Ask Question
6 April, 11:51

In an 8hr day Mark can sell 15 flowers, Lou can sell 6 flowers and Cindy can sell 8 flowers. What is the smallest number of 8hr days needed for each of them to achieve the same number of sales?

+1
Answers (1)
  1. 6 April, 14:26
    0
    Mark = 15 flowers

    Lou = 6 flowers

    Cindy = 8 hours

    Find the Lowest common multiples

    Method 1 (Simple but tedious):

    15: 15, 30, 45, 60, 75, 90, 105, 120

    6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 104, 120

    8: 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120

    Therefore:

    15 x 8 = 120

    6 x 20 = 120

    8 x 15 = 120

    Mark needs 8 8hrs

    Lou needs 20 8hrs

    Cindy needs 15 8hours

    Method 2:

    15 = 5 x 3

    6 = 2 x 3

    8 = 2 x 2 x 2

    LCM = 2 x 2 x 2 x 3 x 5 = 120

    120 : 15 = 8

    120 : 6 = 20

    120 : 8 = 15

    Mark needs 8 8hrs

    Lou needs 20 8hrs

    Cindy needs 15 8hours
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “In an 8hr day Mark can sell 15 flowers, Lou can sell 6 flowers and Cindy can sell 8 flowers. What is the smallest number of 8hr days needed ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers