Before you find the exact value of , can you say about how big it is, and use that to tell whether an answer makes sense?

Grade 5 · 5.NF.A.2 · 7.EE.B.3

Estimating with fractions

Before you find the exact value of \tfrac{9}{10} + \tfrac{3}{7}, can you say about how big it is, and use that to tell whether an answer makes sense?

You need

Benchmark fractionsGrade 4

Where you are 5 in Adding and subtracting fractions

Fractions, ratio and percent3

Fractions15

Adding and subtracting fractions5

Adding and subtracting fractions with like denominators

Estimating with fractions

Adding tenths and hundredths as fractions Adding and subtracting fractions with unlike denominators Word problems that add or subtract fractions

When you reach for it

Bhavin, Emiko and Tunde work in a wood shop. Their boards are measured in inches and fractions of an inch: a strip \tfrac{9}{10} inch thick, a cut 17\tfrac{5}{8} inches long. Many shop questions do not need an exact answer. About how thick is the glued stack? Is this board long enough for both pieces? Does the number on the cut list look right?

Finding an exact sum of fractions with different denominators takes renaming and care (Adding and subtracting fractions with unlike denominators). An estimate takes a few seconds. It answers “about how much” questions, and it catches an answer that cannot be right before any wood is cut.

The tool is a short list of landmark fractions, the benchmarks 0, \tfrac{1}{2} and 1 (Benchmark fractions are familiar fractions used as reference points). Replace each fraction with the benchmark it is nearest, then add or subtract the benchmarks. Those sums are easy: 1 + \tfrac{1}{2} = 1\tfrac{1}{2}, \tfrac{1}{2} + \tfrac{1}{2} = 1.

The procedure

Part 1: rounding fractions to 0, one half or 1

A fraction between 0 and 1 has three benchmarks to choose from. To pick one, look at the top number (the numerator) and set it beside three numbers made from the bottom (the denominator): 0, half the bottom, and the whole bottom.

If the top is closest to 0, the fraction is near 0. In \tfrac{1}{9} the top is 1, and half of 9 is 4\tfrac{1}{2}. The 1 is much closer to 0.
If the top is closest to half the bottom, the fraction is near \tfrac{1}{2}. In \tfrac{3}{7}, half of 7 is 3\tfrac{1}{2}, and 3 is only \tfrac{1}{2} away from it.
If the top is closest to the whole bottom, the fraction is near 1. In \tfrac{9}{10} the top is 9, just 1 short of 10.

Why does this work? A fraction counts pieces of one size. \tfrac{3}{7} is 3 pieces, each a seventh. Half of the whole is 3\tfrac{1}{2} sevenths, so 3 sevenths falls just short of \tfrac{1}{2}. Comparing the top with half the bottom is comparing the number of pieces you have with the number that would make one half.

The number line shows the three stretches. A fraction below \tfrac{1}{4} is closer to 0 than to \tfrac{1}{2}. One between \tfrac{1}{4} and \tfrac{3}{4} is closest to \tfrac{1}{2}. One above \tfrac{3}{4} is closest to 1.

Figure 1The dashed ticks at \tfrac{1}{4} and \tfrac{3}{4} split the line into three stretches. A fraction in the shaded middle is nearest \tfrac{1}{2}; one past \tfrac{3}{4} is nearest 1.

Now estimate a sum. In the lead question, \tfrac{9}{10} is near 1 and \tfrac{3}{7} is near \tfrac{1}{2}, so

\tfrac{9}{10} + \tfrac{3}{7} \approx 1 + \tfrac{1}{2} = 1\tfrac{1}{2}

A difference works the same way. \tfrac{11}{12} is near 1, since 11 is 1 short of 12. \tfrac{1}{9} is near 0. So

\tfrac{11}{12} - \tfrac{1}{9} \approx 1 - 0 = 1

Why is the estimate close? Each fraction moves to a benchmark no more than a quarter away, because no fraction between 0 and 1 is more than \tfrac{1}{4} from its nearest benchmark. Two moves of at most \tfrac{1}{4} each change a sum or difference by at most \tfrac{1}{2}. So the exact value is within \tfrac{1}{2} of the estimate. Usually it is much closer, since most fractions sit nearer their benchmark than a full quarter.

Watching which way each fraction moved also tells you whether the estimate is too big or too small. Friendly numbers and deliberate over and under estimates covers overestimates and underestimates in full. Here is the short form for fractions. In \tfrac{11}{12} - \tfrac{1}{9}, the first fraction moved up to 1 and the second moved down to 0. Starting bigger and taking away less both push the result up, so the exact difference is less than 1.

Part 2: mixed numbers, and judging an answer

A mixed number such as 17\tfrac{5}{8} has a whole-number part and a fraction part. The whole number stays. Only the fraction part moves to a benchmark. Since 5 is closer to 4 (half of 8) than to 8, the fraction \tfrac{5}{8} is near \tfrac{1}{2}, and 17\tfrac{5}{8} is near 17\tfrac{1}{2}. That is rounding to the nearest half.

Figure 217\tfrac{5}{8} is one eighth past 17\tfrac{1}{2} and three eighths short of 18, so its nearest half is 17\tfrac{1}{2}.

For a quicker, rougher estimate, round to the nearest whole number instead. A fraction part of \tfrac{1}{2} or more goes up; less than \tfrac{1}{2} goes down. 17\tfrac{5}{8} is near 18. Rounding to the nearest half keeps each number within \tfrac{1}{4}; rounding to the nearest whole keeps it within \tfrac{1}{2}. Whole numbers are easier to add, halves are closer.

An estimate is also a test. Before trusting an answer, ask whether it sits near the estimate. If it is far away, something went wrong, and you know it without redoing the work. Some answers can be ruled out with no estimate at all, just by asking where a sum must land.

Adding a positive number makes a result bigger. So a sum is more than either of the numbers added.
Taking away a positive number makes a result smaller. So a difference is less than the number you started with.

These facts give a sum or difference a floor or a ceiling. An answer outside it cannot be right.

An estimate has limits. It can show that an answer is wrong. It can never show that an answer is right, only that it is reasonable. And when the estimate lands right on the line you care about, such as the length of the board, it cannot say which side of the line the exact value is on. Then look closer at how far each fraction is from its benchmark, or find the exact value.

Worked examples

Examples 1 to 4 belong to Part 1. Examples 5 to 7 belong to Part 2.

Example 1

Sorting offcuts by size

Bhavin has four thin offcuts (leftover pieces), \tfrac{1}{8}, \tfrac{5}{9}, \tfrac{6}{7} and \tfrac{3}{10} of an inch thick. Round each to 0, \tfrac{1}{2} or 1.

Solution

Set each top beside 0, half the bottom, and the bottom.

Fraction	Compare	Nearest
\tfrac{1}{8}	1 with 0, 4, 8	0
\tfrac{5}{9}	5 with 0, 4\tfrac{1}{2}, 9	\tfrac{1}{2}
\tfrac{6}{7}	6 with 0, 3\tfrac{1}{2}, 7	1
\tfrac{3}{10}	3 with 0, 5, 10	\tfrac{1}{2}

The last row needs care. \tfrac{3}{10} looks small, but 3 is 3 away from 0 and only 2 away from 5. So \tfrac{3}{10} is nearer \tfrac{1}{2}. On the number line it sits just past \tfrac{1}{4}, inside the middle stretch, since \tfrac{1}{4} of 10 is 2\tfrac{1}{2} and 3 is more than that.

Example 2

Gluing two strips

Emiko glues a strip \tfrac{9}{10} inch thick on top of a strip \tfrac{3}{7} inch thick. About how thick is the stack?

Solution

\tfrac{9}{10} is near 1, since 9 is 1 short of 10. \tfrac{3}{7} is near \tfrac{1}{2}, since 3 is close to 3\tfrac{1}{2}. So

\tfrac{9}{10} + \tfrac{3}{7} \approx 1 + \tfrac{1}{2} = 1\tfrac{1}{2}

The stack is about 1\tfrac{1}{2} inches thick. Both strips moved up to reach their benchmarks (\tfrac{9}{10} up to 1, \tfrac{3}{7} up to \tfrac{1}{2}), so the real stack is a little thinner than 1\tfrac{1}{2} inches. It is certainly more than 1 inch, because \tfrac{9}{10} and \tfrac{3}{7} together are more than \tfrac{9}{10} + \tfrac{1}{10} = 1.

Example 3

Planing a board thinner

Tunde has a board \tfrac{11}{12} inch thick. The planer shaves \tfrac{1}{9} inch off. About how thick is the board now, and is the estimate too big or too small?

Solution

\tfrac{11}{12} is near 1 and \tfrac{1}{9} is near 0, so

\tfrac{11}{12} - \tfrac{1}{9} \approx 1 - 0 = 1

The board is now a bit under 1 inch thick. The estimate is too big, and we can say so without the exact value: the board was already thinner than 1 inch before planing, and planing only takes wood away. The estimate tells Tunde the board is close to an inch, which is what he needs to know to pick a screw length.

Example 4

Three steps on one stack

Emiko stacks a piece \tfrac{5}{6} inch thick on one \tfrac{4}{7} inch thick, then sands \tfrac{1}{11} inch off the top. Estimate the final thickness.

Solution

Round each fraction on its own. \tfrac{5}{6}: the top 5 is 1 short of 6, so it is near 1. \tfrac{4}{7}: half of 7 is 3\tfrac{1}{2}, and 4 is close to it, so it is near \tfrac{1}{2}. \tfrac{1}{11}: the top is tiny next to 5\tfrac{1}{2}, so it is near 0.

\tfrac{5}{6} + \tfrac{4}{7} - \tfrac{1}{11} \approx 1 + \tfrac{1}{2} - 0 = 1\tfrac{1}{2}

The stack ends up about 1\tfrac{1}{2} inches thick. With three fractions, the moves can add up, so the estimate may be off by more than with two. Here \tfrac{5}{6} moved up by \tfrac{1}{6}, while \tfrac{4}{7} moved down by only \tfrac{1}{14} (it is half of one seventh past \tfrac{1}{2}), a smaller piece. Rounding the sanding to 0 takes away less than the real \tfrac{1}{11}, which also pushes the estimate up. So the real thickness is under 1\tfrac{1}{2} inches.

Example 5

Two long cuts

Bhavin cuts a piece 24\tfrac{3}{16} inches long and a piece 17\tfrac{5}{8} inches long. About how much board do the two use together? Estimate to the nearest half, then to the nearest whole.

Solution

To the nearest half, keep the whole numbers and move the fraction parts. In \tfrac{3}{16}, the top 3 is closer to 0 than to 8 (half of 16), so 24\tfrac{3}{16} is near 24. In \tfrac{5}{8}, the top 5 is closer to 4 than to 8, so 17\tfrac{5}{8} is near 17\tfrac{1}{2}. The estimate is

24 + 17\tfrac{1}{2} = 41\tfrac{1}{2} \text{ inches}

To the nearest whole, \tfrac{3}{16} is less than \tfrac{1}{2}, so 24\tfrac{3}{16} goes down to 24. \tfrac{5}{8} is more than \tfrac{1}{2}, so 17\tfrac{5}{8} goes up to 18. The estimate is

24 + 18 = 42 \text{ inches}

Both are fair estimates, and together they pin the total down. The first moved both lengths down (\tfrac{3}{16} and \tfrac{1}{8}), so it is under the exact total. The second moved 24\tfrac{3}{16} down by \tfrac{3}{16} but 17\tfrac{5}{8} up by \tfrac{3}{8}, which is \tfrac{6}{16}, a bigger move, so it is over. The total is between 41\tfrac{1}{2} and 42 inches. Adding the mixed numbers exactly is the work of Adding mixed numbers.

Example 6

Checking the cut list

Two lines on the shop’s cut list read

\tfrac{2}{3} + \tfrac{1}{4} = \tfrac{3}{7} \qquad 5\tfrac{1}{3} - 2\tfrac{5}{6} = 2\tfrac{1}{2}

Is each one reasonable?

Solution

First line. Look at \tfrac{2}{3} alone: half of 3 is 1\tfrac{1}{2}, and 2 is more than that, so \tfrac{2}{3} is more than \tfrac{1}{2}. Adding \tfrac{1}{4} makes the sum bigger still. So the sum is more than \tfrac{1}{2}. Now look at \tfrac{3}{7}: half of 7 is 3\tfrac{1}{2}, and 3 is less than that, so \tfrac{3}{7} is less than \tfrac{1}{2}.

Figure 3Adding \tfrac{1}{4} to \tfrac{2}{3} moves right from \tfrac{2}{3} but not as far as 1. The value \tfrac{3}{7} sits left of \tfrac{1}{2}, outside the shaded stretch.

