The L.C.M. of two numbers is 210 and their H.C.F. is 3. If the first number is divided by 3, the quotient is 7 and the remainder is 0. The second number is:
30
Given that the first number divided by 3 gives quotient 7 with remainder 0, the first number is \(3\times 7 = 21\).
For any two numbers, product = LCM × HCF, i.e. \(a\times b = \text{LCM}\times\text{HCF}\).
Substitute: \(21\times b = 210\times 3 = 630\).
Solve: \(b = \dfrac{630}{21} = 30\).
Hence, the second number is 30.
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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?
