Demo Questions in English

(D) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(A) We are given the expression $2^{m} \cdot 5^{n}$ where $m, n \in N$ (natural numbers).
We can rewrite the expression as $2^{m} \cdot 5^{n} = 2^{m-n} \cdot 2^{n} \cdot 5^{n}$ (if $m \ge n$) or $2^{m} \cdot 5^{m} \cdot 5^{n-m}$ (if $n > m$).
Case $1$: If $m = n$,then $2^{m} \cdot 5^{m} = (2 \cdot 5)^{m} = 10^{m}$. The last digit of $10^{m}$ is $0$.
Case $2$: If $m > n$,then $2^{m} \cdot 5^{n} = 2^{m-n} \cdot (2 \cdot 5)^{n} = 2^{m-n} \cdot 10^{n}$. Since $2^{m-n}$ is an even number and $10^{n}$ ends in $0$,the product will end in $0$.
Case $3$: If $n > m$,then $2^{m} \cdot 5^{n} = (2 \cdot 5)^{m} \cdot 5^{n-m} = 10^{m} \cdot 5^{n-m}$. Since $5^{n-m}$ always ends in $5$ for $n > m$ and $10^{m}$ ends in $0$,the product will end in $0$.
In all cases,the last digit is $0$.

(B) The $HCF$ of two numbers always divides their $LCM$ completely.
Given $HCF(a, b) = 16$.
We check each option to see if it is divisible by $16$:
$A) 32 / 16 = 2$ (Divisible)
$B) 72 / 16 = 4.5$ (Not divisible)
$C) 64 / 16 = 4$ (Divisible)
$D) 96 / 16 = 6$ (Divisible)
Since $72$ is not divisible by $16$,it cannot be the $LCM$ of $a$ and $b$.

(C) To find the number of decimal places for a rational number $\frac{p}{q}$ where $q = 2^n \cdot 5^m$,the number of decimal places is given by $\max(n, m)$.
Given the expression $\frac{18}{5^3}$,we can rewrite the denominator as $2^0 \cdot 5^3$.
Here,$n = 0$ and $m = 3$.
The number of decimal places is $\max(0, 3) = 3$.
Alternatively,$\frac{18}{5^3} = \frac{18 \cdot 2^3}{5^3 \cdot 2^3} = \frac{18 \cdot 8}{10^3} = \frac{144}{1000} = 0.144$.
Since there are $3$ digits after the decimal point,the correct option is $C$.

(D) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(C) For distinct prime numbers $a, b, c$,the Highest Common Factor $(HCF)$ is the greatest common divisor,which is $1$ because they share no common factors other than $1$.
The Least Common Multiple $(LCM)$ of distinct prime numbers is the product of the numbers themselves,which is $a \times b \times c = abc$.
Therefore,the ratio of $HCF$ to $LCM$ is $\frac{\text{HCF}}{\text{LCM}} = \frac{1}{abc}$.
This can be expressed as the ratio $1: abc$.

(A) The relationship between two numbers $a$ and $b$ is given by the property: $LCM(a, b)$ must be perfectly divisible by $HCF(a, b)$.
Given $HCF(a, b) = 12$.
We check each option to see if it is divisible by $12$:
$(A)$ $90 / 12 = 7.5$ (Not divisible)
$(B)$ $24 / 12 = 2$ (Divisible)
$(C)$ $48 / 12 = 4$ (Divisible)
$(D)$ $36 / 12 = 3$ (Divisible)
Since $90$ is not divisible by $12$,it cannot be the $LCM$.

(C) For any natural number $n \in N$ (where $n = 1, 2, 3, \dots$):
If $n = 1$,$5^1 = 5$.
If $n = 2$,$5^2 = 25$.
If $n = 3$,$5^3 = 125$.
If $n = 4$,$5^4 = 625$.
It is observed that for any positive integer $n$,the last digit of $5^n$ is always $5$.

(B) First,find the prime factorization of $65$ and $117$.
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
The $\text{HCF}(65, 117)$ is the common factor with the lowest power,which is $13$.
Given the equation: $\text{HCF}(65, 117) = 65m - 117$
Substitute the value of $\text{HCF}$:
$13 = 65m - 117$
Add $117$ to both sides:
$13 + 117 = 65m$
$130 = 65m$
Divide by $65$:
$m = \frac{130}{65} = 2$
Thus,the value of $m$ is $2$.

(C) The $LCM$ (Least Common Multiple) of two numbers is the smallest positive integer that is divisible by both numbers.
Since $p$ and $q$ are distinct prime numbers,they have no common factors other than $1$.
Therefore,the $LCM$ of two distinct prime numbers is simply their product.
Thus,$LCM(p, q) = p \times q = pq$.

