Suppose p and q are the LCM and HCF respectively of two positive numbers. If p : q = 14 : 1 and pq = 1134, then what is the difference between the two numbers?
45
The question provides information about the Least Common Multiple (LCM) and Highest Common Factor (HCF) of two positive numbers. Let the two positive numbers be \(a\) and \(b\). We are given that their LCM is \(p\) and their HCF is \(q\). We are provided with two conditions:
Our goal is to find the absolute difference between the two numbers, which is \(|a - b|\).
A fundamental property connecting two positive numbers, their LCM, and their HCF is:
\[ a \times b = \text{LCM}(a, b) \times \text{HCF}(a, b) \]
In our case, this means:
\[ a \times b = p \times q \]
We are given \(pq = 1134\), so the product of the two numbers is \(a \times b = 1134\).
We have the following system of equations based on the given information:
We can substitute the first equation into the second one to solve for \(q\):
\[ (14q) \times q = 1134 \]
\[ 14q^2 = 1134 \]
Now, we solve for \(q^2\):
\[ q^2 = \frac{1134}{14} \]
\[ q^2 = 81 \]
Since \(q\) is the HCF of positive numbers, it must be positive. Therefore, we take the positive square root:
\[ q = \sqrt{81} = 9 \]
So, the HCF is \(q = 9\).
Now we can find \(p\) using the relation \(p = 14q\):
\[ p = 14 \times 9 \]
\[ p = 126 \]
So, the LCM is \(p = 126\).
We know that the HCF of the two numbers \(a\) and \(b\) is \(q = 9\). This means that \(a\) and \(b\) can be expressed as multiples of 9. Let:
\[ a = 9x \]
\[ b = 9y \]
where \(x\) and \(y\) are positive integers and they must be coprime (their HCF is 1). If \(x\) and \(y\) had a common factor greater than 1, then the HCF of \(a\) and \(b\) would be greater than 9.
We also know that \(a \times b = 1134\). Substituting \(a = 9x\) and \(b = 9y\):
\[ (9x) \times (9y) = 1134 \]
\[ 81xy = 1134 \]
Now, we solve for \(xy\):
\[ xy = \frac{1134}{81} \]
\[ xy = 14 \]
We need to find pairs of positive integers \((x, y)\) such that their product is 14 AND their HCF is 1 (coprime). The pairs of factors for 14 are:
Let's check the HCF for each pair:
All four pairs are coprime. Now we find the corresponding numbers \((a, b)\) using \(a = 9x\) and \(b = 9y\):
The possible pairs of numbers are (9, 126) and (18, 63).
We need to find the difference between the two numbers for each pair:
We check the options provided in the question:
| Option | Difference |
|---|---|
| 1 | 27 |
| 2 | 35 |
| 3 | 45 |
| 4 | cannot be determined due to insufficient data |
Comparing the calculated differences (117 and 45) with the options, we see that 45 is one of the options (Option 3). The difference 117 is not an option.
Therefore, the two numbers must be 18 and 63, and their difference is 45.
By using the properties of LCM and HCF, the given ratio, and the product, we were able to determine the possible pairs of numbers and calculate their differences. The difference that matches one of the given options is 45.
| Step | Action | Details |
|---|---|---|
| 1 | Define variables | Let numbers be \(a, b\), LCM be \(p\), HCF be \(q\). |
| 2 | Note given conditions | Write down ratios, products, etc. (e.g., \(p:q = 14:1\), \(pq = 1134\)). |
| 3 | Use \(ab = pq\) property | Connect the product of numbers to LCM and HCF. |
| 4 | Solve for \(p\) and \(q\) | Use the given conditions to find the values of LCM and HCF. |
| 5 | Express numbers using HCF | Set \(a = qx\) and \(b = qy\) where \(x, y\) are coprime integers. |
| 6 | Use \(ab = pq\) or \(p = qxy\) | Find the value of \(xy\). |
| 7 | Find coprime pairs \((x, y)\) | List pairs whose product is \(xy\) and HCF is 1. |
| 8 | Calculate possible \((a, b)\) pairs | Substitute \((x, y)\) back into \(a=qx, b=qy\). |
| 9 | Calculate the required value | Find the difference, sum, etc., asked in the question for the pairs. |
| 10 | Match with options | Compare results with the given options to find the answer. |
Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?
A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:
Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Calculate the HCF of \(\frac{12}{5}\) , \(\frac{14}{15}\) and \(\frac{16}{17}\) .
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?