How to Tell Which Fraction Is Bigger

Two climbers sit at $\frac{7}{9}$ and $\frac{8}{11}$ of the way up. Eight is more than seven, so the second climber looks ahead, but elevenths are thinner slivers than ninths, so that climber is stacking more pieces that are each smaller. Neither one obviously wins. This lesson gives you three ways to settle a comparison like that, rewriting over a common denominator, cross-multiplying, and leaning on landmarks like $\frac{1}{2}$ , along with a feel for which one is quickest on a given pair.

Problem

North crew:

\frac{5}{8}

of a path. East crew:

\frac{7}{12}

. Rewrite over a common denominator, find which crew leads, then report the gap as a single fraction in simplest form.

Show a hint

A fraction only tells you "how many pieces" once you know the pieces are the same size. Eighths and twelfths are not, so find one denominator that both $8$ and $12$ divide into and rewrite each fraction with it. The smallest one that works is $24$ .
Scale each fraction to twenty-fourths. Multiply $\frac{5}{8}$ by $\frac{3}{3}$ to get $\frac{15}{24}$ , and multiply $\frac{7}{12}$ by $\frac{2}{2}$ to get $\frac{14}{24}$ . Now both are counts of the same twenty-fourths, so the bigger numerator is farther along. Subtract the smaller count from the larger to find the gap.

Show the full solution

Rewrite both over

24

\frac{5}{8} = \frac{5 \times 3}{8 \times 3} = \frac{15}{24}

and

\frac{7}{12} = \frac{7 \times 2}{12 \times 2} = \frac{14}{24}

, so the north crew leads with

15

pieces to

14

. The gap is

\frac{15}{24} - \frac{14}{24} = \frac{1}{24}

, already in simplest form. North is ahead by

1 / 24

of the path. Comparing

5

against

7

directly does not work, since an eighth of the path and a twelfth of the path are different lengths.

Both fractions are laid on one bar sliced into

24

equal cells, so every cell is the same width and comparing turns into plain counting. Rebuilt over

24

, the top strip is

\frac{5}{8} = \frac{15}{24}

with

15

cells filled, and the bottom strip is

\frac{7}{12} = \frac{14}{24}

with

14

filled. The cells line up exactly between the strips, so the longer shaded bar is simply the one with more cells, which makes

\frac{5}{8}

the larger. The single gold cell is the whole difference,

\frac{15}{24} - \frac{14}{24} = \frac{1}{24}

, so

\frac{5}{8}

sits exactly one twenty-fourth past

\frac{7}{12}

Problem

Compare

\frac{4}{7}

and

\frac{5}{8}

via the diagonal products

4 \times 8

and

7 \times 5

. Which is larger, and what is the gap in simplest form?

Show a hint

The two tops are $4 \times 8 = 32$ and $7 \times 5 = 35$ . The larger top wins, because both pieces are now measured in the same $56$ ths. Which fraction does the bigger top belong to?
You have $\frac{32}{56}$ and $\frac{35}{56}$ . The gap is just the difference of the tops over $56$ , so compute $\frac{35 - 32}{56}$ and then reduce.

Show the full solution

Over the shared bottom

56

\frac{4}{7} = \frac{4 \times 8}{56} = \frac{32}{56}

and

\frac{5}{8} = \frac{7 \times 5}{56} = \frac{35}{56}

. Since

35 > 32

\frac{5}{8}

is larger, and the gap is

\frac{35}{56} - \frac{32}{56} = \frac{3}{56}

, already in simplest form. So the gap is

3 / 56

. Those two tops are exactly the diagonal products, which is why the diagonals settle the order without writing the common denominator at all.

Problem

Settle the wall:

\frac{7}{9}

\frac{8}{11}

. Compare

7 \times 11

against

9 \times 8

. Which is higher, and by how much? Gap in simplest form.

