Real Number
Previous Year Solved Questions 2023
(a) 1:2
(b) 2:1
(c) 1:1
(d) 1:3
View Answer
Least composite number = 4
Least prime number = 2
HCF = 2, LCM = 4
Required ratio = HCF/LCM = 2/4 = 1:2
Q2: Find the least number which when divided by 12, 16, and 24 leaves the remainder 7 in each case. (CBSE 2023)
View Answer
Least number = LCM(12,16,24) + 7
LCM(12,16,24) = 48
48 + 7 = 55
Q3: Two numbers are in the ratio 2:3 and their LCM is 180. What is the HCF of these numbers? (CBSE 2023)
View Answer
Let numbers be 2x and 3x
LCM(2x, 3x) = 6x
HCF(2x, 3x) = x
Given: 6x = 180
x = 180/6 = 30
HCF = x = 30
Q4: Prove that \(\sqrt{3}\) is an irrational number. (CBSE 2023)
View Answer
Assume \(\sqrt{3}\) is rational: \(\sqrt{3} = \frac{a}{b}\) where a,b are co-prime integers, b≠0
Squaring: \(3 = \frac{a^2}{b^2}\) ⇒ \(a^2 = 3b^2\)
Thus 3 divides \(a^2\) ⇒ 3 divides a
Let a = 3c
Then: \((3c)^2 = 3b^2\) ⇒ \(9c^2 = 3b^2\) ⇒ \(b^2 = 3c^2\)
Thus 3 divides \(b^2\) ⇒ 3 divides b
Both a and b divisible by 3, contradicting co-prime assumption
Hence \(\sqrt{3}\) is irrational.
5. Check whether \(6^n\) can end with the digit 0 for any natural number n. [NCERT, CBSE 2023]
View Answer
\(6^n = (2 \times 3)^n = 2^n \times 3^n\)
Prime factorization of \(6^n\) does not contain 5 as a factor.
For a number to end with digit 0, its prime factorization must contain both 2 and 5.
Hence \(6^n\) can never end with digit 0.
6. Find the greatest number which divides 85 and 72 leaving remainders 1 and 2 respectively. [CBSE 2023]
View Answer
When number divides 85, remainder 1 ⇒ 85-1=84 divisible
When number divides 72, remainder 2 ⇒ 72-2=70 divisible
So number is HCF of 84 and 70
84 = \(2^2 \times 3 \times 7\)
70 = \(2 \times 5 \times 7\)
HCF = \(2 \times 7 = 14\)
7. Three bells ring at intervals of 6, 12 and 18 minutes. If all the three bells rang at 6 AM, when will they ring together again? [CBSE 2023]
View Answer
Find LCM of 6, 12, 18:
6 = \(2 \times 3\)
12 = \(2^2 \times 3\)
18 = \(2 \times 3^2\)
LCM = \(2^2 \times 3^2 = 4 \times 9 = 36\) minutes
6:00 AM + 36 minutes = 6:36 AM
8. Find the LCM and HCF of 96 and 120 and verify that LCM × HCF = Product of the integers. [CBSE 2023]
View Answer
96 = \(2^5 \times 3\)
120 = \(2^3 \times 3 \times 5\)
HCF = \(2^3 \times 3 = 8 \times 3 = 24\)
LCM = \(2^5 \times 3 \times 5 = 32 \times 3 \times 5 = 480\)
Verification: LCM × HCF = \(480 \times 24 = 11520\)
Product = \(96 \times 120 = 11520\)
Verified: \(11520 = 11520\)
9. Find the LCM and HCF of 26, 65 and 117 by prime factorisation method. [CBSE 2023]
View Answer
26 = \(2 \times 13\)
65 = \(5 \times 13\)
117 = \(3^2 \times 13\)
HCF = 13 (only common factor)
LCM = \(2 \times 3^2 \times 5 \times 13 = 2 \times 9 \times 5 \times 13 = 1170\)
10. Find by prime factorisation the LCM of 18180 and 7575. Also find the HCF. [CBSE 2023]
View Answer
18180 = \(2^2 \times 3^2 \times 5 \times 101\)
7575 = \(3 \times 5^2 \times 101\)
HCF = \(3 \times 5 \times 101 = 1515\)
LCM = \(2^2 \times 3^2 \times 5^2 \times 101 = 4 \times 9 \times 25 \times 101 = 90900\)
11. The traffic lights at three different road crossings change after every 48, 72 and 108 seconds. If they change simultaneously at 7 a.m., at what time will they change together next? [CBSE 2023]
View Answer
LCM of 48, 72, 108:
48 = \(2^4 \times 3\)
72 = \(2^3 \times 3^2\)
108 = \(2^2 \times 3^3\)
LCM = \(2^4 \times 3^3 = 16 \times 27 = 432\) seconds
432 seconds = 7 minutes 12 seconds
7:00 a.m. + 7 min 12 sec = 7:07:12 a.m.
12. Prove that \(\sqrt{2}\) is an irrational number. [NCERT, CBSE 2010, 2023]
View Answer
Assume \(\sqrt{2}\) is rational: \(\sqrt{2} = \frac{a}{b}\) where a,b are co-prime integers, b≠0
Squaring: \(2 = \frac{a^2}{b^2}\) ⇒ \(a^2 = 2b^2\)
Thus 2 divides \(a^2\) ⇒ 2 divides a
Let a = 2c
Then: \((2c)^2 = 2b^2\) ⇒ \(4c^2 = 2b^2\) ⇒ \(b^2 = 2c^2\)
Thus 2 divides \(b^2\) ⇒ 2 divides b
Both a and b divisible by 2, contradicting co-prime assumption
Hence \(\sqrt{2}\) is irrational.