QuanticaPrealgebraOpen in the courseSeven quarter-hops carry you across a creek from \(0\). After \(4\) hops your boot lands on \(1\), and three more put you past \(1\) but short of \(2\). You can call that spot seven quarters, \(\tfrac{7}{4}\), or one whole plank and three quarters more, \(1\tfrac{3}{4}\). Same point, two names. The second name is a mixed number, and this lesson is about writing them and converting between the two forms.
Problem
Your boot is at \(\tfrac{7}{4}\), past \(1\) but short of \(2\). How many whole planks have you completely crossed? Give that count.
Show a hint
- A whole plank only counts as finished if your boot is fully past its far end. You sailed past the mark at \(1\), but you stopped before reaching the mark at \(2\), so the second plank is only partly crossed. How many planks did you finish all the way?
- Count up the quarter-hops. Four quarter-hops carry you across one whole plank to the mark at \(1\). You took seven hops, which is four to reach \(1\) and three more past it. Those three extra hops have not yet reached \(2\), so only the planks behind the mark at \(1\) are truly finished.
Show the full solution
Four quarter-hops make one whole plank, so the marks land at \(0\), \(\tfrac{4}{4}=1\), and \(\tfrac{8}{4}=2\). Seven hops puts your boot at \(\tfrac{7}{4}\), past the mark at \(1\) but short of \(2\), so you have finished \(\boxed{1}\) plank. Those three extra quarter-hops are the \(\tfrac{3}{4}\) in the mixed number \(1\tfrac{3}{4}\).
Problem
Order slip: \(3\tfrac{2}{7}\) bands. Three whole bands = \(21\) beads, then add \(2\) loose. Write \(3\tfrac{2}{7}\) as an improper fraction in sevenths.
Show a hint
- A whole band is not really one thing to the machine, it is 7 beads. So how many single beads hide inside the 3 whole bands before you even touch the loose ones?
- Three whole bands are \(3\times 7=21\) beads. Add the 2 loose beads to get the total bead count, then write that total over 7, since each bead is one seventh of a band.
Show the full solution
Each whole band is 7 beads, so 3 bands hold \(3\times 7=21\) beads. Add the 2 loose beads for \(23\) beads in all, and every bead is one seventh of a band. $$3\tfrac{2}{7}=\frac{3\times 7+2}{7}=\boxed{\dfrac{23}{7}}$$ Going backwards, \(23\div 7\) is 3 with remainder 2, which lands on \(3\tfrac{2}{7}\) again.
Problem
A lookout sits at \(\tfrac{47}{6}\) miles. Convert to a mixed number. Give the nearest post behind the hiker, the largest whole number below \(\tfrac{47}{6}\).
Show a hint
- A mixed number is really a division. How many whole groups of \(6\) fit inside \(47\), and how much is left over? That whole count is the milepost she has already passed.
- Divide \(47\) by \(6\). It goes in \(7\) times because \(6\times 7 = 42\), with \(5\) left over, so \(\frac{47}{6} = 7\tfrac{5}{6}\). The \(7\) is the whole part, so the lookout sits past post \(7\) but not yet at post \(8\).
Show the full solution
Each whole mile is \(\tfrac{6}{6}\), so ask how many \(6\)'s fit into \(47\). Since \(6\times 7=42\) with \(5\) left over, \(\frac{47}{6}=7\tfrac{5}{6}\). She is past post \(7\) and not yet at post \(8\), so the post behind her is \(\boxed{7}\). The whole part of a mixed number is always the largest whole number below it.
Problem
A stretch is \(23\) beads, and every \(5\) beads make one full band (\(\tfrac{23}{5}\)). Divide \(23\) by \(5\): whole bands and leftover. Write the length as a mixed number.
Show a hint
- The bottom number 5 is your group size, so you are really asking how many full fives fit inside 23. Count up in fives, 5, 10, 15, 20, and notice how far you can go before passing 23.
- Divide 23 by 5. The quotient is the number of whole bands, and the remainder is the loose beads left over, which sit on top of 5 in the fraction part.
Show the full solution
\(23\div 5\) gives \(4\) with a remainder of \(3\), since \(5\times 4=20\) and \(23-20=3\). So 23 beads make 4 complete bands with 3 loose beads, and those 3 beads are \(\tfrac{3}{5}\) of a band. $$\frac{23}{5}=\boxed{4\tfrac{3}{5}}$$ The quotient becomes the whole part and the remainder sits over the same bottom number.
Problem
Sensor at \(-7\tfrac{1}{3}\): the minus covers the whole quantity, so the sensor is a third below \(-7\). Give \(-7\tfrac{1}{3}\) as an improper fraction.
Show a hint
- First locate where 7 1/3 lives on the positive side. It sits a third of the way past 7, between 7 and 8. Now reflect that whole spot across 0. Reflecting flips left and right, so whatever was just past 7 going up lands just past -7 going down.
- The minus out front belongs to the entire amount, so -7 1/3 means -(7 1/3) = -(7 + 1/3) = -7 - 1/3. That is a third below -7, landing between -8 and -7. To get the improper fraction, write 7 1/3 = (7 times 3 + 1)/3 = 22/3, then attach the minus.
Show the full solution
The minus covers the whole amount, so \(-7\tfrac{1}{3}=-\left(7+\tfrac{1}{3}\right)\). Convert the positive part, \(7\tfrac{1}{3}=\dfrac{7\times 3+1}{3}=\dfrac{22}{3}\), then attach the minus. $$-7\tfrac{1}{3}=\boxed{-\dfrac{22}{3}}$$ The second sign flips too, so the sensor sits a third below \(-7\), between \(-8\) and \(-7\), not a third above it.
Problem
Potter uses \(2\tfrac{5}{6}+1\tfrac{5}{6}\) scoops. The fraction parts sum past \(1\), so carry a whole. Total scoops as a mixed number?
Show a hint
- Keep the two columns apart for a moment. The whole scoops add to \(2+1=3\). Now add the sixths on their own, \(\tfrac{5}{6}+\tfrac{5}{6}\). That fraction sum is going to be bigger than \(1\), which is the whole point of this problem.
- \(\tfrac{5}{6}+\tfrac{5}{6}=\tfrac{10}{6}\), and \(\tfrac{10}{6}\) is one whole scoop plus \(\tfrac{4}{6}\) left over, which simplifies to \(1\tfrac{2}{3}\). Hand that extra whole scoop up to the whole count. You started with \(3\) wholes, the carry makes it \(4\), and \(\tfrac{2}{3}\) of a scoop stays behind.
Show the full solution
The wholes give \(2+1=3\) and the sixths give \(\tfrac{5}{6}+\tfrac{5}{6}=\tfrac{10}{6}\). Since \(\tfrac{10}{6}=\tfrac{6}{6}+\tfrac{4}{6}=1\tfrac{4}{6}=1\tfrac{2}{3}\), one whole scoop moves over to the whole count, making \(3+1=4\) with \(\tfrac{2}{3}\) left behind. \(\boxed{4\tfrac{2}{3}\text{ scoops}}\) That move is carrying, the same as carrying a ten in column addition, except here a full \(\tfrac{6}{6}\) is what carries.
Problem
Rail is \(5\tfrac{1}{4}\) spans. Cut \(2\tfrac{3}{4}\). Since \(\tfrac{1}{4}<\tfrac{3}{4}\), trade one whole for \(4\) quarters. Finish and give what remains as a mixed number.
Show a hint
- The trade does not change how much rail there is, it just changes how it is written. One whole span is the same as \(\tfrac{4}{4}\), so trading a whole into quarters turns \(5\) wholes and \(\tfrac{1}{4}\) into \(4\) wholes and a bigger pile of quarters. How many quarters are in that pile now?
- After the trade you are looking at \(4\tfrac{5}{4}\) minus \(2\tfrac{3}{4}\). Subtract the quarters first, \(\tfrac{5}{4}-\tfrac{3}{4}\), then subtract the wholes, \(4-2\). Put the two results together and simplify the fraction.
