This page provides comprehensive Chapter 1: Real Numbers - MCQ Worksheet - SJMaths. Multiple Choice Questions (MCQ) worksheet for Class 10 Real Numbers.
Question 1: The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, what is the other number?
(A) 36
(B) 81
(C) 9
(D) 45
Solution: (B) 81 Step 1: Use the formula: Product of two numbers = HCF $\times$ LCM. Step 2: Let the other number be $x$. Given one number is 54, HCF = 27, LCM = 162. Step 3: $54 \times x = 27 \times 162$. Step 4: $x = \frac{27 \times 162}{54} = \frac{4374}{54} = 81$.
Question 2: The ratio of LCM and HCF of the least composite and the least prime numbers is:
(A) 1:2
(B) 2:1
(C) 1:1
(D) 1:3
Solution: (B) 2:1 Step 1: Least composite number is 4. Least prime number is 2. Step 2: LCM(4, 2) = 4. HCF(4, 2) = 2. Step 3: Ratio = LCM : HCF = 4 : 2 = 2 : 1.
Question 3: For some integer $m$, every even integer is of the form:
(A) $m$
(B) $m+1$
(C) $2m$
(D) $2m+1$
Solution: (C) $2m$ Step 1: An even integer is any integer divisible by 2. Step 2: Therefore, it can be written as $2 \times m$ for some integer $m$.
So, the form is $2m$.
Question 4: For some integer $q$, every odd integer is of the form:
(A) $q$
(B) $q+1$
(C) $2q$
(D) $2q+1$
Solution: (D) $2q+1$ Step 1: An odd integer is an integer that is not divisible by 2. Step 2: When divided by 2, it leaves a remainder of 1. Step 3: By Euclid's division lemma, $a = bq + r$. Here $b=2, r=1$.
So, the form is $2q + 1$.
Question 5: $n^2 - 1$ is divisible by 8, if $n$ is:
(A) an integer
(B) a natural number
(C) an odd integer
(D) an even integer
Solution: (C) an odd integer Step 1: Let $n$ be an odd integer, $n = 2k+1$. Step 2: Substitute $n$ in $n^2 - 1$:
$(2k+1)^2 - 1 = 4k^2 + 4k + 1 - 1 = 4k^2 + 4k = 4k(k+1)$. Step 3: Since $k$ and $k+1$ are consecutive integers, their product $k(k+1)$ is always even (divisible by 2). Let $k(k+1) = 2m$. Step 4: $n^2 - 1 = 4(2m) = 8m$.
Thus, $n^2 - 1$ is divisible by 8 if $n$ is an odd integer.
Question 6: If two positive integers $a$ and $b$ are written as $a = x^3y^2$ and $b = xy^3$, where $x, y$ are prime numbers, then HCF$(a, b)$ is:
(A) $xy$
(B) $xy^2$
(C) $x^3y^3$
(D) $x^2y^2$
Solution: (B) $xy^2$ Step 1: Identify the prime factors of $a$ and $b$.
$a = x^3 y^2$ and $b = x y^3$. Step 2: HCF is the product of the smallest power of each common prime factor. Step 3: Common factors are $x$ and $y$.
Smallest power of $x$ is $x^1$ (from $b$).
Smallest power of $y$ is $y^2$ (from $a$). Step 4: HCF = $x \times y^2 = xy^2$.
Question 7: If two positive integers $p$ and $q$ can be expressed as $p = ab^2$ and $q = a^3b$; $a, b$ being prime numbers, then LCM$(p, q)$ is:
(A) $ab$
(B) $a^2b^2$
(C) $a^3b^2$
(D) $a^3b^3$
Solution: (C) $a^3b^2$ Step 1: Identify the prime factors of $p$ and $q$.
$p = a b^2$ and $q = a^3 b$. Step 2: LCM is the product of the greatest power of each prime factor involved. Step 3: Factors involved are $a$ and $b$.
Greatest power of $a$ is $a^3$ (from $q$).
Greatest power of $b$ is $b^2$ (from $p$). Step 4: LCM = $a^3 b^2$.
Question 8: The product of a non-zero rational and an irrational number is:
(A) always irrational
(B) always rational
(C) rational or irrational
(D) one
Solution: (A) always irrational Step 1: Let the non-zero rational number be $a$ and the irrational number be $b$. Step 2: Assume their product $ab$ is rational, say $ab = \frac{p}{q}$ where $p, q$ are integers. Step 3: Then $b = \frac{p}{qa}$. Since $a$ is rational and non-zero, $\frac{p}{qa}$ is rational. Step 4: This implies $b$ is rational, which contradicts the given fact that $b$ is irrational.
Therefore, the product is always irrational.
Question 9: The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is:
Question 10: The total number of factors of a prime number is:
(A) 1
(B) 0
(C) 2
(D) 3
Solution: (C) 2 Step 1: By definition, a prime number is a number greater than 1 that has exactly two factors. Step 2: The factors are 1 and the number itself.