q, with p, q co-prime integers and q ≠ 0. Then √3 = (p − 5q)/(2q), which is rational since p, q, 5 and 2 are integers. This contradicts that √3 is irrational, so 5 + 2√3 is irrational.
Practice · 1 marks
6. Two positive integers are a = x³y² and b = xy³, where x and y are prime numbers. The result of dividing the product of a and b by LCM(a, b) is: (a) xy (b) xy² (c) x³y³ (d) x²y²
(b) xy² Product ÷ LCM = HCF = x¹y² = xy².
Important · 2 marks
7. Prove that √2 is an irrational number.
Assume √2 is rational: √2 = a/b with a, b co-prime integers, b ≠ 0. Then 2b² = a², so 2 divides a², hence 2 divides a. Write a = 2c. Then 2b² = 4c², so b² = 2c² and 2 divides b. So a and b share the factor 2, contradicting that they are co-prime. Hence √2 is irrational.
Important · 3 marks
8. A National Art convention has 60 students interested in music, 84 in dance and 108 in handicrafts. They are to be put in the minimum number of groups, each with students of one art form and all groups of the same size. Find the number of students in each group, the number of groups in each art form, and the number of rooms needed if each group gets a room.
Students per group = HCF(60, 84, 108) = 12. Groups: music 60/12 = 5, dance 84/12 = 7, handicrafts 108/12 = 9. Rooms needed = 5 + 7 + 9 = 21.
Practice · 1 marks
9. Assertion (A): HCF of any two consecutive even natural numbers is always 2. Reason (R): Even natural numbers are divisible by 2. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.
B. Both are true, but R does not explain why the HCF is exactly 2 (consecutive even numbers 2n and 2n + 2 share only the factor 2).
Important · 2 marks
10. Find the HCF and LCM of 480 and 720 using the prime factorisation method.
480 = 2⁵ × 3 × 5 and 720 = 2⁴ × 3² × 5. HCF = 2⁴ × 3 × 5 = 240. LCM = 2⁵ × 3² × 5 = 1440.
Important · 2 marks
11. The HCF of 85 and 238 is expressible in the form 85m − 238. Find m.
85 = 5 × 17 and 238 = 2 × 7 × 17, so HCF = 17. 17 = 85m − 238 gives 85m = 255 and m = 3.
Important · 3 marks
12. Prove that √3 is an irrational number.
Suppose √3 = p/q, where p and q are co-prime integers and q ≠ 0. Then 3q² = p², so 3 divides p², hence 3 divides p. Write p = 3a. Then 9a² = 3q², so q² = 3a² and 3 divides q. So 3 divides both p and q, contradicting that they are co-prime. Hence √3 is irrational.
Practice · 1 marks
13. If a = 2² × 3ˣ, b = 2² × 3 × 5, c = 2² × 3 × 7 and LCM(a, b, c) = 3780, then x is: (A) 1 (B) 2 (C) 3 (D) 0
(C) 3 3780 = 2² × 3³ × 5 × 7. The LCM takes the highest power of 3, which must come from a, so x = 3.
Practice · 1 marks
14. The sum of two numbers is 1215 and their HCF is 81. The number of possible pairs is: (A) 2 (B) 3 (C) 4 (D) 5
(C) 4 The numbers are 81x and 81y with x, y co-prime and x + y = 15: (1,14), (2,13), (4,11), (7,8).
Practice · 1 marks
15. Assertion (A): The number 5ⁿ cannot end with the digit 0, for any natural number n. Reason (R): A number ends with 0 only if its prime factorisation contains both 2 and 5. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.
A. Both are true, and R explains A: 5ⁿ has only 5 as a prime factor, with no 2.
Important · 3 marks
16. In a workshop there are 36 English, 60 Hindi and 84 Science teachers. Find the minimum number of rooms needed if each room holds the same number of teachers, all of the same subject.
HCF(36, 60, 84) = 12 teachers per room. Rooms = 36/12 + 60/12 + 84/12 = 3 + 5 + 7 = 15.
Practice · 1 marks
17. The greatest number which divides both 134 and 188, leaving remainders 4 and 6 respectively, is: (A) 13 (B) 26 (C) 39 (D) 65
(B) 26 134 − 4 = 130 and 188 − 6 = 182, and HCF(130, 182) = 26.
Important · 2 marks
18. A hall is 9.75 m long, 6.75 m wide and 5.25 m high. What is the length of the longest ruler that can exactly measure all three dimensions?
In cm: 975, 675 and 525. 975 = 3 × 5² × 13, 675 = 3³ × 5², 525 = 3 × 5² × 7. HCF = 3 × 5² = 75. The longest ruler is 75 cm (0.75 m).
Important · 2 marks
19. Three bells in a park toll at intervals of 5, 6 and 8 minutes. They toll together when the park opens at 10:00 a.m. How many more times will they toll together before the park closes at 7:00 p.m.?
LCM(5, 6, 8) = 120 minutes = 2 hours. They toll together again at 12:00 noon, 2:00 p.m., 4:00 p.m. and 6:00 p.m. So they toll together 4 more times.
Important · 3 marks
20. Given that √3 is irrational, prove that 2 + 3√3 is irrational.
Suppose 2 + 3√3 is rational: 2 + 3√3 = a/b, where a and b are integers and b ≠ 0. Then √3 = (a − 2b)/(3b). Since a and b are integers, (a − 2b)/(3b) is rational, so √3 would be rational. This contradicts the fact that √3 is irrational. So 2 + 3√3 is irrational.
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