Show a hint

Cross-multiplication sends each top across to the other bottom. Compare $7 \times 11$ with $9 \times 8$ , and remember the bigger product belongs to the bigger fraction. That alone tells you which marker is higher.
Once you know $\frac{7}{9}$ is higher, the gap is $\frac{7}{9} - \frac{8}{11}$ . The common denominator is $9 \times 11 = 99$ , so rewrite both fractions over $99$ and subtract the numerators. Then check whether the result can be reduced.

Show the full solution

Cross-multiply.

7 \times 11 = 77

belongs to

\frac{7}{9}

and

9 \times 8 = 72

belongs to

\frac{8}{11}

, so

\frac{7}{9}

is higher. Those products are the numerators over

9 \times 11 = 99

, so the gap is

\frac{77}{99} - \frac{72}{99} = \frac{5}{99}

. Since

99 = 9 \times 11

shares nothing with

5

, that is simplest form. The left marker beats the right by

5 / 99

of the route.

Problem

\frac{6}{11}

\frac{6}{13}

: same tops. Elevenths are fatter than thirteenths, so six fat slices win. Find the gap in simplest form.

Show a hint

When two fractions share the same top, the one with the smaller bottom is the larger fraction, because the pieces being counted are bigger. So compare $\frac{6}{11}$ and $\frac{6}{13}$ by their bottoms alone, then subtract the smaller from the larger.
To subtract, rewrite both over the same bottom $11 \times 13 = 143$ . That turns $\frac{6}{11}$ into $\frac{78}{143}$ and $\frac{6}{13}$ into $\frac{66}{143}$ . Now subtract the tops and check whether the result can be reduced.

Show the full solution

Both tops are

6

, and an eleventh is a fatter piece than a thirteenth, so

\frac{6}{11}

is the larger share. For the gap use

11 \times 13 = 143

, where

\frac{6}{11} = \frac{78}{143}

and

\frac{6}{13} = \frac{66}{143}

, so

\frac{78}{143} - \frac{66}{143} = \frac{12}{143}

, already in simplest form. The lead is

12 / 143

. The denominators alone settled the direction, and the common bottom was only needed to measure the gap.

Problem

Blue

\frac{4}{15}

, red

\frac{3}{10}

. Match tops to

12

\frac{12}{45}

\frac{12}{40}

. Smaller bottom wins. Which share is larger, and by how much? Gap in simplest form.

Show a hint

You already have the matched-top forms $\frac{12}{45}$ and $\frac{12}{40}$ . Same top, so smaller bottom wins. Which bottom is smaller, $45$ or $40$ ? That tells you the larger fraction. Once you know the order, the gap is just the larger fraction minus the smaller one.
The larger share is $\frac{3}{10}$ . To subtract $\frac{3}{10} - \frac{4}{15}$ , find a common bottom for $10$ and $15$ , which is $30$ . Rewrite both as thirtieths, then subtract the tops. Simplify if you can.

Show the full solution

With the tops matched,

\frac{4}{15} = \frac{12}{45}

and

\frac{3}{10} = \frac{12}{40}

. Both hold

12

pieces, and cutting into

40

parts makes bigger pieces than cutting into

45

, so the red ribbon

\frac{3}{10}

is the larger share. Over

30

\frac{3}{10} = \frac{9}{30}

and

\frac{4}{15} = \frac{8}{30}

, so the gap is

\frac{9}{30} - \frac{8}{30} = \frac{1}{30}

. Red leads by

1 / 30

of a roll. With matched tops the smaller denominator wins, the reverse of the matched-bottom rule.

Problem

Tank 1:

\frac{8}{9}

(missing

\frac{1}{9}

). Tank 2:

\frac{11}{12}

(missing

\frac{1}{12}

). Smaller gap means fuller. Which is fuller, and by how much? Simplest form.

