QuanticaPrealgebraOpen in the courseA baker gives her apprentice \(\tfrac{1}{3}\) of a dough strip, and he uses half of that for a braid. The braid is a part of a part, so it is smaller than either piece. What single fraction of the whole strip is it?
Problem
A garden bed has \(12\) equal squares, and \(\tfrac{2}{3}\) are covered. Split \(12\) into \(3\) groups, keep \(2\). How many squares end up covered?
Show a hint
- The bottom number of \(\tfrac{2}{3}\) tells you how many equal groups to split the 12 squares into. Make the groups first, before you worry about how many to keep.
Show the full solution
The bottom number 3 says split the 12 squares into three equal groups, and \(12 \div 3 = 4\), so each group is 4 squares. The top number 2 says keep two of those groups, which is \(2 \times 4 = 8\) squares. $$\boxed{8}$$
That matches \(\tfrac{2}{3} \times 12 = \tfrac{24}{3} = 8\). The word of works exactly like a multiply sign.
Problem
One batch needs \(\tfrac{5}{8}\) cup of cherries, and \(6\) batches are made. Compute \(6\times\tfrac{5}{8}\). Leave your answer as an improper fraction.
Show a hint
- Six batches means you lay down \(\frac{5}{8}\) of a cup, then another \(\frac{5}{8}\), and so on, 6 times over. The eighths are all the same size, so think about how many eighths you end up with once you have collected 6 of these scoops.
- Write the 6 as \(\frac{6}{1}\) and multiply straight across, tops times tops and bottoms times bottoms. The bottom stays 8 because each scoop is still an eighth of a cup, only the count of eighths changed. So you get \(\frac{6 \times 5}{1 \times 8}\), then simplify.
Show the full solution
Each batch takes 5 eighths of a cup, so 6 batches take \(6 \times 5 = 30\) eighths. Written out, \(\frac{6}{1} \times \frac{5}{8} = \frac{30}{8}\), and both numbers share a factor of 2, so the total is $$\boxed{15/4}$$
The bottom stays 8 the whole way. Collecting eighths changes how many you have, not how big each one is.
Problem
A worker uses \(\tfrac{1}{4}\) of a netting roll, then patches \(\tfrac{1}{3}\) of that panel. The patch is \(1\) cell in the grid of equal cells. Write it as \(a/b\) of the whole roll.
Show a hint
- You do not need to multiply anything yet, just count slivers. Each of the 4 equal panels gets cut into 3 equal slivers. Lay the slivers end to end down the whole roll and count how many there are in total.
- There are \(4 \times 3 = 12\) equal slivers covering the whole roll, and they are all the same size. Her patch is one of those 12 equal pieces, so it is one part out of 12 of the whole.
Show the full solution
She uses \(\tfrac{1}{3}\) of \(\tfrac{1}{4}\) of the roll. Cutting each of the 4 panels into 3 slivers makes \(4 \times 3 = 12\) equal slivers across the whole roll, and her patch is one of them. $$\boxed{1/12}$$
Taking \(\tfrac{1}{a}\) of \(\tfrac{1}{b}\) always gives \(\tfrac{1}{ab}\), because the second cut multiplies the number of pieces.
Problem
A \(3\times 5\) mosaic: pale glass fills \(2\) of \(3\) columns, and gold fills \(3\) of \(5\) rows. What fraction of the panel does the overlap cover? Reduce fully, write as \(a/b\).
Show a hint
- The grid is \(3\) columns wide and \(5\) rows tall, so the whole panel is cut into \(3 \times 5\) equal cells. The overlap is the block that is both in the pale \(2\) columns and in the gold \(3\) rows. How many columns wide is that block, and how many rows tall? Those two counts are the tops of your fractions.
- Multiply tops with tops and bottoms with bottoms. Two thirds of three fifths is \(\frac{2}{3} \times \frac{3}{5} = \frac{2 \times 3}{3 \times 5} = \frac{6}{15}\). Now find \(\gcd(6,15)\) and divide both numbers by it.
