Finding the Other Number Using HCF and LCM
This solution demonstrates how to find an unknown number when its Highest Common Factor (HCF), Least Common Multiple (LCM), and one of the numbers are provided. We utilize the fundamental relationship between these values.
Core Principle: Product Rule
The product of two numbers is always equal to the product of their HCF and LCM.
Let the two numbers be $N_1$ and $N_2$. The relationship is expressed as:
$ N_1 \times N_2 = \text{HCF} \times \text{LCM} $
Applying the Formula
We are given:
- HCF = 16
- LCM = 240
- One number ($N_1$) = 48
- We need to find the other number ($N_2$).
Substitute the known values into the formula:
$ 48 \times N_2 = 16 \times 240 $
Calculation
To find $N_2$, we isolate it:
$ N_2 = \frac{16 \times 240}{48} $
Simplify the calculation:
$ N_2 = \frac{16}{48} \times 240 $
Since $\frac{16}{48}$ simplifies to $\frac{1}{3}$:
$ N_2 = \frac{1}{3} \times 240 $
$ N_2 = 80 $
Thus, the other number is 80.
Similar Questions
- The H.C.F. and the L.C.M. of two numbers are 5 and 495, respectively. If one of the numbers is 55, find the other one.
- The sum of two numbers is 132 and their L.C.M. is 630. What are the two numbers?
- The LCM of the numbers 12.8 and 0.004 is:
- Find the least number which is divisible by first ten natural numbers.
- Find the HCF of ($3^{45} - 1$) and ($3^{35} - 1$).
- The HCF and LCM of two numbers are in the ratio of 1: 30 and the difference between the HCF and LCM is 493. Find the product of LCM and HCF.
- HCF of $\frac{1}{3}, \frac{3}{4}, \frac{4}{5}$ and $\frac{5}{6}$ is:
- Which of the following is the greatest number that, when dividing 183, 127 and 211, leaves the same remainder each time?
- If the LCM of $20x^3y^2$ and $10x^4y^4$ is $20x^4y^4$, find the HCF.
- There are four table clocks. They ring every 10 min, 15 min, 20 min and 25 min respectively. If they all ring together at 10 a.m., then at what time will they ring together again?
Important Questions from LCM and HCF
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?