Show a hint

The empty space in the first tank is $\frac{1}{9}$ , and in the second it is $\frac{1}{12}$ . With the same numerator of 1, the fraction with the bigger bottom number is the smaller piece. So which tank has less empty room, and is therefore fuller?
The fuller tank is the $\frac{11}{12}$ one, since $\frac{1}{12}$ of emptiness is less than $\frac{1}{9}$ . The gap in fill levels is exactly the difference between the two empty pieces, $\frac{1}{9} - \frac{1}{12}$ . Rewrite both over a common bottom of 36 and subtract.

Show the full solution

Each tank is one piece short of full,

\frac{1}{9}

for the first and

\frac{1}{12}

for the second. A twelfth is a thinner sliver than a ninth, so tank 2 has less empty room and

\frac{11}{12} > \frac{8}{9}

. Both missing pieces are measured down from the same full mark, so the fill levels differ by

\frac{1}{9} - \frac{1}{12} = \frac{4}{36} - \frac{3}{36} = \frac{1}{36}

. Tank 2 is fuller by

1 / 36

of a tank.

Both bars run to the same finish line at

1

, so do not bother measuring the long full part. Measure the short empty part instead. The top bar is missing

\frac{1}{9}

and the bottom bar is missing only

\frac{1}{12}

, and lined up against the finish line that thinner gap is plain to see. A ninth of the bar is wider than a twelfth of it, since cutting into more pieces makes each piece smaller. The bar with the smaller gap is the fuller one, so

\frac{11}{12}

wins.

Problem

Bars

\frac{13}{14}

and

\frac{14}{15}

: gaps

\frac{1}{14}

and

\frac{1}{15}

. Smaller gap is closer to done. Report the gap between the two bars in simplest form.

Show a hint

A bar is further along when the piece still left to fill is smaller. Compare the leftover pieces $\frac{1}{14}$ and $\frac{1}{15}$ first, since two unit fractions are easy to rank by their denominators.
The distance between the two bars is exactly the distance between their leftover pieces, because both bars are measured down from the same finish line. So compute $\frac{1}{14} - \frac{1}{15}$ . Over a common denominator of $210$ that is $\frac{15}{210} - \frac{14}{210}$ .

Show the full solution

The top bar has

\frac{1}{14}

left to fill and the bottom has

\frac{1}{15}

. A fifteenth is the smaller piece, so the

\frac{14}{15}

bar is closer to done. Both leftovers are measured down from the same finish line, so the distance between the bars is

\frac{1}{14} - \frac{1}{15} = \frac{15 - 14}{14 \times 15} = \frac{1}{210}

. The bars are

1 / 210

apart. Subtracting the unit fractions this way makes the big denominator appear once, at the end, instead of at the start.

Problem

Six scores:

\frac{5}{9}, \frac{4}{9}, \frac{7}{15}, \frac{9}{16}, \frac{11}{20}, \frac{8}{17}

. Beat

\frac{1}{2}

when top

>

half bottom. How many qualify?

Show a hint

Take one score at a time. For $\frac{5}{9}$ , half of the bottom $9$ is $\frac{9}{2}$ , which sits between $4$ and $5$ . Is the top, $5$ , bigger than that halfway mark? If yes, the score beats $\frac{1}{2}$ . Do the same quick check for the other five.
For an odd bottom like $9$ , $15$ , or $17$ , the top must reach at least the next whole number above half. Half of $9$ is $4 \frac{1}{2}$ , so the top needs to be $5$ or more. For an even bottom like $16$ or $20$ , just double the top and compare to the bottom. Mark each score win or lose, then count the wins.

Show the full solution

A score beats

\frac{1}{2}

when its top is more than half its bottom.

\frac{5}{9}

wins since

5 > 4 \frac{1}{2}

\frac{4}{9}

loses.

\frac{7}{15}

loses since

7 < 7 \frac{1}{2}

\frac{9}{16}

wins since

9 > 8

\frac{11}{20}

wins since

11 > 10

\frac{8}{17}

loses since

8 < 8 \frac{1}{2}

. That is

3

qualifying scores. Half of a whole split into

b

parts is

\frac{b}{2}

parts, so one comparison per score does the job with no common denominators.

Problem

Phone 1:

\frac{7}{13}

. Phone 2:

\frac{9}{20}

. Use

\frac{1}{2}

as landmark, confirm with cross-multiply. By how much is the more-charged phone ahead? Simplest form.

Show a hint

Twice 7 is 14, which clears 13, so $\frac{7}{13}$ sits just above $\frac{1}{2}$ . Twice 9 is 18, which falls short of 20, so $\frac{9}{20}$ sits just below $\frac{1}{2}$ . The one above half is the larger charge. For the exact gap you still need a common denominator.
The winner is $\frac{7}{13}$ . To subtract, put both over $13 \times 20 = 260$ . That turns $\frac{7}{13}$ into $\frac{140}{260}$ and $\frac{9}{20}$ into $\frac{117}{260}$ . Subtract the tops and check whether the result reduces.

Show the full solution

Twice

7

14

, which clears

13

, so

\frac{7}{13}

sits above half. Twice

9

18

, short of

20

, so

\frac{9}{20}

sits below half. Phone 1 is the more charged one. Cross-multiplying agrees,

7 \times 20 = 140

against

9 \times 13 = 117

, and those are the numerators over

13 \times 20 = 260

, so the gap is

\frac{140}{260} - \frac{117}{260} = \frac{23}{260}

. Since

23

is prime and

260 = 2^{2} \times 5 \times 13

, that is lowest terms. Phone 1 leads by

23 / 260

Problem

Temperatures

- \frac{3}{5}

and

- \frac{5}{8}

: both left of

0

. Warmer is nearer

0

. Which is the larger value, and how far apart are they? Positive fraction in simplest form.

Show a hint

Forget the signs for a moment and just compare the sizes $\frac{3}{5}$ and $\frac{5}{8}$ . A common denominator of $40$ lets you stack them up. Whichever size is bigger is the morning that dipped deeper below the baseline, and a deeper dip means a value farther from $0$ .
Rewrite both sizes over $40$ . You get $\frac{3}{5} = \frac{24}{40}$ and $\frac{5}{8} = \frac{25}{40}$ , so $\frac{5}{8}$ is the bigger size and $- \frac{5}{8}$ is the deeper, more negative reading. The larger value is the shallower one, $- \frac{3}{5}$ . For the gap, subtract the two sizes, $\frac{25}{40} - \frac{24}{40}$ .

Show the full solution

Compare the sizes over

40

\frac{3}{5} = \frac{24}{40}

and

\frac{5}{8} = \frac{25}{40}

, so

- \frac{5}{8}

dips deeper below

0

and

- \frac{3}{5}

is the larger, warmer value. The distance between the readings is the same as the distance between the sizes,

\frac{25}{40} - \frac{24}{40} = \frac{1}{40}

, already in simplest form. The two mornings are

1 / 40

of a degree apart. Below

0

the bigger size is the smaller value, so compare as positives and then flip the verdict.

Problem

Three runners are near the finish. Mara is missing

\frac{1}{6}

of the course and Priya is missing

\frac{1}{12}

, so settle those two by the smaller missing piece. Dev is at

\frac{7}{9}

, so place him with a cross-multiply. Order all three, then report the gap between the leader and the runner in last, in simplest form.

Show a hint

Handle Mara and Priya with the missing piece first. Mara still has $\frac{1}{6}$ of the course to go and Priya has only $\frac{1}{12}$ to go. A smaller piece left means more finished, so one of them is clearly ahead of the other. For Dev, compare $\frac{7}{9}$ against the others with a cross-multiply rather than guessing.
Compare Dev to Mara by cross-multiplying $\frac{7}{9}$ and $\frac{5}{6}$ . You get $7 \times 6 = 42$ on Dev's side and $5 \times 9 = 45$ on Mara's side, so $\frac{7}{9} < \frac{5}{6}$ . Together with the missing-piece result, the order from least to most is Dev, then Mara, then Priya. Now the gap you want is leader minus last, which is $\frac{11}{12} - \frac{7}{9}$ . Rewrite both over $36$ and subtract.