Show the full solution
The panel is cut into \(3 \times 5 = 15\) equal cells. The overlap is 2 columns wide and 3 rows tall, so it holds \(2 \times 3 = 6\) cells, giving \(\frac{2}{3} \times \frac{3}{5} = \frac{6}{15}\). Dividing top and bottom by 3 gives $$\boxed{2/5}$$
That is why multiplying fractions multiplies tops and bottoms. The tops count how wide and tall the overlap is, and the bottoms count how the whole square was cut up.
Problem
Compute \(\tfrac{14}{15}\times\tfrac{25}{21}\): prime-factor each number, cancel any prime appearing in a top and a bottom, multiply what remains. Fully reduced result as \(a/b\)?
Show a hint
- Before you multiply anything, look across the two fractions for shared factors. The 14 on top and the 21 on the bottom both hold a 7. The 25 on top and the 15 on the bottom both hold a 5. A factor on any top cancels with the same factor on any bottom, no matter which fraction it sits in.
- Write it as \(\dfrac{2\cdot 7}{3\cdot 5}\times\dfrac{5\cdot 5}{3\cdot 7}\). Cross out the matching \(7\) (from 14 and 21) and one matching \(5\) (from 25 and 15). What survives on top is \(2\cdot 5\) and on the bottom is \(3\cdot 3\). Multiply each side.
Show the full solution
Break each number into primes. \(14=2\cdot 7\), \(25=5\cdot 5\), \(15=3\cdot 5\), and \(21=3\cdot 7\), so the product is $$\frac{2\cdot 7}{3\cdot 5}\times\frac{5\cdot 5}{3\cdot 7}=\frac{2\cdot 7\cdot 5\cdot 5}{3\cdot 5\cdot 3\cdot 7}.$$ The \(7\)s cancel and one \(5\) cancels, leaving \(2\cdot 5=10\) on top and \(3\cdot 3=9\) on the bottom. \(\boxed{10/9}\)
Multiplying first gives \(\tfrac{350}{315}\), the same answer after much more reducing. In a product of fractions every top factor is paired against every bottom factor, so cancel before you multiply.
Problem
\(\tfrac{3}{4}\times\tfrac{4}{5}\times\tfrac{5}{6}\times\tfrac{6}{7}\): each bottom is the next top. Most cancel. What single reduced fraction remains?
Show a hint
- Do not multiply the tops into one giant number and the bottoms into another. Instead, write all four numerators on one line over all four denominators on one line, like \(\dfrac{3\times 4\times 5\times 6}{4\times 5\times 6\times 7}\), and then hunt for numbers that appear in both rows.
- A factor on top cancels the same factor on the bottom. The \(4\) shows up on top and on bottom, so it vanishes. So does the \(5\). So does the \(6\). Cross all three pairs out and read off whatever is still standing on top and on bottom.
Show the full solution
Stack all four tops over all four bottoms, $$\frac{3}{4}\times\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}=\frac{3\times 4\times 5\times 6}{4\times 5\times 6\times 7}.$$ The \(4\), the \(5\), and the \(6\) each sit on top and on the bottom, so all three pairs cancel and leave \(\;\boxed{\dfrac{3}{7}}\;\)
Only the first top and the last bottom have no partner to cancel against. Everything in the middle was both a top and a bottom, so it washed out.
Multiplying by a proper fraction makes things smaller. Division goes the other way and asks how many small pieces fit inside a big one, which is multiplication run backward. The tool that turns a division back into a multiplication is the reciprocal.
Problem
A machine scales by \(\tfrac{7}{12}\), and a second machine must undo it so the combined product is \(1\). What fraction times \(\tfrac{7}{12}\) gives \(1\)? Write as \(a/b\).
Show a hint
- Multiplying fractions multiplies the tops together and the bottoms together. For the result to be \(1\), the top of the product and the bottom of the product have to come out equal. Picture what the second fraction's top and bottom would need to be so that the \(7\) and the \(12\) end up matched.
- The \(7\) sits on top of \(\tfrac{7}{12}\), so to cancel it you need a \(7\) on the bottom of your fraction. The \(12\) sits on the bottom, so to cancel it you need a \(12\) on top. Put those together.
