SEBA Class 10 Maths Miscellaneous Questions Solutions from Chapter 1 New Books
Here you will find all the solutions to SEBA Class 10 Mathematics Miscellaneous Questions from Chapter 1, New Books 2026. For better understanding of this chapter, first visit Class 10 Maths Chapter 1.1 Solutions and Class 10 Maths Chapter 1.2 Solutions
Multiple Choice Questions (MCQs)
1. LCM of $(2^{3} \times 3 \times 5)$ and $(2^{4} \times 5 \times 7)$ is:
(a) 40
(b) 560
(c) 1680
(d) 1120
$\mathbf{Sol^n.}$
LCM is the product of the greatest power of each prime factor involved in the numbers.
$\text{LCM} = 2^{4} \times 3 \times 5 \times 7$
$= 16 \times 3 \times 5 \times 7$
$= 1680$
Correct option is (c) 1680
2. HCF of $(3^{4} \times 2^{4} \times 7^{3})$ and $(3^{2} \times 5 \times 7)$ is:
(a) 630
(b) 63
(c) 729
(d) 567
$\mathbf{Sol^n.}$
HCF is the product of the smallest power of each common prime factor involved in the numbers.
$\text{HCF} = 3^{2} \times 7$
$= 9 \times 7$
$= 63$
Correct option is (b) 63
3. HCF of $(3^{3} \times 5^{2} \times 2)$, $(3^{2} \times 5^{3} \times 2^{2})$ and $(3^{4} \times 5 \times 2^{3})$ is:
(a) 450
(b) 90
(c) 180
(d) 630
$\mathbf{Sol^n.}$
Explanation:-
Smallest power of $2 = 2^{1}$
Smallest power of $3 = 3^{2}$
Smallest power of $5 = 5^{1}$
$\text{HCF} = 2^{1} \times 3^{2} \times 5^{1}$
$= 2 \times 9 \times 5$
$= 90$
Correct option is (b) 90
4. If $\text{HCF}(420, 189) = 21$, then $\text{LCM}(420, 189)$ is:
(a) 420
(b) 1890
(c) 3780
(d) 3680
$\mathbf{Sol^n.}$
We know that:
$\text{LCM} \times \text{HCF} = \text{Product of two numbers}$
$\Rightarrow \text{LCM} \times 21 = 420 \times 189$
$\Rightarrow \text{LCM} = \frac{420 \times 189}{21}$
$\Rightarrow \text{LCM} = 420 \times 9$
$\therefore \text{LCM} = 3780$
Correct option is (c) 3780
Very Short Answer (VSA) Type Questions
5. Using prime factorisation, find HCF and LCM of 96 and 120.
Solution:
By prime factorisation, we get:
$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^{5} \times 3$
$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^{3} \times 3 \times 5$
Therefore,
$\text{HCF}(96, 120) = 2^{3} \times 3 = 8 \times 3 = 24$
$\text{LCM}(96, 120) = 2^{5} \times 3 \times 5 = 32 \times 15 = 480$
6. Find LCM of 576 and 512 by prime factorization.
Solution:
By prime factorisation, we get:
$576 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 = 2^{6} \times 3^{2}$
$512 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{9}$
Therefore,
$\text{LCM}(576, 512) = 2^{9} \times 3^{2} = 512 \times 9 = 4608$
7. Find HCF of 660 and 704 by prime factorization.
Solution:
By prime factorisation, we get:
$660 = 2 \times 2 \times 3 \times 5 \times 11 = 2^{2} \times 3 \times 5 \times 11$
$704 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 11 = 2^{6} \times 11$
Therefore,
$\text{HCF}(660, 704) = 2^{2} \times 11 = 4 \times 11 = 44$
8. Find LCM of 480 and 256 using prime factorization.
Solution:
By prime factorisation, we get:
$480 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^{5} \times 3 \times 5$
$256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^{8}$
Therefore,
$\text{LCM}(480, 256) = 2^{8} \times 3 \times 5 = 256 \times 15 = 3840$
9. Prove that $7 – 2\sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $7 – 2\sqrt{3}$ is rational.
That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:
$7 – 2\sqrt{3} = \frac{a}{b}$
$\Rightarrow 2\sqrt{3} = 7 – \frac{a}{b}$
$\Rightarrow 2\sqrt{3} = \frac{7b – a}{b}$
$\Rightarrow \sqrt{3} = \frac{7b – a}{2b}$
Since $7$, $2$, $a$, and $b$ are integers, $\frac{7b – a}{2b}$ is rational, and so $\sqrt{3}$ is rational.
But this contradicts the fact that $\sqrt{3}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $7 – 2\sqrt{3}$ is rational.
So, we conclude that $7 – 2\sqrt{3}$ is an irrational number.
10. Prove that $8 + 5\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $8 + 5\sqrt{5}$ is rational.
That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:
$8 + 5\sqrt{5} = \frac{a}{b}$
$\Rightarrow 5\sqrt{5} = \frac{a}{b} – 8$
$\Rightarrow 5\sqrt{5} = \frac{a – 8b}{b}$
$\Rightarrow \sqrt{5} = \frac{a – 8b}{5b}$
Since $8$, $5$, $a$, and $b$ are integers, $\frac{a – 8b}{5b}$ is rational, and so $\sqrt{5}$ is rational.