Show the full solution

Priya is

\frac{1}{12}

short of the finish and Mara is

\frac{1}{6}

short, and the smaller leftover means farther along, so Priya is ahead of Mara. For Dev, cross-multiply

\frac{7}{9}

against Mara's

\frac{5}{6}

, giving

7 \times 6 = 42

and

5 \times 9 = 45

, so Dev trails Mara and therefore Priya too. The order from least to most is Dev, Mara, Priya, so the gap from leader to last is

\frac{11}{12} - \frac{7}{9} = \frac{33}{36} - \frac{28}{36} = \frac{5}{36}

, already in simplest form. That is

5 / 36

Problem

\frac{98}{99}

misses

\frac{1}{99}

, and

\frac{100}{101}

misses

\frac{1}{101}

. Smaller missing piece is larger rate. Which rate wins, and what is the gap in simplest form?

Show a hint

Forget the giant denominators for a second. Which missing piece is smaller, $\frac{1}{99}$ or $\frac{1}{101}$ ? Fewer equal parts means bigger parts, so $\frac{1}{99}$ is the larger gap. The rate that is missing the larger gap is further from $1$ , and the rate missing the smaller gap is the winner.
The two rates differ by exactly the difference between their missing pieces, $\frac{1}{99} - \frac{1}{101}$ . Over the common denominator $99 \times 101 = 9999$ , that is $\frac{101}{9999} - \frac{99}{9999}$ . Subtract the tops and check whether the result can be simplified.

Show the full solution

The rate

\frac{98}{99}

is short by

\frac{1}{99}

and

\frac{100}{101}

is short by

\frac{1}{101}

. The bigger miss belongs to

\frac{98}{99}

, so

\frac{100}{101}

is the larger rate. The gap between the rates is the gap between the misses,

\frac{1}{99} - \frac{1}{101} = \frac{101}{9999} - \frac{99}{9999} = \frac{2}{9999}

, and

9999 = 9 \times 11 \times 101

is odd, so that is lowest terms. The rates differ by

2 / 9999

Practice these ideas

Practice

Rosa:

\frac{3}{4}

of her wall. Ben:

\frac{5}{7}

of his. Find the common denominator, compare, give the gap as a fraction in simplest form.

Show the solution

Since

4

and

7

share no factor, the shared denominator is

4 \times 7 = 28

. Rosa has

\frac{3}{4} = \frac{21}{28}

and Ben has

\frac{5}{7} = \frac{20}{28}

, so Rosa has painted more, and the gap is

\frac{21}{28} - \frac{20}{28} = \frac{1}{28}

, already in simplest form. Rosa leads by

1 / 28

of a wall.

Practice

Compare

\frac{8}{11}

and

\frac{7}{10}

by cross-multiplying. Which is larger, and how much greater is it? Report the gap as a fraction in simplest form.

Show the solution

The diagonal products are

8 \times 10 = 80

for

\frac{8}{11}

and

11 \times 7 = 77

for

\frac{7}{10}

, so

\frac{8}{11}

is larger. Those same products are the numerators over

11 \times 10 = 110

, so the gap is

\frac{80}{110} - \frac{77}{110} = \frac{3}{110}

, already in simplest form. The larger beats the smaller by

3 / 110

Practice

Find

k

so that

\frac{6}{k} = \frac{15}{20}

. Set diagonal products equal:

6 \times 20 = k \times 15

. What is

k

Show the solution

Set the diagonal products equal,

6 \times 20 = k \times 15

, so

120 = 15 k

and

k = 120 \div 15 = 8

. Quick check,

\frac{6}{8}

and

\frac{15}{20}

both reduce to

\frac{3}{4}

Practice

Compare

\frac{8}{15}

and

\frac{8}{13}

: same numerator, different denominators. Which is larger? Report how much more that person has as a fraction in simplest form.

