Two unequal pairs of numbers satisfy the following conditions: (i) The product of the two numbers in each pair is 2160 (ii) The HCF of the two numbers in each pair is 12. If x is the mean of the numbers in the first pair and y is the mean of the numbers in the second pair, then what is the mean of x and y?
72
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.
| Pair (p, q) | Numbers (12p, 12q) | Product | HCF | Mean |
|---|---|---|---|---|
| (1, 15) | (12, 180) | \(12 \times 180 = 2160\) | 12 | \((12+180)/2 = 96\) |
| (3, 5) | (36, 60) | \(36 \times 60 = 2160\) | 12 | \((36+60)/2 = 48\) |
The two means are 96 and 48. The mean of these two values is \((96 + 48) / 2 = 144 / 2 = 72\).
Revision Table: Key Concepts
| Concept | Description | Formula/Property |
|---|---|---|
| HCF (Highest Common Factor) | The largest positive integer that divides two or more numbers without leaving a remainder. | - |
| LCM (Least Common Multiple) | The smallest positive integer that is a multiple of two or more numbers. | - |
| Product of two numbers | Equals HCF \(\times\) LCM for those numbers. | \(a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)\) |
| Coprime Numbers | Two numbers are coprime (or relatively prime) if their HCF is 1. | \(\text{HCF}(p, q) = 1\) |
| Mean (Average) | The sum of a set of values divided by the number of values. | \(\text{Mean} = \frac{\text{Sum of values}}{\text{Number of values}}\) |
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.
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