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Grade 5 · 4.NF.B.3C · 5.NF.A.1

Adding mixed numbers

A trail is 3\tfrac{1}{6} miles long and the hill after it is 2\tfrac{1}{2} miles. How long are the two together, when sixths and halves are different pieces?

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Adding and subtracting fractions with unlike denominatorsGrade 5Adding and subtracting fractions with like denominatorsGrade 4
Where you are 4 in Mixed numbers and improper fractions
Fractions, ratio and percent3
Fractions15
Mixed numbers and improper fractions4
Mixed numbers and improper fractions: rewriting and comparingAdding and subtracting mixed numbers with like denominators
Adding mixed numbers
Subtracting mixed numbers

When you reach for it

Hania, Mingyu and Segun train on an obstacle course. It has trail sections measured in miles, and lanes such as a cargo net and a tunnel measured in yards. They also time themselves, in minutes, on each obstacle. Almost none of these amounts is a whole number. A trail is 3\tfrac{1}{6} miles; a rope climb takes 2\tfrac{1}{3} minutes.

A mixed number such as 3\tfrac{1}{6} is a whole number and a fraction less than 1 written side by side, and it means their sum: 3\tfrac{1}{6} = 3 + \tfrac{1}{6} (Mixed numbers and improper fractions). You reach for this lesson whenever two or more mixed numbers are put together: two trail sections laid end to end, two times run one after the other, three lanes in a row. When the fraction parts already share a denominator, the work is simpler and has its own lesson (Adding and subtracting mixed numbers with like denominators). Here the denominators differ, so the fraction parts must first be renamed in shared pieces, which is the work of Adding and subtracting fractions with unlike denominators. This lesson does not reteach that renaming; it uses it.

The lesson has three parts, one for each day:

  • Part 1: add the wholes, add the fractions, and put them back together.
  • Part 2: when the fraction parts make more than one whole, trade a whole’s worth of pieces for a whole number.
  • Part 3: the other route, rewriting each mixed number as an improper fraction; when each route is easier; adding three mixed numbers; and word problems.

The procedure

Part 1: wholes with wholes, parts with parts

Hania’s course is 3\tfrac{1}{6} miles of trail and then 2\tfrac{1}{2} miles of hill. To find the total, picture each length as whole miles and a piece of a mile.

3\tfrac{1}{6}
2\tfrac{1}{2} = 2\tfrac{3}{6}
5\tfrac{4}{6} = 5\tfrac{2}{3}
Figure 1The whole squares are counted with the whole squares, and the sixths with the sixths. 3 wholes and 2 wholes make 5; 1 sixth and 3 sixths make 4 sixths, which is \tfrac{2}{3}.

The picture shows the plan. The whole squares can be counted together: 3 and 2 make 5. The pieces can be counted together too, once they are the same size. A half is 3 sixths, so the pieces are 1 sixth and 3 sixths, which make 4 sixths.

Why are we allowed to split the numbers up and pair them this way? Each mixed number is a sum, so the whole problem is one long sum of four numbers:

3\tfrac{1}{6} + 2\tfrac{1}{2} = 3 + \tfrac{1}{6} + 2 + \tfrac{1}{2}

Addition lets us change the order of the numbers without changing the total. That is the commutative property of addition (properties of addition). It lets us move the 2 in front of the \tfrac{1}{6}. Addition also lets us choose which numbers to add first, by moving the parentheses. That is the associative property of addition. It lets us add the two whole numbers as one group and the two fractions as another.

\begin{aligned} 3 + \tfrac{1}{6} + 2 + \tfrac{1}{2} &= 3 + 2 + \tfrac{1}{6} + \tfrac{1}{2} \\ &= (3 + 2) + \left(\tfrac{1}{6} + \tfrac{1}{2}\right) \end{aligned}

The first line is the commutative property, and the second is the associative property. Now each group is a sum you already know how to find. The wholes give 3 + 2 = 5. For the fractions, 6 is a multiple of 2, so rename the half in sixths, as in Adding and subtracting fractions with unlike denominators:

\tfrac{1}{6} + \tfrac{1}{2} = \tfrac{1}{6} + \tfrac{3}{6} = \tfrac{4}{6} = \tfrac{2}{3}

The last step divides the top and bottom of \tfrac{4}{6} by 2, so the fraction is in simplest form. Put the two groups back together: 5 + \tfrac{2}{3}, which is written 5\tfrac{2}{3}.