Show the solution

Both tops are

8

, and thirteenths are bigger pieces than fifteenths, so

\frac{8}{13} > \frac{8}{15}

and the person with the 13-strand ribbon has more. Over

15 \times 13 = 195

\frac{8}{13} = \frac{120}{195}

and

\frac{8}{15} = \frac{104}{195}

, so the gap is

\frac{120}{195} - \frac{104}{195} = \frac{16}{195}

. Since

16 = 2 \times 2 \times 2 \times 2

and

195 = 3 \times 5 \times 13

, that is already in simplest form, so the gap is

16 / 195

Practice

\frac{6}{35}

and

\frac{9}{40}

each rewritten with numerator

18

\frac{18}{105}

\frac{18}{80}

. Smaller bottom wins. Gap between them in simplest form?

Show the solution

Both matched forms have a top of

18

, and

80

parts are bigger than

105

parts, so

\frac{18}{80} > \frac{18}{105}

and the second recipe,

\frac{9}{40}

, uses more. Both

40

and

35

divide

280

, so

\frac{9}{40} = \frac{63}{280}

and

\frac{6}{35} = \frac{48}{280}

, and the gap is

\frac{63}{280} - \frac{48}{280} = \frac{15}{280} = \frac{3}{56}

after cancelling a

5

. The second recipe uses

3 / 56

more of the jar.

Practice

Scores:

\frac{7}{12}

\frac{5}{11}

\frac{10}{19}

\frac{6}{13}

\frac{9}{17}

. Beat

\frac{1}{2}

when top

>

half the bottom. How many exceed $\frac{1}{2}$ ?

Show the solution

Double the top and compare it with the bottom.

\frac{7}{12}

gives

14 > 12

, a win.

\frac{5}{11}

gives

10 < 11

, a loss.

\frac{10}{19}

gives

20 > 19

, a win.

\frac{6}{13}

gives

12 < 13

, a loss.

\frac{9}{17}

gives

18 > 17

, a win. That is

3

scores above

\frac{1}{2}

. Doubling works because

\frac{1}{2}

is exactly the case where the top equals half the bottom.

Practice

Green trail checkpoint:

\frac{8}{15}

. Orange:

\frac{7}{16}

. Use

\frac{1}{2}

as a landmark to decide which is farther, then find the gap. Give your answer as a fraction in simplest form.

Show the solution

Twice

8

16

, which clears

15

, so

\frac{8}{15} > \frac{1}{2}

. Twice

7

14

, short of

16

, so

\frac{7}{16} < \frac{1}{2}

. Green is the farther checkpoint. For the gap use

15 \times 16 = 240

, where

\frac{8}{15} = \frac{128}{240}

and

\frac{7}{16} = \frac{105}{240}

, so

\frac{128}{240} - \frac{105}{240} = \frac{23}{240}

. Since

23

is prime and does not divide

240

, the gap is

23 / 240

Practice

Cup 1:

\frac{9}{10}

full (gap

\frac{1}{10}

). Cup 2:

\frac{10}{11}

full (gap

\frac{1}{11}

). Smaller gap means fuller. Find the gap between fill levels in simplest form.

Show the solution

Cup 1 leaves

\frac{1}{10}

empty and cup 2 leaves

\frac{1}{11}

. The bigger bottom makes the smaller piece, so cup 2 has less empty space and

\frac{10}{11}

is the fuller one. Both gaps are measured down from the same brim, so the levels differ by

\frac{1}{10} - \frac{1}{11} = \frac{11}{110} - \frac{10}{110} = \frac{1}{110}

. The two levels are

1 / 110

of a cup apart.

Practice

Survey 1: missing

\frac{1}{50}

. Survey 2: missing

\frac{1}{51}

. Smaller missing piece is further along. Gap between completion fractions in simplest form?