Show the full solution
Flip \(\tfrac{7}{12}\) upside down to get \(\tfrac{12}{7}\). Then \(\tfrac{7}{12}\times\tfrac{12}{7}=\tfrac{84}{84}=1\), exactly what the second machine needs. \(\boxed{12/7}\)
A fraction equals \(1\) when its top and bottom match, and swapping the two numbers guarantees that. The flipped partner is called the reciprocal.
Problem
Five builders share a \(3\) m plank: \(3\div 5 = 3\times\tfrac{1}{5}\). How long is each share in meters? Write as \(a/b\).
Show a hint
- Splitting the plank into five equal pieces and keeping one piece is exactly taking one fifth of the plank. So instead of \(3 \div 5\), you can ask what is \(\tfrac{1}{5}\) of \(3\) meters.
- Taking \(\tfrac{1}{5}\) of something means multiplying by \(\tfrac{1}{5}\), so each share is \(3 \times \tfrac{1}{5}\). Multiply the \(3\) by the top of the fraction and keep the \(5\) on the bottom.
Show the full solution
Sharing 3 meters among 5 builders is \(3 \div 5\), and splitting into 5 equal parts and keeping one is taking \(\tfrac{1}{5}\) of the plank. So each share is \(3 \times \tfrac{1}{5}\), which is three fifth-meter pieces. $$\boxed{3/5}$$
Dividing by a whole number is multiplying by its reciprocal, so \(a \div b\) is just the fraction \(\tfrac{a}{b}\).
Problem
A \(5\) L carboy is poured into \(\tfrac{1}{3}\) L jars. One liter fills \(3\) jars, and the carboy holds \(5\) liters. How many \(\tfrac{1}{3}\) L jars does the full carboy fill?
Show a hint
- Forget the division symbol for a second and just count. Each single liter breaks into how many of these little third liter jars? Then you have 5 whole liters to pour, not just one.
- Three thirds fit in every liter, so one liter fills \(3\) jars. With \(5\) liters lined up, you fill \(5\) groups of \(3\) jars. Multiply \(5\times 3\).
Show the full solution
One liter splits into 3 jars, since \(\tfrac{1}{3}+\tfrac{1}{3}+\tfrac{1}{3}=1\). With 5 liters lined up, that is \(5 \times 3 = \boxed{15}\) jars.
The question is \(5\div\tfrac{1}{3}\), and the answer came out bigger than the 5 we started with. Dividing by a number below \(1\) always does that, since many small pieces fit.
Problem
A trim piece is \(\tfrac{5}{6}\) m. You want pieces of \(\tfrac{2}{9}\) m each. Keep the first fraction, flip the divisor, and multiply. What is \(\tfrac{5}{6}\div\tfrac{2}{9}\) in lowest terms?
Show a hint
- Only the divisor flips. The \(\tfrac{5}{6}\) stays exactly as it is, and \(\tfrac{2}{9}\) turns into \(\tfrac{9}{2}\). Now the \(\div\) becomes a \(\times\), so you are really finding \(\tfrac{5}{6} \times \tfrac{9}{2}\).
- Before you multiply across, look for a shared factor. The \(9\) on top and the \(6\) on the bottom both have a factor of \(3\), so the \(9\) becomes \(3\) and the \(6\) becomes \(2\). Then multiply straight across, \(5 \times 3\) over \(2 \times 2\).
Show the full solution
Keep the first fraction and flip the divisor, so \(\tfrac{5}{6} \div \tfrac{2}{9} = \tfrac{5}{6} \times \tfrac{9}{2}\). The \(9\) and the \(6\) share a factor of \(3\), turning them into \(3\) and \(2\), so the product is \(\frac{5 \times 3}{2 \times 2} = \boxed{15/4}\)
That is \(3\tfrac{3}{4}\), so three whole pieces fit with three quarters of another left over.
Problem
A fundraiser raised \(32\) (thousands). That is \(\tfrac{2}{3}\) of the goal. What is the full goal, in thousands?
Show a hint
- You are not asked for two thirds of something here. You already know the two thirds part, it is the 32, and you are hunting for the whole. Dividing by a fraction is how you walk a multiplication backward, so think about \(32 \div \tfrac{2}{3}\).