(A) To find the $LCM(26, 91)$,we first find the prime factorization of each number:
$26 = 2 \times 13$
$91 = 7 \times 13$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(26, 91) = 2 \times 7 \times 13 = 182$
Therefore,the correct option is $A$.

(D) prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $m$ and $n$ are distinct prime integers,they do not share any common factors other than $1$.
The Highest Common Factor $(HCF)$ of two distinct prime numbers is always $1$ because their only common divisor is $1$.
Therefore,the $HCF$ of $m$ and $n$ is $1$.

(B) For any two positive integers $a$ and $b$,the $LCM(a, b)$ must be perfectly divisible by their $HCF(a, b)$.
Given that $HCF(a, b) = 25$,the $LCM(a, b)$ must be a multiple of $25$.
Checking the given options:
$(A)$ $50 = 25 \times 2$ (Possible)
$(B)$ $105$ is not divisible by $25$ $(105 / 25 = 4.2)$ (Impossible)
$(C)$ $100 = 25 \times 4$ (Possible)
$(D)$ $25 = 25 \times 1$ (Possible)
Therefore,$105$ cannot be the $LCM$ of $a$ and $b$.

(C) An irrational number is a number that cannot be expressed as a fraction $p/q$ where $p$ and $q$ are integers and $q \neq 0$. Its decimal expansion is non-terminating and non-recurring.
$1$. $\sqrt{4} = 2$,which is a rational number.
$2$. $0.010101...$ is a recurring decimal,which can be expressed as $1/99$,so it is a rational number.
$3$. $\pi$ (pi) is a well-known irrational number because its decimal expansion is non-terminating and non-recurring.
$4$. $5$ can be written as $5/1$,which is a rational number.
Therefore,the correct option is $C$.

(A) We know that for any two positive integers $a$ and $b$,the relationship $a \times b = LCM(a, b) \times GCD(a, b)$ holds true.
Given that $LCM(a, b) = a \times b$.
Substituting this into the formula,we get $a \times b = (a \times b) \times GCD(a, b)$.
Dividing both sides by $a \times b$ (assuming $a, b \neq 0$),we get $GCD(a, b) = 1$.
Therefore,the correct answer is $1$.

(C) To find the last digit of $5^n$ for any positive integer $n$,let us observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
In each case,the last digit (units digit) is $5$.
Therefore,for any positive integer $n$,the last digit of $5^n$ is always $5$.

(A) To find the $HCF$ (Highest Common Factor) of $12$,$15$,and $21$,we first find the prime factorization of each number:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
The $HCF$ is the product of the smallest power of each common prime factor present in the numbers.
The only common prime factor is $3$,and its smallest power is $3^1$.
Therefore,$HCF$ $(12, 15, 21) = 3$.

(A) To find the $HCF$ of $6$ and $20$,we first find the prime factorization of each number:
$6 = 2 \times 3$
$20 = 2^2 \times 5$
The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $2$,and its smallest power is $2^1 = 2$.
Therefore,the $HCF$ of $6$ and $20$ is $2$.

(C) The Least Common Multiple $(LCM)$ of two numbers is the smallest positive integer that is divisible by both numbers.
For any two prime numbers $p$ and $q$,since they are distinct,they have no common factors other than $1$.
Therefore,the $LCM$ of two distinct prime numbers is simply their product.
$LCM(p, q) = p \times q = pq$.

(A) We are given the expression $2^5 \cdot 5^5$.
Using the law of exponents $a^n \cdot b^n = (a \cdot b)^n$,we can rewrite the expression as:
$2^5 \cdot 5^5 = (2 \cdot 5)^5$
$= 10^5$
$= 100000$
The last digit of $100000$ is $0$.

(B) The number $\pi$ is defined as the ratio of the circumference of a circle to its diameter.
It is a non-terminating and non-repeating decimal, which means it cannot be expressed in the form $p/q$ where $p$ and $q$ are integers and $q \neq 0$.
Therefore, $\pi$ is an irrational number.

(C) To find the $LCM$ of $7, 11,$ and $17$,we first observe that all three numbers are prime numbers.
Since $7, 11,$ and $17$ are prime numbers,they have no common factors other than $1$.
The $LCM$ of a set of prime numbers is simply the product of those numbers.
Therefore,$LCM(7, 11, 17) = 7 \times 11 \times 17 = 1309$.