3\tfrac{1}{6} + 2\tfrac{1}{2} = 5\tfrac{2}{3}

Hania’s course is 5\tfrac{2}{3} miles long.

Many people set the work out in a column, wholes under wholes and fractions under fractions, the way you add whole numbers by place value. The column is the same two groups written one above the other:

whole partfraction part
trail3\tfrac{1}{6} = \tfrac{1}{6}
hill2\tfrac{1}{2} = \tfrac{3}{6}
total5\tfrac{4}{6} = \tfrac{2}{3}

A quick check. The answer should be close to what you get from easy numbers. 3\tfrac{1}{6} is a little more than 3, and 2\tfrac{1}{2} is exactly 2\tfrac{1}{2}, so the total should be a little more than 5\tfrac{1}{2}. It is. Rounding mixed numbers to the nearest half or whole for a check like this is the subject of Estimating with fractions.

Part 2: when the parts make more than a whole

Mingyu runs the course in two pieces. The first takes 2\tfrac{3}{4} minutes and the second takes 1\tfrac{2}{3} minutes. Group the wholes and the fractions as before:

2\tfrac{3}{4} + 1\tfrac{2}{3} = (2 + 1) + \left(\tfrac{3}{4} + \tfrac{2}{3}\right)

The wholes make 3. For the fractions, neither 4 nor 3 is a multiple of the other, and they share no factor except 1, so the shared piece is 4 \times 3 = 12 twelfths:

\tfrac{3}{4} + \tfrac{2}{3} = \tfrac{9}{12} + \tfrac{8}{12} = \tfrac{17}{12}

So the total is 3 and \tfrac{17}{12}, which could be written 3\tfrac{17}{12}. That names the right amount, but it is not finished. The fraction part is more than 1, because 17 twelfths is more than the 12 twelfths that fill one whole.

\tfrac{3}{4} = \tfrac{9}{12}
\tfrac{2}{3} = \tfrac{8}{12}
\tfrac{17}{12} = 1\tfrac{5}{12}
Figure 2Nine twelfths and eight twelfths are 17 twelfths. Twelve of them fill one whole, and 5 twelfths are left, so \tfrac{17}{12} = 1\tfrac{5}{12}.

The picture shows what to do. Out of the 17 twelfths, 12 fill a whole, and 5 are left over. So \tfrac{17}{12} = 1 + \tfrac{5}{12} = 1\tfrac{5}{12}. In numbers, divide the top by the bottom: 17 \div 12 = 1 remainder 5. Trading those 12 twelfths for 1 whole is regrouping. It is the same move as trading ten ones for one ten when you add whole numbers (Regrouping in addition): a place holds only what fits, and a full set moves up.

The traded whole joins the other wholes. That step uses the associative property again:

\begin{aligned} 3 + \tfrac{17}{12} &= 3 + \left(1 + \tfrac{5}{12}\right) \\ &= (3 + 1) + \tfrac{5}{12} \\ &= 4\tfrac{5}{12} \end{aligned}

So the whole run is

2\tfrac{3}{4} + 1\tfrac{2}{3} = 3\tfrac{17}{12} = 4\tfrac{5}{12}

Mingyu ran the course in 4\tfrac{5}{12} minutes. A check: each piece of his run is a little under the next whole number (2\tfrac{3}{4} is a little under 3, and 1\tfrac{2}{3} is a little under 2), so the total should be a little under 5. And 4\tfrac{5}{12} is.

How much can carry over? Each fraction part of a mixed number is less than 1. Two of them together are less than 1 + 1 = 2. So two fraction parts make at most one extra whole, and the whole part of the answer is either the sum of the whole parts or one more than that. With three mixed numbers, the fraction parts are less than 3, so up to two wholes can carry over. Part 3 has an example.

Simplify the leftover. After the trade, look at the fraction part again. If its top and bottom share a factor, divide both by it, so the final answer has its fraction part in simplest form. Example 3 needs this.

