A craftsperson makes jewelry to sell at art fairs. For Christmas fairs, the craftsperson bought 5 7/10 ounces of gold, 4 4/5 ounces of silver, and 7 5/6 ounces of turquoise. What was the total weight that she purchased?
Group of answer choices
18 1/3 ounces
18 1/30 ounces
16 29/30 ounces
10 1/2 ounces
To find the total weight, we need to add the weights of gold, silver, and turquoise.
5 7/10 ounces + 4 4/5 ounces + 7 5/6 ounces
To add these mixed numbers, we need to find a common denominator. The common denominator for 10, 5, and 6 is 30.
5 7/10 ounces + 4 4/5 ounces + 7 5/6 ounces
= (50/10) + (24/5) + (35/6) ounces
= 50/10 + 48/10 + 35/6 ounces
Now, we can add the fractions:
= (50 + 48 + (35 * 5))/(10) = (50 + 48 + 175)/10
= 273/10 ounces
Since the fraction cannot be simplified any further, the total weight the craftsperson purchased is 273/10 ounces.
To express this as a mixed number, divide the numerator (273) by denominator (10):
273 รท 10 = 27 remainder 3
So, the total weight is 27 3/10 ounces.
Therefore, the correct answer is 27 3/10 ounces, which is closest to 27 1/3 ounces.