A sum that must be more than \tfrac{1}{2} cannot equal a number less than \tfrac{1}{2}. The first line is wrong. We can also say where the sum must land. It is more than \tfrac{2}{3}, and since \tfrac{1}{4} is less than \tfrac{1}{3}, it is less than \tfrac{2}{3} + \tfrac{1}{3} = 1. The exact sum, \tfrac{11}{12}, found in Adding and subtracting fractions with unlike denominators, does sit between \tfrac{2}{3} and 1.

Second line. Round to the nearest half: \tfrac{1}{3} is near \tfrac{1}{2} (the top 1 is closer to 1\tfrac{1}{2} than to 0), so 5\tfrac{1}{3} is near 5\tfrac{1}{2}. And \tfrac{5}{6} is near 1, so 2\tfrac{5}{6} is near 3. The estimate is

5\tfrac{1}{2} - 3 = 2\tfrac{1}{2}

The answer 2\tfrac{1}{2} matches the estimate, so it is reasonable. That is not proof that it is exactly right. A wrong answer such as 2\tfrac{2}{5} would also sit near the estimate. To be sure, find the exact difference (Subtracting mixed numbers); here it does come out to 2\tfrac{1}{2}.

Example 7

When the estimate lands on the line

Tunde needs a piece 15\tfrac{4}{9} inches long and a piece 14\tfrac{5}{8} inches long. His board is 30 inches long. Ignoring the wood lost to the saw, is the board long enough?

Solution

To the nearest half, \tfrac{4}{9} is near \tfrac{1}{2} and \tfrac{5}{8} is near \tfrac{1}{2}. The estimate is

15\tfrac{1}{2} + 14\tfrac{1}{2} = 30

That is exactly the length of the board, so the estimate cannot decide. The exact total could be a hair under 30 or a hair over.

Look closer at how far each fraction is from \tfrac{1}{2}. Half of 9 ninths is 4\tfrac{1}{2} ninths, so \tfrac{4}{9} is half of one ninth short of \tfrac{1}{2}, which is \tfrac{1}{18} short. Half of 8 eighths is 4 eighths, so \tfrac{5}{8} is \tfrac{1}{8} over \tfrac{1}{2}. One fraction is short by \tfrac{1}{18}, the other is over by \tfrac{1}{8}. An eighth is a bigger piece than an eighteenth, since cutting a whole into 8 parts makes bigger parts than cutting it into 18. So the overshoot wins, and the exact total is more than 30 inches. The board is too short.

The exact total is 30\tfrac{5}{72} inches, just over 30, as the comparison said.

Boundaries

The benchmarks 0, \tfrac{1}{2} and 1 suit fractions between 0 and 1. For a fraction larger than 1, such as \tfrac{17}{8}, first write it as a mixed number, 2\tfrac{1}{8}, then round the fraction part.
A fraction exactly halfway between two benchmarks, such as \tfrac{1}{4} or \tfrac{3}{4}, is equally near both. Either choice is a fair estimate; say which one you used.
Each fraction can move up to \tfrac{1}{4} to reach the nearest benchmark, so an estimate built from several fractions can drift: \tfrac{1}{4} for each fraction, at most. With many terms, the estimate may be too rough to be useful.
An estimate can show an answer is wrong, never that it is right. A reasonable answer still needs an exact check when exactness matters, as with a cut that must fit.
When the estimate equals the very value a decision turns on (a board length, a limit), it cannot decide. A closer look or the exact value is needed.

Where this goes next

Exact sums and differences of fractions with different denominators come in Adding and subtracting fractions with unlike denominators, and every exact answer there can be held up against a benchmark estimate first. Mixed numbers come next in Adding mixed numbers and Subtracting mixed numbers, where rounding to the nearest half or whole is the quick check on a longer calculation. Comparing two fractions exactly, as in Comparing by finding a common denominator, settles the cases an estimate leaves open. The same habit, round first and then compute, runs through all estimation, from Estimating by rounding with whole numbers to the over and under reasoning of Friendly numbers and deliberate over and under estimates.

Exercises

Part 1: rounding to a benchmark

Round each fraction to 0, \tfrac{1}{2} or 1.

\tfrac{2}{11}
\tfrac{7}{12}
\tfrac{13}{15}
\tfrac{5}{11}

Part 1: estimating sums and differences

Estimate each sum or difference using the benchmarks 0, \tfrac{1}{2} and 1.