Part 3: the improper-fraction route, and choosing a route

The other route. Any mixed number can be rewritten as an improper fraction, a fraction whose top is at least its bottom. Cut each whole into pieces of the fraction’s size and count all the pieces (Mixed numbers and improper fractions: rewriting and comparing). For Mingyu’s run, 2\tfrac{3}{4} is 2 wholes of 4 quarters each plus 3 quarters, which is 8 + 3 = 11 quarters. 1\tfrac{2}{3} is 3 + 2 = 5 thirds.

2\tfrac{3}{4} = \tfrac{11}{4} \qquad 1\tfrac{2}{3} = \tfrac{5}{3}

Now the problem is a sum of two fractions with unlike denominators, so rename both in twelfths and add:

\tfrac{11}{4} + \tfrac{5}{3} = \tfrac{33}{12} + \tfrac{20}{12} = \tfrac{53}{12}

To read the answer as a mixed number, divide: 53 \div 12 = 4 remainder 5, so \tfrac{53}{12} = 4\tfrac{5}{12}. That is the same 4\tfrac{5}{12} minutes as in Part 2. It has to be: \tfrac{11}{4} and 2\tfrac{3}{4} are two names for one number, and so are \tfrac{5}{3} and 1\tfrac{2}{3}, so the two routes add the same numbers. They differ only in when the wholes are cut into pieces and when the pieces are packed back into wholes. On this route the regrouping is the final division.

Which route is easier? Here are the two routes side by side for Mingyu’s run.

wholes and partsimproper fractions
rewrite2 + 1 and \tfrac{3}{4} + \tfrac{2}{3}\tfrac{11}{4} + \tfrac{5}{3}
rename\tfrac{9}{12} + \tfrac{8}{12}\tfrac{33}{12} + \tfrac{20}{12}
add3 + \tfrac{17}{12}\tfrac{53}{12}
finish3 + 1\tfrac{5}{12} = 4\tfrac{5}{12}53 \div 12 gives 4\tfrac{5}{12}

Both are fine when the whole numbers are small. As the whole numbers grow, the improper fractions grow with them. 18\tfrac{5}{6} becomes \tfrac{113}{6}, and renaming that in twenty-fourths gives a top of 452. The wholes-and-parts route never makes the wholes bigger: they are added as whole numbers, and only the small fraction parts are renamed. So for large whole parts, the wholes-and-parts route keeps the numbers small. The improper-fraction route has one step fewer to remember, since the regrouping happens in the last division, and it is the route that multiplying mixed numbers needs later, so it is worth knowing well.

Three mixed numbers. The wholes-and-parts route works for any number of mixed numbers. The commutative and associative properties let you gather every whole part into one group and every fraction part into another. Then add the whole parts, rename all the fraction parts with one denominator that every denominator divides, add them, and regroup. With three fraction parts the sum can pass 1 or even 2.

Word problems. In this lesson, every word problem puts amounts together: lengths laid end to end, times run one after another, a total over several days. Those are sums. Problems where you must read the story to decide whether to add or subtract are in Word problems that add or subtract fractions. Whatever the story, two rules hold. Add only amounts of the same kind, measured in the same unit: miles with miles, minutes with minutes. And end with a check on the size of the answer.

Worked examples

Examples 1 and 2 belong to Part 1, Examples 3 and 4 to Part 2, and Examples 5, 6 and 7 to Part 3.

Example 1

Two obstacles, one denominator a multiple of the other

Segun climbs the wall in 4\tfrac{1}{3} minutes and then crosses the rope bridge in 2\tfrac{1}{6} minutes. How long do the two obstacles take him?

Solution

The times are added, since one obstacle comes right after the other. Gather the wholes and the fractions, using the commutative and associative properties:

4\tfrac{1}{3} + 2\tfrac{1}{6} = (4 + 2) + \left(\tfrac{1}{3} + \tfrac{1}{6}\right)

The wholes make 6. For the fractions, 6 \div 3 = 2, a whole number, so 6 is a multiple of 3 and sixths can be the shared piece. Cut each third into 2 sixths:

\tfrac{1}{3} + \tfrac{1}{6} = \tfrac{2}{6} + \tfrac{1}{6} = \tfrac{3}{6} = \tfrac{1}{2}

The fraction \tfrac{3}{6} simplifies, since 3 divides both its top and its bottom. So the two obstacles take 6\tfrac{1}{2} minutes.

