CBSE Class 10 Mathematics Real Numbers VBQs Set B

Question. The sum of exponents of prime factors in the prime-factorisation of 196 is:
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b)
Explanation: The prime factorisation of 196 is: \( 196 = 2^2 \times 7^2 \). So, sum of the exponents of prime factors 2 and 7 is \( 2 + 2 \) i.e., 4.

Question. The total number of factors of a prime number is
(a) 1
(b) 0
(c) 2
(d) 3
Answer: (c)
Explanation: Factors of a prime number are 1 and the number itself.

Question. The HCF and the LCM of 12, 21, 15 respectively are
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 420, 3
Answer: (c)
Explanation: Here, \( 12 = 2^2 \times 3 \), \( 21 = 3 \times 7 \), \( 15 = 3 \times 5 \). So, HCF = 3; LCM = \( 2^2 \times 3 \times 7 \times 5 \) i.e., 420.

Question. The decimal representation of \( \frac{11}{2^3 \times 5} \) will:
(a) terminate after 1 decimal place
(b) terminate after 2 decimal places
(c) terminate after 3 decimal places
(d) not terminate
Answer: (c)
Explanation: The decimal representation of \( \frac{11}{2^3 \times 5} \) will terminate after 3 decimal places because the highest power of 2 or 5 in the denominator is 3.

Question. The LCM of smallest two digit composite number and smallest composite number is:
(a) 12
(b) 4
(c) 20
(d) 44
Answer: (c)
Explanation: Smallest two-digit composite number is 10. Smallest composite number is 4. LCM(10, 4) = 20.

Question. If two positive integers \( a \) and \( b \) are written as \( a = x^3y^2 \) and \( b = xy^3 \), where \( x \) and \( y \) are prime numbers, then the HCF \( (a, b) \) is:
(a) \( xy \)
(b) \( xy^2 \)
(c) \( x^3y^3 \)
(d) \( x^2y^2 \)
Answer: (b)
Explanation: Given that \( a = x^3y^2 = x \times x \times x \times y \times y \) and \( b = xy^3 = x \times y \times y \times y \). \( \Rightarrow \) HCF of \( a \) and \( b = \text{HCF}(x^3y^2, xy^3) = x \times y \times y = xy^2 \). We know that HCF is the product of the smallest power of each common prime factor involved in the numbers.

Question. If two positive integers \( p \) and \( q \) can be expressed as \( p = ab^2 \) and \( q = a^3b \) where \( a \) and \( b \) are prime numbers, then the LCM \( (p, q) \) is:
(a) \( ab \)
(b) \( a^2b^2 \)
(c) \( a^3b^2 \)
(d) \( a^3b^3 \)
Answer: (c)
Explanation: Given that \( p = ab^2 = a \times b \times b \) and \( q = a^3b = a \times a \times a \times b \). We know that LCM is the product of the greatest power of each prime factor of the numbers. \( \Rightarrow \) LCM of \( p \) and \( q = \text{LCM}(ab^2, a^3b) = a \times a \times a \times b \times b = a^3b^2 \).

Question. \( 7 \times 11 \times 13 \times 15 + 15 \) is a:
(a) Composite number
(b) Whole number
(c) Prime number
(d) (a) and (b) both
Answer: (d)
Explanation: \( 7 \times 11 \times 13 \times 15 + 15 = 15(7 \times 11 \times 13 + 1) = 15 \times 1002 \). Also \( 15 \times 1002 \) is a whole number. The number having factors more than two therefore, this is composite number and whole number.

Question. LCM of \( (2^3 \times 3 \times 5) \) and \( (2^4 \times 5 \times 7) \) is
(a) 40
(b) 560
(c) 1120
(d) 1680
Answer: (d)
 Explanation: \( (2^3 \times 3 \times 5) \) and \( (2^4 \times 5 \times 7) \). LCM = \( 2^4 \times 3 \times 5 \times 7 = 1680 \).

Question. 1.23451326... is
(a) an integer
(b) an irrational number
(c) a rational number
(d) none of these
Answer: (b)
Explanation: Number neither terminating nor repeated, therefore this is an irrational number.

Question. If the LCM of \( a \) and 18 is 36 and the HCF of \( a \) and 18 is 2, then \( a = \)
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d)
Explanation: We know that, \( \text{LCM}(a, b) \times \text{HCF}(a, b) = a \times b \). \( \Rightarrow 36 \times 2 = a \times 18 \Rightarrow a = \frac{36 \times 2}{18} \Rightarrow a = 4 \).

