CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 2 Marks - Part 1
Deepa Maths Academy • Global Examination Portal
CBSE Class 10 Maths Chapter 1 Real Numbers Model Questions - 2 Marks - Part 1
Secure University-Grade Repository for Model Assessments, Board Examinations, and Step-by-Step Solutions.
Portal ID: DMA-EXAM-2026• Security: Encrypted• Status: Active Examination
SECTION B — Very Short Answer Type Questions
[2 Marks Each]
Q1. Find the HCF and LCM of 120 and 144 using the prime factorization method.
Answer: #
To find the HCF and LCM of 120 and 144 using prime factorization:
1. Find the prime factorizations:
$120 = 2^3 \times 3 \times 5$
$144 = 2^4 \times 3^2$
2. HCF (Product of the smallest power of each common prime factor):
$\text{HCF}(120, 144) = 2^3 \times 3^1 = 8 \times 3 = 24$
3. LCM (Product of the greatest power of each prime factor involved):
$\text{LCM}(120, 144) = 2^4 \times 3^2 \times 5 = 16 \times 9 \times 5 = 720$
Result: $\text{HCF} = 24$, $\text{LCM} = 720$
Q2. Given that $\text{HCF}(306, 657) = 9$, find $\text{LCM}(306, 657)$.
Answer:
We know that for any two positive integers $a$ and $b$:
$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$
Given numbers are $a = 306$ and $b = 657$, and $\text{HCF}(306, 657) = 9$.
Since the given expression can be expressed as a product of prime factors other than 1 and itself (it has factors other than 1 and 13), it is a composite number.
Result: Proved that it is a composite number.
Q4. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.
Answer:
No, two numbers cannot have 15 as their HCF and 175 as their LCM.
Reason:
We know that the HCF of two numbers must always be a exact factor (divisor) of their LCM.
Let us check if 175 is completely divisible by 15:
$175 \div 15 = 11$ with a remainder of $10$ (or $11.66...$ which is not an integer).
Since the LCM is not an exact multiple of the HCF, two such numbers do not exist.
Result: No, because the HCF must divide the LCM exactly.
Q5. Find the smallest number which when divided by 28 and 32 leaves remainders 8 and 12 respectively.
Answer:
First, let us observe the difference between the divisors and their respective remainders:
$28 - 8 = 20$
$32 - 12 = 20$
Since the difference is the same ($20$) in both cases, the required number will be the LCM of 28 and 32 minus this common difference ($20$).
1. Find the prime factorizations of 28 and 32:
$28 = 2^2 \times 7$
$32 = 2^5$
3. Subtract the common difference:
$\text{Required Number} = 224 - 20 = 204$
Result: The smallest number is $204$.
Q6. Check whether $12^n$ can end with the digit 0 for any natural number $n$.
Answer:
No, $12^n$ cannot end with the digit 0 for any natural number $n$.
Reason:
1. If any number ends with the digit 0, it must be divisible by $10$. In other words, its prime factorization must contain both $2$ and $5$ as prime factors (since $10 = 2 \times 5$).
2. Let us find the prime factorization of $12$:
$12 = 2^2 \times 3$
So, $12^n = (2^2 \times 3)^n = 2^{2n} \times 3^n$
3. The prime factors of $12^n$ are only $2$ and $3$. It does not contain $5$ as a prime factor.
By the Fundamental Theorem of Arithmetic, the prime factorization of a number is unique. Since $5$ is missing from the prime factors, $12^n$ can never end with the digit $0$.
Result: No, it cannot end with the digit 0.
Q7. Find the HCF of 96 and 404 by the prime factorization method. Hence, find their LCM.
Answer:
1. Find the prime factorizations of 96 and 404:
$96 = 2^5 \times 3^1$
$404 = 2^2 \times 101^1$
2. Find the HCF (Product of the smallest power of each common prime factor):
$\text{HCF}(96, 404) = 2^2 = 4$
3. Find the LCM using the relation $\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$:
$4 \times \text{LCM}(96, 404) = 96 \times 404$
$\text{LCM}(96, 404) = \frac{96 \times 404}{4}$
$\text{LCM}(96, 404) = 96 \times 101$
$\text{LCM}(96, 404) = 9,696$
Result: $\text{HCF} = 4$, $\text{LCM} = 9,696$
Q8. Two bells toll at intervals of 12 and 18 minutes respectively. If they toll together at 12:00 PM, when will they toll together again?