But this contradicts the fact that $\sqrt{5}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $8 + 5\sqrt{5}$ is rational.
So, we conclude that $8 + 5\sqrt{5}$ is an irrational number.
11. Prove that $11 + 3\sqrt{2}$ is an irrational number, given that $\sqrt{2}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $11 + 3\sqrt{2}$ is rational.
That is, we can find coprime integers $a$ and $b$ ($b \ne 0$) such that:
$11 + 3\sqrt{2} = \frac{a}{b}$
$\Rightarrow 3\sqrt{2} = \frac{a}{b} – 11$
$\Rightarrow 3\sqrt{2} = \frac{a – 11b}{b}$
$\Rightarrow \sqrt{2} = \frac{a – 11b}{3b}$
Since $11$, $3$, $a$, and $b$ are integers, $\frac{a – 11b}{3b}$ is rational, and so $\sqrt{2}$ is rational.
But this contradicts the fact that $\sqrt{2}$ is irrational.
This contradiction has arisen because of our incorrect assumption that $11 + 3\sqrt{2}$ is rational.
So, we conclude that $11 + 3\sqrt{2}$ is an irrational number.
12. (a) Find the H.C.F and L.C.M of 480 and 720 using the Prime factorization method.
Solution:
By prime factorisation, we get:
$480 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 5 = 2^{5} \times 3 \times 5$
$720 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 2^{4} \times 3^{2} \times 5$
Therefore,
$\text{HCF}(480, 720) = 2^{4} \times 3 \times 5 = 16 \times 3 \times 5 = 240$
$\text{LCM}(480, 720) = 2^{5} \times 3^{2} \times 5 = 32 \times 9 \times 5 = 1440$
OR
(b) The H.C.F of 85 and 238 is expressible in the form $85m – 238$. Find the value of $m$?
Solution:
First, we find the prime factors of $85$ and $238$:
$85 = 5 \times 17$
$238 = 2 \times 7 \times 17$
$\text{HCF}(85, 238) = 17$
According to the question:
$85m – 238 = \text{HCF}(85, 238)$
$\Rightarrow 85m – 238 = 17$
$\Rightarrow 85m = 17 + 238$
$\Rightarrow 85m = 255$
$\Rightarrow m = \frac{255}{85}$
$\therefore m = 3$
So, the value of $m$ is $3$.
Short Answer (SA) Type Questions
13. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change together next?
Solution:
The time after which the traffic lights change together will be the LCM of $48$, $72$, and $108$.
By prime factorisation:
$48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^{4} \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^{3} \times 3^{2}$
$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^{2} \times 3^{3}$
Now,
$\text{LCM}(48, 72, 108) = 2^{4} \times 3^{3} = 16 \times 27 = 432$
So, the lights will change together again after $432\text{ seconds}$.
Converting seconds into minutes:
$432\text{ seconds} = \frac{432}{60}\text{ minutes} = 7\text{ minutes } 12\text{ seconds}$
Therefore, the lights will change together next at $7\text{ hr } 7\text{ min } 12\text{ sec a.m.}$ (or $7:07:12\text{ a.m.}$).
14. Prove that $\sqrt{5}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $\sqrt{5}$ is rational.
Let $\sqrt{5} = \frac{a}{b}$, where $a$ and $b$ are coprime and $b \ne 0$.
So, $b\sqrt{5} = a$.
Squaring on both sides and rearranging, we get:
$5b^2 = a^2 \quad \text{— (1)}$
$\Rightarrow b^2 = \frac{a^2}{5}$
$\therefore a^2$ is divisible by $5$.
So, $a$ is divisible by $5$.
Let $a = 5c$ for some integer $c$. Substituting the value of $a$ in equation (1), we get:
$5b^2 = (5c)^2$
$\Rightarrow 5b^2 = 25c^2$
$\Rightarrow b^2 = 5c^2$
$\Rightarrow \frac{b^2}{5} = c^2$
This means $b^2$ is divisible by $5$, and so $b$ is also divisible by $5$.
Therefore, $a$ and $b$ have at least $5$ as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{5}$ is rational.
So, we conclude that $\sqrt{5}$ is irrational.
15. Prove that $\sqrt{3}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $\sqrt{3}$ is rational.
Let $\sqrt{3} = \frac{a}{b}$, where $a$ and $b$ are coprime and $b \ne 0$.
So, $b\sqrt{3} = a$.
Squaring on both sides and rearranging, we get:
$3b^2 = a^2 \quad \text{— (1)}$
$\Rightarrow b^2 = \frac{a^2}{3}$
$\therefore a^2$ is divisible by $3$.
So, $a$ is divisible by $3$.
Let $a = 3c$ for some integer $c$. Substituting the value of $a$ in equation (1), we get:
$3b^2 = (3c)^2$
$\Rightarrow 3b^2 = 9c^2$
$\Rightarrow b^2 = 3c^2$
$\Rightarrow \frac{b^2}{3} = c^2$
This means $b^2$ is divisible by $3$, and so $b$ is also divisible by $3$.
Therefore, $a$ and $b$ have at least $3$ as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational.
So, we conclude that $\sqrt{3}$ is irrational.