Practice Questions
The product of two numbers is 960 and their HCF is 8. Find their LCM.
$\text{LCM} = \text{Product} / \text{HCF} = 960 / 8 = 120$.
The LCM of two numbers is 420 and their HCF is 14. If one number is 70, find the other.
$\text{Number}_2 = (\text{HCF} \times \text{LCM}) / \text{Number}_1 = (14 \times 420) / 70 = 14 \times 6 = 84$.
Find the least number divisible by 9, 12 and 15.
$\text{LCM}(9, 12, 15) = 180$.
Find the greatest number that divides 1053 and 1443 leaving the same remainder.
Subtract the numbers: $1443 - 1053 = 390$. The greatest such number divides the difference. 390 divides both and leaves a remainder of 273.
Find the smallest number which when divided by 12 and 18 leaves remainder 7 in each case.
Smallest number = $\text{LCM}(12, 18) + 7 = 36 + 7 = 43$.
Find the least number which when divided by 10, 15 and 20 leaves remainders 6, 11 and 16 respectively.
Common difference: $10-6=4, 15-11=4, 20-16=4$. Result = $\text{LCM}(10, 15, 20) - 4 = 60 - 4 = 56$.
Two events occur every 18 days and 24 days. If they occur together today, after how many days will they coincide again?
They will coincide at intervals of $\text{LCM}(18, 24) = 72$ days.
Find the smallest number which when divided by 8 leaves remainder 3 and by 12 leaves remainder 7.
Common difference: $8-3=5, 12-7=5$. Result = $\text{LCM}(8, 12) - 5 = 24 - 5 = 19$.
Find the HCF of 144 and 198.
$144 = 2^4 \times 3^2$, $198 = 2 \times 3^2 \times 11$. HCF = $2 \times 3^2 = 18$.
Three bells ring at intervals of 6, 9 and 15 seconds. If they ring together now, after how many seconds will they ring together again?
$\text{LCM}(6, 9, 15) = 90$ seconds.
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