Chapter 7 ยท Add and subtract fractions

Fraction Word Problems: Estimate, Solve, and Check

Estimate fraction sums and differences with benchmarks, solve word problems about parts of the same whole, and use the estimate to check your answer.

15 min read

Nia pours 25\frac{2}{5} liter of juice and 12\frac{1}{2} liter of water into an empty jug. A classmate says the jug now holds 37\frac{3}{7} liter. Can that be right?

You can tell in a few seconds, before you work out the exact answer. The tool is an estimate: a quick answer that is close to the exact answer, the one you get by working the problem out fully.

In this lesson you will estimate sums and differences of fractions. A sum is the result of adding. A difference is the result of subtracting. Then you will solve word problems with fractions, and you will use your estimate to check that each answer makes sense.

Start with what a fraction means. A fraction tells you an amount only when you know its whole, the one full thing it is a part of. Half of a small jug of water and half of a big jug of water are both written 12\frac{1}{2}, but they are different amounts.

So here is the rule for every problem in this lesson: compare, add or subtract fractions as amounts only when they refer to the same whole, or to wholes of equal size.

In a word problem, look at the unit, such as liter, hour or yard. In Nia's problem, both amounts are parts of 1 liter. They share a whole, so you can add them.

Benchmark fractions are easy amounts that you compare other fractions with. Your benchmarks in this lesson are 00, 12\frac{1}{2} and 11 of the same whole. Every fraction you estimate in this lesson is from 00 to 11.

To find the nearest benchmark for 58\frac{5}{8}, think of a whole cut into 8 equal pieces. The denominator (bottom number) is 8, the number of pieces in the whole. The numerator (top number) is 5, the number of pieces you have.

  • No pieces is 00, which is 08\frac{0}{8}.
  • Half of the pieces is 12\frac{1}{2}, which is 48\frac{4}{8}.
  • All of the pieces is 11, which is 88\frac{8}{8}.

So the benchmark numerators are 0, half of the denominator, and the denominator itself. Here they are 0, 4 and 8. Which one is closest to your numerator, 5? It is 1 away from 4, 3 away from 8 and 5 away from 0. So 58\frac{5}{8} is nearest 12\frac{1}{2}.

You can write 58โ‰ˆ12\frac{5}{8} \approx \frac{1}{2}. The sign โ‰ˆ\approx means is about. The table shows the same steps for more fractions.

Nearest benchmarks for fractions of the same whole
FractionBenchmark numeratorsClosest to your numeratorNearest benchmark
110\frac{1}{10}
0, 5, 10
0 (1 is 1 away from 0 and 4 away from 5)
00
712\frac{7}{12}
0, 6, 12
6 (7 is 1 away from 6 and 5 away from 12)
12\frac{1}{2}
910\frac{9}{10}
0, 5, 10
10 (9 is 1 away from 10 and 4 away from 5)
11
25\frac{2}{5}
0, 2122\frac{1}{2}, 5
2122\frac{1}{2} (2 is 12\frac{1}{2} away from 2122\frac{1}{2} and 2 away from 0)
12\frac{1}{2}
45\frac{4}{5}
0, 2122\frac{1}{2}, 5
5 (4 is 1 away from 5 and 1121\frac{1}{2} away from 2122\frac{1}{2})
11

Worked example

Estimate a sum and a difference

Omar does homework for 56\frac{5}{6} of an hour and then reads for 512\frac{5}{12} of an hour. About how much time does he spend in all? About how much longer does he spend on homework than on reading?

  1. Check the whole first. Both times are parts of 1 hour, so they refer to the same whole. That means you may add them, subtract them and compare them.

  2. Find the nearest benchmark for 5 sixths. With sixths, the benchmark numerators are 0, 3 and 6. The numerator 5 is 1 away from 6 and 2 away from 3. So 5 sixths is nearest 1.