(D) rational number is a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
$1$. $\sqrt{7}$ is an irrational number because $7$ is not a perfect square.
$2$. $3\pi$ is an irrational number because $\pi$ is irrational.
$3$. $0.101000...$ is a non-terminating and non-repeating decimal,which makes it an irrational number.
$4$. $0.25$ can be written as $\frac{25}{100} = \frac{1}{4}$,which is in the form $\frac{p}{q}$.
Therefore,$0.25$ is a rational number.

(D) To find the $HCF(510, 92)$,we perform prime factorization of both numbers:
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2 \times 2 \times 23 = 2^2 \times 23$
The common prime factor with the lowest exponent is $2^1 = 2$.
Therefore,$HCF(510, 92) = 2$.

(A) We know that for any two positive integers $a$ and $b$,the relationship between their $HCF$ and $LCM$ is given by:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Given:
$\text{HCF} = 9$
$\text{Product of numbers} (a \times b) = 288$
Substituting the values in the formula:
$9 \times \text{LCM} = 288$
$\text{LCM} = \frac{288}{9}$
$\text{LCM} = 32$
Therefore,the $LCM$ of the two numbers is $32$.

(A) To find the $LCM$ of $6$ and $20$,we first find their prime factorizations:
$6 = 2 \times 3$
$20 = 2^2 \times 5$
The $LCM$ is the product of the highest powers of all prime factors involved:
$LCM = 2^2 \times 3^1 \times 5^1$
$LCM = 4 \times 3 \times 5 = 60$
Therefore,the $LCM$ of $6$ and $20$ is $60$.

(C) According to the Fundamental Theorem of Arithmetic,every composite number greater than $1$ can be expressed as a product of prime numbers. This representation is unique,apart from the order in which the prime factors occur. This theorem is also known as the Unique Factorization Theorem.

(B) To find the last digit of $5^{25}$,we observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
It is observed that for any positive integer $n$,the last digit of $5^n$ is always $5$.
Therefore,the last digit of $5^{25}$ is $5$.

(B) For any two positive integers $a$ and $b$,the fundamental theorem of arithmetic states that the product of the two numbers is equal to the product of their Highest Common Factor $(HCF)$ and Least Common Multiple $(LCM)$.
Mathematically,this is expressed as: $HCF(a, b) \times LCM(a, b) = a \times b$.

(B) To find the $LCM$ of $6, 72,$ and $120$,we first find the prime factorization of each number:
$6 = 2^1 \times 3^1$
$72 = 8 \times 9 = 2^3 \times 3^2$
$120 = 8 \times 15 = 2^3 \times 3^1 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM$ $= 2^3 \times 3^2 \times 5^1$
$LCM$ $= 8 \times 9 \times 5$
$LCM$ $= 72 \times 5 = 360$
Therefore,the $LCM$ of $6, 72,$ and $120$ is $360$.

(D) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.
$a \times b = HCF(a, b) \times LCM(a, b)$
Given $HCF(a, b) = 8$ and $LCM(a, b) = 64$,we have:
$a \times b = 8 \times 64 = 512$
Since $8$ is the $HCF$,both $a$ and $b$ must be multiples of $8$. Let $a = 8x$ and $b = 8y$,where $x$ and $y$ are coprime and $x > y$ (since $a > b$).
Substituting these into the product equation:
$(8x) \times (8y) = 512$
$64xy = 512$
$xy = 8$
Possible pairs $(x, y)$ such that $x$ and $y$ are coprime and $x > y$ are $(8, 1)$.
If $x = 8$ and $y = 1$,then $a = 8 \times 8 = 64$ and $b = 8 \times 1 = 8$.
Since $a > b$,$a = 64$ is the correct value.

(D) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(C) To find the last digit of $5^n$ for any positive integer $n$,let us observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
As we can see,for any positive integer $n$,the result of $5^n$ always ends in the digit $5$. Therefore,the last digit is always $5$.

(A) The given number is $0.01111...$,which can be written as $0.0\bar{1}$.
Since the decimal expansion is non-terminating and repeating,it can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Let $x = 0.0111...$ (Equation $1$).
Multiply by $10$: $10x = 0.1111...$ (Equation $2$).
Multiply by $100$: $100x = 1.1111...$ (Equation $3$).
Subtract Equation $2$ from Equation $3$: $100x - 10x = 1.1111... - 0.1111...$.
$90x = 1$.
$x = \frac{1}{90}$.
Since the number can be expressed as a fraction of two integers,it is a rational number.

(C) To find the prime factorization of $30$,we divide it by the smallest prime numbers:
$30 \div 2 = 15$
$15 \div 3 = 5$
$5 \div 5 = 1$
Therefore,the prime factorization of $30$ is $2 \times 3 \times 5$.

