CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 2 Marks - Part 2
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CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 2 Marks - Part 2
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SECTION B — Very Short Answer Type Questions
[2 Marks Each]
Q26. If $\frac{p}{q}$ is a rational number, what can you say about $q$ if its decimal expansion is non-terminating repeating?
Answer:
Let $\frac{p}{q}$ be a rational number in its simplest form (where $p$ and $q$ are co-prime).
1. If the decimal expansion of $\frac{p}{q}$ is non-terminating repeating, then the prime factorization of the denominator $q$ is **not** of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.
2. This means that $q$ must have at least one prime factor other than $2$ and $5$.
Result: The prime factorization of $q$ contains prime factors other than $2$ or $5$.
Q27. Determine if $\frac{17}{8}$ has a terminating decimal expansion. If yes, find it.
Answer:
A rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.
1. Find the prime factorization of the denominator $8$:
$8 = 2^3 = 2^3 \times 5^0$
2. Since the prime factorization of the denominator is of the form $2^n \times 5^m$ (where $n = 3$ and $m = 0$), the rational number has a terminating decimal expansion.
3. To find the decimal expansion without actual division, make the powers of $2$ and $5$ equal by multiplying the numerator and the denominator by $5^3$:
$\frac{17}{2^3} = \frac{17 \times 5^3}{2^3 \times 5^3}$
$= \frac{17 \times 125}{(2 \times 5)^3}$
$= \frac{2125}{10^3}$
$= \frac{2125}{1000}$
$= 2.125$
Result: $\frac{17}{8}$ has a terminating decimal expansion, and its value is $2.125$.
Q28. After how many decimal places will the decimal expansion of $\frac{23}{2^4 \times 5^3}$ terminate?
Answer:
The number of decimal places after which the decimal expansion of a terminating rational number $\frac{p}{2^n \times 5^m}$ will terminate is equal to the maximum of the exponents of $2$ and $5$ in the denominator (i.e., $\max(n, m)$).
1. Identify the exponents of $2$ and $5$ in the denominator of $\frac{23}{2^4 \times 5^3}$:
Exponent of $2$ ($n$) $= 4$
Exponent of $5$ ($m$) $= 3$
2. Find the maximum of the two exponents:
$\max(4, 3) = 4$
Result: The decimal expansion of $\frac{23}{2^4 \times 5^3}$ will terminate after $4$ decimal places.
Q29. Write down the prime factorization of the denominator of the rational number $0.125$
Answer:
1. First, convert the given decimal number into a fraction in its simplest form:
$0.125 = \frac{125}{1000}$
2. Simplify the fraction by dividing the numerator and the denominator by their highest common factor ($125$):
$\frac{125 \div 125}{1000 \div 125} = \frac{1}{8}$
3. Find the prime factorization of the denominator ($8$):
$8 = 2^3$
Result: The prime factorization of the denominator is $2^3$ (or $2^3 \times 5^0$).
Q30. Is $\frac{6}{15}$ a terminating decimal? Justify your answer.
Answer:
To determine if a rational number has a terminating decimal expansion, we first express it in its simplest form (where the numerator and denominator are co-prime).
1. Simplify the given fraction $\frac{6}{15}$ by dividing the numerator and denominator by their HCF ($3$):
$\frac{6}{15} = \frac{6 \div 3}{15 \div 3} = \frac{2}{5}$
2. Examine the denominator of the simplified fraction ($5$):
$5 = 2^0 \times 5^1$
3. Since the prime factorization of the denominator is of the form $2^n \times 5^m$ (where $n = 0$ and $m = 1$), it satisfies the condition for a terminating decimal.
Result: Yes, $\frac{6}{15}$ is a terminating decimal.
Q31. Find the HCF of 180, 252, and 324 using prime factorization.
Answer:
To find the HCF using prime factorization, we find the prime factors of each number and take the product of the smallest powers of each common prime factor.
