There were live pieces of cloth of lengths 15 m.
1 m. 36 m, 42 m, 48 m. But all of them could
se measured in whole units of a measuring rod.
what could be the largest length of the rod?
Share
There were live pieces of cloth of lengths 15 m.
1 m. 36 m, 42 m, 48 m. But all of them could
se measured in whole units of a measuring rod.
what could be the largest length of the rod?
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
Once you get rid of the extraneous details, this question is asking for a common factor of each of these numbers. The way I would go about this is start with fifteen and try out each of its factors and take it from there. As it only has four (1, 15, 3, & 5) it was quite an easy process. I would start at the largest factor but it was obvious 5 could not be the answer so I began with three.
15/3=5
21/3=7
36/3=12
42/3= 14
48/3=16
All of these are whole numbers, all of these are factors of the previous numbers, any larger number could not work, the rod is 3m.This should take a person all of about ten seconds to figure out at most. This is merely explaining the exact process.
To be honest a reasonably intelligent second grader could have figured this out so I am led to question why the asker questioned this in the first place.