Check in seconds. A third of a minute is 20 seconds and a sixth is 10 seconds, so the times are 4 minutes 20 seconds and 2 minutes 10 seconds. Together that is 6 minutes 30 seconds, and 30 seconds is half a minute.

Example 2

Two lanes, neither denominator a multiple of the other

The cargo net lane is 5\tfrac{2}{5} yards long and the tunnel is 3\tfrac{1}{4} yards long. The crew lays them end to end. How long are the two together?

Solution

Laid end to end, the lengths add. The wholes make 5 + 3 = 8. For the fractions, neither 5 nor 4 divides the other, and they share no factor except 1, so use 5 \times 4 = 20:

\tfrac{2}{5} + \tfrac{1}{4} = \tfrac{8}{20} + \tfrac{5}{20} = \tfrac{13}{20}

13 has no factor except 1 and itself, and it does not divide 20, so \tfrac{13}{20} is in simplest form. It is less than 1, so there is nothing to regroup.

5\tfrac{2}{5} + 3\tfrac{1}{4} = 8\tfrac{13}{20}

The two lanes are 8\tfrac{13}{20} yards long. Check: rounding each length down to a whole gives 5 + 3 = 8, and both fraction parts were dropped, so the exact total must be a little more than 8. It is.

Example 3

Regrouping, then simplifying

Hania runs 3\tfrac{5}{6} miles on Monday and 2\tfrac{1}{2} miles on Tuesday. How far does she run on the two days in all?

Solution

The wholes make 3 + 2 = 5. For the fractions, rename the half in sixths:

\tfrac{5}{6} + \tfrac{1}{2} = \tfrac{5}{6} + \tfrac{3}{6} = \tfrac{8}{6}

So far the total is 5\tfrac{8}{6}. The fraction part is more than 1, so regroup: six of the eight sixths make a whole, and 2 sixths are left, so \tfrac{8}{6} = 1\tfrac{2}{6}. Then simplify the leftover, since 2 divides both 2 and 6: \tfrac{2}{6} = \tfrac{1}{3}.

\begin{aligned} 3\tfrac{5}{6} + 2\tfrac{1}{2} &= 5\tfrac{8}{6} \\ &= 5 + 1\tfrac{2}{6} \\ &= 6\tfrac{1}{3} \end{aligned}

Hania runs 6\tfrac{1}{3} miles. Simplifying first works just as well: \tfrac{8}{6} = \tfrac{4}{3}, and \tfrac{4}{3} = 1\tfrac{1}{3}, so the total is again 5 + 1\tfrac{1}{3} = 6\tfrac{1}{3}.

Example 4

Regrouping with a least common denominator

Mingyu crawls under the net in 6\tfrac{5}{6} minutes and then swings across the monkey bars in 1\tfrac{3}{4} minutes. What is his time for the two obstacles?

Solution

The wholes make 6 + 1 = 7. For the fractions, 6 and 4 share the factor 2, so the least common denominator is smaller than the product 24. Multiples of 6 are 6 and 12, and 4 divides 12. Rename both parts in twelfths:

\tfrac{5}{6} + \tfrac{3}{4} = \tfrac{10}{12} + \tfrac{9}{12} = \tfrac{19}{12}

The fraction part is more than 1. Divide: 19 \div 12 = 1 remainder 7, so \tfrac{19}{12} = 1\tfrac{7}{12}. The traded whole joins the 7 wholes:

6\tfrac{5}{6} + 1\tfrac{3}{4} = 7\tfrac{19}{12} = 8\tfrac{7}{12}

7 and 12 share no factor except 1, so Mingyu’s time is 8\tfrac{7}{12} minutes.

Check in seconds. \tfrac{5}{6} of a minute is 50 seconds and \tfrac{3}{4} of a minute is 45 seconds. So the times are 6 minutes 50 seconds and 1 minute 45 seconds. Together they are 7 minutes 95 seconds, which is 8 minutes 35 seconds. And \tfrac{7}{12} of 60 seconds is 35 seconds.

Example 5

Both routes for one sum

Segun jogs the warm-up loop, 1\tfrac{4}{5} miles, and then the course itself, 2\tfrac{1}{2} miles. Find the total distance by rewriting both numbers as improper fractions. Then find it again by adding wholes and parts.

