QuanticaPrealgebraOpen in the courseOne ranger cuts a trail into \(4\) equal stretches and marks \(\tfrac{3}{4}\) of the way out. Another cuts the same trail into \(12\) stretches and marks \(\tfrac{9}{12}\). They walk out and find their ribbons on the same patch of dirt. Two different fractions can name the same amount, and there is a rule for when they do.
Problem
Trail cut into \(4\) stretches, then into \(12\). Every third stake of the second ranger lands on a stake of the first. Which stake sits on \(\tfrac{3}{4}\)? Write as \(a/b\).
Show a hint
- Her ribbon at \(\tfrac{3}{4}\) is the third of her four marks, so three full stretches of hers lie behind it. Each one of those stretches is worth how many of his little pieces?
- Three of her stretches sit before the ribbon, and each one is worth \(3\) of his pieces, so \(3 \times 3 = 9\) of his pieces bring you exactly to the ribbon. His ninth stake is at \(\tfrac{9}{12}\).
Show the full solution
Her ribbon sits after \(3\) of her \(4\) stretches, and each of her stretches holds \(3\) of his pieces, so \(3 \times 3 = 9\) of his pieces reach the ribbon. His ninth stake, \(\boxed{9/12}\), stands on the same patch of dirt. That is the same as \(\tfrac{3}{4} \times \tfrac{3}{3}\). Multiplying top and bottom by the same number renames the point without moving it.
Problem
A fudge tray has \(3\) strips, and \(2\) go to an order (\(\tfrac{2}{3}\)). Each strip is re-scored into \(5\) mini strips. Write the new fraction name for the order as \(a/b\).
Show a hint
- The fudge in the order did not change at all, only how it is sliced. Count two things in the new mini strips, how many mini strips make up the whole tray, and how many make up the order.
- Each of the 3 strips becomes 5 mini strips, so the tray holds \(3 \times 5\) of them. Each of the 2 order strips also becomes 5 mini strips, so the order is \(2 \times 5\). Put the order count over the tray count.
Show the full solution
The tray had \(3\) strips and each becomes \(5\) mini strips, so the tray holds \(3 \times 5 = 15\). The order was \(2\) strips, which becomes \(2 \times 5 = 10\). So the order is \(\boxed{10/15}\). Slicing every piece into \(k\) smaller pieces multiplies the top and the bottom by the same \(k\), which is why the amount of fudge never changes.
Problem
A coach multiplies top and bottom of \(\tfrac{5}{6}\) by \(4\). The extra \(4\) appears on both floors and cancels, leaving the same value. Write the bigger name as \(\tfrac{a}{b}\).
Show a hint
- You are not changing the value, you are just renaming it with larger numbers. Multiply the top by \(4\) to get the new top, and multiply the bottom by \(4\) to get the new bottom. The factor of \(4\) on top and the factor of \(4\) on the bottom cancel in the single division, which is why the split stays the same.
- Top is \(5\times 4\). Bottom is \(6\times 4\). Compute each one and stack the new top over the new bottom as \(\frac{a}{b}\).
Show the full solution
Multiply the top by \(4\) and the bottom by \(4\). The top becomes \(5 \times 4 = 20\) and the bottom becomes \(6 \times 4 = 24\), so the bigger name is \(\boxed{20/24}\). A fraction is one division, so a factor of \(4\) on top and a matching \(4\) on the bottom cancel, leaving \(5\) divided by \(6\) again.
Problem
A counter must display fractions over \(21\). A runner is \(\tfrac{2}{3}\) done. Find \(k\) where \(3k=21\), scale the top. How many of \(21\) lights should show?
Show a hint
- The bottom of the fraction has to go from 3 to 21 without changing the actual amount. The only honest way to do that is to multiply the bottom by some whole number \(k\). Ask yourself what you multiply 3 by to land exactly on 21, since that same \(k\) is how many small pieces each third gets cut into.
- You need \(3 \times k = 21\), so \(k = 21 \div 3 = 7\). Cutting each third into 7 means you must also multiply the top by 7. Compute \(2 \times 7\).
Show the full solution
You need \(3 \times k = 21\), so \(k = 21 \div 3 = 7\). Multiply the top by that same \(7\), giving \(2 \times 7 = 14\), so $$\frac{2}{3} = \frac{2 \times 7}{3 \times 7} = \frac{14}{21}.$$ The counter should light \(\boxed{14}\) of its \(21\) lights. Aim at the bottom first, then copy that same multiplier onto the top, or the value shifts.