- Dividing by \(\tfrac{2}{3}\) is the same as multiplying by the flipped fraction \(\tfrac{3}{2}\). So compute \(32 \times \tfrac{3}{2}\), and notice the 32 and the 2 are ready to cancel before you multiply.
Show the full solution
The forward story is goal \(\times\,\tfrac{2}{3} = 32\), so the goal is \(32 \div \tfrac{2}{3} = 32 \times \tfrac{3}{2}\). Cancel the 2 against the 32, since \(32 = 16 \times 2\), which leaves \(16 \times 3 = \boxed{48}\) thousand.
Checking forward, \(48 \div 3 = 16\) and \(16 \times 2 = 32\). Multiplying by the reciprocal is how you recover a whole from a part.
Problem
Feed \(12\) into a machine: first multiply by \(\tfrac{4}{9}\), then divide by \(\tfrac{8}{15}\). What is the result?
Show a hint
- Dividing by a fraction is the same as multiplying by its reciprocal. So the second setting, divide by \(\tfrac{8}{15}\), is really multiply by \(\tfrac{15}{8}\). Once both steps are multiplications, they collapse into one, because two multiplications in a row are just one multiplication by their product.
- The single fraction is \(\tfrac{4}{9}\times\tfrac{15}{8}\). Before multiplying straight across, cancel across the fractions. The \(4\) and the \(8\) share a factor of \(4\), and the \(15\) and the \(9\) share a factor of \(3\). That leaves \(\tfrac{1}{3}\times\tfrac{5}{2}=\tfrac{5}{6}\). Now compute \(12\times\tfrac{5}{6}\).
Show the full solution
Dividing by \(\tfrac{8}{15}\) is multiplying by \(\tfrac{15}{8}\), so the machine does \(12 \times \tfrac{4}{9} \times \tfrac{15}{8}\). Cancel first. The \(4\) and the \(8\) share a factor of \(4\), and the \(15\) and the \(9\) share a factor of \(3\), so $$\frac{4}{9}\times\frac{15}{8}=\frac{1\times 5}{3\times 2}=\frac{5}{6}.$$ Then \(12\times\tfrac{5}{6}=\tfrac{60}{6}=\boxed{10}\)
A multiply and a divide folded into one fraction. Chain as many of them as you like and the result is still just a fraction.
Practice these ideas
Practice
\(\tfrac{3}{4}\) of a parking lot is shaded, and \(\tfrac{2}{9}\) of shaded spaces are reserved. What fraction of the whole lot is shaded-reserved? Write in lowest terms as \(a/b\).
Show the solution
Two ninths of three quarters is \(\tfrac{2}{9}\times\tfrac{3}{4}\). Cancel first. The \(3\) on top and the \(9\) on the bottom share a \(3\), and the \(2\) on top and the \(4\) on the bottom share a \(2\), leaving \(\frac{1\times 1}{3\times 2}=\boxed{1/6}\)
Without cancelling you get \(\tfrac{6}{36}\), which reduces to the same thing with more work.
Practice
Three quarters of \(20\) students ride bikes. Split \(20\) into \(4\) equal groups, keep three. How many students ride bikes?
Show the solution
Taking \(\tfrac{3}{4}\) of 20 means splitting 20 into four groups of \(20 \div 4 = 5\) and keeping three of them, so \(3 \times 5 = \boxed{15}\) students.
As one multiplication it is \(\tfrac{3}{4} \times 20 = \tfrac{60}{4} = 15\), the same answer either way.
Practice
Light through two panes: outer passes \(\tfrac{2}{3}\), and inner passes \(\tfrac{6}{7}\). Cancel \(3\) and \(6\) first. What fraction gets through?
Show the solution
The light through both panes is \(\tfrac{2}{3}\times\tfrac{6}{7}\). The \(6\) on top and the \(3\) on the bottom both divide by \(3\), giving \(\tfrac{2}{1}\times\tfrac{2}{7}\), and multiplying across gives \(\boxed{4/7}\)
Cancelling first kept the numbers small and left nothing to reduce at the end.