Solution

Improper fractions first. 1\tfrac{4}{5} is 5 fifths plus 4 fifths, which is \tfrac{9}{5}. 2\tfrac{1}{2} is 4 halves plus 1 half, which is \tfrac{5}{2}. Rename both in tenths and add:

\tfrac{9}{5} + \tfrac{5}{2} = \tfrac{18}{10} + \tfrac{25}{10} = \tfrac{43}{10}

Divide: 43 \div 10 = 4 remainder 3, so \tfrac{43}{10} = 4\tfrac{3}{10}.

Now wholes and parts. The wholes make 1 + 2 = 3. The fractions make \tfrac{8}{10} + \tfrac{5}{10} = \tfrac{13}{10} = 1\tfrac{3}{10}, and 3 + 1\tfrac{3}{10} = 4\tfrac{3}{10}.

Both routes give 4\tfrac{3}{10} miles, as they must. Here the numbers stay small either way. The improper route needed one division at the end; the other route needed a regroup in the middle.

Example 6

Three sections of the course

The full course has three trail sections: 1\tfrac{1}{2} miles, 2\tfrac{2}{3} miles and 1\tfrac{5}{12} miles. How long is the trail in all?

Solution

By the commutative and associative properties, gather the three wholes into one group and the three fractions into another. The wholes make 1 + 2 + 1 = 4. For the fractions, 2, 3 and 12 all divide 12, so rename each in twelfths:

\tfrac{1}{2} + \tfrac{2}{3} + \tfrac{5}{12} = \tfrac{6}{12} + \tfrac{8}{12} + \tfrac{5}{12} = \tfrac{19}{12}

Regroup: 19 \div 12 = 1 remainder 7, so \tfrac{19}{12} = 1\tfrac{7}{12}. Add the traded whole to the 4:

1\tfrac{1}{2} + 2\tfrac{2}{3} + 1\tfrac{5}{12} = 4\tfrac{19}{12} = 5\tfrac{7}{12}

The trail is 5\tfrac{7}{12} miles long.

Check by rounding to the nearest half. 1\tfrac{1}{2} is already a half. \tfrac{2}{3} is closer to \tfrac{1}{2} than to 1, so 2\tfrac{2}{3} is near 2\tfrac{1}{2}. \tfrac{5}{12} is just under \tfrac{6}{12}, so 1\tfrac{5}{12} is near 1\tfrac{1}{2}. The estimate is 1\tfrac{1}{2} + 2\tfrac{1}{2} + 1\tfrac{1}{2} = 5\tfrac{1}{2}, and 5\tfrac{7}{12} is just above it.

Example 7

Choosing the route when the wholes are large

To mark the edge of the course, the crew stretches one rope 18\tfrac{5}{6} yards long and a second rope 24\tfrac{3}{8} yards long, end to end. How much ground do the ropes cover? Choose the easier route and say why.

Solution

The whole parts are large. On the improper route, 18\tfrac{5}{6} = \tfrac{113}{6} and 24\tfrac{3}{8} = \tfrac{195}{8}, and renaming both in twenty-fourths makes tops of 452 and 585. The wholes-and-parts route keeps 18 and 24 as whole numbers, so use it.

The wholes make 18 + 24 = 42. For the fractions, multiples of 8 are 8, 16 and 24, and 6 divides 24, so rename in twenty-fourths:

\tfrac{5}{6} + \tfrac{3}{8} = \tfrac{20}{24} + \tfrac{9}{24} = \tfrac{29}{24}

Regroup: 29 \div 24 = 1 remainder 5, so \tfrac{29}{24} = 1\tfrac{5}{24}, and

18\tfrac{5}{6} + 24\tfrac{3}{8} = 42\tfrac{29}{24} = 43\tfrac{5}{24}

The ropes cover 43\tfrac{5}{24} yards.

The improper route gives the same answer with bigger numbers: \tfrac{452}{24} + \tfrac{585}{24} = \tfrac{1037}{24}, and 1037 \div 24 = 43 remainder 5. As a check, round to the nearest whole: 18\tfrac{5}{6} is near 19 and 24\tfrac{3}{8} is near 24, so the total should be near 43. It is.

