90295
The number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\)
1 )an integer
2 )not a real number
3 )an irrational number
4 )a rational number
Explanation:
Exp: C an irrational number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\) \(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\)\(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\) \(=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\) \(=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}\) \(=\frac{5+2+2\sqrt{10}}{3}\) \(=\frac{7+2\sqrt{10}}{3}\) Here\(\sqrt{10}=\sqrt{2}\times\sqrt{5}\) since\(\sqrt{2}\) and \(\sqrt{5}\) both are an irrational number Therefore, \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}\) is an irrational number. Applying
REAL NUMBERS
90296
On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n - 1) is divided by 9?
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 On dividing n by 9 the remainder is 7 \(\Rightarrow\)n = 9q + 7, where q is the quotient \(\Rightarrow\)3n = 3(9q + 7) \(\Rightarrow\)3n = 27q + 21 \(\Rightarrow\)3n - 1 = 27q + 21 - 1 \(\Rightarrow\)3n - 1 = 27q + 20 \(\Rightarrow\)3n - 1 = 27q + 18 + 2 \(\Rightarrow\)3n - 1 = 9(3q + 2) + 2 So, the remainder will be 2 Never Active Hard **[Real Numbers]**
REAL NUMBERS
90299
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
1 )2
2 )3
3 )4
4 )1
Explanation:
Exp: C 4 LCM (a, 18) = 36 HCF (a, 18) = 2 We know that the product of numbers is equal to the product of their HCF and LCM. Therefore, 18a = 2(36) \(\text{a}=\frac{2(36)}{18}\) a = 4 Hence the correct choice is (c).
REAL NUMBERS
90301
The HCF and the LCM of 12, 21, 15 respectively are:
90295
The number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\)
1 )an integer
2 )not a real number
3 )an irrational number
4 )a rational number
Explanation:
Exp: C an irrational number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\) \(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\)\(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\) \(=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\) \(=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}\) \(=\frac{5+2+2\sqrt{10}}{3}\) \(=\frac{7+2\sqrt{10}}{3}\) Here\(\sqrt{10}=\sqrt{2}\times\sqrt{5}\) since\(\sqrt{2}\) and \(\sqrt{5}\) both are an irrational number Therefore, \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}\) is an irrational number. Applying
REAL NUMBERS
90296
On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n - 1) is divided by 9?
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 On dividing n by 9 the remainder is 7 \(\Rightarrow\)n = 9q + 7, where q is the quotient \(\Rightarrow\)3n = 3(9q + 7) \(\Rightarrow\)3n = 27q + 21 \(\Rightarrow\)3n - 1 = 27q + 21 - 1 \(\Rightarrow\)3n - 1 = 27q + 20 \(\Rightarrow\)3n - 1 = 27q + 18 + 2 \(\Rightarrow\)3n - 1 = 9(3q + 2) + 2 So, the remainder will be 2 Never Active Hard **[Real Numbers]**
REAL NUMBERS
90299
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
1 )2
2 )3
3 )4
4 )1
Explanation:
Exp: C 4 LCM (a, 18) = 36 HCF (a, 18) = 2 We know that the product of numbers is equal to the product of their HCF and LCM. Therefore, 18a = 2(36) \(\text{a}=\frac{2(36)}{18}\) a = 4 Hence the correct choice is (c).
REAL NUMBERS
90301
The HCF and the LCM of 12, 21, 15 respectively are:
90295
The number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\)
1 )an integer
2 )not a real number
3 )an irrational number
4 )a rational number
Explanation:
Exp: C an irrational number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\) \(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\)\(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\) \(=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\) \(=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}\) \(=\frac{5+2+2\sqrt{10}}{3}\) \(=\frac{7+2\sqrt{10}}{3}\) Here\(\sqrt{10}=\sqrt{2}\times\sqrt{5}\) since\(\sqrt{2}\) and \(\sqrt{5}\) both are an irrational number Therefore, \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}\) is an irrational number. Applying
REAL NUMBERS
90296
On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n - 1) is divided by 9?
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 On dividing n by 9 the remainder is 7 \(\Rightarrow\)n = 9q + 7, where q is the quotient \(\Rightarrow\)3n = 3(9q + 7) \(\Rightarrow\)3n = 27q + 21 \(\Rightarrow\)3n - 1 = 27q + 21 - 1 \(\Rightarrow\)3n - 1 = 27q + 20 \(\Rightarrow\)3n - 1 = 27q + 18 + 2 \(\Rightarrow\)3n - 1 = 9(3q + 2) + 2 So, the remainder will be 2 Never Active Hard **[Real Numbers]**
REAL NUMBERS
90299
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
1 )2
2 )3
3 )4
4 )1
Explanation:
Exp: C 4 LCM (a, 18) = 36 HCF (a, 18) = 2 We know that the product of numbers is equal to the product of their HCF and LCM. Therefore, 18a = 2(36) \(\text{a}=\frac{2(36)}{18}\) a = 4 Hence the correct choice is (c).
REAL NUMBERS
90301
The HCF and the LCM of 12, 21, 15 respectively are:
90295
The number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\)
1 )an integer
2 )not a real number
3 )an irrational number
4 )a rational number
Explanation:
Exp: C an irrational number \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\) \(=\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}-\sqrt{2}}\times\)\(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\) \(=\frac{\big(\sqrt{5}+\sqrt{2}\big)^2}{\big(\sqrt{5}\big)^2-\big(\sqrt{2}\big)^2}\) \(=\frac{\big(\sqrt{5}\big)^2+\big(\sqrt{2}\big)^2+2\times\sqrt{5}\times\sqrt{2}}{5-2}\) \(=\frac{5+2+2\sqrt{10}}{3}\) \(=\frac{7+2\sqrt{10}}{3}\) Here\(\sqrt{10}=\sqrt{2}\times\sqrt{5}\) since\(\sqrt{2}\) and \(\sqrt{5}\) both are an irrational number Therefore, \(\frac{\sqrt{5}+\sqrt{2}}{\sqrt{5-\sqrt{2}}}\) is an irrational number. Applying
REAL NUMBERS
90296
On dividing a positive integer n by 9, we get 7 as remainder. What will be the remainder if (3n - 1) is divided by 9?
1 )1
2 )2
3 )3
4 )4
Explanation:
Exp: B 2 On dividing n by 9 the remainder is 7 \(\Rightarrow\)n = 9q + 7, where q is the quotient \(\Rightarrow\)3n = 3(9q + 7) \(\Rightarrow\)3n = 27q + 21 \(\Rightarrow\)3n - 1 = 27q + 21 - 1 \(\Rightarrow\)3n - 1 = 27q + 20 \(\Rightarrow\)3n - 1 = 27q + 18 + 2 \(\Rightarrow\)3n - 1 = 9(3q + 2) + 2 So, the remainder will be 2 Never Active Hard **[Real Numbers]**
REAL NUMBERS
90299
If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =
1 )2
2 )3
3 )4
4 )1
Explanation:
Exp: C 4 LCM (a, 18) = 36 HCF (a, 18) = 2 We know that the product of numbers is equal to the product of their HCF and LCM. Therefore, 18a = 2(36) \(\text{a}=\frac{2(36)}{18}\) a = 4 Hence the correct choice is (c).
REAL NUMBERS
90301
The HCF and the LCM of 12, 21, 15 respectively are:
