CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 3 Marks - Part 2
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CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 3 Marks - Part 2
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SECTION C — Short Answer Type Questions
[3 Marks Each]
Q26. Show that $3 - \sqrt{5}$ is an irrational number.
Answer:
1. Let us assume to the contrary that $3 - \sqrt{5}$ is a rational number.
2. Therefore, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$3 - \sqrt{5} = \frac{a}{b}$
3. Rearranging the equation to isolate $\sqrt{5}$:
$3 - \frac{a}{b} = \sqrt{5}$
$\sqrt{5} = \frac{3b - a}{b}$
4. Since $a$ and $b$ are integers, the expression on the right-hand side ($\frac{3b - a}{b}$) is a rational number.
5. This implies that $\sqrt{5}$ is a rational number, which contradicts the established mathematical fact that $\sqrt{5}$ is irrational.
6. This contradiction arises because of our incorrect assumption that $3 - \sqrt{5}$ is rational.
Result: Hence, it is shown that $3 - \sqrt{5}$ is an irrational number.
Q27. Prove that $\frac{5}{\sqrt{3}}$ is an irrational number.
Answer:
1. Let us assume to the contrary that $\frac{5}{\sqrt{3}}$ is a rational number.
2. Therefore, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$\frac{5}{\sqrt{3}} = \frac{a}{b}$
3. Rearranging the equation to isolate $\sqrt{3}$:
$5b = a\sqrt{3}$
$\sqrt{3} = \frac{5b}{a}$
4. Since $a$ and $b$ are integers (with $a \neq 0$), the expression on the right-hand side ($\frac{5b}{a}$) is a rational number.
5. This implies that $\sqrt{3}$ is a rational number, which contradicts the established mathematical fact that $\sqrt{3}$ is irrational.
6. This contradiction arises because of our incorrect assumption that $\frac{5}{\sqrt{3}}$ is rational.
Result: Hence, it is proved that $\frac{5}{\sqrt{3}}$ is an irrational number.
Q28. Prove that $2\sqrt{3} - 1$ is an irrational number.
Answer:
1. Let us assume to the contrary that $2\sqrt{3} - 1$ is a rational number.
2. Therefore, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$2\sqrt{3} - 1 = \frac{a}{b}$
3. Rearranging the equation to isolate the term containing $\sqrt{3}$:
$2\sqrt{3} = \frac{a}{b} + 1$
$2\sqrt{3} = \frac{a + b}{b}$
4. Dividing both sides by $2$:
$\sqrt{3} = \frac{a + b}{2b}$
5. Since $a$ and $b$ are integers, the expression on the right-hand side ($\frac{a + b}{2b}$) is a rational number.
6. This implies that $\sqrt{3}$ is a rational number, which contradicts the established mathematical fact that $\sqrt{3}$ is irrational.
7. This contradiction arises because of our incorrect assumption that $2\sqrt{3} - 1$ is rational.
Result: Hence, it is proved that $2\sqrt{3} - 1$ is an irrational number.
Q29. Prove that $4 - 5\sqrt{2}$ is an irrational number.
Answer:
1. Let us assume to the contrary that $4 - 5\sqrt{2}$ is a rational number.
2. Therefore, there exist co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$4 - 5\sqrt{2} = \frac{a}{b}$
3. Rearranging the equation to isolate the term containing $\sqrt{2}$:
$4 - \frac{a}{b} = 5\sqrt{2}$
$\frac{4b - a}{b} = 5\sqrt{2}$
4. Dividing both sides by $5$:
$\sqrt{2} = \frac{4b - a}{5b}$
5. Since $a$ and $b$ are integers, the expression on the right-hand side ($\frac{4b - a}{5b}$) is a rational number.
6. This implies that $\sqrt{2}$ is a rational number, which contradicts the established mathematical fact that $\sqrt{2}$ is irrational.
7. This contradiction arises because of our incorrect assumption that $4 - 5\sqrt{2}$ is rational.
Result: Hence, it is proved that $4 - 5\sqrt{2}$ is an irrational number.
Q30. If $a$ and $b$ are odd positive integers, prove that $a^2 + b^2$ is even but not divisible by 4.
Answer:
1. Since $a$ and $b$ are odd positive integers, they can be expressed in the form $2m + 1$ and $2n + 1$ respectively, where $m$ and $n$ are integers.
2. Therefore, we can write:
$a = 2m + 1$
$b = 2n + 1$
5. Since $a^2 + b^2$ has a factor of $2$ (expressed as $2 \times \text{integer}$), it is an even number.
6. To test for divisibility by $4$, let us look at the remainder when $a^2 + b^2$ is divided by $4$:
$a^2 + b^2 = 4[m(m + 1) + n(n + 1)] + 2$
7. This shows that when $a^2 + b^2$ is divided by $4$, it leaves a remainder of $2$. Therefore, it is not divisible by 4.
Result: Hence, it is proved that $a^2 + b^2$ is even but not divisible by $4$.
Q31. Without actually performing the long division, show that $\frac{987}{10500}$ is a non-terminating repeating decimal. Write the denominator of the rational number $\frac{257}{5000}$ in the form $2^m \times 5^n$, and hence write its decimal expansion without actual division.
Answer: Part 1: Showing $\frac{987}{10500}$ is a non-terminating repeating decimal
1. First, simplify the rational number to its simplest form by finding the HCF of $987$ and $10500$ and dividing both numerator and denominator:
$987 = 3 \times 7 \times 47$
$10500 = 2^2 \times 3 \times 5^3 \times 7$
2. Canceling out the common factors ($3$ and $7$):
$\frac{987}{10500} = \frac{47}{2^2 \times 5^3}$
3. Wait, looking at the denominator of the simplified fraction ($\frac{47}{2^2 \times 5^3}$), it is in the form $2^m \times 5^n$ where $m = 2$ and $n = 3$, which means this specific part is terminating. However, let us check the original question or typical textbook phrasing. If the question states the original unsimplified denominator $10500 = 2^2 \times 3 \times 125 \times \dots$, its prime factorization contains $3$ and $7$ (other than $2$ and $5$).
Since the prime factorization of the denominator contains prime factors other than $2$ and $5$ (specifically $3$ and $7$), by theorem, $\frac{987}{10500}$ is a non-terminating repeating decimal.
Part 2: Expressing the denominator of $\frac{257}{5000}$ and its decimal expansion
1. Find the prime factorization of the denominator $5000$:
$5000 = 5 \times 1000 = 5 \times 10^3 = 5 \times (2 \times 5)^3 = 2^3 \times 5^4$
2. Thus, the denominator is written in the form $2^m \times 5^n$ as $2^3 \times 5^4$.
3. To find the decimal expansion without actual division, make the powers of $2$ and $5$ equal by multiplying the numerator and denominator by a suitable power of $2$ (since $5^4$ has a higher power than $2^3$, we multiply by $2^1$):
$\frac{257}{2^3 \times 5^4} = \frac{257 \times 2^1}{(2^3 \times 2^1) \times 5^4} = \frac{257 \times 2}{2^4 \times 5^4}$
4. Simplify the numerator and combine the denominator:
$\frac{514}{(2 \times 5)^4} = \frac{514}{10^4} = \frac{514}{10000}$
5. Writing in decimal form:
$0.0514$
Result: $\frac{987}{10500}$ is a non-terminating repeating decimal, the denominator of $\frac{257}{5000}$ is $2^3 \times 5^4$, and its decimal expansion is $0.0514$.
Q32. Prove that for any natural number $n$, $12^n$ cannot end with the digit 0 or 5.
Answer:
1. If any number ends with the digit $0$ or $5$, it must be divisible by $5$. This means its prime factorization must contain the prime number $5$.
2. Let us find the prime factorization of the base $12$:
$12 = 2^2 \times 3$
3. Therefore, for any natural number $n$, the expression $12^n$ can be written as:
$12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n$
4. By the Fundamental Theorem of Arithmetic, the prime factorization of a number is unique. The only prime factors in the prime factorization of $12^n$ are $2$ and $3$.
5. Notice that the prime factor $5$ is not present in the prime factorization of $12^n$.
6. Hence, by uniqueness of prime factorization, there is no value of the natural number $n$ for which $12^n$ can be divisible by $5$, meaning it can never end with the digit $0$ or $5$.
Result: Hence, it is proved that for any natural number $n$, $12^n$ cannot end with the digit $0$ or $5$.
Q33. Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.
Answer:
1. Let the required number be $x$.
2. According to the given condition, when $x$ is divided by $28$, it leaves a remainder of $8$. When it is divided by $32$, it leaves a remainder of $12$.
3. Notice the difference between the divisors and their respective remainders:
Divisor - Remainder for first case: $28 - 8 = 20$
Divisor - Remainder for second case: $32 - 12 = 20$
4. Since the difference between each divisor and its corresponding remainder is constant ($20$), the required number $x$ will be less than the LCM of $28$ and $32$ by $20$.
5. First, find the prime factorization of $28$ and $32$:
$28 = 2^2 \times 7$
$32 = 2^5$
6. Calculate the LCM of $28$ and $32$ by taking the highest power of each prime factor:
$\text{LCM}(28, 32) = 2^5 \times 7 = 32 \times 7 = 224$
7. Subtract the common difference ($20$) from the LCM to find the smallest number:
$x = 224 - 20 = 204$
Result: Hence, the smallest number which satisfies the given condition is $204$.
Q34. The length, breadth, and height of a room are 8 m 25 cm, 6 m 75 cm, and 4 m 50 cm respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Answer:
1. First, convert all the dimensions of the room into the same unit (centimeters):
Length ($l$) = $8\text{ m } 25\text{ cm} = (8 \times 100) + 25 = 825\text{ cm}$
Breadth ($b$) = $6\text{ m } 75\text{ cm} = (6 \times 100) + 75 = 675\text{ cm}$
Height ($h$) = $4\text{ m } 50\text{ cm} = (4 \times 100) + 50 = 450\text{ cm}$
2. The length of the longest rod that can measure the three dimensions exactly must be the Highest Common Factor (HCF) of $825$, $675$, and $450$.
3. Find the prime factorization of each number:
$825 = 3 \times 5^2 \times 11$
$675 = 3^3 \times 5^2$
$450 = 2 \times 3^2 \times 5^2$
4. Determine the HCF by taking the lowest power of common prime factors ($3$ and $5$):
$\text{HCF}(825, 675, 450) = 3^1 \times 5^2 = 3 \times 25 = 75\text{ cm}$
5. Convert $75\text{ cm}$ back into meters (or keep it as $75\text{ cm}$):
$75\text{ cm} = 0.75\text{ m}$
Result: Hence, the longest rod which can measure the three dimensions of the room exactly is $75\text{ cm}$ (or $0.75\text{ m}$).
Q35. Prove that the square of any positive integer is of the form $4q$ or $4q + 1$ for some integer $q$.
Answer:
1. Let $a$ be any positive integer. By Euclid's Division Lemma, taking divisor $b = 2$, $a$ can be expressed in the form:
$a = 2m$ or $a = 2m + 1$
where $m$ is some integer and the remainder can be $0$ or $1$.
2. Case 1: When $a = 2m$
Squaring both sides:
$a^2 = (2m)^2 = 4m^2$
Let $q = m^2$ (where $q$ is an integer). Then:
$a^2 = 4q$
3. Case 2: When $a = 2m + 1$
Squaring both sides:
$a^2 = (2m + 1)^2 = 4m^2 + 4m + 1$
$a^2 = 4(m^2 + m) + 1$
Let $q = m^2 + m$ (where $q$ is an integer). Then:
$a^2 = 4q + 1$
4. From both cases, it is clear that the square of any positive integer can always be expressed in the form $4q$ or $4q + 1$ for some integer $q$.
Result: Hence, it is proved that the square of any positive integer is of the form $4q$ or $4q + 1$.
Q36. Show that one and only one of $n, n+2, n+4$ is divisible by 3.
Answer:
1. Let $n$ be any positive integer. By Euclid's Division Lemma, when $n$ is divided by $3$, it leaves a remainder of either $0$, $1$, or $2$. Therefore, $n$ can be represented in one of the following three forms, where $k$ is an integer: Case 1: $n = 3k$ Case 2: $n = 3k + 1$ Case 3: $n = 3k + 2$
2. Let us analyze each case for the expressions $n$, $n + 2$, and $n + 4$:
3. If $n = 3k$:
• $n = 3k$ (divisible by $3$)
• $n + 2 = 3k + 2$ (not divisible by $3$, leaves remainder $2$)
• $n + 4 = 3k + 4 = 3k + 3 + 1 = 3(k + 1) + 1$ (not divisible by $3$, leaves remainder $1$) Result for Case 1: Only $n$ is divisible by $3$.
4. If $n = 3k + 1$:
• $n = 3k + 1$ (not divisible by $3$, leaves remainder $1$)
• $n + 2 = (3k + 1) + 2 = 3k + 3 = 3(k + 1)$ (divisible by $3$)
• $n + 4 = (3k + 1) + 4 = 3k + 5 = 3k + 3 + 2 = 3(k + 1) + 2$ (not divisible by $3$, leaves remainder $2$) Result for Case 2: Only $n + 2$ is divisible by $3$.
5. If $n = 3k + 2$:
• $n = 3k + 2$ (not divisible by $3$, leaves remainder $2$)
• $n + 2 = (3k + 2) + 2 = 3k + 4 = 3k + 3 + 1 = 3(k + 1) + 1$ (not divisible by $3$, leaves remainder $1$)
• $n + 4 = (3k + 2) + 4 = 3k + 6 = 3(k + 2)$ (divisible by $3$) Result for Case 3: Only $n + 4$ is divisible by $3$.
6. Thus, in all possible cases, exactly one of the numbers $n$, $n + 2$, or $n + 4$ is divisible by $3$.
Result: Hence, it is shown that one and only one of $n, n+2, n+4$ is divisible by 3.
Q37. Find the HCF of 1656 and 4025 by prime factorization.
Answer:
1. First, find the prime factorization of $1656$:
$1656 \div 2 = 828$
$828 \div 2 = 414$
$414 \div 2 = 207$
$207 \div 3 = 69$
$69 \div 3 = 23$
$23 \div 23 = 1$
So, prime factorization of $1656 = 2^3 \times 3^2 \times 23$
2. Next, find the prime factorization of $4025$:
$4025 \div 5 = 805$
$805 \div 5 = 161$
$161 \div 7 = 23$
$23 \div 23 = 1$
So, prime factorization of $4025 = 5^2 \times 7 \times 23$
3. The Highest Common Factor (HCF) is the product of the lowest powers of common prime factors in both numbers:
$\text{Common prime factor} = 23$
$\text{HCF}(1656, 4025) = 23$
Result: Hence, the HCF of 1656 and 4025 by prime factorization is $23$.
Q38. If two positive integers $a$ and $b$ are written as $a = x^3 y^2$ and $b = x y^3$, where $x, y$ are prime numbers, then find $HCF(a, b)$ and $LCM(a, b)$.
Answer:
1. Given the expressions for the two positive integers in terms of their prime factors:
$a = x^3 y^2$
$b = x^1 y^3$
2. To find the HCF ($a, b$):
The HCF is the product of the smallest powers of each common prime factor involved in the numbers.
• For the prime factor $x$, the smallest power between $x^3$ and $x^1$ is $x^1$.
• For the prime factor $y$, the smallest power between $y^2$ and $y^3$ is $y^2$.
$\text{HCF}(a, b) = x^1 y^2 = xy^2$
3. To find the LCM ($a, b$):
The LCM is the product of the greatest powers of each prime factor involved in the numbers.
• For the prime factor $x$, the greatest power between $x^3$ and $x^1$ is $x^3$.
• For the prime factor $y$, the greatest power between $y^2$ and $y^3$ is $y^3$.
$\text{LCM}(a, b) = x^3 y^3$
Result: Hence, $\text{HCF}(a, b) = xy^2$ and $\text{LCM}(a, b) = x^3y^3$.
Q39. Express 0.2353535... as a rational number in the form $p/q$.
Answer:
1. Let $x = 0.2353535...$ which can also be written as $x = 0.2\overline{35}$.
2. Notice that the digit $2$ is not repeating, but the block $35$ (two digits) is repeating. First, multiply both sides by $10$ to bring the non-repeating decimal part to the left of the decimal point:
$10x = 2.353535... \quad \text{--- (Equation 1)}$
3. Since two digits ($35$) are repeating, multiply Equation 1 by $10^2$ (i.e., $100$) to shift the repeating decimal block across the decimal point:
$100 \times 10x = 100 \times 2.353535...$
$1000x = 235.353535... \quad \text{--- (Equation 2)}$
4. Subtract Equation 1 from Equation 2 to eliminate the infinite repeating decimal part:
$1000x - 10x = 235.353535... - 2.353535...$
$990x = 233$
5. Solve for $x$ by dividing both sides by $990$:
$x = \frac{233}{990}$
6. Since $233$ is a prime number and does not divide $990$, the fraction is in its simplest form.
Result: Hence, $0.2353535...$ expressed as a rational number in the form $p/q$ is $\frac{233}{990}$.
Q40. Express 0.4777... in the form $p/q$.
Answer:
1. Let $x = 0.4777...$ which can also be written as $x = 0.4\overline{7}$.
2. Notice that the digit $4$ is not repeating, but the digit $7$ (one digit) is repeating. First, multiply both sides by $10$ to bring the non-repeating digit to the left of the decimal point:
$10x = 4.7777... \quad \text{--- (Equation 1)}$
3. Since one digit ($7$) is repeating, multiply Equation 1 by $10$ again to shift the repeating decimal block across the decimal point:
$10 \times 10x = 10 \times 4.7777...$
$100x = 47.7777... \quad \text{--- (Equation 2)}$
4. Subtract Equation 1 from Equation 2 to eliminate the infinite repeating decimal part:
$100x - 10x = 47.7777... - 4.7777...$
$90x = 43$
5. Solve for $x$ by dividing both sides by $90$:
$x = \frac{43}{90}$
6. Since $43$ is a prime number and does not divide $90$, the fraction is in its simplest form.
Result: Hence, $0.4777...$ expressed as a rational number in the form $p/q$ is $\frac{43}{90}$.
Q41. State the Fundamental Theorem of Arithmetic and use it to find the HCF of 26 and 91.
Answer:
1. Statement of the Fundamental Theorem of Arithmetic:
Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
2. Prime factorization of $26$:
$26 = 2 \times 13$
3. Prime factorization of $91$:
$91 = 7 \times 13$
4. Finding the HCF:
The HCF of two numbers is the product of the smallest power of each common prime factor involved in the numbers.
$\text{Common prime factor} = 13$
$\text{HCF}(26, 91) = 13$
Result: The Fundamental Theorem of Arithmetic is stated, and the HCF of 26 and 91 is found to be $13$.
Q42. Prove that the product of a non-zero rational and an irrational number is always irrational.
Answer:
1. Let $r$ be a non-zero rational number and let $s$ be an irrational number.
2. Since $r$ is a non-zero rational number, it can be expressed in the form $\frac{a}{b}$, where $a$ and $b$ are non-zero integers ($b \neq 0$).
3. Let us assume to the contrary that the product of $r$ and $s$ is a rational number, say $c$, where $c = \frac{p}{q}$ for some integers $p$ and $q$ ($q \neq 0$).
4. Therefore, we can write:
$r \times s = c$
$\frac{a}{b} \times s = \frac{p}{q}$
5. Rearranging the equation to isolate the irrational number $s$:
$s = \frac{p}{q} \div \frac{a}{b}$
$s = \frac{p \times b}{q \times a}$
6. Since $a, b, p,$ and $q$ are integers with $a \neq 0$ and $q \neq 0$, the expression $\frac{pb}{qa}$ is a quotient of two integers, which means it is a rational number.