Show the solution

Survey 1 is

\frac{1}{50}

short of done and survey 2 is

\frac{1}{51}

short. A fifty-first is the smaller leftover, so survey 2 is further along. Both leftovers are measured from the same finished whole, so the completion fractions differ by

\frac{1}{50} - \frac{1}{51} = \frac{51}{2550} - \frac{50}{2550} = \frac{1}{2550}

, already in simplest form. Survey 2 leads by

1 / 2550

Practice

Temperatures

- \frac{5}{9}

and

- \frac{4}{7}

. Warmer is closer to

0

. Which reading is warmer? Report how far apart the readings are as a positive fraction in simplest form.

Show the solution

Compare the sizes over

9 \times 7 = 63

\frac{5}{9} = \frac{35}{63}

and

\frac{4}{7} = \frac{36}{63}

, so

\frac{4}{7}

is the bigger size, which puts

- \frac{4}{7}

farther below

0

and makes

- \frac{5}{9}

the warmer reading. The distance between them is the difference of the sizes,

\frac{36}{63} - \frac{35}{63} = \frac{1}{63}

. The readings are

1 / 63

apart.

Practice

Four sprinters:

\frac{7}{12}

\frac{5}{8}

\frac{11}{18}

\frac{2}{3}

. Use the fastest comparison tool for each pair. Write the single largest fraction in its original form.

Show the solution

All four denominators divide

72

. Rewriting,

\frac{7}{12} = \frac{42}{72}

\frac{5}{8} = \frac{45}{72}

\frac{11}{18} = \frac{44}{72}

, and

\frac{2}{3} = \frac{48}{72}

. The pieces are now the same size, so the biggest numerator wins. That is

48

, which came from

2 / 3

Practice

Sand: gains

\frac{3}{4}

, loses

\frac{1}{3}

. Find what remains, compare to

\frac{1}{2}

. By how much does it differ from

\frac{1}{2}

? Simplest form.

Show the solution

Over

12

, the leftover sand is

\frac{3}{4} - \frac{1}{3} = \frac{9}{12} - \frac{4}{12} = \frac{5}{12}

of a scoop. Half a scoop is

\frac{6}{12}

, so what remains is a little less than half, short by

\frac{6}{12} - \frac{5}{12} = \frac{1}{12}

. The sand misses the halfway mark by

1 / 12

of a scoop.

Practice

Maya:

\frac{13}{16}

of her goal. Theo:

\frac{17}{20}

. Both past halfway, so use cross-multiplication. By what fraction, in simplest form, is the leader ahead?

Show the solution

Both

16

and

20

divide

80

. Maya has

\frac{13}{16} = \frac{65}{80}

and Theo has

\frac{17}{20} = \frac{68}{80}

, so Theo leads. The gap is

\frac{68}{80} - \frac{65}{80} = \frac{3}{80}

, and since

80 = 2 \times 2 \times 2 \times 2 \times 5

has no factor of

3

, that is already in simplest form. Theo is ahead by

3 / 80

Practice

Mosaic 1:

\frac{123}{124}

done (missing

\frac{1}{124}

). Mosaic 2:

\frac{199}{200}

(missing

\frac{1}{200}

). Smaller missing piece wins. Report the gap in simplest form.

Show the solution

Each mosaic is one tile from done,

\frac{1}{124}

missing for the first and

\frac{1}{200}

for the second. The bigger denominator makes the smaller hole, so mosaic 2 is the more complete one. The completions differ by the difference of the holes. Here

124 = 4 \times 31

and

200 = 8 \times 25

share only a factor of

4

, so the least common denominator is

\frac{124 \times 200}{4} = 6200

, and

\frac{1}{124} - \frac{1}{200} = \frac{50}{6200} - \frac{31}{6200} = \frac{19}{6200}

, lowest terms since

19

is prime. Mosaic 2 leads by

19 / 6200

Comparing Fractions

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