Problem
Scoreboard shows \(\tfrac{18}{24}\). Divide by \(2\) to get \(\tfrac{9}{12}\), then divide by \(3\). Stop when top and bottom share no factor above \(1\). Write the result as \(a/b\).
Show a hint
- You already did the first cut, going from \(\tfrac{18}{24}\) to \(\tfrac{9}{12}\). The seed tells you the next shared factor is 3. Apply that one division to both the top and the bottom and see where you stop.
- Take \(\tfrac{9}{12}\) and divide both numbers by 3. The top becomes \(9\div 3\) and the bottom becomes \(12\div 3\). Work out those two numbers and write the fraction they form.
Show the full solution
From \(\tfrac{9}{12}\), divide both numbers by \(3\). That gives \(\tfrac{9\div 3}{12\div 3} = \tfrac{3}{4}\), and \(3\) and \(4\) share no factor above \(1\), so this is where you stop. The smallest name is \(\boxed{3/4}\). Reducing is building up run backward, dividing out a shared factor instead of multiplying one in.
Problem
A coach writes a split as \(\tfrac{84}{120}\). Find \(\gcd(84,120)\), divide both top and bottom by it in one stroke, and write \(\tfrac{84}{120}\) in lowest terms as \(\tfrac{a}{b}\).
Show a hint
- The idea is that you do not have to peel factors one by one. One number can strip them all at once. What is the largest number that divides both 84 and 120 cleanly? Break each into its prime factors and grab everything they share.
- Both numbers are even, both are divisible by 4, and both are divisible by 3, so \(4\times 3\) divides each. That makes \(\gcd(84,120)=12\). Now compute \(84\div 12\) for the new top and \(120\div 12\) for the new bottom.
Show the full solution
Factor both numbers. \(84 = 2 \times 2 \times 3 \times 7\) and \(120 = 2 \times 2 \times 2 \times 3 \times 5\), so they share two \(2\)s and a \(3\), making \(\gcd(84,120) = 12\). Then \(84 \div 12 = 7\) and \(120 \div 12 = 10\), so lowest terms is \(\boxed{7/10}\). One pass is enough because the gcd already holds every prime the two numbers share.
Problem
Shared primes of \(126\) and \(350\) are \(2\) and \(7\). Multiply them for the common factor, divide both by it. Write \(\tfrac{126}{350}\) in lowest terms as \(a/b\).
Show a hint
- Lay the two prime lists side by side. A prime can only cancel if it appears in both lists. Which primes show up in the top list and the bottom list?
- The shared primes are \(2\) and \(7\), so the common factor is \(2 \times 7 = 14\). Divide both \(126\) and \(350\) by \(14\).
Show the full solution
The top is \(126 = 2 \times 3 \times 3 \times 7\) and the bottom is \(350 = 2 \times 5 \times 5 \times 7\). Only the \(2\) and the \(7\) appear in both lists, so the common factor is \(2 \times 7 = 14\). Dividing gives \(126 \div 14 = 9\) and \(350 \div 14 = 25\), so the gears in lowest terms are \(\boxed{9/25}\). A prime cancels only when it sits on both floors, which is why the \(3\)s and \(5\)s stay put.
Problem
An app stores \(\tfrac{2}{5}\), but the display only allows bottom \(\leq 40\). Allowed bottoms are \(5k\leq 40\). How many different names fit?
Show a hint
- Forget the top number for a moment. The only thing that decides whether a name fits is its bottom, and the bottom is always \(5k\). So you are really asking how many whole numbers \(k\) keep \(5k\) from going past \(40\). That is just counting multiples of \(5\), the kind of thing you did back in Chapter 3.
- List the bottoms in order. With \(k = 1\) the bottom is \(5\), with \(k = 2\) it is \(10\), and so on. Keep going as long as the bottom stays at or below \(40\), then count how many you wrote down.
Show the full solution
Every allowed name is \(\tfrac{2k}{5k}\), so the only rule is \(5k \le 40\). Dividing, \(40 \div 5 = 8\), so \(k\) runs \(1\) through \(8\), since \(k = 9\) would push the bottom to \(45\). That is \(\boxed{8}\) names. The whole family of names for a fraction is just the multiples of its bottom, so counting names is counting multiples.