Show the full solution
One whole span is \(\tfrac{4}{4}\), so trade a whole out of the \(5\) and \(5\tfrac{1}{4}\) becomes \(4\tfrac{5}{4}\), the same length written differently. Now the columns subtract, \(\tfrac{5}{4}-\tfrac{3}{4}=\tfrac{2}{4}\) and \(4-2=2\), leaving \(2\tfrac{2}{4}\). Halving top and bottom, $$2\tfrac{2}{4}=\boxed{2\tfrac{1}{2}}$$ spans. Borrowing is carrying run backwards, one whole broken into denominator-many pieces.
Problem
Legs: \(1\tfrac{1}{2}+2\tfrac{1}{3}\) laps. Halves and thirds differ, so find lcd, rename fractions, then add. Check for carry. Total as a mixed number?
Show a hint
- Split each amount into its whole part and its fraction part. The wholes are \(1\) and \(2\), so those give \(3\) laps right away. The fractions \(\tfrac{1}{2}\) and \(\tfrac{1}{3}\) cannot be added as they stand because the pieces are different sizes, so first find a denominator that both \(2\) and \(3\) divide into.
- Use \(6\) as the shared bottom. Rename \(\tfrac{1}{2}=\tfrac{3}{6}\) and \(\tfrac{1}{3}=\tfrac{2}{6}\), then add to get \(\tfrac{3}{6}+\tfrac{2}{6}=\tfrac{5}{6}\). Since \(\tfrac{5}{6}\) is less than one whole lap, there is nothing to carry, so just attach it to the \(3\) whole laps.
Show the full solution
The wholes give \(1+2=3\) laps. The fractions need a shared bottom, and \(6\) is the smallest number both \(2\) and \(3\) divide into, so \(\tfrac{1}{2}=\tfrac{3}{6}\) and \(\tfrac{1}{3}=\tfrac{2}{6}\), which add to \(\tfrac{5}{6}\). That is under one whole lap, so nothing carries, and the total is \(\boxed{3\tfrac{5}{6}}\) laps. As a single improper fraction that is \(\tfrac{23}{6}\), a quick way to check the work.
Problem
\(4\) planters each need \(2\tfrac{5}{8}\) bags. Convert to an improper fraction, multiply by \(4\), convert back. Total bags as a mixed number?
Show a hint
- Rewrite \(2\tfrac{5}{8}\) as a single improper fraction before you touch the \(4\). Eighths times eighths stays in eighths, so the multiply step is clean.
- \(2\tfrac{5}{8}=\tfrac{21}{8}\). Now \(4\times\tfrac{21}{8}=\tfrac{84}{8}\). Simplify that fraction, then split it into a whole number and a remaining fraction.
Show the full solution
\(2\tfrac{5}{8}=\tfrac{16}{8}+\tfrac{5}{8}=\tfrac{21}{8}\), so $$4\times\frac{21}{8}=\frac{84}{8}=\frac{21}{2}.$$ Half of \(21\) is \(10\) with \(1\) left over, so the four boxes hold \(\boxed{10\tfrac{1}{2}}\) bags. Keeping the parts separate is what causes slips here, since \(4\times\tfrac{5}{8}=\tfrac{20}{8}\) is itself more than two whole bags.
Problem
Full day: \(3\tfrac{3}{4}\) km, and a half-staffed day gets \(\tfrac{2}{3}\) of that. Convert, multiply by \(\tfrac{2}{3}\), convert back. How many km, as a mixed number?
Show a hint
- You cannot reliably take \(\tfrac{2}{3}\) of \(3\) and \(\tfrac{2}{3}\) of \(\tfrac{3}{4}\) separately and trust it. Turn \(3\tfrac{3}{4}\) into a single improper fraction first. Three wholes is \(\tfrac{12}{4}\), so \(3\tfrac{3}{4}=\tfrac{15}{4}\). Now you just owe yourself \(\tfrac{2}{3}\) of \(\tfrac{15}{4}\).
- Multiply straight across, top times top and bottom times bottom: \(\tfrac{2}{3}\times\tfrac{15}{4}=\tfrac{2\times 15}{3\times 4}=\tfrac{30}{12}\). Now reduce \(\tfrac{30}{12}\) by its biggest common factor, then split that improper fraction into a whole number and a leftover fraction.