Answer:
To find the time when the two bells will toll together again, we need to find the LCM of their individual tolling intervals ($12$ minutes and $18$ minutes).
1. Find the prime factorizations:
$12 = 2^2 \times 3^1$
$18 = 2^1 \times 3^2$
2. Find the $\text{LCM}(12, 18)$ (Product of the greatest power of each prime factor):
$\text{LCM}(12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36 \text{ minutes}$
3. Calculate the next tolling time:
Since they toll together at $12:00\text{ PM}$, they will toll together again after $36$ minutes.
$12:00\text{ PM} + 36\text{ minutes} = 12:36\text{ PM}$
Result: They will toll together again at $12:36\text{ PM}$.
Q9. Find the HCF of 81 and 237 and express it in the form $81x + 237y$.
Answer:
1. Use Euclidean's Division Algorithm to find the HCF of 237 and 81:
$237 = 81 \times 2 + 75$
$81 = 75 \times 1 + 6$
$75 = 6 \times 12 + 3$
$6 = 3 \times 2 + 0$
The last non-zero remainder is $3$. So, $\text{HCF}(81, 237) = 3$.
Result: $\text{HCF} = 3$, expressed as $81(-38) + 237(13)$ (where $x = -38$ and $y = 13$).
Q10. Is it possible for the HCF and LCM of two numbers to be 18 and 380 respectively? Justify.
Answer:
No, it is not possible for the HCF and LCM of two numbers to be 18 and 380 respectively.
Justification:
We know that the HCF of two numbers must always be an exact factor (divisor) of their LCM.
Let us check if $380$ is completely divisible by $18$:
$380 \div 18 = 21.11...$ (which is not an integer, or leaves a remainder of $2$).
Since the LCM is not an exact multiple of the HCF, no such pair of numbers can exist.
Result: No, because the HCF ($18$) does not divide the LCM ($380$) exactly.
Q11. Prove that $\sqrt{2}$ is irrational.
Answer:
Let us assume to the contrary that $\sqrt{2}$ is a rational number.
1. Then, we can find two integers $a$ and $b$ ($b \neq 0$) such that:
$\sqrt{2} = \frac{a}{b}$
Let $a$ and $b$ be co-prime (i.e., their HCF is 1).
2. Squaring on both sides:
$2 = \frac{a^2}{b^2}$
$a^2 = 2b^2$
This means that $2$ divides $a^2$, and by extension, $2$ also divides $a$.
3. Let $a = 2c$ for some integer $c$. Substituting this back:
$(2c)^2 = 2b^2$
$4c^2 = 2b^2$
$b^2 = 2c^2$
This means that $2$ divides $b^2$, and therefore $2$ also divides $b$.
4. Thus, both $a$ and $b$ have at least $2$ as a common factor. This contradicts our assumption that $a$ and $b$ are co-prime.
Result: This contradiction arises due to our incorrect assumption that $\sqrt{2}$ is rational. Hence, $\sqrt{2}$ is irrational.
Q12. Prove that $\sqrt{3}$ is irrational.
Answer:
Let us assume to the contrary that $\sqrt{3}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$\sqrt{3} = \frac{a}{b}$
2. Squaring on both sides:
$3 = \frac{a^2}{b^2}$
$a^2 = 3b^2$
This means that $3$ divides $a^2$, and therefore $3$ also divides $a$.
3. Let $a = 3c$ for some integer $c$. Substituting this value:
$(3c)^2 = 3b^2$
$9c^2 = 3b^2$
$b^2 = 3c^2$
This means that $3$ divides $b^2$, and therefore $3$ also divides $b$.
4. Thus, both $a$ and $b$ have at least $3$ as a common factor. This contradicts our assumption that $a$ and $b$ are co-prime.
Result: This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence, $\sqrt{3}$ is irrational.
Q13. Show that $5 - \sqrt{3}$ is an irrational number.