To compare two fractions, give them the same denominator first. Then both are counted in pieces of one size, so the bigger top wins. The same shared-denominator move is what makes fraction addition work later on, so it is worth learning here.
Problem
Slopes \(\tfrac{5}{6}\) and \(\tfrac{7}{8}\) over shared bottom \(24\): sixths \(\times 4\), eighths \(\times 3\). Compare tops. Write the steeper slope as \(\tfrac{a}{b}\).
Show a hint
- You cannot compare the tops while the bottoms disagree. Force both fractions onto the same bottom of \(24\) first, then the bigger top wins. Multiplying top and bottom by the same number does not change a fraction's value, it just renames it.
- Multiply \(\frac{5}{6}\) by \(\frac{4}{4}\) to get \(\frac{20}{24}\), and multiply \(\frac{7}{8}\) by \(\frac{3}{3}\) to get \(\frac{21}{24}\). Now both bottoms are \(24\), so compare the tops \(20\) and \(21\). The bigger top names the steeper ramp, and your answer should be its original unbuilt fraction.
Show the full solution
Put both slopes over \(24\). $$\frac{5}{6} = \frac{5\times 4}{6\times 4} = \frac{20}{24}, \qquad \frac{7}{8} = \frac{7\times 3}{8\times 3} = \frac{21}{24}$$ Since \(21 > 20\), the west ramp is steeper, with slope \(\boxed{7/8}\). Tops are only comparable once the bottoms match, because only then are you counting pieces of the same size.
Problem
Is \(\tfrac{4}{6}=\tfrac{6}{9}\)? Compute \(4\times 9\) and \(6\times 6\). Equal products mean equal fractions. Enter the common product if equal, else \(0\).
Show a hint
- Multiplying \(\tfrac{4}{6}\) by \(6\times 9\) cancels the bottom \(6\) and leaves \(4\times 9\). Multiplying \(\tfrac{6}{9}\) by the same \(6\times 9\) cancels the bottom \(9\) and leaves \(6\times 6\). Now you only have to compare two whole-number products, no fractions left.
- Work out \(4\times 9\) and \(6\times 6\) separately. If they land on the same number, that number is your answer. If they differ, the answer is \(0\).
Show the full solution
\(4 \times 9 = 36\) and \(6 \times 6 = 36\). The two products agree, so \(\tfrac{4}{6}\) and \(\tfrac{6}{9}\) are the same number and the common product is \(\boxed{36}\). Multiplying both fractions by \(6 \times 9\) clears the bottoms and leaves exactly those crossed products, so \(\tfrac{a}{b} = \tfrac{c}{d}\) exactly when \(a \times d = b \times c\).
Problem
Master fraction \(\tfrac{8}{14}\). Three locks: \(\tfrac{12}{21}\), \(\tfrac{20}{35}\), \(\tfrac{16}{30}\). Cross-product test each. How many equal \(\tfrac{8}{14}\)?
Show a hint
- You do not need to simplify a single fraction here. For each stamped lock \(\tfrac{c}{d}\), build the two products \(8 \times d\) and \(14 \times c\) and just ask whether they land on the same number. Equal products mean the lock belongs, different products mean it does not.
- Take them one at a time. For \(\tfrac{12}{21}\), compare \(8 \times 21\) with \(14 \times 12\). For \(\tfrac{20}{35}\), compare \(8 \times 35\) with \(14 \times 20\). For \(\tfrac{16}{30}\), compare \(8 \times 30\) with \(14 \times 16\). Count up how many of the three pairs come out matching.
Show the full solution
Each lock \(\tfrac{c}{d}\) matches when \(8 \times d = 14 \times c\). For \(\tfrac{12}{21}\), \(8 \times 21 = 168\) and \(14 \times 12 = 168\), a match. For \(\tfrac{20}{35}\), \(8 \times 35 = 280\) and \(14 \times 20 = 280\), a match. For \(\tfrac{16}{30}\), \(8 \times 30 = 240\) but \(14 \times 16 = 224\), so that one fails. \(\boxed{2}\) locks equal \(\tfrac{8}{14}\). The near miss gets caught without reducing anything.
Problem
Ticks \(\tfrac{n}{36}\) for \(n=1\ldots 36\). Gold plate only where \(\gcd(n,36)=1\). Since \(36=2^2\times 3^2\), \(n\) must miss multiples of \(2\) and \(3\). How many ticks are gold plated?
