What Does Evaluate Imply in Math Problems?

  • Thread starter Thread starter Gringo123
  • Start date Start date
  • Tags Tags
    Fraction
AI Thread Summary
The term "evaluate" in math problems typically means to perform the necessary calculations to find an answer. In the context of multiplying mixed numbers like 3 1/3 and 1 1/5, it involves converting them into improper fractions before multiplying. The discussion clarifies that "evaluate" does not imply a sum but rather the arithmetic operation needed to solve the problem. Participants note that "sum" can be a British term used broadly for any arithmetic operation. Ultimately, "evaluate" is understood as a directive to complete the calculation.
Gringo123
Messages
141
Reaction score
0
What does the word mean when you are presented with a question like this?

Evaluate 3 1/3 x 1 1/5

Is it simply asking us to work out the answer to the sum?
 
Physics news on Phys.org
A sum? Isn't this a product of fractions? In words, three wholes and one third times one whole and one fifth?
 
Gringo123 said:
What does the word mean when you are presented with a question like this?

Evaluate 3 1/3 x 1 1/5

Is it simply asking us to work out the answer to the sum?
As Radou noted, this is a product, not a sum. The two numbers are mixed numbers (a whole number plus a fraction), which are not very useful for calculations. Convert each of the mixed numbers to an improper fraction, and then multiply, using the rules of fraction multiplication.
 
Hi Mark
Thanks once again fior helping out.
Does that mean the solution is as follows?
10/3 x 6/5 = 60/15 = 4
When the question says, "Evaluate", I guess it just means "work out the answer"?
 
Yes, "evaluate" simply means "do the arithmetic".

And, radou and Mark44, "sum" is a common Britishism for any arithmetic problem.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top