Real Numbers  Class 10  Mathematics
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Exercise 1.1  Real Numbers  NCERT  Mathematics  Class 10
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Here are the prime factorizations of the given numbers:
(i) $140 = 2^2 \times 5 \times 7$
(ii) $156 = 2^2 \times 3 \times 13$
(iii) $3825 = 3^2 \times 5^2 \times 17$
(iv) $5005 = 5 \times 7 \times 11 \times 13$
(v) $7429 = 17 \times 19 \times 23$
Find the LCM and HCF of the following pairs of integers and verify that $\mathrm{LCM} \times \mathrm{HCF}=$ product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
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Sign up nowFind the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8,9 and 25
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Sign up nowGiven that $\operatorname{HCF}(306,657)=9$, find $\operatorname{LCM}(306,657)$.
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Sign up nowCheck whether $6^{n}$ can end with the digit 0 for any natural number $n$.
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Sign up nowExplain why $7 \times 11 \times 13+13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1+5$ are composite numbers.
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Sign up nowThere is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?
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Sign up nowExercise 1.2  Real Numbers  NCERT  Mathematics  Class 10
Prove that $\sqrt{5}$ is irrational.
To prove that (\sqrt{5}) is irrational, we'll use a proof by contradiction. Here are the steps:

Assume the opposite: Assume that (\sqrt{5}) is rational. This means it can be written as a fraction (\frac{a}{b}), where (a) and (b) are coprime integers (they have no common positive divisor other than 1), and (b \neq 0).

Expressing the Assumption: Based on the assumption, (\sqrt{5} = \frac{a}{b}). Squaring both sides gives us (5 = \frac{a^2}{b^2}), which implies (a^2 = 5b^2).

Contradiction on Integrality: The last equation implies that (a^2) is divisible by 5. Therefore, (a) must also be divisible by 5 (because the square of a nonmultiple of 5 is never a multiple of 5). Let (a = 5k) for some integer (k).

Substituting Back: Substitute (a = 5k) back into (a^2 = 5b^2). We get ((5k)^2 = 5b^2), which simplifies to (25k^2 = 5b^2), and further to (5k^2 = b^2).

Contradiction on (b): The equation (5k^2 = b^2) implies that (b^2), and hence (b), is divisible by 5. This is a contradiction because we initially assumed that (a) and (b) are coprime, meaning they cannot both be divisible by 5.

Conclusion: Since assuming that (\sqrt{5}) is rational leads to a contradiction, (\sqrt{5}) must be irrational.
This proof by contradiction confirms the irrationality of (\sqrt{5}).
Prove that $3+2 \sqrt{5}$ is irrational.
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Sign up nowProve that the following are irrationals :
(i) $\frac{1}{\sqrt{2}}$
(ii) $7 \sqrt{5}$
(iii) $6+\sqrt{2}$
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Sign up nowExtra Questions  Real Numbers  NCERT  Mathematics  Class 10
For any integers $a$ and $3$, there exists unique integers $q$ and $r$ such that $a = 3q + r$. Find the possible value of $r$.
By applying Euclid's division lemma, which states that for any two integers $ a $ and $ b $, there exist unique integers $ q $ (the quotient) and $ r $ (the remainder) such that:
$$ a = bq + r $$
and
$$ 0 \leq r < b $$
For the given question, we can substitute $ b = 3 $ into the lemma:
$$ a = 3q + r $$
Based on the condition provided by Euclid's division lemma, the value of $ r $ must satisfy:
$$ 0 \leq r < 3 $$
Therefore, the possible values of $ r $ are:
$0$
$1$
$2$
The number of positive integers satisfying the equation $x + \log_{10}\left(2^{x} + 1\right) = x \log_{10} 5 + \log_{10} 6$ is
A) 0
B) 1
C) 2
D) infinite
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Sign up nowWhich one of the following is not correct for the features of the exponential function given by $f(x)=b^{x}$, where $b>1$?
A. For very large negative values of $x$, the function is very close to 0.
B. The domain of the function is $\mathbb{R}$, the set of real numbers.
C. The point $(1,0)$ is always on the graph of the function.
D. The range of the function is the set of all positive real numbers.
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