    56โ‰ˆ1\frac{5}{6} \approx 1
  3. Find the nearest benchmark for 5 twelfths. With twelfths, the benchmark numerators are 0, 6 and 12. The numerator 5 is 1 away from 6 and 5 away from 0. So 5 twelfths is nearest one half.

    512โ‰ˆ12\frac{5}{12} \approx \frac{1}{2}
  4. To estimate the time in all, add. Put the benchmarks in place of the fractions, because benchmarks are easy to add in your head.

    56+512โ‰ˆ1+12=112\frac{5}{6} + \frac{5}{12} \approx 1 + \frac{1}{2} = 1\frac{1}{2}
  5. To estimate how much longer, subtract. Use the same two benchmarks.

    56โˆ’512โ‰ˆ1โˆ’12=12\frac{5}{6} - \frac{5}{12} \approx 1 - \frac{1}{2} = \frac{1}{2}

Answer

Omar spends about 1121\frac{1}{2} hours in all, and he spends about 12\frac{1}{2} hour longer on homework than on reading.

An estimate can be 00. Suppose one piece of string is 58\frac{5}{8} of a foot long and another is 512\frac{5}{12} of a foot long. Both lengths are parts of 1 foot, the same whole, so you can compare them. About how much longer is the first piece?

You already found that 58\frac{5}{8} and 512\frac{5}{12} are each nearest 12\frac{1}{2}. So the estimate is 12โˆ’12=0\frac{1}{2} - \frac{1}{2} = 0. This does not say the pieces are the same length. An estimate is a rough guide, and this one tells you the exact difference is not far from 00. Check it. Each eighth is 3 twenty-fourths, so 58=1524\frac{5}{8} = \frac{15}{24}. Each twelfth is 2 twenty-fourths, so 512=1024\frac{5}{12} = \frac{10}{24}. The exact difference is 1524โˆ’1024=524\frac{15}{24} - \frac{10}{24} = \frac{5}{24} of a foot.

An exact answer can be 00 as well. For example, 36โˆ’12=0\frac{3}{6} - \frac{1}{2} = 0, because 36\frac{3}{6} and 12\frac{1}{2} of the same whole are the same amount.

Key concept

Estimate, solve, then check

Use these three moves when a word problem adds or subtracts fractions from 00 to 11 that refer to the same whole, or to wholes of equal size.

  1. Estimate. Replace each fraction with its nearest benchmark: 00, 12\frac{1}{2} or 11. Then add or subtract the benchmarks.
  2. Solve. Rewrite the fractions with a common denominator (the same denominator for both). Then add or subtract.
  3. Check. Compare your exact answer with your estimate. If they are close, your answer is reasonable: it makes sense. If they are far apart, look for a mistake.

If the fractions refer to wholes of different sizes, do not add, subtract or compare them as amounts.

Worked example

Add two amounts of the same whole

Go back to Nia's jug. She pours 25\frac{2}{5} liter of juice and 12\frac{1}{2} liter of water into an empty jug. How much liquid is in the jug?

  1. Check the whole and decide what to do. Both amounts are parts of 1 liter, so they refer to the same whole. The two amounts are put together, so add.

    25+12\frac{2}{5} + \frac{1}{2}
  2. Estimate. With fifths, the benchmark numerators are 0, 2 and a half, and 5. The numerator 2 is only a half away from 2 and a half, so 2 fifths is nearest one half. The other amount is already the benchmark one half.

    25+12โ‰ˆ12+12=1\frac{2}{5} + \frac{1}{2} \approx \frac{1}{2} + \frac{1}{2} = 1
  3. Solve the way you already know. Fifths and halves are pieces of different sizes, so rename both as tenths. Cut each fifth into 2 equal pieces, and 2 fifths become 4 tenths. Cut each half into 5 equal pieces, and 1 half becomes 5 tenths. The amounts stay the same.

    25=2ร—25ร—2=41012=1ร—52ร—5=510\frac{2}{5} = \frac{2 \times 2}{5 \times 2} = \frac{4}{10} \qquad \frac{1}{2} = \frac{1 \times 5}{2 \times 5} = \frac{5}{10}
  4. Now the pieces are the same size, so add the numerators. Together, 4 tenths and 5 tenths make 9 tenths.