Practice
Each of \(8\) ornaments uses \(\tfrac{3}{10}\) m of wire. Total wire is \(8\times\tfrac{3}{10}\). How many meters are needed? Give as an improper fraction \(a/b\) in lowest terms.
Show the solution
Eight copies of \(\tfrac{3}{10}\) is \(\tfrac{8\times 3}{10}=\tfrac{24}{10}\). Both numbers are even, so dividing each by 2 gives \(\boxed{12/5}\) meters.
Multiplying a fraction by a whole number scales the top only. The pieces stay tenths, you just have more of them.
Practice
Compute \(\dfrac{9}{14}\times\dfrac{7}{12}\). Cancel \(7\) from \(7\) and \(14\), and \(3\) from \(9\) and \(12\). What fraction \(a/b\) in lowest terms do you get?
Show the solution
Cancel the crossing pairs first. The \(7\) and the \(14\) share a \(7\), so they become \(1\) and \(2\). The \(9\) and the \(12\) share a \(3\), so they become \(3\) and \(4\). The problem turns into $$\frac{3}{2}\times\frac{1}{4}.$$ Now multiply straight across. The tops give \(3\times1=3\) and the bottoms give \(2\times4=8\), so the product is \(\tfrac{3}{8}\). Since \(\gcd(3,8)=1\), it is already in lowest terms. $$\boxed{3/8}$$
Practice
A jug is \(\tfrac{5}{8}\) full. You pour off \(\tfrac{2}{5}\) of the juice inside. The two \(5\)s cancel. What fraction of the full jug did you pour out?
Show the solution
Pouring off \(\tfrac{2}{5}\) of what is in the jug, while \(\tfrac{5}{8}\) is in there, is \(\tfrac{2}{5}\times\tfrac{5}{8}\). Instead of multiplying straight across into \(\tfrac{10}{40}\) and reducing later, cancel first. The 5 on the bottom of \(\tfrac{2}{5}\) and the 5 on the top of \(\tfrac{5}{8}\) cancel to 1, leaving \(\tfrac{2}{1}\times\tfrac{1}{8}=\tfrac{2}{8}\). One more pair, the 2 against the 8, gives \(\tfrac{1}{4}\). So you poured out \(\boxed{1/4}\) of the full jug.
Practice
Chain: \(\tfrac{4}{5}\times\tfrac{5}{6}\times\tfrac{6}{7}\times\tfrac{7}{8}\). Most tops and bottoms cancel in pairs. What single fraction \(a/b\) does the chain equal?
Show the solution
Collect all the tops over all the bottoms, $$\frac{4}{5}\times\frac{5}{6}\times\frac{6}{7}\times\frac{7}{8}=\frac{4\times5\times6\times7}{5\times6\times7\times8}.$$ The \(5\), the \(6\), and the \(7\) each appear on top and on the bottom, so those pairs cancel and leave \(\tfrac{4}{8}\), which reduces to $$\boxed{1/2}$$
Only the first top and the last bottom have no partner, which is what makes a chain like this telescope.
Practice
A stretch setting multiplies by \(\tfrac{9}{4}\). What single fraction \(a/b\) multiplies with \(\tfrac{9}{4}\) to give exactly \(1\)?
Show the solution
Flip \(\tfrac{9}{4}\) to get \(\tfrac{4}{9}\). Checking, \(\frac{9}{4}\times\frac{4}{9}=\frac{36}{36}=1\), so the flip really does undo the stretch. \(\boxed{4/9}\)
Two numbers that multiply to \(1\) are reciprocals, and for a fraction the reciprocal is always the same two numbers swapped.
Practice
A \(4\) L drum is emptied with a \(\tfrac{1}{3}\) L dispenser. How many full pumps empty the drum?
Show the solution
One liter holds 3 of the third-liter pumps, and there are 4 liters, so \(4 \times 3 = \boxed{12}\) pumps.
That count is \(4 \div \tfrac{1}{3}\), and dividing by \(\tfrac{1}{3}\) is the same as multiplying by \(3\).
Practice
A tank is \(\tfrac{7}{10}\) full. Each truck carries \(\tfrac{1}{2}\) of a full tank. How many truckloads does the current water fill? Give as \(a/b\) in lowest terms.