Question. The product of a non-zero rational and an irrational number is:
(a) always irrational
(b) always rational
(c) rational or irrational
(d) one
Answer: (a)
Explanation: Product of a non-zero rational and an irrational number is always irrational. For example: \( \frac{7}{9} \) is rational and \( \sqrt{2} \) is irrational numbers. Their product is \( \frac{7\sqrt{2}}{9} \), which is an irrational number.

Question. The number of decimal places after which the decimal expansion of the rational number \( \frac{9}{2^4 \times 5} \) will terminate, is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer: (d)
Explanation: Number is \( \frac{9}{2^4 \times 5} = \frac{9 \times 5^3}{2^4 \times 5^4} = \frac{1125}{10^4} = 0.1125 \). Therefore, number terminate after 4 decimal places.

Question. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is:
(a) 10
(b) 100
(c) 504
(d) 2520
Answer: (d)
Explanation: As we require the least number, the problem is based on finding the LCM. Factors of 1 to 10 numbers are as follows: \( 1 = 1 \), \( 2 = 1 \times 2 \), \( 3 = 1 \times 3 \), \( 4 = 1 \times 2 \times 2 \), \( 5 = 1 \times 5 \), \( 6 = 1 \times 2 \times 3 \), \( 7 = 1 \times 7 \), \( 8 = 1 \times 2 \times 2 \times 2 \), \( 9 = 1 \times 3 \times 3 \), \( 10 = 1 \times 2 \times 5 \). LCM of number 1 to 10 = \( \text{LCM}(1, 2, 3, 4, 5, 6, 7, 8, 9, 10) = 1 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 2520 \).

Question. The decimal expansion of the rational number \( \frac{14587}{1250} \) will terminate after:
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal places
Answer: (d)
Explanation: \( \frac{14587}{1250} = \frac{14587}{5^4 \times 2} \). Simplifying the given fraction: \( \Rightarrow \frac{14587}{5^4 \times 2} \times \frac{2^3}{2^3} = \frac{14587 \times 8}{5^4 \times 2^4} = \frac{116696}{10^4} = 11.6696 \). Hence, the given rational number will terminate after four decimal places.

Question. If \( \text{HCF}(a, b) = 45 \) and \( a \times b = 30375 \), then \( \text{LCM}(a, b) \) is:
(a) 1875
(b) 1350
(c) 625
(d) 675
Answer: (d)
We know that, \( \text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} \). So, \( \text{LCM}(a, b) = \frac{30375}{45} = 675 \).

Question. The cube of any positive integer is not of the form:
(a) \( 9q \)
(b) \( 9q + 1 \)
(c) \( 9q + 3 \)
(d) \( 9q + 8 \)
Answer: (c)
The cube of any positive integer is of the form \( 9q \) or \( 9q + 1 \) or \( 9q + 8 \). So, \( 9q + 3 \) is incorrect.

Question. 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75, what is the HCF of (525, 3000)?
(a) 25
(b) 125
(c) 75
(d) 15
Answer: (c)
Explanation: Since 3, 5, 15, 25 and 75 are the only common factors of 525 and 3000, 75 is the HCF.

Question. If HCF of two numbers is 1, the numbers are called relatively ________ or ________.
(a) Prime, co-prime
(b) Composite, prime
(c) Both (a) and (b)
(d) None of the above
Answer: (a)
Explanation: Prime numbers are those numbers which have only two factors i.e., 1 and itself. Example, 3, 5, 11 etc. Co-prime numbers: Two numbers that have only 1 as a common factor. Example, 35 and 39. \( 35 = 1 \times 5 \times 7 \), \( 39 = 1 \times 3 \times 13 \). Here, common factor is 1.

Fill in the Blanks

Question. \( \left( \frac{2 + \sqrt{5}}{3} \right) \) is ________ number.
Answer: irrational. Explanation: As \( \sqrt{5} \) is irrational, \( 2 + \sqrt{5} \) is irrational.

Question. The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, the other number is ________.
Answer: 81. Explanation: HCF of two numbers is 27 and their LCM is 162. Let the other number be \( x \). Product of two numbers = \( \text{HCF} \times \text{LCM} = 27 \times 162 \Rightarrow 54x = 27 \times 162 \Rightarrow x = 81 \).

Question. If \( a = (2^2 \times 3^3 \times 5^4) \) and \( b = (2^3 \times 3^2 \times 5) \) then \( \text{HCF}(a, b) = \) ________.
Answer: 180. Explanation: \( a = (2^2 \times 3^3 \times 5^4) \), \( b = (2^3 \times 3^2 \times 5) \). \( \text{HCF}(a, b) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180 \).

Question. A decimal number \( 0.\bar{8} \) can be expressed in its simplest form as ________.
Answer: \( \frac{8}{9} \). Explanation: Let \( x = 0.88888... \). \( 10x = 8.88888... \Rightarrow 10x - x = 8 \Rightarrow 9x = 8 \Rightarrow x = \frac{8}{9} \).

Question. Product of two numbers is 18144 and their HCF is 6, then their LCM is ________.
Answer: 324. Explanation: Product of two numbers = 18144. HCF of two numbers is 6. Product of two numbers = \( \text{HCF} \times \text{LCM} = 18144 \Rightarrow 6 \times \text{LCM} = 18144 \Rightarrow \text{LCM} = \frac{18144}{6} = 324 \).

Question. The decimal expression of the rational number \( \frac{23}{2^2 \times 5} \) will terminate after ________ decimal place(s).
Answer: 2. Explanation: Here the power of 2 is 2 and the power of 5 is 1. \( 2 > 1 \). Hence, \( \frac{23}{2^2 \times 5} \) has terminating decimal expansion which terminates after 2 places of decimals.

Question. The HCF of smallest composite number and the smallest prime number is ________.
Answer: 2. Explanation: Smallest prime number = 2. Smallest composite number = 4. HCF(2, 4) = 2.

Question. If \( a \) and \( b \) are positive integers, then \( \frac{\text{HCF}(a, b) \times \text{LCM}(a, b)}{ab} = \) ________.
Answer: 1. Explanation: \( \text{HCF}(a, b) \times \text{LCM}(a, b) = ab \Rightarrow \frac{\text{HCF}(a, b) \times \text{LCM}(a, b)}{ab} = 1 \).

Question. ________ is the H.C.F. of two consecutive even numbers.
Answer: 2. Explanation: All even numbers are divisible by 2. Therefore, HCF of two consecutive even numbers is 2.

Question. If two positive integers \( p \) and \( q \) can be expressed as \( p = a^2b^3 \) and \( q = a^4b \); \( a, b \) being prime numbers, then \( \text{LCM}(p, q) \) is ________.
Answer: \( a^4b^3 \).

Very Short Answer Type Questions

Question. The LCM of two numbers is 182 and their HCF is 13. If one of the number is 26, find the other.
Answer: We know that \( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \). So, \( 13 \times 182 = 26 \times b \Rightarrow b = \frac{13 \times 182}{26} = 91 \). Thus, the other number is 91.

Question. Given that \( \text{HCF}(135, 225) = 45 \), find the \( \text{LCM}(135, 225) \).
Answer: We know that \( \text{LCM} \times \text{HCF} = \text{Product of two numbers} \). \( \therefore \text{LCM}(135, 225) = \frac{\text{Product of 135 and 225}}{\text{HCF}(135, 225)} = \frac{135 \times 225}{45} = 675 \).

Question. After how many decimal places will the decimal representation of the rational number \( \frac{229}{2^2 \times 5^7} \) terminate?
Answer: Here, \( \frac{229}{2^2 \times 5^7} = \frac{229 \times 2^5}{2^7 \times 5^7} = \frac{229 \times 32}{(10)^7} \). Hence, the given rational number will terminate after 7 decimal places.

Question. Are the smallest prime and the smallest composite numbers co-prime? Justify.
Answer: No. We know that, smallest prime number is 2 and smallest composite number is 4. HCF of (2, 4) = 2. Since, there is a common factor 2, they are not co-prime.

Question. The HCF of two numbers \( a \) and \( b \) is 5 and their LCM is 200. Find the product \( ab \).
Answer: Given, \( \text{HCF}(a, b) = 5 \), \( \text{LCM}(a, b) = 200 \). \( \text{HCF} \times \text{LCM} = \text{Product of the numbers} \Rightarrow a \times b = 5 \times 200 \Rightarrow ab = 1000 \). Hence, the product \( ab \) is 1000.

Question. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons.
Answer: No. We know that: “The HCF of any two numbers must be a factor of the LCM of those numbers.” So, two numbers cannot have their HCF 18 and LCM 380, as 18 does not divide 380.

