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?

Question

Group of answer choices

18 1/3 ounces

18 1/30 ounces

16 29/30 ounces

10 1/2 ounces

Answer 1

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 ÷ 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.