Answer:
Let us assume to the contrary that $5 - \sqrt{3}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$5 - \sqrt{3} = \frac{a}{b}$
2. Rearranging the terms to isolate the root on one side:
$\sqrt{3} = 5 - \frac{a}{b}$
$\sqrt{3} = \frac{5b - a}{b}$
3. Since $a$ and $b$ are integers, $\frac{5b - a}{b}$ is a rational number. This would imply that $\sqrt{3}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{3}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $5 - \sqrt{3}$ is rational. Hence, $5 - \sqrt{3}$ is irrational.
Q14. Prove that $3 + 2\sqrt{5}$ is irrational.
Answer:
Let us assume to the contrary that $3 + 2\sqrt{5}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$3 + 2\sqrt{5} = \frac{a}{b}$
2. Rearranging the terms to isolate the root term:
$2\sqrt{5} = \frac{a}{b} - 3$
$2\sqrt{5} = \frac{a - 3b}{b}$
$\sqrt{5} = \frac{a - 3b}{2b}$
3. Since $a$ and $b$ are integers, $\frac{a - 3b}{2b}$ is a rational number. This would imply that $\sqrt{5}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{5}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $3 + 2\sqrt{5}$ is rational. Hence, $3 + 2\sqrt{5}$ is irrational.
Q15. Show that $\frac{1}{\sqrt{2}}$ is irrational.
Answer:
Let us assume to the contrary that $\frac{1}{\sqrt{2}}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$\frac{1}{\sqrt{2}} = \frac{a}{b}$
2. Taking the reciprocal on both sides:
$\sqrt{2} = \frac{b}{a}$
3. Since $a$ and $b$ are integers, $\frac{b}{a}$ is a rational number. This would imply that $\sqrt{2}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{2}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $\frac{1}{\sqrt{2}}$ is rational. Hence, $\frac{1}{\sqrt{2}}$ is irrational.
Q16. Prove that $7\sqrt{5}$ is irrational.
Answer:
Let us assume to the contrary that $7\sqrt{5}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$7\sqrt{5} = \frac{a}{b}$
2. Rearranging the terms to isolate the root term:
$\sqrt{5} = \frac{a}{7b}$
3. Since $a$ and $b$ are integers, $\frac{a}{7b}$ is a rational number. This would imply that $\sqrt{5}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{5}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $7\sqrt{5}$ is rational. Hence, $7\sqrt{5}$ is irrational.
Q17. Show that $2 - 3\sqrt{5}$ is irrational.
Answer:
Let us assume to the contrary that $2 - 3\sqrt{5}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$2 - 3\sqrt{5} = \frac{a}{b}$
2. Rearranging the terms to isolate the root term:
$3\sqrt{5} = 2 - \frac{a}{b}$
$3\sqrt{5} = \frac{2b - a}{b}$
$\sqrt{5} = \frac{2b - a}{3b}$
3. Since $a$ and $b$ are integers, $\frac{2b - a}{3b}$ is a rational number. This would imply that $\sqrt{5}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{5}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $2 - 3\sqrt{5}$ is rational. Hence, $2 - 3\sqrt{5}$ is irrational.
Q18. Prove that $\sqrt{p} + \sqrt{q}$ is irrational, where $p$ and $q$ are primes.
Answer:
Let us assume to the contrary that $\sqrt{p} + \sqrt{q}$ is a rational number, say $r$.
1. Then we can write:
$\sqrt{p} + \sqrt{q} = r$
$\sqrt{p} = r - \sqrt{q}$
2. Squaring on both sides:
$p = (r - \sqrt{q})^2$
$p = r^2 + q - 2r\sqrt{q}$
4. Since $r$ is rational and $p, q$ are primes (whose square roots are irrational), the right-hand side is a rational number, which implies that $\sqrt{q}$ is rational.
5. However, this contradicts the fact that $\sqrt{q}$ is irrational (since $q$ is a prime).
Result: This contradiction has arisen due to our incorrect assumption. Hence, $\sqrt{p} + \sqrt{q}$ is irrational.
Q19. Prove that $5 - 2\sqrt{3}$ is an irrational number.