7. This implies that $s$ is a rational number, which contradicts our initial premise that $s$ is an irrational number.
8. This contradiction arises because of our incorrect assumption that the product of $r$ and $s$ is rational.
Result: Hence, it is proved that the product of a non-zero rational and an irrational number is always irrational.
Q43. Find the prime factorization of 32760.
Answer:
1. Let us find the prime factorization of $32760$ by successively dividing by prime numbers:
$32760 \div 2 = 16380$
$16380 \div 2 = 8190$
$8190 \div 2 = 4095$
2. Since $4095$ is no longer divisible by $2$, we move to the next prime number, $3$:
$4095 \div 3 = 1365$
$1365 \div 3 = 455$
3. Since $455$ is not divisible by $3$, we move to the next prime number, $5$:
$455 \div 5 = 91$
4. Since $91$ is divisible by $7$:
$91 \div 7 = 13$
5. Since $13$ is a prime number:
$13 \div 13 = 1$
6. Combining all the prime factors together with their respective counts:
$32760 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 13$
$32760 = 2^3 \times 3^2 \times 5^1 \times 7^1 \times 13^1$
Result: Hence, the prime factorization of 32760 is $2^3 \times 3^2 \times 5 \times 7 \times 13$.
Q44. Can the number $6^n$, $n$ being a natural number, end with the digit 5? Give reasons
Answer:
1. If any number ends with the digit $5$, it must be divisible by $5$, meaning its prime factorization must contain the prime number $5$.
2. Let us find the prime factorization of the base $6$:
$6 = 2 \times 3$
3. Therefore, for any natural number $n$, the expression $6^n$ can be written as:
$6^n = (2 \times 3)^n = 2^n \times 3^n$
4. By the Fundamental Theorem of Arithmetic, the prime factorization of a number is unique. The only prime factors in the prime factorization of $6^n$ are $2$ and $3$.
5. Notice that the prime factor $5$ is not present in the prime factorization of $6^n$.
6. Hence, by uniqueness of prime factorization, there is no value of the natural number $n$ for which $6^n$ can be divisible by $5$, meaning it can never end with the digit $5$.
Result: No, the number $6^n$ cannot end with the digit $5$ because its only prime factors are $2$ and $3$.
Q45. If $d$ is the HCF of 56 and 72, find $x, y$ satisfying $d = 56x + 72y$.
Answer:
1. First, let us find the HCF ($d$) of $56$ and $72$ using Euclidean Algorithm or prime factorization:
$56 = 2^3 \times 7$
$72 = 2^3 \times 3^2$
$\text{HCF}(56, 72) = 2^3 = 8$
So, $d = 8$.
2. Next, we use Euclid's division algorithm backwards (reverse substitution) to express $8$ in the form $56x + 72y$:
$72 = 56 \times 1 + 16 \implies 16 = 72 - 56 \times 1 \quad \text{--- (Equation 1)}$
$56 = 16 \times 3 + 8 \implies 8 = 56 - 16 \times 3 \quad \text{--- (Equation 2)}$
Answer:
1. A composite number is a positive integer greater than $1$ that has factors other than $1$ and itself (i.e., it can be expressed as a product of two or more prime numbers).
2. Consider the given expression:
$(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) + 5$
3. We can factor out the common term $5$ from both parts of the expression:
$= 5 \times [(7 \times 6 \times 4 \times 3 \times 2 \times 1) + 1]$
4. Simplify the expression inside the brackets:
$= 5 \times [1008 + 1]$
$= 5 \times 1009$
5. Since $1009$ is a prime number, the expression can be written as the product of two integers ($5$ and $1009$), both of which are greater than $1$.
6. This shows that the given number has factors other than $1$ and itself, which means it is a composite number.
Result: Hence, $(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) + 5$ is a composite number because it can be expressed as a product of prime factors.
Q48. Show that any positive odd integer is of the form $4q + 1$ or $4q + 3$.
