A Grandmother buys one strawberry doughnut and one chocolate doughnut for her two grandchildren, Jessy and James.
Jessy eats of the strawberry doughnut, while James eats
of the chocolate doughnut.
How much of the doughnuts do they eat in total?
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A Grandmother buys one strawberry doughnut and one chocolate doughnut for her two grandchildren, Jessy and James.
Jessy eats of the strawberry doughnut, while James eats
of the chocolate doughnut.
How much of the doughnuts do they eat in total?
To determine how much of the doughnuts they eat in total, let's find the sum of the fractions that represent their consumption.
First, consider Jessy's consumption of the strawberry doughnut: .
Next, consider James's consumption of the chocolate doughnut: .
To add these fractions, we need a common denominator. The denominators are 6 and 3. The least common multiple of these is 6.
Convert to an equivalent fraction with a denominator of 6:
Now we have the fractions and .
We can add them since they have the same denominator:
Therefore, in total, Jessy and James eat:
of the doughnuts.
The correct answer choice is the one that corresponds to , which is Choice 2.
Thus, the solution to this problem is that they eat of the doughnuts in total.
\( \frac{2}{4}+\frac{1}{4}= \)\( \)
Because fractions represent parts of a whole. Adding means combining different-sized pieces. You need the same size pieces (common denominator) to add them correctly!
Look at your denominators: 6 and 3. Since 6 is already a multiple of 3, the LCD is 6. For other problems, list multiples of each denominator until you find the smallest one they share.
It means they ate 3 out of 6 equal pieces total. Think of it as combining Jessy's 1 piece with James's 2 pieces, giving you 3 pieces altogether!
You can simplify it, but in this problem is the correct answer choice. Both forms are mathematically equivalent - choose based on what the question asks for.
Use the multiplication method: multiply the denominators together, then adjust both numerators accordingly. For example, would need LCD of 20.
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