1. Find the prime factorization of each number:
$180 = 2^2 \times 3^2 \times 5$
$252 = 2^2 \times 3^2 \times 7$
$324 = 2^2 \times 3^4$
2. Identify the common prime factors with the smallest exponents:
Common prime factors are $2$ and $3$.
Smallest power of $2$ is $2^2$.
Smallest power of $3$ is $3^2$.
Q32. The product of two numbers is 1600 and their HCF is 5. Find their LCM.
Answer:
For any two positive integers $a$ and $b$, the relationship between their HCF, LCM, and product is given by:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = \text{Product of two numbers}$
1. Substitute the given values into the formula:
$5 \times \text{LCM} = 1600$
2. Solve for $\text{LCM}$:
$\text{LCM} = \frac{1600}{5}$
$\text{LCM} = 320$
Result: The LCM of the two numbers is $320$.
Q33. Can the product of two irrational numbers be rational? Give an example.
Answer:
Yes, the product of two irrational numbers can be a rational number.
1. Let us consider two irrational numbers, say $\sqrt{3}$ and $2\sqrt{3}$ (or simply $\sqrt{3}$ and $\sqrt{3}$).
3. Since $3$ is a rational number, this demonstrates that the product of two irrational numbers can indeed be rational.
Result: Yes. For example, $\sqrt{3} \times \sqrt{3} = 3$, which is a rational number.
Q34. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.
Answer:
To find the largest number that divides two given numbers leaving specific remainders, we subtract the respective remainders from each number and then find the HCF of the resulting numbers.
1. Subtract the remainders from the given numbers:
$2053 - 5 = 2048$
$967 - 7 = 960$
2. Find the prime factorization of $2048$ and $960$:
$2048 = 2^{11}$
$960 = 2^6 \times 3 \times 5$
3. Find the HCF by taking the lowest power of common prime factors:
$\text{HCF}(2048, 960) = 2^6 = 64$
Result: The largest number that divides 2053 and 967 leaving remainders 5 and 7 respectively is $64$.
Q35. If $n$ is an odd integer, show that $n^2 - 1$ is divisible by 8.
Answer:
Since $n$ is an odd integer, it can be expressed in the form $n = 2k + 1$, where $k$ is an integer.
2. Expand the square using the identity $(a + b)^2 = a^2 + 2ab + b^2$:
$= (4k^2 + 4k + 1) - 1$
$= 4k^2 + 4k$
3. Factor out $4k$ from the expression:
$= 4k(k + 1)$
4. Since $k$ and $k + 1$ are consecutive integers, one of them must be even. Therefore, the product $k(k + 1)$ is always divisible by $2$ (let $k(k + 1) = 2m$ for some integer $m$).
5. Substitute this back into our expression:
$n^2 - 1 = 4(2m) = 8m$
Result: Since $n^2 - 1$ is a multiple of $8$, it is divisible by $8$.
Q36. Find the HCF of the smallest prime number and the smallest composite number.
Answer:
1. Identify the smallest prime number and the smallest composite number:
The smallest prime number is $2$.
The smallest composite number is $4$.
2. Find the prime factorization of both numbers:
$2 = 2^1$
$4 = 2^2$
3. Determine the HCF by taking the lowest power of the common prime factor:
$\text{HCF}(2, 4) = 2^1 = 2$
Result: The HCF of the smallest prime number and the smallest composite number is $2$.
Q37. Prove that the product of a non-zero rational and an irrational number is irrational.
Answer:
Let us prove this by the method of contradiction.
1. Let $r$ be a non-zero rational number and let $x$ be an irrational number. We need to show that their product $rx$ is irrational.
2. Assume to the contrary that $rx$ is a rational number, say $rx = q$, where $q$ is a rational number.
3. Since $r$ is a non-zero rational number, its multiplicative inverse $\frac{1}{r}$ is also a rational number.
4. Multiply both sides of our assumed equation by $\frac{1}{r}$:
$x = q \times \frac{1}{r}$
5. Since the product of two rational numbers ($q$ and $\frac{1}{r}$) is always rational, this implies that $x$ must be a rational number.