Boundaries

  • A mixed number means a sum. 3\tfrac{1}{6} means 3 + \tfrac{1}{6}, not 3 \times \tfrac{1}{6}. Splitting it into wholes and parts is allowed only because of that plus sign, and the regrouping rests on the commutative and associative properties of addition.
  • The same whole and the same unit. Wholes add to wholes and pieces to pieces only when every number counts the same thing. Miles add to miles. A length in miles and a length in yards must be put in one unit first, and a length never adds to a time.
  • Fraction parts less than one. In a mixed number the fraction part is less than 1. That is why two fraction parts carry at most one whole. A sum such as 3\tfrac{17}{12} names the right amount but is not a mixed number in this sense until it is regrouped.
  • No zero denominators. A denominator says how many equal pieces make a whole, so it is never 0.
  • Positive numbers, and adding only. Every number here is positive. Taking one mixed number from another can need a whole broken into pieces first, which is Subtracting mixed numbers. Mixed numbers with negative signs, where the whole and the fraction parts must keep the same sign, belong to Operations with rational numbers.
  • Adding, not multiplying. Adding wholes to wholes and parts to parts works for addition only. For a product such as 2\tfrac{1}{2} \times 3\tfrac{1}{3}, multiplying the wholes and the parts separately misses the cross pieces, which is why Multiplying mixed numbers uses improper fractions.

Where this goes next

Subtracting mixed numbers comes next (Subtracting mixed numbers). It runs the same plan in reverse: wholes from wholes and parts from parts. Where this lesson traded a full set of pieces up for a whole, subtraction sometimes has to trade a whole down into pieces. Story problems that ask you to decide whether to join or take away are in Word problems that add or subtract fractions.

Multiplying and dividing mixed numbers (Multiplying mixed numbers, Dividing mixed numbers) lean on the improper-fraction route from Part 3. And the same idea of grouping like with like, wholes with wholes and pieces with pieces, returns in algebra, where 3x + 2 + 5x + 4 is gathered into 8x + 6 by the same two properties of addition.

Exercises

Part 1: wholes and parts, no regrouping

Add. Write each answer as a mixed number with its fraction part in simplest form.

  1. 2\tfrac{1}{4} + 3\tfrac{1}{2}
  2. 1\tfrac{1}{3} + 4\tfrac{1}{2}
  3. 6\tfrac{1}{5} + 2\tfrac{3}{10}
  4. 3\tfrac{2}{9} + 5\tfrac{1}{3}
  5. 7\tfrac{1}{4} + 1\tfrac{3}{5}
  6. 1\tfrac{1}{6} + 3\tfrac{1}{8}
  7. Segun runs 1\tfrac{3}{10} miles to the first obstacle and then 2\tfrac{1}{2} miles more to the finish. How far does he run in all?

Part 2: regrouping

Add. Write each answer as a mixed number with its fraction part in simplest form.

  1. 1\tfrac{2}{3} + 2\tfrac{1}{2}
  2. 3\tfrac{3}{4} + 2\tfrac{1}{2}
  3. 4\tfrac{4}{5} + 1\tfrac{3}{10}
  4. 2\tfrac{5}{6} + 3\tfrac{4}{9}
  5. 5\tfrac{7}{10} + 2\tfrac{3}{4}
  6. 1\tfrac{5}{8} + 1\tfrac{5}{12}
  7. Hania runs one lap of the obstacle course in 3\tfrac{2}{3} minutes and a second lap in 3\tfrac{4}{5} minutes. How long are the two laps together? Check your answer in minutes and seconds.
  8. Mingyu adds 4\tfrac{5}{6} + 1\tfrac{2}{3} and writes 5\tfrac{9}{6}. Is the sum correct? Is it finished? Write it as a mixed number with a proper fraction part in simplest form.

Part 3: improper fractions, three mixed numbers, word problems

  1. Add by rewriting both as improper fractions: 1\tfrac{1}{2} + 2\tfrac{2}{5}.
  2. Add by rewriting both as improper fractions: 2\tfrac{1}{3} + 1\tfrac{4}{5}.
  3. Find 3\tfrac{5}{8} + 1\tfrac{1}{2} by both routes and show that the answers agree.
  4. 1\tfrac{1}{2} + 2\tfrac{1}{4} + 3\tfrac{1}{8}
  5. 2\tfrac{2}{3} + 1\tfrac{5}{6} + 4\tfrac{1}{2}
  6. A training route has three parts: a \tfrac{3}{4}-mile warm-up, a 2\tfrac{2}{5}-mile trail and a 1\tfrac{1}{2}-mile hill. How long is the whole route?
  7. Segun times three obstacles: 1\tfrac{1}{4} minutes on the tires, 2\tfrac{1}{3} minutes on the wall and \tfrac{3}{4} minute on the balance beam. What is his total time on the three?