Answer:
Let us assume to the contrary that $5 - 2\sqrt{3}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$5 - 2\sqrt{3} = \frac{a}{b}$
2. Rearranging the terms to isolate the root term:
$2\sqrt{3} = 5 - \frac{a}{b}$
$2\sqrt{3} = \frac{5b - a}{b}$
$\sqrt{3} = \frac{5b - a}{2b}$
3. Since $a$ and $b$ are integers, $\frac{5b - a}{2b}$ is a rational number. This would imply that $\sqrt{3}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{3}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $5 - 2\sqrt{3}$ is rational. Hence, $5 - 2\sqrt{3}$ is irrational.
Q20. Show that $\frac{3}{\sqrt{7}}$ is not a rational number.
Answer:
Let us assume to the contrary that $\frac{3}{\sqrt{7}}$ is a rational number.
1. Then, we can find two co-prime integers $a$ and $b$ ($b \neq 0$) such that:
$\frac{3}{\sqrt{7}} = \frac{a}{b}$
2. Rearranging the terms to isolate the root term:
$\sqrt{7} = \frac{3b}{a}$
3. Since $a$ and $b$ are integers, $\frac{3b}{a}$ is a rational number. This would imply that $\sqrt{7}$ is also a rational number.
4. However, this contradicts the fact that $\sqrt{7}$ is an irrational number.
Result: This contradiction has arisen due to our incorrect assumption that $\frac{3}{\sqrt{7}}$ is rational. Hence, $\frac{3}{\sqrt{7}}$ is not a rational number.
Q21. Without actually performing the long division, state whether $\frac{13}{3125}$ has a terminating or non-terminating repeating decimal expansion.
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 $3125$:
$3125 = 5^5 = 2^0 \times 5^5$
2. Since the prime factorization of the denominator is of the form $2^n \times 5^m$ (where $n = 0$ and $m = 5$), the rational number has a terminating decimal expansion.
Result: $\frac{13}{3125}$ has a terminating decimal expansion.
Q22. Write the decimal expansion of $\frac{7}{80}$ without actual division.
Answer:
To write the decimal expansion without actual division, we can express the denominator in the form $2^n \times 5^m$ and make the powers of $2$ and $5$ equal.
1. Find the prime factorization of the denominator $80$:
$80 = 2^4 \times 5^1$
2. To make the exponents of $2$ and $5$ equal, multiply the numerator and the denominator by $5^3$ (since $4 - 1 = 3$):
$\frac{7}{2^4 \times 5^1} = \frac{7 \times 5^3}{2^4 \times 5^1 \times 5^3}$
Result: The decimal expansion of $\frac{7}{80}$ is $0.0875$.
Q23. State the condition on the denominator $q$ such that a rational number $\frac{p}{q}$ has a terminating decimal expansion.
Answer:
Let $\frac{p}{q}$ be a rational number, where $p$ and $q$ are co-prime. The rational number $\frac{p}{q}$ has a terminating decimal expansion if and only 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.
Result: The prime factors of $q$ must be only $2$ and/or $5$ (i.e., of the form $2^n 5^m$).
Q24. Without dividing, show that $\frac{129}{2^2 \times 5^7 \times 7^5}$ is a non-terminating repeating decimal.
Answer:
A rational number $\frac{p}{q}$ has a terminating decimal expansion if and only 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. Observe the denominator of the given rational number:
Denominator $q = 2^2 \times 5^7 \times 7^5$
2. Along with powers of $2$ and $5$, the prime factorization also contains $7$ (i.e., $7^5$), which is not of the form $2^n \times 5^m$.
3. Since the prime factors of the denominator include a prime number other than $2$ and $5$, the decimal expansion cannot terminate.
Result: $\frac{129}{2^2 \times 5^7 \times 7^5}$ has a non-terminating repeating decimal expansion.
Q25. Find the decimal representation of $\frac{14587}{1250}$.
Answer:
To find the decimal representation without long division, we can express the denominator in the form $2^n \times 5^m$ and make the powers of $2$ and $5$ equal.
1. Find the prime factorization of the denominator $1250$:
$1250 = 125 \times 10 = 5^3 \times (2 \times 5) = 2^1 \times 5^4$
2. To make the exponents of $2$ and $5$ equal, multiply the numerator and the denominator by $2^3$ (since $4 - 1 = 3$):
$\frac{14587}{2^1 \times 5^4} = \frac{14587 \times 2^3}{2^1 \times 5^4 \times 2^3}$
Result: The decimal representation of $\frac{14587}{1250}$ is $11.6696$.
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.
