Answer  Hint:Terminating Decimal: A terminating decimal is usually defined as a decimal number that contains a finite number of digits after the decimal point.If x is a rational number whose simplest form is $\dfrac{p}{q}$, where p and q are integers and $q \ne 0$. Then,(i) x is terminating only when q is of the form $(2^m \times 5^n)$ for some non-negative integers m and n.(ii) x is non-terminating and repeating, if $q \neq (2^m \times 5^n)$. Complete step by step solution: $(i) \dfrac{33}{375}$If we do prime factorization of denominator, we get,$\dfrac{33}{375}=\dfrac{33}{3 \times 5^3}$The denominator of the fraction contains factors of 3 and 5. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.$(ii) \dfrac{15}{28}$If we do prime factorization of denominator, we get,$\dfrac{15}{28}=\dfrac{15}{2^2 \times 7}$The denominator of the fraction contains factors of 2 and 7. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.$(iii) \dfrac{16}{45}$If we do prime factorization of denominator, we get,$\dfrac{16}{45}=\dfrac{16}{3^2 \times 5}$The denominator of the fraction contains factors of 3 and 5. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.$(iv) \dfrac{12}{35}$If we do prime factorization of denominator, we get,$\dfrac{12}{35}=\dfrac{12}{5 \times 7}$The denominator of the fraction contains factors of 5 and 7. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.$(v) \dfrac{80}{27}$If we do prime factorization of denominator, we get,$\dfrac{80}{27}=\dfrac{80}{3^3}$The denominator of the fraction contains factors of 3. Since the denominator of given fraction does not have only factors of 2 and factors of 5, i.e. denominator is not of the form $(2^m \times 5^n)$, thus, the given fraction does not have terminating decimal representation.$(vi) \dfrac{123}{1250}$If we do prime factorization of denominator, we get,$\dfrac{123}{1250}=\dfrac{123}{2 \times 5^4}$The denominator of the fraction contains factors of 2 and 5. Since the denominator of given fraction has only factors of 2 and factors of 5, i.e. denominator is of the form $(2^m \times 5^n)$, thus, the given fraction have terminating decimal representation.Hence, only (vi) fractions have terminating decimal representation. Note: In these types of questions, we just check whether the prime factorization of the denominator of the fraction has only factors of 2 and 5. If it has only factors of 2 and 5, we can conclude that it is a terminating decimal.No worries! We‘ve got your back. Try BYJU‘S free classes today! No worries! We‘ve got your back. Try BYJU‘S free classes today! Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses No worries! We‘ve got your back. Try BYJU‘S free classes today! Open in App Suggest Corrections 1 Q. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. Which of the following rational numbers have terminating decimal?(i) 16225 (ii) 518 (iii) 221 (iv) 7250 (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (i) and (iv) (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is (a) 203400 (b) 194400 (c) 198400(d) 205400 If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n = (a) 1 (b) 2 (c) 3(d) 4 The exponent of 2 in the prime factorisation of 144, is (a) 4 (b) 5 (c) 6(d) 3 If n is any natural number, then 6n − 5n always ends with (a) 1 (b) 3 (c) 5 (d) 7[Hint: For any n ∈ N, 6n and 5n end with 6 and 5 respectively. Therefore, 6n − 5n always ends with 6 − 5 = 1.] Q. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. Which of the following rational numbers have terminating decimal?(i) 16225 (ii) 518 (iii) 221 (iv) 7250 (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (i) and (iv) (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is (a) 203400 (b) 194400 (c) 198400(d) 205400 If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n = (a) 1 (b) 2 (c) 3(d) 4 The exponent of 2 in the prime factorisation of 144, is (a) 4 (b) 5 (c) 6(d) 3 If n is any natural number, then 6n − 5n always ends with (a) 1 (b) 3 (c) 5 (d) 7[Hint: For any n ∈ N, 6n and 5n end with 6 and 5 respectively. Therefore, 6n − 5n always ends with 6 − 5 = 1.] Q. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. Which of the following rational numbers have terminating decimal?(i) 16225 (ii) 518 (iii) 221 (iv) 7250 (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (i) and (iv) (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is (a) 203400 (b) 194400 (c) 198400(d) 205400 If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n = (a) 1 (b) 2 (c) 3(d) 4 The exponent of 2 in the prime factorisation of 144, is (a) 4 (b) 5 (c) 6(d) 3 If n is any natural number, then 6n − 5n always ends with (a) 1 (b) 3 (c) 5 (d) 7[Hint: For any n ∈ N, 6n and 5n end with 6 and 5 respectively. Therefore, 6n − 5n always ends with 6 − 5 = 1.] Q. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. Which of the following rational numbers have terminating decimal?(i) 16225 (ii) 518 (iii) 221 (iv) 7250 (a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (i) and (iv) (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. (i) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (ii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iii) We have, Theorem states: Let Then, x has a decimal expression which does not have terminating decimal. (iv) We have, Theorem states: Let Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n. Then, x has a decimal expression which will have terminating decimal after 3 places of decimal. Hence the (iv) option will have terminating decimal expansion. There is no correct option. If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is (a) 203400 (b) 194400 (c) 198400(d) 205400 If a = 23 ✕ 3, b = 2 ✕ 3 ✕ 5, c = 3n ✕ 5 and LCM (a, b, c) = 23 ✕ 32 ✕ 5, then n = (a) 1 (b) 2 (c) 3(d) 4 The exponent of 2 in the prime factorisation of 144, is (a) 4 (b) 5 (c) 6(d) 3 If n is any natural number, then 6n − 5n always ends with (a) 1 (b) 3 (c) 5 (d) 7[Hint: For any n ∈ N, 6n and 5n end with 6 and 5 respectively. Therefore, 6n − 5n always ends with 6 − 5 = 1.] |
Which two of the following rational numbers have a terminating decimal representation?
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