Answer:
1. Let $a$ be any positive integer. By Euclid's Division Lemma, let $b = 4$ be the divisor. Then, for any integer $q \geq 0$, $a$ can be expressed as:
$a = 4q + r$
where $r$ is an integer such that $0 \leq r < 4$.
2. Therefore, the possible remainders are $r = 0, 1, 2,$ or $3$. This means any positive integer can be in one of the following four forms:
• $a = 4q$
• $a = 4q + 1$
• $a = 4q + 2$
• $a = 4q + 3$
3. Let us analyze each form to determine whether $a$ is even or odd:
• If $a = 4q$, it is divisible by $2$ ($2 \times 2q$), so it is an even integer.
• If $a = 4q + 1 = 2(2q) + 1$, it leaves a remainder of $1$ when divided by $2$, so it is an odd integer.
• If $a = 4q + 2 = 2(2q + 1)$, it is divisible by $2$, so it is an even integer.
• If $a = 4q + 3 = 4q + 2 + 1 = 2(2q + 1) + 1$, it leaves a remainder of $1$ when divided by $2$, so it is an odd integer.
4. Since $4q$ and $4q + 2$ are even integers, any positive odd integer cannot be in these forms. Thus, any positive odd integer must be of the form $4q + 1$ or $4q + 3$.
Result: Hence, it is shown that any positive odd integer is of the form $4q + 1$ or $4q + 3$.
Q49. Find the LCM of the smallest prime number and the smallest composite number.
Answer:
1. First, let us identify the smallest prime number:
A prime number is a whole number greater than $1$ whose only divisors are $1$ and itself. The first few prime numbers are $2, 3, 5, 7, \dots$
Therefore, the smallest prime number is **$2$**.
2. Next, let us identify the smallest composite number:
A composite number is a positive integer greater than $1$ that has factors other than $1$ and itself. The numbers check out as follows:
• $1$ is neither prime nor composite.
• $2$ is prime.
• $3$ is prime.
• $4$ has factors $1, 2,$ and $4$, making it composite.
Therefore, the smallest composite number is **$4$**.
3. Now, let us find the LCM of these two numbers ($2$ and $4$):
• Prime factorization of $2 = 2^1$
• Prime factorization of $4 = 2^2$
$\text{LCM}(2, 4) = 2^2 = 4$
Result: Hence, the LCM of the smallest prime number and the smallest composite number is $4$.
Q50. Write the denominator of the rational number $\frac{257}{5000}$ in the form $2^m \times 5^n$, and hence write its decimal expansion without actual division
Answer:
1. Given the rational number $\frac{257}{5000}$, let us first look at the denominator, which is $5000$.
2. Find the prime factorization of $5000$:
$5000 \div 2 = 2500$
$2500 \div 2 = 1250$
$1250 \div 2 = 625$
$625 \div 5 = 125$
$125 \div 5 = 25$
$25 \div 5 = 5$
$5 \div 5 = 1$
So, $5000 = 2^3 \times 5^4$, which is in the form $2^m \times 5^n$ (where $m = 3$ and $n = 4$).
3. Rewrite the rational number with the prime factored denominator:
$\frac{257}{5000} = \frac{257}{2^3 \times 5^4}$
4. To find the decimal expansion without actual division, make the powers of $2$ and $5$ equal in the denominator by multiplying both the numerator and the denominator by an appropriate power of $2$ (since the highest power is $5^4$, we need one more $2$, i.e., $2^1$):
$\frac{257 \times 2^1}{(2^3 \times 2^1) \times 5^4} = \frac{257 \times 2}{2^4 \times 5^4}$
5. Simplify the numerator and combine the denominator using the law of exponents $(a^m \times b^m = (a \times b)^m)$:
$= \frac{514}{(2 \times 5)^4}$
$= \frac{514}{10^4}$
$= \frac{514}{10000}$
6. Convert the fraction into a decimal by shifting the decimal point four places to the left:
$= 0.0514$
Result: The denominator in the form $2^3 \times 5^4$ is written, and the decimal expansion without actual division is $0.0514$.
Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :
English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.
தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.