Show the full solution
\(3\tfrac{3}{4}=\tfrac{12}{4}+\tfrac{3}{4}=\tfrac{15}{4}\), so take \(\tfrac{2}{3}\) of it by multiplying straight across, \(\tfrac{2}{3}\times\tfrac{15}{4}=\tfrac{30}{12}=\tfrac{5}{2}\). Five halves is two wholes and one half, so the half-staffed day clears $$\boxed{2\tfrac{1}{2}}$$ kilometres. The multiply rule only works on a plain fraction, so clear the mixed form first and fold it back at the end.
Problem
Ribbon \(4\tfrac{1}{2}\) m, and bows are \(\tfrac{3}{4}\) m each. Convert, divide using flip-and-multiply. How many whole bows?
Show a hint
- Before you can divide, the mixed number has to become a single fraction. Four and a half metres is the same as \(\tfrac{9}{2}\) metres, since two halves make each metre and there are nine halves in all. Now you are dividing \(\tfrac{9}{2}\) by \(\tfrac{3}{4}\).
- Dividing by \(\tfrac{3}{4}\) means multiplying by its flip, \(\tfrac{4}{3}\). So compute \(\tfrac{9}{2}\times\tfrac{4}{3}\). Multiply the tops, multiply the bottoms, then simplify and read off the whole number of pieces.
Show the full solution
\(4\tfrac{1}{2}=\tfrac{9}{2}\), since each whole metre is two halves. Dividing by \(\tfrac{3}{4}\) is the same as multiplying by \(\tfrac{4}{3}\), so $$\frac{9}{2}\div\frac{3}{4}=\frac{9}{2}\times\frac{4}{3}=\frac{36}{6}=\boxed{6}.$$ The count comes out whole, so the ribbon splits into six \(\tfrac{3}{4}\)-metre bows with nothing left over. Checking the other way, \(6\times\tfrac{3}{4}=\tfrac{18}{4}=4\tfrac{1}{2}\) metres.
Problem
A sign maker uses \(1\tfrac{5}{8}\) m of ribbon per letter for a \(4\)-letter word, starting from a \(7\) m roll. Convert, multiply, subtract. How many metres of ribbon are left on the roll?
Show a hint
- Turn \(1\tfrac{5}{8}\) into eighths first, since every letter uses the same amount. Four equal letters means you can multiply that single fraction by \(4\) instead of adding it four times.
- One letter is \(\tfrac{13}{8}\) metres, so four letters use \(4\times\tfrac{13}{8}=\tfrac{52}{8}\) metres. Simplify that, then write \(7\) as a fraction over the same denominator so you can subtract cleanly.
Show the full solution
One letter uses \(1\tfrac{5}{8}=\tfrac{13}{8}\) metres, so four letters use \(4\times\tfrac{13}{8}=\tfrac{52}{8}=\tfrac{13}{2}=6\tfrac{1}{2}\) metres. Write the roll as \(7=\tfrac{14}{2}\) and subtract. $$\frac{14}{2}-\frac{13}{2}=\boxed{\tfrac{1}{2}}\text{ metre}$$ Half a metre is well short of the \(1\tfrac{5}{8}\) another letter would need, so no fifth letter fits.
Practice these ideas
Practice
An anchor leg is logged as \(\tfrac{29}{6}\) of a segment. Rewrite \(\tfrac{29}{6}\) as a mixed number in simplest form.
Show the solution
Each whole is \(\tfrac{6}{6}\), so ask how many \(6\)'s fit into \(29\). Since \(6\times 4=24\) and \(29-24=5\), the quotient \(4\) is the whole part and the remainder \(5\) stays over \(6\). That gives \(\boxed{4\tfrac{5}{6}}\), already in simplest form since \(5\) and \(6\) share no factor.
Practice
A staircase rises \(6\tfrac{2}{5}\) full turns. Rewrite \(6\tfrac{2}{5}\) as a single improper fraction in fifths.
