junita is making necklace to give present.she plans to put15 beads in each necklace . Beads are sold in packages of 20.what is the least number of packages she can buy to make necklaces and have no beads left over?
What is the least common multiple of 15 and 20?
Divide that by 15.
15, 30,45,60
20,40,60,80
LCM= 60
60 divided by 15= 4
To find the least number of packages Junita needs to buy, we first need to determine the total number of beads required for making necklaces.
Junita plans to put 15 beads in each necklace, and she wants to make necklaces without any beads left over. So the total number of beads required for making necklaces would be a multiple of 15.
To find the least number of packages she can buy, we need to divide the total number of beads required by the number of beads in each package (20), and round the result up to the nearest whole number.
Let's calculate it step by step:
1. Calculate the total number of beads required:
Total Beads Required = Number of Necklaces × Beads per Necklace
Since Junita wants to make necklaces without any beads left over, the total number of beads required would be a multiple of 15. We can assume the number of necklaces as 'n' and calculate the total beads required as:
Total Beads Required = n × 15
2. Calculate the number of packages required:
Number of Packages = Total Beads Required ÷ Beads per Package
Since each package contains 20 beads, we can calculate the number of packages required as:
Number of Packages = (n × 15) ÷ 20
3. Round up the result to the nearest whole number:
Least Number of Packages = Round up(Number of Packages)
Now, let's put all the steps together and calculate the least number of packages Junita needs to buy:
Total Beads Required = n × 15
Number of Packages = (n × 15) ÷ 20
Least Number of Packages = Round up(Number of Packages)
Note: To round up the result, we can use the ceiling function or round the result up to the nearest whole number.
This way, we can find the least number of packages Junita needs to buy to make necklaces without any beads left over.