Reasoning

  1. Explain why, when two mixed numbers are added, the whole part of the answer is always either the sum of the two whole parts or one more than that sum.
  2. Mingyu says 2\tfrac{1}{2} + 1\tfrac{3}{5} = 3\tfrac{4}{7}. Without finding the exact sum, explain why he must be wrong. Then find the exact sum.
  3. Explain, naming each property you use, why 4\tfrac{2}{5} + 3\tfrac{1}{3} may be found as (4 + 3) + \left(\tfrac{2}{5} + \tfrac{1}{3}\right).
Answers
  1. 5\tfrac{3}{4}. Wholes 2 + 3 = 5; fractions \tfrac{1}{4} + \tfrac{2}{4} = \tfrac{3}{4}.
  2. 5\tfrac{5}{6}. Wholes 1 + 4 = 5; fractions \tfrac{2}{6} + \tfrac{3}{6} = \tfrac{5}{6}.
  3. 8\tfrac{1}{2}. Wholes 6 + 2 = 8; fractions \tfrac{2}{10} + \tfrac{3}{10} = \tfrac{5}{10}, and dividing top and bottom by 5 gives \tfrac{1}{2}.
  4. 8\tfrac{5}{9}. Wholes 3 + 5 = 8; fractions \tfrac{2}{9} + \tfrac{3}{9} = \tfrac{5}{9}.
  5. 8\tfrac{17}{20}. Wholes 7 + 1 = 8; fractions \tfrac{5}{20} + \tfrac{12}{20} = \tfrac{17}{20}.
  6. 4\tfrac{7}{24}. Wholes 1 + 3 = 4; the least common denominator of 6 and 8 is 24, and \tfrac{4}{24} + \tfrac{3}{24} = \tfrac{7}{24}.
  7. 3\tfrac{4}{5} miles. Wholes 1 + 2 = 3; fractions \tfrac{3}{10} + \tfrac{5}{10} = \tfrac{8}{10}, which simplifies to \tfrac{4}{5}.
  8. 4\tfrac{1}{6}. Wholes 3; fractions \tfrac{4}{6} + \tfrac{3}{6} = \tfrac{7}{6} = 1\tfrac{1}{6}, so 3\tfrac{7}{6} = 4\tfrac{1}{6}.
  9. 6\tfrac{1}{4}. Wholes 5; fractions \tfrac{3}{4} + \tfrac{2}{4} = \tfrac{5}{4} = 1\tfrac{1}{4}, so 5\tfrac{5}{4} = 6\tfrac{1}{4}.
  10. 6\tfrac{1}{10}. Wholes 5; fractions \tfrac{8}{10} + \tfrac{3}{10} = \tfrac{11}{10} = 1\tfrac{1}{10}, so 5\tfrac{11}{10} = 6\tfrac{1}{10}.
  11. 6\tfrac{5}{18}. Wholes 5; the least common denominator is 18, and \tfrac{15}{18} + \tfrac{8}{18} = \tfrac{23}{18} = 1\tfrac{5}{18}, so 5\tfrac{23}{18} = 6\tfrac{5}{18}.
  12. 8\tfrac{9}{20}. Wholes 7; \tfrac{14}{20} + \tfrac{15}{20} = \tfrac{29}{20} = 1\tfrac{9}{20}, so 7\tfrac{29}{20} = 8\tfrac{9}{20}.
  13. 3\tfrac{1}{24}. Wholes 2; the least common denominator is 24, and \tfrac{15}{24} + \tfrac{10}{24} = \tfrac{25}{24} = 1\tfrac{1}{24}, so 2\tfrac{25}{24} = 3\tfrac{1}{24}.
  14. 7\tfrac{7}{15} minutes. Wholes 6; \tfrac{10}{15} + \tfrac{12}{15} = \tfrac{22}{15} = 1\tfrac{7}{15}. Check: 3 minutes 40 seconds and 3 minutes 48 seconds make 7 minutes 28 seconds, and \tfrac{7}{15} of 60 seconds is 28 seconds.
  15. Correct but not finished: 6\tfrac{1}{2}. The wholes make 5 and \tfrac{5}{6} + \tfrac{4}{6} = \tfrac{9}{6}, so 5\tfrac{9}{6} is the right amount. But \tfrac{9}{6} is more than 1: \tfrac{9}{6} = 1\tfrac{3}{6} = 1\tfrac{1}{2}, so the sum is 6\tfrac{1}{2}.
