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

Exercise 1.2

Step 1: Recall the division algorithm.

When any integer \(n\) is divided by 4, the remainder can only be 0, 1, 2, or 3.

Therefore, every integer can be expressed in one of the following forms:

\(n = 4q\),

\(n = 4q + 1\),

\(n = 4q + 2\),

\(n = 4q + 3\).

Step 2: Check the required form.

The question asks whether every positive integer can be written in the form \(4q + 2\).

But this is only true for those numbers which leave remainder 2 when divided by 4 (like 2, 6, 10, 14, …).

Step 3: Counterexample.

Consider \(n = 5\).

Dividing by 4: \(5 = 4 \times 1 + 1\).

This is of the form \(4q + 1\), not \(4q + 2\).

Conclusion.

Since not all integers fit the form \(4q + 2\), the statement is not true.

Step 1: Represent consecutive integers.

Let the first positive integer be \(n\).

Then the next consecutive integer is \(n + 1\).

Step 2: Write their product.

The product is:

\(n(n + 1)\).

Step 3: Reason about even and odd numbers.

Among any two consecutive integers, one is always even and the other is odd.

For example: \(2,3\) or \(5,6\) or \(10,11\).

Step 4: Conclude divisibility.

If one factor is even, the product is always even.

Therefore, \(n(n+1)\) is always divisible by 2.

Final Answer: The statement is True.

Step 1: Write the three consecutive integers.

Let the integers be \(n, n+1, n+2\).

Step 2: Check divisibility by 2 (even number).

Any three consecutive numbers must include at least one even number,

because every second number is even.

So, \(n(n+1)(n+2)\) is divisible by 2.

Step 3: Check divisibility by 3.

Every third number is divisible by 3, so among \(n, n+1, n+2\),

at least one will be a multiple of 3.

So, \(n(n+1)(n+2)\) is divisible by 3.

Step 4: Combine the results.

If a number is divisible by both 2 and 3, it is divisible by 6.

Hence, the product \(n(n+1)(n+2)\) is always divisible by 6.

Conclusion: The statement is True.

Step 1: Expressing any integer in terms of 3.

Every integer \(n\) can be written in one of the following forms:

\(n = 3q\), or \(n = 3q + 1\), or \(n = 3q + 2\),

where \(q\) is some integer (quotient on division by 3).

Step 2: Squaring each case.

Case A: If \(n = 3q\)

Then \(n^2 = (3q)^2 = 9q^2\).

This can be written as \(n^2 = 3(3q^2)\), i.e. of the form \(3m\).

Case B: If \(n = 3q + 1\)

Then \(n^2 = (3q + 1)^2\).

Expanding: \(n^2 = 9q^2 + 6q + 1\).

This is \(n^2 = 3(3q^2 + 2q) + 1\).

So \(n^2\) is of the form \(3m + 1\).

Case C: If \(n = 3q + 2\)

Then \(n^2 = (3q + 2)^2\).

Expanding: \(n^2 = 9q^2 + 12q + 4\).

This can be written as \(n^2 = 3(3q^2 + 4q + 1) + 1\).

So \(n^2\) is again of the form \(3m + 1\).

Step 3: Combine results.

Thus, a square of any integer can only be of the form:

• \(3m\), or
• \(3m + 1\).

It can never be of the form \(3m + 2\).

Conclusion: The square of any positive integer cannot be of the form \(3m + 2\).

Step 1: Express the number in the given form.

Let the number be \(n = 3q + 1\), where \(q\) is a natural number.

Step 2: Find its square.

\(n^2 = (3q + 1)^2\)

Expand: \(n^2 = 9q^2 + 6q + 1\)

Step 3: Factorize the expression.

\(n^2 = 3(3q^2 + 2q) + 1\)

Here, \(3q^2 + 2q\) is an integer. Let \(m = 3q^2 + 2q\).

Step 4: Final form.

So, \(n^2 = 3m + 1\).

Conclusion. The square of a number of the form \(3q+1\) is always of the form \(3m+1\). It can never be written as \(3m\) or \(3m+2\).

Step 1: Prime factorisation of 525.

