*Last Updated on April 9, 2024.*

Here, we’ll talk about Word Problems that are using fractions. We’ll let you practice while keeping it quite simple.

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**In Addition:**

The Question is: A grocer gives you \( \frac{5}{9}\) of a full bunch of bananas. Earlier, you already got \( \frac{2}{9}\) of a banana bunch in the bag. So now, do you have a whole banana bunch in your shopping bag?

The Strategy: The presented problem includes two parts and you must determine if you have a full bunch now and then you’ll have to figure out if that’s less, greater, or equal to a full bunch.

So how much is it what you have?

\( \frac{5}{9} + \frac{2}{9}= ?\)

• Do we have common denominators: Yes, 9.

• Add the numerators: \( 5 + 2 = 7\)

• Our new fraction is: \( \frac{7}{9}\)

• Simplifying needed? No.

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So the answer is: You’ve got \( \frac{7}{9}\) of a full bunch.

So are you having a whole banana bunch?

Well, is \(\frac{7}{9} \) greater or equal to one bunch?

The answer is No. \( \frac{7}{9} < 1\)

**In Subtraction:**

The question is: You’ve got a bag that has \(8\) marbles. Your bag can maximally hold \(12\) marbles. Your friend borrows \( 3\) of the marbles in your bag. Now, what fraction of your bag is still filled?

The Strategy: In this problem, you didn’t get a direct fraction in this problem. But do you see that you’ve got \( \frac{8}{12}\) of your bag, and that you gave up \( \frac{3}{12}\) of that bag? You got two (2) values: \(8\) marbles and \( 3\) marbles and the possible (or maximum) number is 12 and this applies to both values.

So how many marbles (or rather, how much of your bag) have you got left?

\( \frac{8}{12} – \frac{3}{12} = ?\)

• Do we have common denominators: Yes we do.

• So we just subtract the numerators: \( 8 – 3 = 5\)

• Our new fraction is: \( \frac{5}{12}\)

• Simplifying needed? No.

The answer is: You’ve got \( \frac{5}{12}\) of your marbles bag left.

**In Multiplication:**

The question is: When a carpenter fixes a house, he must cut some beams. He’ll need to saw off \( \frac{2}{3}\) of a 6.5-foot beam. Now, how much will the carpenter saw off?

The Strategy: Keep in mind that in multiplication word problems, we usually see the word “of.” You may see expressions like “1 half of 6” or \( “\frac{2}{3}\, of\, 9”\). When you see that, you may figure out that it’s about multiplication.

Often, we also see words used rather than numerals in these problems. So you’ll have to figure out what these numbers are.

The question here asks you to figure out how much the carpenter will be sawing off. So we’ve got 1 fraction that’s multiplied by 1 mixed number.

So, how many feet is the carpenter sawing off the beam?

\( \frac{2}{3} * 6\, \frac{1}{2} = ?\)

• Fisrt, we have to convert to improper fractions, so: \( 6\, \frac{1}{2} = 6 + \frac{1}{2} = \frac{12}{2} + \frac{1}{2}= \frac{13}{2}\)

• Then, we need to rewrite the problem: \( \frac{2}{3} * \frac{13}{2} = ?\)

• Then, multiply the numerators: \( 2*13 = 26\)

• And multiply the denominators: \( 3*2 = 6\)

• Our new fraction is: \( \frac{26}{6}\)

• Now, convert this to mixed number: \( \frac{26}{6} = 26\div 6 = 4r2 = 4 \,\frac{2}{6}\)

• Now simplify: \(2\) and \(6\) are having the common factor of \( 2\). So we divide both the top and bottom number by \( 2\). \( \frac{2}{6} = \frac{1}{3}\)

The answer: The carpenter will be sawing off \( 4 \,\frac{1}{3}\) feet of the beam.

**In Division:**

The question is: At a birthday party, there are \(12\) kids. In total, we have three and a half (3,5) pies, and all the kids will get an equal share of the pies. Now, how much pie exactly does each of the kids get?

The Strategy: Here, we have to deal with a division problem. We have a rather big amount, and we’ll have to divide it up between a bunch of kids. But here, we’ve got to have a mixed number (3.5) that we must divide by the whole number 12.

So we’ll have to convert our whole number (12) as well as our mixed number (3.5) into 2 improper fractions. When we’ve done that, we can go through our normal steps again of division for fractions.

So how much pie will each of the kids get?

So: \( 3 \,\frac{1}{2}\div 12 = ?\)

• First, convert:

\( 3 \,\frac{1}{2} = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2}\)
\( 12 = \frac{12}{1}\)

• Now, rewrite: \( \frac{7}{2} \div \frac{12}{1}= ?\)

• Then, apply Reciprocal of Divisor: \( \frac{1}{12} \)

• So, rewrite this as a multiplication problem: \( \frac{7}{2} * \frac{1}{12} =?\)

• We multiply our numerators: \( 7*1 = 7\)

• And also multiply our denominators: \( 2*12 = 24\)

• Our new fraction is: \( \frac{7}{24} \)

• Simplifying? Not needed. There aren’t any common factors.

So our answer is: Each of the kids gets \( \frac{7}{24}\) of a pie.

Sure, \( \frac{7}{24}\) is the correct answer, but do you notice that this number is quite close to \( \frac{8}{24}\,?\)

Well, \( \frac{8}{24}\) is the equivalent of \( \frac{1}{3}\). And this means each of the kids gets almost one-third of a pie. And sometimes, it’s easier to round a number off. It’s so much easier to think about 1/3, or one-third, of a pie.

Do you see that 7/24 (seven twenty-fourths) is so much harder to visualize than 1/3 (one-third)? When you’re going to cut up those pies, you may just as easily cut the pies in thirds, right? Of course, that isn’t exact, but don’t forget that we’re having a party! Everyone will understand. [/spoiler]