    410+510=910\frac{4}{10} + \frac{5}{10} = \frac{9}{10}
  5. Check. The estimate was 1 liter. The exact answer is 9 tenths of a liter, only 1 tenth less than 1. The two are close, so the answer is reasonable.

Answer

The jug holds 910\frac{9}{10} liter of liquid.

Worked example

Subtract two amounts of the same whole

Kai has 56\frac{5}{6} yard of ribbon. He cuts off 38\frac{3}{8} yard to wrap a gift. How much ribbon is left?

  1. Both lengths are parts of 1 yard, the same whole. A piece is cut away from the ribbon, so subtract.

    56โˆ’38\frac{5}{6} - \frac{3}{8}
  2. Estimate. The amount 5 sixths is nearest 1, because the numerator 5 is 1 away from 6 and 2 away from 3. The amount 3 eighths is nearest one half, because half of 8 is 4, and 3 is 1 away from 4 and 3 away from 0.

    56โˆ’38โ‰ˆ1โˆ’12=12\frac{5}{6} - \frac{3}{8} \approx 1 - \frac{1}{2} = \frac{1}{2}
  3. Solve. Rename both fractions as twenty-fourths, because 6 times 4 is 24 and 8 times 3 is 24. Cut each sixth into 4 equal pieces, and 5 sixths become 20 twenty-fourths. Cut each eighth into 3 equal pieces, and 3 eighths become 9 twenty-fourths.

    56=5ร—46ร—4=202438=3ร—38ร—3=924\frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24} \qquad \frac{3}{8} = \frac{3 \times 3}{8 \times 3} = \frac{9}{24}
  4. The pieces are the same size now, so subtract the numerators. Taking 9 twenty-fourths from 20 twenty-fourths leaves 11 twenty-fourths.

    2024โˆ’924=1124\frac{20}{24} - \frac{9}{24} = \frac{11}{24}
  5. Check. Half of 24 is 12, so 11 twenty-fourths is just 1 twenty-fourth less than one half. That is close to the estimate. It is also less than the 5 sixths of a yard Kai started with, which fits cutting a piece off.

Answer

Kai has 1124\frac{11}{24} yard of ribbon left.

How close should your answer be to your estimate?

Neighboring benchmarks are 12\frac{1}{2} apart. A fraction exactly in the middle of two neighbors is 14\frac{1}{4} from each, and no fraction from 00 to 11 is farther than that from its nearest benchmark. So swapping a fraction for its nearest benchmark changes it by 14\frac{1}{4} at most. With two fractions, the two changes add up to 12\frac{1}{2} at most.

This gives you a check. Suppose you estimate with two fractions of the same whole, each from 00 to 11, and you use the nearest benchmark for each. Then a correct answer is never more than 12\frac{1}{2} away from your estimate. If your answer is farther away, look for a mistake.

Try it on Omar's total time. Each sixth is 2 twelfths, so 56=1012\frac{5}{6} = \frac{10}{12}. The exact sum is 1012+512=1512=1312=114\frac{10}{12} + \frac{5}{12} = \frac{15}{12} = 1\frac{3}{12} = 1\frac{1}{4} hours. His estimate was 1121\frac{1}{2} hours. The two are 14\frac{1}{4} hour apart, so the answer passes.

Passing means the answer is reasonable. It does not prove the answer is exactly right, so still compute with care.

The big idea: estimate before you solve, and compare after you solve. Replace each fraction with its nearest benchmark, 00, 12\frac{1}{2} or 11, to get a quick estimate. Find the exact answer with a common denominator. Then ask whether the two are close. This works when the fractions refer to the same whole, or to wholes of equal size.

Next, you will solve word problems that take more than one step, with fractions and mixed numbers. You will name each step and the whole each fraction refers to. You will also find and fix mistakes in solved problems.