(A) To find the Greatest Common Divisor $(GCD)$ of $13, 23,$ and $31$,we first identify the prime factorization of each number:
$13 = 13 \times 1$
$23 = 23 \times 1$
$31 = 31 \times 1$
Since $13, 23,$ and $31$ are all prime numbers and they have no common factors other than $1$,their $GCD$ is $1$.

(B) number is called rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
$\pi$ is a mathematical constant defined as the ratio of a circle's circumference to its diameter.
Its decimal representation is non-terminating and non-repeating.
Therefore, $\pi$ is an irrational number.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $HCF$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
The $HCF$ is the product of the smallest power of each common prime factor,which is $2^1 = 2$.
Therefore,the $HCF$ of the smallest prime number and the smallest composite number is $2$.

(A) To find the remainder when $(5k+1)^2$ is divided by $5$,we expand the expression using the identity $(a+b)^2 = a^2 + 2ab + b^2$.
$(5k+1)^2 = (5k)^2 + 2(5k)(1) + (1)^2$
$= 25k^2 + 10k + 1$
Now,we can rewrite this expression as:
$= 5(5k^2 + 2k) + 1$
Since $5(5k^2 + 2k)$ is clearly divisible by $5$,the remainder when the entire expression is divided by $5$ is $1$.

(B) number is called rational if it can be expressed in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
Since $\sqrt{2}$ cannot be expressed as a ratio of two integers, it is an irrational number.
Therefore, $\sqrt{2}$ is an irrational number.

(B) The given expression is $\sqrt{1+1}$.
Calculating the sum inside the square root: $1+1 = 2$.
Therefore,the expression becomes $\sqrt{2}$.
Since $2$ is not a perfect square,$\sqrt{2}$ is an irrational number.
Thus,the correct option is $B$.

(B) We know the fundamental property of two positive integers $a$ and $b$:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Given that $\text{HCF}(a, b) = 7$ and $\text{LCM}(a, b) = 385$.
Therefore,$a \times b = 7 \times 385$.
Calculating the product: $7 \times 385 = 2695$.
Thus,the product of the two integers is $2695$.

(A) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(A) Given expression is $3 + \sqrt{16}$.
We know that $\sqrt{16} = 4$.
Substituting this value into the expression: $3 + 4 = 7$.
Since $7$ can be written in the form $p/q$ where $p=7$ and $q=1$ (both integers and $q \neq 0$),it is a rational number.

(B) To find the last digit of $5^{25}$,we observe the powers of $5$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
Since the last digit of any positive integer power of $5$ is always $5$,the last digit of $5^{25}$ is $5$.

(B) For any two positive integers $a$ and $b$,the product of their Highest Common Factor $(HCF)$ and Least Common Multiple $(LCM)$ is always equal to the product of the two numbers themselves.
Mathematically,this is expressed as: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.

(B) To find the $LCM$ of $6$,$72$,and $120$,we first find the prime factorization of each number:
$6 = 2^1 \times 3^1$
$72 = 8 \times 9 = 2^3 \times 3^2$
$120 = 8 \times 15 = 2^3 \times 3^1 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM = 2^3 \times 3^2 \times 5^1$
$LCM = 8 \times 9 \times 5$
$LCM = 72 \times 5 = 360$
Therefore,the $LCM$ of $6$,$72$,and $120$ is $360$.

(D) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.
$a \times b = HCF(a, b) \times LCM(a, b)$
Given $HCF(a, b) = 8$ and $LCM(a, b) = 64$,we have:
$a \times b = 8 \times 64 = 512$
Since $HCF(a, b) = 8$,both $a$ and $b$ must be multiples of $8$. Let $a = 8x$ and $b = 8y$,where $x$ and $y$ are coprime and $x > y$ (since $a > b$).
Substituting these into the product equation:
$(8x) \times (8y) = 512$
$64xy = 512$
$xy = 8$
The possible pairs of $(x, y)$ such that $x > y$ and $gcd(x, y) = 1$ are $(8, 1)$.
If $x = 8$ and $y = 1$,then $a = 8 \times 8 = 64$ and $b = 8 \times 1 = 8$.
Checking the $LCM$: $LCM(64, 8) = 64$,which matches the given condition.
Thus,$a = 64$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM(2, 4)$:
Prime factorization of $2 = 2^1$.
Prime factorization of $4 = 2^2$.
$LCM$ is the product of the highest power of each prime factor involved,which is $2^2 = 4$.
Therefore,the $LCM$ is $4$.

(C) To find the last digit of $5^n$ for any positive integer $n$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
As observed,for any positive integer $n$,the product of $5$ multiplied by itself $n$ times will always result in a number ending in $5$. Therefore,the last digit of $5^n$ is always $5$.