Show a hint
- You are really counting how many numbers from 1 to 36 are coprime to 36, meaning they have no factor of 2 and no factor of 3. It is faster to count the ones that get thrown away than to check all 36 by hand. How many multiples of 2 are in 1 to 36, and how many multiples of 3?
- There are 18 multiples of 2 and 12 multiples of 3 in the range, but the multiples of 6 got counted in both groups, so subtract them back once. That removes \(18 + 12 - 6 = 24\) numbers. Whatever is left out of 36 is your answer.
Show the full solution
Count the tops that fail and subtract. From \(1\) to \(36\) there are \(36 \div 2 = 18\) multiples of \(2\) and \(36 \div 3 = 12\) multiples of \(3\), and the \(36 \div 6 = 6\) multiples of \(6\) got counted in both groups, so \(18 + 12 - 6 = 24\) tops fail. That leaves \(36 - 24 = \boxed{12}\) gold ticks. Counting failures beats testing all \(36\), as long as you subtract the multiples of \(6\) once to fix the double count.
Practice these ideas
Practice
A window box is split into \(3\) equal sections, and flowers fill the first section (\(\tfrac{1}{3}\)). Re-scoring into \(12\) cells: how many cells do the flowers cover?
Show the solution
The \(3\) sections become \(3 \times 4 = 12\) cells, so the bottom grew by a factor of \(4\) and the top has to grow by \(4\) as well. $$\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}$$ The flowers cover \(\boxed{4}\) cells. Re-scoring does not move the flowers, it just counts the same patch in smaller pieces.
Practice
Five sections report harmony fractions: \(\tfrac{8}{12}\), \(\tfrac{6}{10}\), \(\tfrac{10}{15}\), \(\tfrac{14}{21}\), \(\tfrac{9}{12}\). How many equal \(\tfrac{2}{3}\)?
Show the solution
Cross multiply against \(\tfrac{2}{3}\), so \(\tfrac{a}{b}\) matches when \(3a = 2b\). For \(\tfrac{8}{12}\), \(24 = 24\), a match. For \(\tfrac{6}{10}\), \(18\) against \(20\), no. For \(\tfrac{10}{15}\), \(30 = 30\), a match. For \(\tfrac{14}{21}\), \(42 = 42\), a match. For \(\tfrac{9}{12}\), \(27\) against \(24\), no. Three of the five pass, so \(\boxed{3}\).
Practice
A hose covers \(\tfrac{2}{7}\) of a garden row (\(2\) of \(7\) beds). Every bed is re-cut into \(4\) patches. After the re-cut, how many equal patches make up the whole row?
Show the solution
Each of the \(7\) beds splits into \(4\) patches, so the row now holds \(7 \times 4 = \boxed{28}\) equal patches. The watered share becomes \(2 \times 4 = 8\) of them, and \(\tfrac{8}{28}\) reduces back to \(\tfrac{2}{7}\), so the coverage never changed, only the piece size.
Practice
A jar is \(\tfrac{10}{25}\) full. Reduce to lowest terms by dividing by the common divisor. Write the simplest name as \(a/b\).
Show the solution
Both \(10\) and \(25\) are multiples of \(5\), and \(5\) is the largest number dividing both. Dividing gives \(10 \div 5 = 2\) on top and \(25 \div 5 = 5\) on the bottom, so the simplest name is \(\boxed{2/5}\). Since \(2\) and \(5\) share no factor above \(1\), it cannot shrink any further.
Practice
\(30\) yes votes out of \(42\) total. Find \(\gcd(30,42)\), divide top and bottom by it. Write \(\tfrac{30}{42}\) in lowest terms as \(\tfrac{a}{b}\).
Show the solution
\(30 = 6 \times 5\) and \(42 = 6 \times 7\), and nothing larger divides both, so \(\gcd(30,42) = 6\). Divide the top and bottom by that \(6\) in one step. $$\frac{30}{42} = \frac{30 \div 6}{42 \div 6} = \frac{5}{7}$$ Lowest terms is \(\boxed{5/7}\), since \(5\) and \(7\) share no factor bigger than \(1\).
Practice
A label reads \(\tfrac{15}{28}\). Before reducing, check: is it already in lowest terms? A fraction is simplest when \(\gcd(\text{top},\text{bottom})=1\). Find \(\gcd(15,28)\).