6. However, this contradicts our initial premise that $x$ is an irrational number. Our assumption that $rx$ is rational must therefore be false.
Result: The product of a non-zero rational and an irrational number is always irrational.
1. Since there is $1$ repeating digit after the decimal point, multiply both sides of Equation 1 by $10$:
$10x = 6.666...$ $\text{--- (Equation 2)}$
2. Subtract Equation 1 from Equation 2:
$10x - x = 6.666... - 0.666...$
$9x = 6$
3. Solve for $x$:
$x = \frac{6}{9}$
4. Simplify the fraction to its lowest terms by dividing the numerator and denominator by their HCF ($3$):
$x = \frac{2}{3}$
Result: $0.666...$ expressed in the form $p/q$ is $\frac{2}{3}$.
Q39. Write the condition for a rational number to have a non-terminating repeating decimal expansion.
Answer:
Let $\frac{p}{q}$ be a rational number in its simplest form (where $p$ and $q$ are co-prime integers and $q \neq 0$).
1. The condition for the rational number to have a non-terminating repeating decimal expansion is that the prime factorization of the denominator $q$ is **not** of the form $2^n \times 5^m$, where $n$ and $m$ are non-negative integers.
2. In other words, $q$ must contain at least one prime factor other than $2$ or $5$.
Result: The prime factorization of $q$ is not of the form $2^n \times 5^m$.
Q40. Check if $6^n$ can end with the digit 0 for any natural number $n$.
Answer:
For any number to end with the digit $0$, it must be divisible by $10$, which means its prime factorization must contain both $2$ and $5$ as prime factors.
1. Consider the prime factorization of $6^n$:
$6^n = (2 \times 3)^n = 2^n \times 3^n$
2. Observe the prime factors of $6^n$:
The only prime factors in the prime factorization of $6^n$ are $2$ and $3$. The prime factor $5$ is absent.
3. By the Fundamental Theorem of Arithmetic, this prime factorization is unique. Since $5$ is not a prime factor of $6^n$, $6^n$ can never be divisible by $5$ for any natural number $n$.
Result: No, $6^n$ cannot end with the digit $0$ for any natural number $n$.
Q41. Find the LCM of $2^3 \times 3^2$ and $2^2 \times 3^3$.
Answer:
To find the LCM of two numbers given in their prime factorized form, we take the product of the greatest powers of each prime factor involved.
1. Identify the prime factors present in both numbers ($2$ and $3$).
2. Find the highest power of each prime factor:
For $2$, the highest power is $2^3$.
For $3$, the highest power is $3^3$.
Result: The LCM of $2^3 \times 3^2$ and $2^2 \times 3^3$ is $216$.
Q42. State the Fundamental Theorem of Arithmetic.
Answer:
The Fundamental Theorem of Arithmetic (also known as the Unique Factorization Theorem) states that:
1. Every composite number can be expressed (or factorized) as a product of prime numbers.
2. This factorization is unique, apart from the order in which the prime factors occur.
Result: Every composite number can be uniquely factored as a product of primes, ignoring the order of the factors.
Q43. If $a = x^3 y^2$ and $b = x y^3$, find the HCF of $a$ and $b$.
Answer:
To find the HCF of two numbers given in their prime factorized form, we take the product of the smallest powers of each common prime factor.
1. Identify the given expressions in prime factorized form:
$a = x^3 y^2$
$b = x^1 y^3$
2. Identify the common prime factors ($x$ and $y$) and find their lowest powers:
For $x$, the smallest power between $x^3$ and $x^1$ is $x^1$ (or $x$).
For $y$, the smallest power between $y^2$ and $y^3$ is $y^2$.
3. Calculate the HCF:
$\text{HCF}(a, b) = x^1 y^2 = xy^2$
Result: The HCF of $a$ and $b$ is $xy^2$.