Show the solution
Counting truckloads is \(\tfrac{7}{10} \div \tfrac{1}{2}\). Flip the divisor and multiply, \(\tfrac{7}{10} \times \tfrac{2}{1} = \tfrac{14}{10}\), and dividing top and bottom by \(2\) gives $$\boxed{7/5}$$
A little more than one half-tank should come out, and \(\tfrac{7}{5}\) sits just past \(1\).
Practice
Tray A holds \(\tfrac{10}{21}\) of a ream, and Tray B holds \(\tfrac{5}{14}\). Compute \(\tfrac{10}{21}\div\tfrac{5}{14}\) and give the result as \(a/b\) in lowest terms.
Show the solution
Flip the divisor, so \(\tfrac{10}{21}\div\tfrac{5}{14}=\tfrac{10}{21}\times\tfrac{14}{5}\). The \(10\) and the \(5\) share a factor of \(5\), and the \(14\) and the \(21\) share a factor of \(7\), which leaves \(\tfrac{2}{3}\times\tfrac{2}{1}\). $$\boxed{4/3}$$
Straight across it would be \(\tfrac{140}{105}\), and since \(\gcd(140,105)=35\) that reduces to the same thing.
Practice
Priya is \(\tfrac{3}{5}\) through her novel, which is \(27\) pages. Since \(\tfrac{3}{5}\) of the whole \(=27\), multiply \(27\) by \(\tfrac{5}{3}\). How many pages total?
Show the solution
Three fifths of the total is 27, so the total is \(27 \div \tfrac{3}{5} = 27 \times \tfrac{5}{3}\). The \(3\) on the bottom divides the \(27\), leaving \(9 \times 5 = \boxed{45}\) pages.
Checking, \(\tfrac{3}{5}\times 45 = 27\). Multiplying by the reciprocal is how you get a whole back from a part.
Practice
A spool holds \(\tfrac{15}{4}\) m, and each bow needs \(\tfrac{3}{8}\) m. Compute \(\tfrac{15}{4}\div\tfrac{3}{8}\). How many whole bows does the ribbon make?
Show the solution
Flip \(\tfrac{3}{8}\) to \(\tfrac{8}{3}\), so \(\tfrac{15}{4}\div\tfrac{3}{8}=\tfrac{15}{4}\times\tfrac{8}{3}\). The \(8\) and the \(4\) share a factor of \(4\), and the \(15\) and the \(3\) share a factor of \(3\), leaving \(\tfrac{5}{1}\times\tfrac{2}{1}=\boxed{10}\)
The quotient is a whole number, so the ribbon divides evenly with nothing left over.
Practice
Five eighths of a \(24\) g block is measured out and split into jars of \(\tfrac{3}{4}\) g each. How many jars does the merchant fill?
Show the solution
She measures out \(24 \times \tfrac{5}{8}\) grams. Since \(24 \div 8 = 3\), that is \(3 \times 5 = 15\) grams. Splitting 15 grams into \(\tfrac{3}{4}\) gram jars is \(15 \div \tfrac{3}{4} = 15 \times \tfrac{4}{3} = \tfrac{60}{3} = \boxed{20}\) jars.
The two steps pull opposite ways. Multiplying by a fraction under \(1\) shrank the amount, then dividing by one grew the count back up.
Practice
Three filters: \(\tfrac{5}{6}\times\tfrac{6}{7}\times\tfrac{7}{8}\). The chain telescopes. Divide that result by \(\tfrac{5}{8}\). What single number is the answer?
Show the solution
The three filters give \(\dfrac{5\times 6\times 7}{6\times 7\times 8}\), and the \(6\) and the \(7\) each cancel top against bottom, leaving \(\tfrac{5}{8}\). Dividing by \(\tfrac{5}{8}\) means multiplying by \(\tfrac{8}{5}\), and \(\tfrac{5}{8}\times\tfrac{8}{5}=\tfrac{40}{40}\). $$\boxed{1}$$
Any nonzero fraction times its own reciprocal is \(1\), and here the surviving signal was exactly the baseline.