Question. Find a rational number between \( \sqrt{2} \) and \( \sqrt{7} \).
Answer: \( \sqrt{2} \approx 1.414 \) and \( \sqrt{7} \approx 2.6 \). Let the rational number be \( x \). \( \sqrt{2} < x < \sqrt{7} \) or \( 1.4 < x < 2.6 \). Hence, any rational number like 1.5, 2.0, 2.5, can be the answer.

Question. Write the number of zeroes in the end of a number whose prime factorization is \( 2^2 \times 5^3 \times 3^2 \times 17 \).
Answer: Given, \( 2^2 \times 5^2 \times 5 \times 3^2 \times 17 = (2 \times 5)^2 \times 5 \times 3^2 \times 17 = (10)^2 \times 5 \times 3^2 \times 17 \). The power of 10 in the given expression is 2. Hence, the number of zeroes in the end will be 2.

Question. If the HCF of \( (336, 54) = 6 \), find the LCM \( (336, 54) \).
Answer: The HCF of (336, 54) = 6. We know that: \( \text{LCM} \times \text{HCF} = \text{Product of two numbers} \Rightarrow \text{LCM} = \frac{336 \times 54}{6} = 336 \times 9 = 3024 \). Hence, the LCM of the two numbers is 3024.

Question. Find a rational number between \( \sqrt{2} \) and \( \sqrt{3} \).
Answer: Rational number between \( \sqrt{2} \) (1.41 approx) and \( \sqrt{3} \) (1.73 approx) can be 1.5, 1.6, 1.63 etc. So, a required rational number may be 1.5.

Question. Write one rational and one irrational number lying between 0.25 and 0.32.
Answer: Rational number = 0.30. Irrational number = 0.3010203040… Or any other correct rational and irrational number.

Question. Write the exponent of 3 in the prime factorization of 144.
Answer: Prime factorization of \( 144 = 2^4 \times 3^2 \). So, exponent of 3 = 2.

SHORT ANSWER Type Questions

Question. Check whether \( 12^n \) can end with the digit 0 for any natural number \( n \).
Answer: Let, if possible, \( 12^n \) have a value which ends with the digit 0. \( \Rightarrow \) 10 is a factor of \( 12^n \Rightarrow 5 \) is a prime factor of \( 12^n \) i.e., \( 12^n = 5 \times q \), where \( q \) is some natural number \( \Rightarrow (2^2 \times 3)^n = 5 \times q \) or \( 2^{2n} \times 3^n = 5 \times q \). The assumption, 5 is a prime factor of \( 2^{2n} \times 3^n \), is not possible because \( 2^{2n} \times 3^n \) can have only 2 and 3 as prime factors. Hence, our assumption is wrong and \( 12^n \) cannot end with the digit 0.

Question. The product of the LCM and HCF of two natural numbers is 24. The difference of two numbers is 2. Find the numbers.
Answer: Let the natural numbers be \( p \) and \( q \). According to question, \( p \times q = 24 \) ...(i) and \( p - q = 2 \Rightarrow p = 2 + q \) ...(ii). From (i) and (ii), \( (q + 2) \times q = 24 \Rightarrow q^2 + 2q - 24 = 0 \Rightarrow q^2 + 6q - 4q - 24 = 0 \Rightarrow (q + 6)(q - 4) = 0 \Rightarrow q = -6, 4 \). \( q = 4 \) [Since -6 is not a natural number]. So, the numbers are 4, 6.

Question. Two alarm clocks ring their alarms at regular intervals of 72 seconds and 50 seconds if they first beep together at 12 noon, at what time will they beep again for the first time?
Answer: Here, we need to find the LCM of 72 and 50. \( 72 = 2 \times 2 \times 2 \times 3 \times 3 \), \( 50 = 2 \times 5 \times 5 \). LCM of 72 and 50 = \( 2^3 \times 3^2 \times 5^2 = 1800 \). So, 1800 sec = 30 min. Hence, alarm clocks will beep again for the first time at 12.30 pm.

Question. Find the HCF of 612 and 1314 using prime factorisation.
Answer: Prime factors of 612 and 1314. \( 612 = 2 \times 2 \times 3 \times 3 \times 17 \), \( 1314 = 2 \times 3 \times 3 \times 73 \). HCF is the product of common prime factors with smallest power: \( 2^1 \times 3^2 = 2 \times 9 = 18 \).