QuanticaPrealgebraOpen in the courseA robot at \(0\) steps a third of the way to \(1\), then half of the way, and stops past the middle. It is tempting to add the tops and add the bottoms and read off the landing spot, but the two steps are different lengths, and that breaks the shortcut. Adding fractions is stepping right along the line and subtracting is stepping left. Either way the counting only works when both steps are built from the same size piece, so when they are not, you rebuild them until they are.
Problem
A counter on a strip notched into ninths slides right \(2\) notches, then \(5\) more. Every notch is the same width. What is \(\tfrac{2}{9}+\tfrac{5}{9}\), written as a fraction?
Show a hint
- The two slides are made of the same ninth-sized notches, so you never have to change the notch size partway through. You only need to count how many of those equal notches the counter has crossed by the end.
- Add the number of notches, 2 then 5, to get the total notches crossed, and keep the notch size the same. That total goes on top, and the 9 that names the notch size stays on the bottom.
Show the full solution
The notches are all the same width, so just count them. Two notches then five more is \(2+5=7\) notches, and each notch is \(\tfrac{1}{9}\) of the strip. So \(\tfrac{2}{9}+\tfrac{5}{9}=\tfrac{7}{9}\), which is \(\boxed{7/9}\). The 9 on the bottom names the size of one notch, and that size does not change as the counter slides, so only the count on top moves.
Problem
A marker at \(\tfrac{7}{8}\) steps left by \(\tfrac{3}{8}\). Count remaining eighth-notches, then reduce. What is \(\tfrac{7}{8}-\tfrac{3}{8}\) in simplest form?
Show a hint
- Both fractions are built from the same eighth-sized notches, so subtracting them is really just counting notches. Start at 7 notches and step back 3. How many notches remain, and over what bottom number?
- Seven eighths minus three eighths leaves \(\tfrac{4}{8}\). Now do not stop there. The top is 4 and the bottom is 8, and they share a factor of 4. Divide both by 4 and write what you get.
Show the full solution
Both fractions count eighth-sized notches, so start at 7 notches and step back 3. That leaves \(7-3=4\) notches, so \(\tfrac{7}{8}-\tfrac{3}{8}=\tfrac{4}{8}\). Top and bottom both divide by 4, which gives \(\boxed{1/2}\). A sum or difference often is not in simplest form yet, so check for a shared factor before you stop. Landing on the fourth notch out of eight is the halfway point, which confirms it.
Problem
A friend adds \(\tfrac{1}{3}+\tfrac{1}{2}=\tfrac{2}{5}\) (tops and bottoms). But \(\tfrac{2}{5}<\tfrac{1}{2}\), less than one step. Find the correct sum \(\tfrac{1}{3}+\tfrac{1}{2}\) over the lcd.
Show a hint
- The two fractions are different sizes, so you cannot add them yet. Find a denominator that both 3 and 2 fit into evenly. The smallest one is the lcd of 2 and 3.
- The lcd is \(6\). Rename each fraction over 6, since the pieces have to match before you count them: \(\tfrac{1}{3}=\tfrac{2}{6}\) and \(\tfrac{1}{2}=\tfrac{3}{6}\). Now the denominators agree, so keep the \(6\) on the bottom and add only the top numbers.
Show the full solution
The lcd of \(2\) and \(3\) is \(6\), so rewrite both over sixths. \(\tfrac{1}{3}=\tfrac{2}{6}\) and \(\tfrac{1}{2}=\tfrac{3}{6}\). Now the steps are the same size, so add the counts, \(2+3=5\), giving \(\tfrac{2}{6}+\tfrac{3}{6}=\boxed{5/6}\). Stacking tops and bottoms fails because a third and a half are different sizes. You can only add counts of pieces that match.
Problem
A glider burns \(\tfrac{1}{4}\) then \(\tfrac{1}{6}\) of a charge. Find \(\operatorname{lcm}(4,6)\), rewrite, add. How much burned total? Write in simplest form.
Show a hint
- You can only add numerators once the bottoms match, so first find the lcd of \(4\) and \(6\). The smallest number both divide into is \(12\). Now think about what each fraction becomes when its denominator is \(12\).
- Use the build-up move from 4.4. Since \(4\times 3=12\), multiply top and bottom of \(\tfrac{1}{4}\) by \(3\) to get \(\tfrac{3}{12}\). Since \(6\times 2=12\), multiply top and bottom of \(\tfrac{1}{6}\) by \(2\) to get \(\tfrac{2}{12}\). Add the numerators over the common bottom \(12\), then check whether the result can be reduced.