Show the solution
Each full turn is \(\tfrac{5}{5}\), so 6 turns hold \(6\times 5=30\) fifths, and the extra \(\tfrac{2}{5}\) brings the total to \(32\) fifths. So \(6\tfrac{2}{5}=\boxed{\tfrac{32}{5}}\). That is the \(\text{whole}\times\text{bottom}+\text{top}\) shortcut, with the denominator left alone.
Practice
A runner covers \(\tfrac{38}{7}\) laps. \(\tfrac{38}{7}\) falls between two consecutive whole numbers. What is the smaller of those two whole numbers?
Show the solution
Since \(7\times 5=35\) and \(7\times 6=42\), five whole sevens fit inside \(38\) with \(3\) left over, so \(\tfrac{38}{7}=5\tfrac{3}{7}\). That sits between \(5\) and \(6\), and the smaller of those is \(\boxed{5}\). The runner has finished 5 full laps and is partway through her sixth.
Practice
A diver's tag reads \(-3\tfrac{2}{5}\) m: the minus covers the whole amount, so this is \(-3-\tfrac{2}{5}\). Rewrite \(-3\tfrac{2}{5}\) as a single improper fraction.
Show the solution
Convert the positive part first. \(3\tfrac{2}{5}=\dfrac{3\times 5+2}{5}=\dfrac{17}{5}\), and the minus covers the whole amount, so \(-3\tfrac{2}{5}=\boxed{-\tfrac{17}{5}}\). Since \(\tfrac{17}{5}\) sits between \(3\) and \(4\), the negative sits between \(-4\) and \(-3\), about \(-3.4\) metres, which fits a depth reading.
Practice
A flag is planted at \(\frac{23}{4}\) on a number line. What is the largest whole number to the left of the flag?
Show the solution
\(23\div 4=5\) with a remainder of \(3\), so \(\frac{23}{4}=5\tfrac{3}{4}\). The flag lands three quarters past \(5\), so the largest whole number to its left is \(\boxed{5}\).
Practice
A runner starts at a trail marker \(3\tfrac{4}{5}\) km from start, then jogs \(2\tfrac{3}{5}\) km more. How far from start is the runner? Give as a mixed number in simplest form.
Show the solution
The wholes give \(3+2=5\) and the fifths give \(\tfrac{4}{5}+\tfrac{3}{5}=\tfrac{7}{5}\). Seven fifths is more than one whole, so trade it, \(\tfrac{7}{5}=1\tfrac{2}{5}\), and carry that whole across. The wholes become \(6\) with \(\tfrac{2}{5}\) left, so the runner is \(\boxed{6\tfrac{2}{5}}\) kilometers from the start. Improper fractions agree, \(\tfrac{19}{5}+\tfrac{13}{5}=\tfrac{32}{5}\).
Practice
Evaluate \(6\tfrac{1}{3}-2\tfrac{2}{3}\) and write the result as a mixed number. The thirds will not subtract cleanly, so you will need to borrow one whole from the \(6\). What is the value?
Show the solution
\(\tfrac{1}{3}\) is smaller than \(\tfrac{2}{3}\), so borrow. Take one whole from the \(6\), leaving \(5\), and turn it into \(\tfrac{3}{3}\), which makes \(6\tfrac{1}{3}=5\tfrac{4}{3}\). Now the wholes give \(5-2=3\) and the thirds give \(\tfrac{4}{3}-\tfrac{2}{3}=\tfrac{2}{3}\), so the value is \(\boxed{3\tfrac{2}{3}}\). Improper fractions check out, \(\tfrac{19}{3}-\tfrac{8}{3}=\tfrac{11}{3}\).
Practice
Two legs: \(1\tfrac{1}{2}\) miles and \(2\tfrac{1}{4}\) miles. Line up fraction parts over the lcd before adding. What is the total distance as a mixed number?
Show the solution
Give both fraction parts the same bottom, \(\tfrac{1}{2}=\tfrac{2}{4}\). The wholes give \(1+2=3\) and the fourths give \(\tfrac{2}{4}+\tfrac{1}{4}=\tfrac{3}{4}\), which is under one whole, so nothing carries. The total is \(\boxed{3\tfrac{3}{4}}\) miles, or \(\tfrac{15}{4}\) as an improper fraction.