Divide step by step:

\(525 \div 3 = 175\) ⇒ so one factor is \(3\).

\(175 \div 5 = 35\) ⇒ factor \(5\).

\(35 \div 5 = 7\) ⇒ another factor \(5\).

So, \(525 = 3 \times 5^2 \times 7\).

Step 2: Prime factorisation of 3000.

Divide step by step:

\(3000 \div 2 = 1500\)

\(1500 \div 2 = 750\)

\(750 \div 2 = 375\)

\(375 \div 3 = 125\)

\(125 \div 5 = 25\)

\(25 \div 5 = 5\)

\(5 \div 5 = 1\)

So, \(3000 = 2^3 \times 3 \times 5^3\).

Step 3: Identify common prime factors.

From 525: \(3, 5^2, 7\).

From 3000: \(2^3, 3, 5^3\).

Common factors are:

\(3\) and \(5^2\).

Step 4: Multiply common factors to get HCF.

\(HCF = 3 \times 5^2 = 3 \times 25 = 75\).

Conclusion. The highest common factor (HCF) of 525 and 3000 is 75.

Step 1: Simplify the given expression.

We are asked to check if \(3 \times 5 \times 7 + 7\) is composite.

First calculate the product: \(3 \times 5 \times 7 = 105\).

Add 7: \(105 + 7 = 112\).

Step 2: Factorize the number.

Notice that we can factor 7 from the original expression:

\(3 \times 5 \times 7 + 7 = 7(3 \times 5 + 1)\).

Simplify inside the bracket: \(3 \times 5 + 1 = 15 + 1 = 16\).

So the number becomes \(7 \times 16\).

Step 3: Check for factors.

Since \(112 = 7 \times 16\), it is divisible by 7 and 16 in addition to 1 and 112.

This means it has more than two factors.

Conclusion. A number with more than two factors is a composite number. Therefore, 112 is composite.

Step 1: Recall the property of HCF and LCM.

For any two positive integers \(a\) and \(b\):

\( \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b \).

From this property, it follows that the HCF of two numbers must always divide their LCM.

Step 2: Check if the given HCF divides the given LCM.

Given HCF = \(18\), LCM = \(380\).

Check \(\dfrac{380}{18}\).

\(380 \div 18 = 21.11...\) (not an integer).

Step 3: Reasoning.

Since \(18\) does not divide \(380\), the pair of numbers cannot exist.

Conclusion: No two numbers can have HCF = \(18\) and LCM = \(380\).

Step 1: Start with the given fraction: \(\dfrac{987}{10500}\).

Step 2: Divide numerator and denominator by their common factor 3: \(\dfrac{987}{10500} = \dfrac{329}{3500}\).

Step 3: Now, divide numerator and denominator again by 7:

\(\dfrac{329}{3500} = \dfrac{47}{500}\).

Step 4: Factorize the denominator 500. We get \(500 = 2^2 \times 5^3\).

Step 5: Since the denominator in lowest terms has only the prime factors 2 and 5, the decimal expansion must be terminating.

Final Conclusion: The fraction \(\dfrac{987}{10500}\) has a terminating decimal expansion.

Step 1: Recall the property of terminating decimals.

A rational number \(\dfrac{p}{q}\) in lowest terms has a terminating decimal expansion if and only if the denominator \(q\) has no prime factors other than 2 or 5.

Step 2: Connect with powers of 10.

Any terminating decimal can be written as a fraction with denominator \(10^k\) for some integer \(k\).

For example, \(327.7081 = \dfrac{3277081}{10000}\).

Since \(10000 = 10^4 = 2^4 \times 5^4\), the denominator has only 2 and 5 as prime factors.

Step 3: Simplify and check denominator form.

Even if we reduce \(\dfrac{3277081}{10000}\) to lowest terms, the denominator will still be of the form \(2^m 5^n\).

This is because cancelling common factors between numerator and denominator cannot introduce new primes in the denominator.

Conclusion.

The prime factors of \(q\) are only 2 and 5, i.e., \(q = 2^m 5^n\) for some non-negative integers \(m,n\).