Show the solution
\(15 = 3 \times 5\) and \(28 = 2 \times 2 \times 7\). The two lists share no prime at all, so the only whole number dividing both is \(\boxed{1}\). A gcd of \(1\) is exactly what lowest terms means, so \(\tfrac{15}{28}\) has nothing left to cancel.
Practice
A reading of \(\tfrac{7}{12}\) must be rewritten over \(48\). Find \(k\) where \(12\times k=48\), then scale the top by \(k\). What top number is stored over \(48\)?
Show the solution
The bottom grows from \(12\) to \(48\), so \(k = 48 \div 12 = 4\). Multiply the top by that same \(4\), giving \(7 \times 4 = \boxed{28}\), so \(\tfrac{7}{12} = \tfrac{28}{48}\). Scaling only one floor would change the value, which is why the same \(k\) has to go on both.
Practice
A battery shows \(\tfrac{56}{64}\) charged. Divide top and bottom by their gcd. What is \(\tfrac{56}{64}\) in lowest terms as \(a/b\)?
Show the solution
\(56 = 8 \times 7\) and \(64 = 8 \times 8\), so \(\gcd(56,64) = 8\). Dividing both parts by \(8\) gives $$\frac{56}{64} = \frac{56 \div 8}{64 \div 8} = \frac{7}{8}.$$ The battery is \(\boxed{7/8}\) charged. Since \(7\) is prime and does not divide \(8\), nothing is left to cancel.
Practice
Sensor 1: \(\tfrac{8}{14}\). Sensor 2: \(\tfrac{12}{21}\). Cross-multiply: compare \(8\times 21\) against \(14\times 12\). Do they name the same point? Enter yes or no.
Show the solution
\(8 \times 21 = 168\) and \(14 \times 12 = 168\). The two diagonals match, so the readings name the same point, \(\boxed{yes}\). Reducing agrees, since \(\tfrac{8}{14}\) and \(\tfrac{12}{21}\) both come down to \(\tfrac{4}{7}\).
Practice
A dial stores \(\tfrac{5}{8}\) as \(\tfrac{x}{40}\). Cross-multiplying gives \(5\times 40=8\times x\). What is \(x\)?
Show the solution
Cross-multiplying \(\dfrac{x}{40} = \dfrac{5}{8}\) gives \(8 \times x = 5 \times 40 = 200\), so \(x = 200 \div 8 = \boxed{25}\). Checking, \(\dfrac{25}{40}\) reduces by \(5\) back to \(\dfrac{5}{8}\).
Practice
A scanner counted \(162\) clean jars out of \(252\) total. Reduce \(\tfrac{162}{252}\) all the way to simplest form \(\tfrac{a}{b}\).
Show the solution
Factor both numbers. \(162 = 2 \times 3^4\) and \(252 = 2^2 \times 3^2 \times 7\), so the shared part is one \(2\) and two \(3\)s, which makes \(\gcd(162,252) = 2 \times 3 \times 3 = 18\). Dividing, $$\frac{162}{252} = \frac{162 \div 18}{252 \div 18} = \frac{9}{14}.$$ The clean-jar share is \(\boxed{9/14}\). Since \(9 = 3^2\) and \(14 = 2 \times 7\) share no prime, it is fully reduced.
Practice
A machine stamps fractions equal to \(\tfrac{3}{8}\) with bottom \(\leq 80\). Allowed bottoms are multiples of \(8\). How many such fractions fit?
Show the solution
Every fraction equal to \(\tfrac{3}{8}\) is \(\tfrac{3k}{8k}\), so its bottom is \(8k\). Needing \(8k \le 80\) gives \(k \le 10\), so the bottoms run \(8, 16, 24, \dots, 80\). That is \(\boxed{10}\) fractions. Because \(\tfrac{3}{8}\) is already in lowest terms, every equal name has a bottom that is a multiple of \(8\).
Practice
Prize wheel: \(30\) slots labelled \(\tfrac{n}{30}\). Sticker when \(\gcd(n,30)=1\). \(30=2\times 3\times 5\), so \(n\) must miss multiples of \(2\), \(3\), \(5\). How many slots get stickers?
Show the solution
A slot wins when \(n\) misses every prime factor of \(30 = 2 \times 3 \times 5\). Cross out the evens, the multiples of \(3\), and the multiples of \(5\), and what survives from \(1\) to \(30\) is \(1, 7, 11, 13, 17, 19, 23, 29\). That is \(\boxed{8}\) slots. With only \(30\) numbers in play, listing the survivors is faster than any formula.