  16. 3\tfrac{9}{10}. \tfrac{3}{2} + \tfrac{12}{5} = \tfrac{15}{10} + \tfrac{24}{10} = \tfrac{39}{10}, and 39 \div 10 = 3 remainder 9.
  17. 4\tfrac{2}{15}. \tfrac{7}{3} + \tfrac{9}{5} = \tfrac{35}{15} + \tfrac{27}{15} = \tfrac{62}{15}, and 62 \div 15 = 4 remainder 2.
  18. 5\tfrac{1}{8} both ways. Wholes and parts: 4 + \tfrac{5}{8} + \tfrac{4}{8} = 4\tfrac{9}{8} = 5\tfrac{1}{8}. Improper fractions: \tfrac{29}{8} + \tfrac{12}{8} = \tfrac{41}{8}, and 41 \div 8 = 5 remainder 1.
  19. 6\tfrac{7}{8}. Wholes 1 + 2 + 3 = 6; fractions \tfrac{4}{8} + \tfrac{2}{8} + \tfrac{1}{8} = \tfrac{7}{8}.
  20. 9. Wholes 2 + 1 + 4 = 7; fractions \tfrac{4}{6} + \tfrac{5}{6} + \tfrac{3}{6} = \tfrac{12}{6} = 2, exactly two wholes, so the sum is 7 + 2 = 9.
  21. 4\tfrac{13}{20} miles. Wholes 0 + 2 + 1 = 3; in twentieths, \tfrac{15}{20} + \tfrac{8}{20} + \tfrac{10}{20} = \tfrac{33}{20} = 1\tfrac{13}{20}, so the route is 3 + 1\tfrac{13}{20} = 4\tfrac{13}{20} miles.
  22. 4\tfrac{1}{3} minutes. Wholes 1 + 2 = 3; in twelfths, \tfrac{3}{12} + \tfrac{4}{12} + \tfrac{9}{12} = \tfrac{16}{12} = 1\tfrac{4}{12} = 1\tfrac{1}{3}, so the total is 4\tfrac{1}{3}. A shortcut by the commutative property: put 1\tfrac{1}{4} next to \tfrac{3}{4}, which makes exactly 2, and 2 + 2\tfrac{1}{3} = 4\tfrac{1}{3}.
  23. Each fraction part is at least 0 and less than 1, so the two fraction parts together are at least 0 and less than 2. If their sum is less than 1, nothing regroups and the whole part of the answer is the sum of the whole parts. If their sum is 1 or more but less than 2, exactly one whole is traded up, and the whole part is one more. A sum of 2 or more is not possible, so two or more extra wholes never happen.
  24. Both fraction parts are at least a half: \tfrac{1}{2} is a half and \tfrac{3}{5} is more than a half. So the fraction parts make more than 1, and the sum is more than 2 + 1 + 1 = 4. But 3\tfrac{4}{7} is less than 4, so it cannot be the sum. (It came from adding tops and adding bottoms.) The exact sum: wholes 3, and \tfrac{5}{10} + \tfrac{6}{10} = \tfrac{11}{10} = 1\tfrac{1}{10}, so the sum is 4\tfrac{1}{10}.
  25. A mixed number means a sum, so the problem is 4 + \tfrac{2}{5} + 3 + \tfrac{1}{3}. The commutative property of addition lets the 3 and the \tfrac{2}{5} trade places, giving 4 + 3 + \tfrac{2}{5} + \tfrac{1}{3}. The associative property lets us choose which numbers to add first, so we may group the wholes and the fractions: (4 + 3) + \left(\tfrac{2}{5} + \tfrac{1}{3}\right). Neither step changes the total. (The sum is 7 + \tfrac{11}{15} = 7\tfrac{11}{15}.)
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0