Practice
Six planters each need \(1\tfrac{2}{3}\) bags of mulch. Convert to an improper fraction, then multiply by \(6\). How many bags total?
Show the solution
One whole is \(\tfrac{3}{3}\), so \(1\tfrac{2}{3}=\tfrac{3}{3}+\tfrac{2}{3}=\tfrac{5}{3}\). Then \(6\times\tfrac{5}{3}=\tfrac{30}{3}=10\), so the row of planters needs \(\boxed{10}\) bags of mulch. The improper fraction is the shortcut here, since the \(6\) and the \(3\) cancel instead of you adding \(1\tfrac{2}{3}\) six separate times.
Practice
One press pass uses \(2\tfrac{2}{3}\) oz of ink, and a proof run needs \(\tfrac{3}{4}\) of a pass. Convert, then multiply. How many ounces does the proof run use?
Show the solution
The whole number 2 is \(\tfrac{6}{3}\), so \(2\tfrac{2}{3}=\tfrac{8}{3}\) and the problem is \(\tfrac{3}{4}\times\tfrac{8}{3}=\tfrac{24}{12}\). Since \(24\div 12=2\), the proof run uses \(\boxed{2}\) ounces of ink.
Practice
A ribbon is \(3\tfrac{3}{4}\) m. Cut into \(1\tfrac{1}{2}\) m pieces. Convert both to improper fractions, divide, write the result as a mixed number. How many pieces fit?
Show the solution
\(3\tfrac{3}{4}=\tfrac{15}{4}\) and \(1\tfrac{1}{2}=\tfrac{3}{2}\), so flip the second one and multiply. $$\frac{15}{4}\div\frac{3}{2}=\frac{15}{4}\times\frac{2}{3}=\frac{30}{12}=\frac{5}{2}$$ Two goes into five twice with one left over, so \(\boxed{2\tfrac{1}{2}}\) pieces fit. Two full pieces use \(3\) meters and the leftover \(\tfrac{3}{4}\) meter is exactly half of another piece.
Practice
One loop lap is \(4\tfrac{2}{3}\) km, and two runners split it equally. How long is each leg? Give as a mixed number in km.
Show the solution
Three thirds make one whole, so four wholes hold twelve thirds and \(4\tfrac{2}{3}=\tfrac{14}{3}\). Dividing by 2 is the same as multiplying by \(\tfrac{1}{2}\), so \(\tfrac{14}{3}\cdot\tfrac{1}{2}=\tfrac{14}{6}=\tfrac{7}{3}\), and three goes into seven twice with one left over. Each leg is \(\boxed{2\tfrac{1}{3}\text{ km}}\). Two legs of \(2\tfrac{1}{3}\) add back to \(4\tfrac{2}{3}\), the whole lap.
Practice
A full batch needs \(2\tfrac{1}{4}\) cups of flour. You make \(\tfrac{1}{3}\) of a batch. How many cups do you need? Give as a fraction in simplest form.
Show the solution
A third of a batch means \(\tfrac{1}{3}\times 2\tfrac{1}{4}\), and the word "of" is the signal to multiply. Convert first, \(2\tfrac{1}{4}=\tfrac{8}{4}+\tfrac{1}{4}=\tfrac{9}{4}\), then multiply straight across. $$\frac{1}{3}\times\frac{9}{4}=\frac{9}{12}=\boxed{\tfrac{3}{4}}$$ Three quarters of a cup fits the story, since a third of a batch should use well under the full \(2\tfrac{1}{4}\) cups.
Practice
A board is \(5\tfrac{1}{2}\) ft. Two side brackets, each \(1\tfrac{3}{4}\) ft, are cut off. How many feet of board remain?
Show the solution
Each bracket is \(1\tfrac{3}{4}=\tfrac{7}{4}\) feet, so the two together use \(2\times\tfrac{7}{4}=\tfrac{14}{4}=\tfrac{7}{2}=3\tfrac{1}{2}\) feet. Subtracting from the board, \(5\tfrac{1}{2}-3\tfrac{1}{2}\), the halves cancel and \(5-3=2\), so \(\boxed{2}\) feet of board remain.