Demo Questions in English

(A) The given number is $0.01111...$,which can be written as $0.0overline{1}$.
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 $3$).
Subtracting 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 $\frac{1}{90}$,it is a rational number.

(A) Let the three consecutive positive integers be $n, (n+1)$,and $(n+2)$.
The product of these integers is $P = n(n+1)(n+2)$.
This expression represents the product of three consecutive integers.
We know that in any set of $k$ consecutive integers,there is always at least one multiple of $k!$ (k factorial).
For $k=3$,the product is always divisible by $3! = 3 \times 2 \times 1 = 6$.
For example,if $n=1$,$1 \times 2 \times 3 = 6$ (divisible by $6$).
If $n=2$,$2 \times 3 \times 4 = 24$ (divisible by $6$).
If $n=3$,$3 \times 4 \times 5 = 60$ (divisible by $6$).
Thus,the product of any three consecutive positive integers is always divisible by $6$.

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

(A) To find the $HCF$ (Highest Common Factor) of $13$,$23$,and $31$:
$1$. Prime factorization of $13 = 13 \times 1$.
$2$. Prime factorization of $23 = 23 \times 1$.
$3$. Prime factorization of $31 = 31 \times 1$.
Since $13$,$23$,and $31$ are all prime numbers,they have no common factors other than $1$.
Therefore,the $HCF$ of $13$,$23$,and $31$ is $1$.

(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 $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Therefore,$\pi$ is an irrational number.

(D) To find the remainder when $27$ is divided by $4$,we perform the division:
$27 \div 4 = 6$ with a remainder.
$4 \times 6 = 24$.
Remainder $= 27 - 24 = 3$.
Therefore,the remainder is $3$.

(C) According to the Fundamental Theorem of Arithmetic,every composite number greater than $1$ can be uniquely expressed as a product of primes,except for the order in which the prime factors occur. This means that for any given composite number,the set of prime factors is fixed,regardless of the order in which they are written. Therefore,the factorization is unique.

(B) We are given the expression $(5k+1)^2$.
Expanding this using the identity $(a+b)^2 = a^2 + 2ab + b^2$,we get:
$(5k+1)^2 = (5k)^2 + 2(5k)(1) + (1)^2$
$= 25k^2 + 10k + 1$
$= 5(5k^2 + 2k) + 1$
Here,the expression is in the form $5q + 1$,where $q = 5k^2 + 2k$ is an integer.
Therefore,when $(5k+1)^2$ is divided by $5$,the remainder 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) Given expression is $\sqrt{1+1}$.
First,calculate the sum inside the square root: $1+1 = 2$.
So,the expression becomes $\sqrt{2}$.
Since $2$ is not a perfect square,$\sqrt{2}$ is an irrational number.
Therefore,the correct option is $B$.

(C) To find the number of decimal places,we express the denominator in the form $2^n \times 5^m$.
Given fraction: $\frac{35457}{6250}$.
Prime factorization of $6250 = 625 \times 10 = 5^4 \times (2 \times 5) = 2^1 \times 5^5$.
The number of decimal places is determined by the maximum of the exponents of $2$ and $5$,which is $\max(1, 5) = 5$.
Therefore,the decimal representation terminates after $5$ decimal places.

(B) The decimal expansion $0.9999...$ can be written as $0.\bar{9}$.
Since the digit $9$ repeats infinitely after the decimal point,it is a non-terminating repeating decimal expansion.
Furthermore,$0.9999... = 1$,which is a rational number,and all rational numbers have either terminating or non-terminating repeating decimal expansions.

(D) The number of decimal places in the decimal expansion of a rational number $\frac{p}{q}$ (where $q$ is of the form $2^n \cdot 5^m$) is given by $\max(n, m)$.
However,for a fraction $\frac{p}{q}$ where $q$ is not of the form $2^n \cdot 5^m$,the decimal expansion is non-terminating and repeating.
For $\frac{3}{113}$,the denominator $113$ is a prime number and is not of the form $2^n \cdot 5^m$.
Therefore,the decimal expansion is non-terminating and repeating.
If the question implies the length of the period of the repeating decimal,for a prime $p$,the period length divides $p-1$.
For $p = 113$,the period length is $112$.