\tfrac{8}{9} + \tfrac{6}{13}
\tfrac{1}{10} + \tfrac{6}{11}
\tfrac{12}{13} + \tfrac{14}{15}
\tfrac{10}{11} - \tfrac{7}{15}
\tfrac{11}{20} - \tfrac{1}{16}
\tfrac{3}{5} + \tfrac{11}{13} + \tfrac{1}{15}

Part 1: in the shop

Bhavin glues a piece \tfrac{15}{16} inch thick to a piece \tfrac{3}{8} inch thick. About how thick is the result?
Tunde planes \tfrac{4}{11} inch off a board \tfrac{17}{18} inch thick. About how thick is the board now?

Part 2: mixed numbers

Round 6\tfrac{3}{10} to the nearest half and to the nearest whole number.
Round 8\tfrac{5}{7} to the nearest half and to the nearest whole number.
Estimate 12\tfrac{1}{8} + 9\tfrac{14}{15} by rounding to the nearest half.
Estimate 20\tfrac{3}{5} - 7\tfrac{1}{6} by rounding to the nearest half.
Emiko cuts 10\tfrac{2}{7} inches off a board 48\tfrac{1}{16} inches long. Estimate, to the nearest half, how much board is left.

Part 2: is it reasonable?

Is \tfrac{3}{5} + \tfrac{1}{3} = \tfrac{4}{8} reasonable? Explain without finding the exact sum.
Is \tfrac{9}{14} - \tfrac{1}{2} = \tfrac{1}{7} reasonable?
Is 4\tfrac{1}{5} + 2\tfrac{5}{6} = 6\tfrac{1}{30} reasonable?
Is 7\tfrac{1}{9} - 3\tfrac{2}{3} = 3\tfrac{4}{9} reasonable?

Part 2: when an estimate cannot decide, and why it works

Bhavin needs pieces 20\tfrac{3}{8} inches and 9\tfrac{6}{11} inches long from a 30 inch board. Ignoring saw waste, is the board long enough? Estimate first, then decide.
Emiko lays a strip \tfrac{2}{5} inch wide beside a strip \tfrac{1}{2} inch wide in a groove 1 inch wide. Her estimate of the total width is 1 inch. Does that tell her whether the strips fit? Decide whether they fit.
Explain why rounding each of two fractions between 0 and 1 to the nearest of 0, \tfrac{1}{2} and 1 always gives an estimate within \tfrac{1}{2} of the exact sum.
Tunde says that an answer matching its estimate must be correct. Use the claim \tfrac{6}{11} + \tfrac{7}{16} = 1 to show he is wrong.