Q44. Find the smallest number which when divided by 15, 20, and 30 leaves a remainder of 4 in each case.
Answer:
To find the smallest number that leaves a constant remainder when divided by multiple numbers, we first find the LCM of those divisors and then add the remainder.
1. Find the prime factorization of $15$, $20$, and $30$:
$15 = 3 \times 5$
$20 = 2^2 \times 5$
$30 = 2 \times 3 \times 5$
2. Determine the LCM by taking the highest power of each prime factor:
$\text{LCM}(15, 20, 30) = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$
3. Add the common remainder ($4$) to the LCM to find the required number:
$\text{Required number} = 60 + 4 = 64$
Result: The smallest number is $64$.
Q45. Prove that $n^2 - n$ is divisible by 2 for every positive integer $n$.
Answer:
Let us factorize the given expression $n^2 - n$:
$n^2 - n = n(n - 1)$
Notice that $n$ and $n - 1$ are two consecutive positive integers.
1. For any two consecutive integers, one of them is always even and the other is odd.
2. Since the product of an even integer and any integer is always even, the product $n(n - 1)$ must be an even integer.
3. An even integer is always divisible by $2$.
Result: Therefore, $n^2 - n$ is divisible by $2$ for every positive integer $n$.
Q46. Can $\frac{x}{y}$ be an irrational number if $x$ and $y$ are both irrational?
Answer:
Yes, the quotient of two irrational numbers can be an irrational number.
1. Let us consider two irrational numbers, say $x = \sqrt{6}$ and $y = \sqrt{2}$ (both are irrational).
3. Since $\sqrt{3}$ is an irrational number, this demonstrates that $\frac{x}{y}$ can indeed be irrational.
Result: Yes. For example, if $x = \sqrt{6}$ and $y = \sqrt{2}$, then $\frac{x}{y} = \sqrt{3}$, which is irrational.
Q47. Find the HCF of $2^3 \times 3^2 \times 5$ and $2^2 \times 3^3 \times 5^2$
Answer:
To find the HCF of two numbers given in their prime factorized form, we take the product of the smallest powers of each common prime factor.
1. Identify the common prime factors in both numbers ($2$, $3$, and $5$).
2. Find the lowest power of each common prime factor:
For $2$, the smallest power between $2^3$ and $2^2$ is $2^2$.
For $3$, the smallest power between $3^2$ and $3^3$ is $3^2$.
For $5$, the smallest power between $5^1$ and $5^2$ is $5^1$ (or $5$).
Q48. Explain the steps to represent $0.\bar{7}$ as a rational number.
Answer:
To express the repeating decimal $0.\bar{7}$ as a rational number in the form $\frac{p}{q}$, follow these steps:
1. Let $x = 0.\bar{7}$, which means $x = 0.777...$ $\text{--- (Equation 1)}$
2. Since there is $1$ repeating digit ($7$) after the decimal point, multiply both sides of Equation 1 by $10$:
$10x = 7.777...$ $\text{--- (Equation 2)}$
3. Subtract Equation 1 from Equation 2 to eliminate the repeating decimal part:
$10x - x = 7.777... - 0.777...$
$9x = 7$
4. Solve for $x$ by dividing both sides by $9$:
$x = \frac{7}{9}$
Result: $0.\bar{7}$ represented as a rational number is $\frac{7}{9}$.
Q49. Find the prime factorization of 3825.
Answer:
To find the prime factorization of $3825$, we successively divide the number by prime numbers starting from the smallest:
1. Check divisibility by $3$ (sum of digits $3 + 8 + 2 + 5 = 18$, which is divisible by $3$):
$3825 \div 3 = 1275$
2. Check divisibility by $3$ again ($1 + 2 + 7 + 5 = 15$, which is divisible by $3$):
$1275 \div 3 = 425$
3. Check divisibility by $5$ (ends in $5$):
$425 \div 5 = 85$
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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.
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