Understanding the Problem: Finding Number Pairs

The question asks us to find two different pairs of numbers that meet specific conditions: their product is 2160, and their Highest Common Factor (HCF) is 12. Once we find these two pairs, we need to calculate the mean (average) of the numbers in each pair. Let's call these means x and y. Finally, we need to find the mean of x and y.

Finding the Number Pairs

Let a and b be a pair of numbers. We are given:

• \(a \times b = 2160\)
• \(\text{HCF}(a, b) = 12\)

We know that for any two positive integers a and b, the product of the numbers is equal to the product of their HCF and Least Common Multiple (LCM). That is, \(a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)\).

Using this property, we can find the LCM of the numbers in each pair:

\[ \text{LCM}(a, b) = \frac{a \times b}{\text{HCF}(a, b)} = \frac{2160}{12} = 180 \]Therefore, the LCM for both pairs of numbers is 180.

Now, since the HCF of a and b is 12, we can write \(a = 12p\) and \(b = 12q\), where p and q are positive integers and their HCF is 1 (i.e., p and q are coprime).

Substitute these into the product equation:

\[ (12p) \times (12q) = 2160 \]\[ 144pq = 2160 \]\[ pq = \frac{2160}{144} = 15 \]

So, we need to find pairs of coprime integers (p, q) whose product is 15. The possible pairs (p, q) such that \(pq = 15\) and \(\text{HCF}(p, q) = 1\) are:

• (1, 15)
• (3, 5)

These two pairs of (p, q) will give us the two unequal pairs of numbers.

Determining the Two Number Pairs

Using the pairs of (p, q) we found, we can find the corresponding pairs of numbers (a, b) where \(a = 12p\) and \(b = 12q\):

• Pair 1: If (p, q) = (1, 15), the numbers are \(a = 12 \times 1 = 12\) and \(b = 12 \times 15 = 180\). The first pair is (12, 180).
• Pair 2: If (p, q) = (3, 5), the numbers are \(a = 12 \times 3 = 36\) and \(b = 12 \times 5 = 60\). The second pair is (36, 60).

Let's verify these pairs:

• Pair (12, 180): Product \(12 \times 180 = 2160\). HCF(12, 180) = 12. Correct.
• Pair (36, 60): Product \(36 \times 60 = 2160\). HCF(36, 60) = 12. Correct.

These are indeed two unequal pairs of numbers satisfying the given conditions.

Calculating the Means of the Pairs

The question states that x is the mean of the numbers in the first pair and y is the mean of the numbers in the second pair. We can assign the pairs in either order.

Let the first pair be (12, 180) and the second pair be (36, 60).

• Mean of the first pair (x): \(x = \frac{12 + 180}{2} = \frac{192}{2} = 96\)
• Mean of the second pair (y): \(y = \frac{36 + 60}{2} = \frac{96}{2} = 48\)

Alternatively, let the first pair be (36, 60) and the second pair be (12, 180).

• Mean of the first pair (x): \(x = \frac{36 + 60}{2} = \frac{96}{2} = 48\)
• Mean of the second pair (y): \(y = \frac{12 + 180}{2} = \frac{192}{2} = 96\)

In both scenarios, the two means we get are 96 and 48.

Calculating the Mean of x and y

Finally, we need to find the mean of x and y.

\[ \text{Mean of x and y} = \frac{x + y}{2} \]Using the values x = 96 and y = 48:

\[ \text{Mean} = \frac{96 + 48}{2} = \frac{144}{2} = 72 \]

If we used x = 48 and y = 96, the result would be the same:

\[ \text{Mean} = \frac{48 + 96}{2} = \frac{144}{2} = 72 \]

So, the mean of x and y is 72.

The two means are 96 and 48. The mean of these two values is \((96 + 48) / 2 = 144 / 2 = 72\).

Revision Table: Key Concepts

Additional Information: HCF and LCM Properties

Understanding HCF and LCM is crucial for solving problems involving factors and multiples of numbers. The relationship \(a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)\) is fundamental.

When two numbers a and b have an HCF of H, they can be expressed as \(a = Hp\) and \(b = Hq\), where p and q are coprime integers. This is a very useful way to approach problems like the one we just solved, where the HCF is given. The product \(ab = (Hp)(Hq) = H^2pq\). Also, their LCM is \(Hpq\). Substituting these into the property: \(H^2pq = H \times Hpq\), which shows the consistency of this representation.

In our problem, H = 12, \(ab = 2160\). So, \(12^2 pq = 2160\), which means \(144pq = 2160\), leading to \(pq = 15\). Finding coprime factors of 15 (which are (1, 15) and (3, 5)) directly gives us the structure of the two pairs of numbers.