NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 1: Real Numbers

Exercise 1.1

An even integer is any number that can be divided by 2 without leaving a remainder.

If we take an integer \(m\), then \(2m\) is always even because it has a factor of 2.

The form \(m\) or \(m+1\) can represent any integer, not specifically even ones. The form \(2m+1\) is the general form of odd integers.

Therefore, every even integer can be written in the form \(2m\).

Step 1: Recall that even integers are always multiples of 2.

So we can write any even integer as:

\(2q\), where \(q\) is some integer.

Step 2: Odd integers are always one more than an even integer.

Therefore, if an even number is \(2q\), then the next odd number is:

\(2q + 1\).

Step 3: This shows that every odd integer can be expressed in the form:

\(2q + 1\).

Answer: Option D (\(2q + 1\))

Let \(n\) be an odd integer, so we can write \(n = 2k+1\) for some integer \(k\).

Now, \(n^2 - 1 = (2k+1)^2 - 1 = 4k(k+1)\).

Since \(k\) and \(k+1\) are consecutive numbers, one of them is always even. Therefore, their product \(k(k+1)\) is divisible by 2.

That makes the whole expression divisible by \(4 \times 2 = 8\).

So, \(n^2 - 1\) is divisible by 8 when \(n\) is odd.

We first find HCF(65, 117) using Euclid’s division algorithm:

\(117 = 65 \times 1 + 52\)

\(65 = 52 \times 1 + 13\)

\(52 = 13 \times 4 + 0\)

So the HCF is 13.

Now, it is given that HCF can be expressed as \(65m - 117\).

So, \(65m - 117 = 13 \Rightarrow 65m = 130 \Rightarrow m = 2\).

Thus, the required value of \(m\) is 2.

If the number is \(d\), then:

When 70 is divided by \(d\), remainder = 5. This means \(d\) divides \(70 - 5 = 65\).

When 125 is divided by \(d\), remainder = 8. This means \(d\) divides \(125 - 8 = 117\).

So \(d\) must be a common divisor of 65 and 117.

The HCF of 65 and 117 is 13. Hence, the required number is 13.

The HCF is found by taking the smallest power of each prime factor in both numbers.

For \(x\), the powers are \(3\) (in a) and \(1\) (in b). The minimum is \(1\).

For \(y\), the powers are \(2\) (in a) and \(3\) (in b). The minimum is \(2\).

So, the HCF is \(x^1y^2 = xy^2\).

The LCM is found by taking the highest power of each prime factor from both numbers.

For \(a\), the powers are \(1\) (in p) and \(3\) (in q). The maximum is \(3\).

For \(b\), the powers are \(2\) (in p) and \(1\) (in q). The maximum is \(2\).

So, the LCM is \(a^3b^2\).

Let \(r \neq 0\) be a rational number and \(\alpha\) an irrational number.

If their product \(r\alpha\) were rational, then dividing it by the non-zero rational \(r\) would give \(\alpha = \dfrac{r\alpha}{r}\), which would make \(\alpha\) rational. This is a contradiction.

Hence, the product is always irrational.

The required number is the least common multiple (LCM) of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

The highest powers of primes within this range are: \(2^3, 3^2, 5, 7\).

So, \(\text{LCM} = 2^3 \times 3^2 \times 5 \times 7 = 2520\).

Therefore, the least number divisible by all the numbers from 1 to 10 is 2520.

First, factorize the denominator: \(1250 = 2 \times 5^4\).

Since the denominator is of the form \(2^m5^n\), the decimal expansion will terminate.

The number of decimal places equals the higher of the powers of 2 or 5 in the denominator.

Here, we have \(m = 1\) and \(n = 4\). The maximum is 4.

Therefore, the decimal expansion will terminate after 4 decimal places.