Maya’s school held an aluminum collection program in which they collected aluminum and brought it to be recycled. In the first week, Maya collected four over twenty lb of aluminum and her friend Abigail collected seven over eight lb.
a. Which girl collected more aluminum?
b. How much aluminum did the two girls collect?
a. 4/20 < 7/8
b. 4/20 + 7/8 = ?
1 3/40
hm but I feel like that's not the answer
4/20 + 7/8 = 1 3/40
meh I use it anyways
To determine which girl collected more aluminum, we need to compare the amounts collected by Maya and Abigail.
First, let's convert the fractions to a common denominator. The fraction "four over twenty" can be simplified as "one over five" by dividing the numerator and denominator by 4. Similarly, "seven over eight" remains the same since there are no common factors to simplify.
Now, let's compare the two fractions:
- Maya collected "one over five" lb of aluminum.
- Abigail collected "seven over eight" lb of aluminum.
To compare these two fractions, we need to find a common denominator. The least common multiple (LCM) of 5 and 8 is 40.
Now, let's convert both fractions to have a denominator of 40:
- Maya's fraction: (1/5) * (8/8) = 8/40
- Abigail's fraction: (7/8) * (5/5) = 35/40
Now we can see that Maya collected 8/40 lb of aluminum, and Abigail collected 35/40 lb of aluminum.
a. Maya collected less aluminum than Abigail, since 8/40 lb is less than 35/40 lb. Therefore, Abigail collected more aluminum.
To calculate the total amount of aluminum collected by the two girls, we can simply add their respective amounts:
- Maya collected 8/40 lb of aluminum.
- Abigail collected 35/40 lb of aluminum.
Adding these fractions together:
8/40 + 35/40 = 43/40 lb
b. Therefore, the two girls collected a total of 43/40 lb (or approximately 1.075 lb) of aluminum.