(A) To find the greatest positive integer that divides $70$ and $125$ leaving remainders $5$ and $8$ respectively,we need to find the Highest Common Factor $(HCF)$ of $(70 - 5)$ and $(125 - 8)$.
Step $1$: Subtract the remainders from the given numbers.
$70 - 5 = 65$
$125 - 8 = 117$
Step $2$: Find the $HCF$ of $65$ and $117$.
Prime factorization of $65 = 5 \times 13$.
Prime factorization of $117 = 3^2 \times 13$.
Step $3$: The common factor with the lowest power is $13$.
Therefore,the greatest positive integer is $13$.

(B) First,find the $\text{HCF}$ of $65$ and $117$ using prime factorization:
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
Therefore,$\text{HCF}(65, 117) = 13$.
Given the equation: $\text{HCF}(65, 117) = 65m - 117$.
Substitute the value of $\text{HCF}$:
$13 = 65m - 117$
$13 + 117 = 65m$
$130 = 65m$
$m = \frac{130}{65} = 2$.
Thus,the value of $m$ is $2$.

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

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

(C) By Euclid's division lemma,for any positive integer $a$ and divisor $b=4$,we have $a = 4m + r$,where $r$ can be $0, 1, 2, 3$.
If $r=0$,$a = 4m = 2(2m)$,which is even.
If $r=1$,$a = 4m+1$,which is odd.
If $r=2$,$a = 4m+2 = 2(2m+1)$,which is even.
If $r=3$,$a = 4m+3$,which is odd.
Therefore,any odd positive integer is of the form $4m+1$ or $4m+3$.

(C) The largest $4$-digit number is $9999$.
To find the largest $4$-digit number divisible by $95$,we divide $9999$ by $95$.
$9999 \div 95 = 105$ with a remainder.
$9999 = 95 \times 105 + 14$.
The remainder is $14$.
To find the largest $4$-digit number divisible by $95$,we subtract the remainder from the largest $4$-digit number:
$9999 - 14 = 9985$.
Thus,$9985$ is the largest $4$-digit integer divisible by $95$.

(C) Euclid's division lemma states that for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \le r < b$.
In this problem,$b = 5$.
Therefore,the possible values for the remainder $r$ are $0, 1, 2, 3, 4$.
Since $r$ must be less than $5$,the value $r = 6$ is not possible.

(B) Let $a$ be any positive integer. Any integer $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$ for some integer $k ge 0$.
We calculate $a^2 \pmod{6}$ for each case:
$1$. If $a = 6k$,then $a^2 = 36k^2 = 6(6k^2)$,so $a^2 \equiv 0 \pmod{6}$.
$2$. If $a = 6k+1$,then $a^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1$,so $a^2 \equiv 1 \pmod{6}$.
$3$. If $a = 6k+2$,then $a^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4$,so $a^2 \equiv 4 \pmod{6}$.
$4$. If $a = 6k+3$,then $a^2 = 36k^2 + 36k + 9 = 36k^2 + 36k + 6 + 3 = 6(6k^2 + 6k + 1) + 3$,so $a^2 \equiv 3 \pmod{6}$.
$5$. If $a = 6k+4$,then $a^2 = 36k^2 + 48k + 16 = 36k^2 + 48k + 12 + 4 = 6(6k^2 + 8k + 2) + 4$,so $a^2 \equiv 4 \pmod{6}$.
$6$. If $a = 6k+5$,then $a^2 = 36k^2 + 60k + 25 = 36k^2 + 60k + 24 + 1 = 6(6k^2 + 10k + 4) + 1$,so $a^2 \equiv 1 \pmod{6}$.
The possible remainders are $0, 1, 3, 4$. The remainders $2$ and $5$ are not possible. Among the given options,$2$ is the correct answer.

(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 $2$.

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

(A) To find the $LCM(26, 91)$,we first find the prime factorization of both numbers:
$26 = 2 \times 13$
$91 = 7 \times 13$
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) By definition,a 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 $1$.
Therefore,the Highest Common Factor $(HCF)$ of two distinct prime numbers is always $1$.

(B) The $LCM$ of two numbers must always be divisible by their $HCF$.
Given $HCF(a, b) = 25$.
We check the divisibility of the given options by $25$:
$1$. $50 / 25 = 2$ (Divisible)
$2$. $105 / 25 = 4.2$ (Not divisible)
$3$. $100 / 25 = 4$ (Divisible)
Since $105$ is not divisible by $25$,it cannot be the $LCM$ of two numbers whose $HCF$ is $25$.

(C) Let $d = HCF(a - b, a + b)$.
Then $d$ must divide their sum and their difference.
$d$ divides $(a + b) + (a - b) = 2a$.
$d$ divides $(a + b) - (a - b) = 2b$.
Since $d$ divides $2a$ and $2b$,$d$ must divide $HCF(2a, 2b) = 2 \times HCF(a, b)$.
Given $HCF(a, b) = 1$,so $d$ must divide $2 \times 1 = 2$.
Therefore,$d$ can be either $1$ or $2$.
For example,if $a = 3, b = 2$,$HCF(3, 2) = 1$,then $HCF(3-2, 3+2) = HCF(1, 5) = 1$.
If $a = 3, b = 1$,$HCF(3, 1) = 1$,then $HCF(3-1, 3+1) = HCF(2, 4) = 2$.
Thus,the result is $1$ or $2$.

(A) For any two positive integers $a$ and $b$,the relationship between their $LCM$ and $GCD$ is given by: $a \times b = LCM(a, b) \times GCD(a, b)$.
Given that $LCM(a, b) = a \times b$.
Substituting this into the formula: $a \times b = (a \times b) \times GCD(a, b)$.
Dividing both sides by $(a \times b)$,we get: $GCD(a, b) = 1$.
Therefore,the $GCD$ of the two numbers is $1$.

(B) The decimal expansion $0.9999...$ can be written as $0.\bar{9}$.
Since the digit $9$ repeats infinitely,it is a non-terminating repeating decimal expansion.
Furthermore,mathematically,$0.9999... = 1$,which is a rational number,and all rational numbers have either terminating or non-terminating repeating decimal expansions.

(B) Any natural number $a$ can be expressed in the form $6k, 6k+1, 6k+2, 6k+3, 6k+4,$ or $6k+5$.
Calculating $a^2 \pmod{6}$ for each case:
$1) (6k)^2 = 36k^2 = 6(6k^2) \equiv 0 \pmod{6}$
$2) (6k+1)^2 = 36k^2 + 12k + 1 = 6(6k^2 + 2k) + 1 \equiv 1 \pmod{6}$
$3) (6k+2)^2 = 36k^2 + 24k + 4 = 6(6k^2 + 4k) + 4 \equiv 4 \pmod{6}$
$4) (6k+3)^2 = 36k^2 + 36k + 9 = 6(6k^2 + 6k + 1) + 3 \equiv 3 \pmod{6}$
$5) (6k+4)^2 = 36k^2 + 48k + 16 = 6(6k^2 + 8k + 2) + 4 \equiv 4 \pmod{6}$
$6) (6k+5)^2 = 36k^2 + 60k + 25 = 6(6k^2 + 10k + 4) + 1 \equiv 1 \pmod{6}$
The possible remainders are $0, 1, 3, 4$.
Thus,the remainders $2$ and $5$ are not possible.

(C) To find the last digit of $5^n$ for any positive integer $n$,let's 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 last digit of $5^n$ is always $5$.

(A) To find the $HCF$ 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.
The only common prime factor is $3$,and its smallest power is $3^1$.
Therefore,$\text{HCF}(12, 15, 21) = 3$.

(C) Let the four consecutive positive integers be $n, (n+1), (n+2),$ and $(n+3)$.
The product of these integers is $P = n(n+1)(n+2)(n+3)$.
This expression is equivalent to $24 \times \binom{n+3}{4}$.
Since $\binom{n+3}{4}$ is always an integer,the product $P$ must be divisible by $24$.
For example,if $n=1$,the product is $1 \times 2 \times 3 \times 4 = 24$,which is divisible by $24$.
If $n=2$,the product is $2 \times 3 \times 4 \times 5 = 120$,which is $24 \times 5$,also divisible by $24$.

(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 power of each prime factor present in the numbers:
$LCM = 2^2 \times 3^1 \times 5^1$
$LCM = 4 \times 3 \times 5 = 60$
Therefore,the $LCM$ of $6$ and $20$ is $60$.

(A) We are given the expression $2^m \cdot 5^n$ where $m, n \in \mathbb{N}$.
We can rewrite this as $2^m \cdot 5^n = 2^{m-n} \cdot (2^n \cdot 5^n)$ assuming $m \ge n$,or $2^m \cdot 5^n = 5^{n-m} \cdot (2^m \cdot 5^m)$ assuming $n > m$.
In either case,we can factor out $10^k$ where $k = \min(m, n)$.
Since $2 \cdot 5 = 10$,the expression will always contain at least one factor of $10$ because $m, n \ge 1$.
Any integer multiplied by $10$ ends in the digit $0$.
Therefore,the last digit of $2^m \cdot 5^n$ is $0$.

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

(B) 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 $HCF$ 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,$HCF(510, 92) = 2$.

(B) 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 equal to $\max(n, m)$.
Given the fraction is $\frac{15}{2^2}$.
Here,the denominator is $2^2 \cdot 5^0$.
Comparing this with $2^n \cdot 5^m$,we get $n = 2$ and $m = 0$.
The maximum value of $n$ and $m$ is $\max(2, 0) = 2$.
Therefore,the decimal expansion terminates after $2$ decimal places.
Verification: $\frac{15}{2^2} = \frac{15}{4} = 3.75$,which has $2$ decimal places.

(B) We are given the expression $(6k+1)^2$.
Expanding this using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$,we get:
$(6k+1)^2 = (6k)^2 + 2(6k)(1) + 1^2$
$= 36k^2 + 12k + 1$
We can factor out $6$ from the first two terms:
$= 6(6k^2 + 2k) + 1$
Let $m = 6k^2 + 2k$,where $m$ is an integer.
Then the expression becomes $6m + 1$.
When $6m + 1$ is divided by $6$,the quotient is $m$ and the remainder is $1$.

(A) To determine if a rational number $p/q$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$. If $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,then the decimal expansion is terminating.
Here,the fraction is $44/625$.
The denominator is $625 = 5^4$.
Since the denominator is in the form $2^0 \times 5^4$,it satisfies the condition for a terminating decimal expansion.
Therefore,$44/625$ is a terminating decimal.

(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.
$4$. $0$ can be written as $\frac{0}{1}$,which satisfies the condition of a rational number.
Therefore,$0$ is the correct rational number.

(D) To find the $LCM$ of prime numbers,we multiply them together because they have no common factors other than $1$.
Since $7$,$11$,and $17$ are all prime numbers,their $LCM$ is their product.
$LCM = 7 \times 11 \times 17$
$LCM = 77 \times 17$
$LCM = 1309$
Therefore,the correct option is $D$.

(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 $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Therefore,$\pi$ is an irrational number.

(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:
$2^5 \cdot 5^5 = (2 \cdot 5)^5$
$= 10^5$
$= 100,000$
The last digit of $100,000$ is $0$.

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

(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 $2$.

(A) The $LCM$ of two numbers must always be divisible by their $HCF$.
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)
$(D) 60 / 12 = 5$ (Divisible)
Since $90$ is not divisible by $12$,it cannot be the $LCM$ of two numbers whose $HCF$ is $12$.

(C) For any natural number $n \in N$,the expression $5^n$ represents the product of $5$ multiplied by itself $n$ times.
For $n = 1$,$5^1 = 5$.
For $n = 2$,$5^2 = 25$.
For $n = 3$,$5^3 = 125$.
For $n = 4$,$5^4 = 625$.
As observed,the last digit of $5^n$ is always $5$ for any natural number $n$.

(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 equal to $\max(n, m)$.
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 $\max(0, 3) = 3$.
Thus,the decimal expansion terminates after $3$ decimal places.

(B) 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 $4$.