Show the full solution
The lcd of \(4\) and \(6\) is \(12\). Since \(4\times 3=12\), \(\tfrac{1}{4}=\tfrac{3}{12}\), and since \(6\times 2=12\), \(\tfrac{1}{6}=\tfrac{2}{12}\). Both are twelfths now, so add the tops, \(3+2=5\). The glider burned \(\boxed{5/12}\) of a charge. Since \(\gcd(5,12)=1\), that is already in simplest form.
Nothing about subtraction is new. If adding is a step right along the line, subtracting is a step left, and the rule about sizes is the same whichever way you walk. You still cannot count pieces until they match, so a difference takes the same three moves as a sum. Match the sizes, count the pieces, simplify. The next problem is \(\tfrac{5}{6}-\tfrac{3}{8}\).
Problem
Timer is \(\tfrac{5}{6}\) used, and a shortcut saves \(\tfrac{3}{8}\). Find \(\operatorname{lcm}(6,8)\), rename, subtract. Find \(\tfrac{5}{6}-\tfrac{3}{8}\) in simplest form.
Show a hint
- The two slices are different sizes, so first give them a common bottom. List multiples of \(6\) and of \(8\) and find the smallest number both reach. That smallest shared multiple is your lcd, and every sixth and every eighth can be re-cut into pieces of that size.
- The lcd is \(24\). Rename \(\tfrac{5}{6}=\tfrac{20}{24}\) since \(6\times 4=24\) and \(5\times 4=20\), and rename \(\tfrac{3}{8}=\tfrac{9}{24}\) since \(8\times 3=24\) and \(3\times 3=9\). Now the bottoms match, so subtract the tops: \(20-9\) over \(24\). Check whether the result can be reduced.
Show the full solution
The multiples of \(8\) are \(8, 16, 24\), and \(24\) is the first one \(6\) also reaches, so the lcd is \(24\). Rename both. Since \(6\times 4=24\), \(\tfrac{5}{6}=\tfrac{20}{24}\), and since \(8\times 3=24\), \(\tfrac{3}{8}=\tfrac{9}{24}\). Subtract the tops, \(20-9=11\), so \(\tfrac{5}{6}-\tfrac{3}{8}=\boxed{11/24}\). The number \(11\) is prime and does not divide \(24\), so the answer is already in lowest terms.
Problem
Mixer pours \(\tfrac{5}{12}\) then \(\tfrac{1}{4}\). The lcd is \(12\). Add the tops, then reduce, since the sum shares a factor. What fraction was poured in simplest form?
Show a hint
- The lcd of \(12\) and \(4\) is \(12\), since \(12\) is already a multiple of \(4\). The \(\tfrac{5}{12}\) is fine as is. Rename \(\tfrac{1}{4}\) so its bottom is \(12\) by multiplying top and bottom by \(3\).
- You should have \(\tfrac{5}{12}+\tfrac{3}{12}=\tfrac{8}{12}\). That is correct but not reduced. Both \(8\) and \(12\) divide by \(4\), so split both by \(4\) to land on your final fraction.
Show the full solution
Since \(12\) is already a multiple of \(4\), the lcd is \(12\) and only \(\tfrac{1}{4}\) needs renaming. Multiply its top and bottom by \(3\) to get \(\tfrac{3}{12}\), then add over the shared bottom, \(\tfrac{5}{12}+\tfrac{3}{12}=\tfrac{8}{12}\). Both \(8\) and \(12\) divide by \(4\), so the mixer poured \(\boxed{2/3}\) of the can. When one denominator is already a multiple of the other, that one is the lcd and only one fraction has to be recut.
Problem
Dial reads \(\tfrac{3}{8}\). Stepping left by \(\tfrac{5}{6}\) crosses \(0\). Common denominator, subtract in order, keep the sign. What is \(\tfrac{3}{8}-\tfrac{5}{6}\) in simplest form?
Show a hint
- The two denominators are \(8\) and \(6\). Their least common denominator is \(24\), since \(24\) is the smallest number both \(8\) and \(6\) divide into. Rewrite each fraction with \(24\) on the bottom before you try to subtract anything.
- Multiply top and bottom of \(\tfrac{3}{8}\) by \(3\) to get \(\tfrac{9}{24}\), and multiply \(\tfrac{5}{6}\) by \(4\) to get \(\tfrac{20}{24}\). Now the subtraction is \(\tfrac{9}{24}-\tfrac{20}{24}\). Since \(9\) is smaller than \(20\), the count \(9-20\) comes out negative. Work out \(9-20\), keep it over \(24\), and check whether the result can be reduced.
Show the full solution
The lcd of \(8\) and \(6\) is \(24\). Rename each fraction, \(\tfrac{3}{8}=\tfrac{9}{24}\) and \(\tfrac{5}{6}=\tfrac{20}{24}\). Taking \(20\) twenty-fourths away from only \(9\) of them drops the count below zero, \(9-20=-11\), so \(\tfrac{3}{8}-\tfrac{5}{6}=\boxed{-11/24}\). Order matters in a subtraction. The dial walks left past \(0\) and stops on the negative side of the same line you used in 4.1.
Problem
Spool holds \(4\) m. Cut \(\tfrac{2}{3}\) m off. Rewrite \(4=\tfrac{12}{3}\), subtract. How much cord remains? Find \(4-\tfrac{2}{3}\) as a single fraction.
Show a hint
- You cannot take thirds away from something that is still written in whole meters. First turn the \(4\) into thirds. Since \(4=\tfrac{4}{1}\) and you want a denominator of \(3\), multiply the top and bottom by \(3\). How many thirds are in 4 whole meters?
- Four whole meters is \(\tfrac{4\times 3}{1\times 3}=\tfrac{12}{3}\), so you have 12 thirds on the spool. Now both amounts are measured in the same size pieces, and the subtraction becomes \(\tfrac{12}{3}-\tfrac{2}{3}\). Keep the denominator and subtract the tops, \(12-2\).
Show the full solution
Write the whole number as a fraction and give it thirds, \(4=\tfrac{4}{1}=\tfrac{4\times 3}{1\times 3}=\tfrac{12}{3}\), since each meter holds \(3\) thirds. Both amounts are in thirds now, so \(\tfrac{12}{3}-\tfrac{2}{3}=\tfrac{10}{3}\), and \(\boxed{10/3}\) of a meter is left. Any whole number is already a fraction over \(1\), which is all you need to rename it over any denominator.
Problem
Spring: \(\tfrac{3}{4}\) planted, path takes \(\tfrac{2}{3}\), \(\tfrac{1}{6}\) replanted. Find lcd for \(4,3,6\), combine in one sweep. What fraction ends up in use?
Show a hint
- The smallest number that \(4\), \(3\), and \(6\) all divide into is \(12\). Rewrite every fraction with \(12\) on the bottom before you touch the top numbers, and keep the plus and minus signs attached to the right pieces.
- Over \(12\) the three fractions are \(\tfrac{9}{12}\), \(\tfrac{8}{12}\), and \(\tfrac{2}{12}\). The expression \(\tfrac{3}{4}-\tfrac{2}{3}+\tfrac{1}{6}\) becomes \(\tfrac{9-8+2}{12}\). Work left to right on top, then simplify the fraction you get.
Show the full solution
All of \(4\), \(3\), and \(6\) divide \(12\), so put every fraction over \(12\). That gives \(\tfrac{3}{4}=\tfrac{9}{12}\), \(\tfrac{2}{3}=\tfrac{8}{12}\), and \(\tfrac{1}{6}=\tfrac{2}{12}\), so the expression becomes $$\frac{9-8+2}{12}=\frac{3}{12}.$$ Divide top and bottom by \(3\), and \(\boxed{1/4}\) of the plot ends up in use. Converting all three at once and then sweeping the top left to right keeps each sign attached to its own number.
Problem
Rider: \(+\tfrac{7}{10}-\tfrac{1}{5}+\tfrac{1}{2}\). lcd for \(10,5,2\) is \(10\). Combine in one pass. What fraction of the loop is her total progress?
Show a hint
- You need one denominator that all three fractions can share. Ask yourself the smallest number that \(10\), \(5\), and \(2\) all divide into evenly. Since \(5\) and \(2\) both already divide \(10\), that shared denominator is just \(10\).
- Rewrite each fraction over \(10\). The first is already \(\tfrac{7}{10}\). For \(\tfrac{1}{5}\), multiply top and bottom by \(2\) to get \(\tfrac{2}{10}\). For \(\tfrac{1}{2}\), multiply top and bottom by \(5\) to get \(\tfrac{5}{10}\). Now combine the numerators in order, keeping the minus sign on the second one, so \(7-2+5\) over \(10\).
Show the full solution
Both \(5\) and \(2\) divide evenly into \(10\), so \(10\) is already the lcd. Rename the other two terms, \(\tfrac{1}{5}=\tfrac{2}{10}\) and \(\tfrac{1}{2}=\tfrac{5}{10}\), then combine the tops in one pass with their signs, \(7-2+5=10\). That is \(\tfrac{10}{10}\), one full loop, so her progress is \(\boxed{1}\). Check the denominators you already have before hunting for a new one, since the largest is often the lcd.
Practice these ideas
Practice
A counter on a sevenths number line steps right \(\tfrac{1}{7}\) then \(\tfrac{3}{7}\). Count the notches. Write \(\tfrac{1}{7}+\tfrac{3}{7}\) in simplest form.
Show the solution
The bottoms already match, so just count notches. One notch then \(3\) more is \(1+3=4\) sevenths, so \(\tfrac{1}{7}+\tfrac{3}{7}=\boxed{4/7}\). Since \(\gcd(4,7)=1\), nothing cancels.
Practice
A counter at \(\tfrac{9}{10}\) slides left by \(\tfrac{3}{10}\). Subtract and reduce. What is \(\tfrac{9}{10}-\tfrac{3}{10}\) in simplest form?
Show the solution
Both fractions count tenths, so the subtraction happens only on top. \(9-3=6\), giving \(\tfrac{9}{10}-\tfrac{3}{10}=\tfrac{6}{10}\). Divide top and bottom by \(2\) to get \(\boxed{3/5}\). Same-size pieces subtract straight across, but the result still needs a simplicity check.
Practice
A classmate wrote \(\tfrac{1}{3}+\tfrac{1}{4}=\tfrac{2}{7}\) by stacking tops and bottoms. But \(\tfrac{2}{7}<\tfrac{1}{3}\), so a sum came out smaller than a part, which is impossible. Find the correct sum over the lcd, in simplest form.
Show the solution
The lcd of \(3\) and \(4\) is \(12\). Rename both, \(\tfrac{1}{3}=\tfrac{4}{12}\) and \(\tfrac{1}{4}=\tfrac{3}{12}\), then add the tops, \(4+3=7\). The correct sum is \(\boxed{7/12}\). Stacking tops and bottoms treats thirds and fourths as the same size, which is how it produced a sum smaller than one of the parts.
Practice
"Sunset" blend: \(\tfrac{2}{5}\) mango, \(\tfrac{1}{3}\) pineapple. Find \(\operatorname{lcm}(5,3)\), rename, add. Total fruit share \(\tfrac{2}{5}+\tfrac{1}{3}\) in simplest form?
Show the solution
The lcd of \(5\) and \(3\) is \(15\). Rename each share, \(\tfrac{2}{5}=\tfrac{6}{15}\) and \(\tfrac{1}{3}=\tfrac{5}{15}\), then add the tops, \(6+5=11\). The blend is \(\boxed{11/15}\) fruit by volume. Since \(11\) is prime and does not divide \(15\), that is already in lowest terms.
Practice
A candle is \(\tfrac{7}{8}\) of its original height, and it burns \(\tfrac{1}{3}\) more. Find the lcd of \(8\) and \(3\), rename, subtract. What is \(\tfrac{7}{8}-\tfrac{1}{3}\) in simplest form?
Show the solution
The denominators \(8\) and \(3\) share no factor, so the lcd is their product, \(8\times 3=24\). Rename both, \(\tfrac{7}{8}=\tfrac{21}{24}\) and \(\tfrac{1}{3}=\tfrac{8}{24}\), then subtract the tops, \(21-8=13\). The candle keeps \(\boxed{13/24}\) of its original height. When two denominators share no common factor, multiplying them always gives the lcd.
Practice
A bottle is \(\tfrac{7}{10}\) full, and the hiker drinks \(\tfrac{1}{2}\) of a bottle. Find the lcd, rename, subtract. What fraction of a bottle is left, in simplest form?
Show the solution
Since \(10\) is already a multiple of \(2\), the lcd is \(10\) and only the half gets renamed, \(\tfrac{1}{2}=\tfrac{5}{10}\). Subtract the tops, \(7-5=2\), giving \(\tfrac{2}{10}\). Both divide by \(2\), so \(\boxed{1/5}\) of a bottle is left. A correct difference is not always reduced, so hunt for a shared factor last.
Practice
A probe is at \(+\tfrac{1}{6}\) then sinks by \(\tfrac{3}{4}\), crossing \(0\). Compute \(\tfrac{1}{6}-\tfrac{3}{4}\) over a common denominator. What is the result, with sign, in simplest form?
Show the solution
The lcd of \(6\) and \(4\) is \(12\). Rename each one, \(\tfrac{1}{6}=\tfrac{2}{12}\) and \(\tfrac{3}{4}=\tfrac{9}{12}\), then subtract the tops, \(2-9=-7\). The reading is \(\boxed{-7/12}\). The probe started only \(2\) twelfths above zero and dropped \(9\), so it ends below the surface and the answer comes out negative.
Practice
A pitcher holds \(5\) cups, and \(\tfrac{2}{3}\) of a cup is added. Rewrite \(5=\tfrac{5}{1}\) over denominator \(3\), then combine. Find \(5+\tfrac{2}{3}\) as a single fraction.
Show the solution
Write the whole number as \(5=\tfrac{5}{1}\), then multiply top and bottom by \(3\) to get \(\tfrac{15}{3}\). Both amounts are thirds now, so add the tops, \(15+2=17\). The pitcher holds \(\boxed{17/3}\) cups. Since \(17\) is prime and shares no factor with \(3\), nothing cancels.
Practice
A ribbon is \(2\) m, and \(\tfrac{3}{8}\) m is cut off. Rewrite \(2=\tfrac{2}{1}\) over eighths, then subtract. What is \(2-\tfrac{3}{8}\) as a single fraction?
Show the solution
Write \(2=\tfrac{2}{1}\), then multiply top and bottom by \(8\) to get \(\tfrac{16}{8}\), because \(16\) eighths really is two whole meters. Subtract the tops, \(16-3=13\), so \(2-\tfrac{3}{8}=\boxed{13/8}\) of a meter. That is a little over one and a half meters, which fits, since only a small piece was cut off.
Practice
Three bursts: \(\tfrac{1}{2}+\tfrac{1}{4}+\tfrac{1}{8}\) of a fuel cell. lcd for \(2,4,8\) is \(8\). Total fraction used, in simplest form?
Show the solution
Both \(2\) and \(4\) divide \(8\), so \(8\) works for all three fractions at once. Rename, \(\tfrac{1}{2}=\tfrac{4}{8}\), \(\tfrac{1}{4}=\tfrac{2}{8}\), and \(\tfrac{1}{8}\) is already there, then add the tops in one sweep, \(4+2+1=7\). The bursts use \(\boxed{7/8}\) of a cell. With three fractions you never combine them in pairs. Put them all over one denominator and add once.
Practice
Kite: \(+\tfrac{1}{2}+\tfrac{1}{3}-\tfrac{1}{5}\). lcd for \(2,3,5\) is \(30\). How far above ground, in simplest form?
Show the solution
The denominators \(2\), \(3\), and \(5\) are primes with nothing in common, so the lcd is their product, \(2\times 3\times 5=30\). Rename each move, $$\tfrac{1}{2}=\tfrac{15}{30}, \qquad \tfrac{1}{3}=\tfrac{10}{30}, \qquad \tfrac{1}{5}=\tfrac{6}{30},$$ then combine the tops with their signs, \(15+10-6=19\). The kite sits \(\boxed{19/30}\) of the way to the marker. Here the lcd is the product of the denominators, not simply the largest one.
Practice
A tank starts \(\tfrac{11}{12}\) full, and pipes draw \(\tfrac{1}{2}\) then \(\tfrac{1}{4}\). Find \(\tfrac{11}{12}-\tfrac{1}{2}-\tfrac{1}{4}\) over the lcd. What fraction remains, in simplest form?
Show the solution
Twelve is a multiple of \(2\) and of \(4\), so put everything over \(12\). The \(\tfrac{11}{12}\) stays, \(\tfrac{1}{2}=\tfrac{6}{12}\), and \(\tfrac{1}{4}=\tfrac{3}{12}\), so the chain reads \(\tfrac{11}{12}-\tfrac{6}{12}-\tfrac{3}{12}\). Sweep the tops in one go, \(11-6-3=2\), giving \(\tfrac{2}{12}\). Divide both by \(2\), so \(\boxed{1/6}\) of the tank remains.