Answers

0. The top 2 is closer to 0 than to 5\tfrac{1}{2}, half of 11.
\tfrac{1}{2}. Half of 12 is 6, and 7 is 1 away from it but 5 away from 12.
1. The top 13 is 2 short of 15 and far from 7\tfrac{1}{2}.
\tfrac{1}{2}. Half of 11 is 5\tfrac{1}{2}, and 5 is only \tfrac{1}{2} away.
1\tfrac{1}{2}. \tfrac{8}{9} is near 1 and \tfrac{6}{13} is near \tfrac{1}{2} (half of 13 is 6\tfrac{1}{2}), so 1 + \tfrac{1}{2} = 1\tfrac{1}{2}.
\tfrac{1}{2}. \tfrac{1}{10} is near 0 and \tfrac{6}{11} is near \tfrac{1}{2}, so 0 + \tfrac{1}{2} = \tfrac{1}{2}.
2. Both are 1 short of their bottoms, so each is near 1, and 1 + 1 = 2.
\tfrac{1}{2}. \tfrac{10}{11} is near 1 and \tfrac{7}{15} is near \tfrac{1}{2} (half of 15 is 7\tfrac{1}{2}), so 1 - \tfrac{1}{2} = \tfrac{1}{2}.
\tfrac{1}{2}. \tfrac{11}{20} is near \tfrac{1}{2} (half of 20 is 10) and \tfrac{1}{16} is near 0, so \tfrac{1}{2} - 0 = \tfrac{1}{2}.
1\tfrac{1}{2}. \tfrac{3}{5} is near \tfrac{1}{2} (3 is close to 2\tfrac{1}{2}), \tfrac{11}{13} is near 1, and \tfrac{1}{15} is near 0: \tfrac{1}{2} + 1 + 0 = 1\tfrac{1}{2}.
About 1\tfrac{1}{2} inches. \tfrac{15}{16} is near 1, and \tfrac{3}{8} is near \tfrac{1}{2} because 3 is closer to 4 than to 0.
About \tfrac{1}{2} inch. \tfrac{17}{18} is near 1 and \tfrac{4}{11} is near \tfrac{1}{2} (4 is closer to 5\tfrac{1}{2} than to 0), so 1 - \tfrac{1}{2} = \tfrac{1}{2}.
Nearest half 6\tfrac{1}{2}; nearest whole 6. In \tfrac{3}{10}, 3 is closer to 5 than to 0, but \tfrac{3}{10} is less than \tfrac{1}{2}, so it rounds down to the whole.
Nearest half 8\tfrac{1}{2}; nearest whole 9. \tfrac{5}{7} is closer to \tfrac{1}{2} than to 1 (5 is 1\tfrac{1}{2} from 3\tfrac{1}{2} and 2 from 7), and it is more than \tfrac{1}{2}, so it rounds up to the whole.
22. 12\tfrac{1}{8} is near 12 and 9\tfrac{14}{15} is near 10, so 12 + 10 = 22.
13\tfrac{1}{2}. 20\tfrac{3}{5} is near 20\tfrac{1}{2} and 7\tfrac{1}{6} is near 7 (1 is closer to 0 than to 3), so 20\tfrac{1}{2} - 7 = 13\tfrac{1}{2}.
About 37\tfrac{1}{2} inches. 48\tfrac{1}{16} is near 48; in \tfrac{2}{7}, 2 is closer to 3\tfrac{1}{2} than to 0, so 10\tfrac{2}{7} is near 10\tfrac{1}{2}, and 48 - 10\tfrac{1}{2} = 37\tfrac{1}{2}.
Not reasonable. \tfrac{4}{8} = \tfrac{1}{2}, but \tfrac{3}{5} alone is more than \tfrac{1}{2} (3 is more than 2\tfrac{1}{2}), and adding \tfrac{1}{3} makes the sum larger still. The estimate \tfrac{1}{2} + \tfrac{1}{2} = 1 is nearer; the exact sum is \tfrac{14}{15}.
Reasonable. \tfrac{9}{14} is a little more than \tfrac{1}{2} (half of 14 is 7), so taking away \tfrac{1}{2} leaves a small amount, and \tfrac{1}{7} is small. In fact \tfrac{9}{14} is \tfrac{2}{14} more than \tfrac{7}{14}, and \tfrac{2}{14} = \tfrac{1}{7}, so it is exactly right.
Not reasonable. 4\tfrac{1}{5} is near 4 and 2\tfrac{5}{6} is near 3, so the sum is near 7. The answer 6\tfrac{1}{30} is about 6, a whole off. The exact sum is 7\tfrac{1}{30}.
Reasonable. 7\tfrac{1}{9} is near 7 and 3\tfrac{2}{3} is near 3\tfrac{1}{2}, so the estimate is 3\tfrac{1}{2}, and 3\tfrac{4}{9} is very close to it. (It is also exact.)
Yes. Estimate: 20\tfrac{1}{2} + 9\tfrac{1}{2} = 30, which cannot decide. Closer: \tfrac{3}{8} is \tfrac{1}{8} short of \tfrac{4}{8} = \tfrac{1}{2}, and \tfrac{6}{11} is \tfrac{1}{22} over \tfrac{1}{2} (half of 11 is 5\tfrac{1}{2}, and 6 is half a piece more). An eighth is bigger than a twenty-second, so the total is under 30. Exactly, it is 29\tfrac{81}{88} inches.
The estimate cannot tell her, since it equals the groove width. But \tfrac{2}{5} is less than \tfrac{1}{2} (2 is less than 2\tfrac{1}{2}), so the total is less than \tfrac{1}{2} + \tfrac{1}{2} = 1. They fit; the exact width is \tfrac{9}{10} inch.
Every fraction between 0 and 1 lies within \tfrac{1}{4} of one of 0, \tfrac{1}{2}, 1, because those three split the line into pieces \tfrac{1}{2} long, and no point is more than half a piece from an end. Each of the two fractions moves at most \tfrac{1}{4}, so the sum moves at most \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{1}{2}.
The estimate is \tfrac{1}{2} + \tfrac{1}{2} = 1, matching the claim. But \tfrac{7}{16} is \tfrac{1}{16} short of \tfrac{8}{16} = \tfrac{1}{2}, while \tfrac{6}{11} is only \tfrac{1}{22} over \tfrac{1}{2}. The shortfall is bigger, so the sum is less than 1. Exactly, it is \tfrac{173}{176}.