QuanticaAlgebra IOpen in the courseChapter 5 removed letters from a system one at a time. Chapter 6 opens by putting one in. Prealgebra chapter 7 settled what \(a:b\) says. What is new here is algebraic. Naming the size of one part with a letter turns a ratio statement into an equation, and chapters 4 and 5 already solve equations. Start with a ratio and a total.
Problem
A bindery finishes 552 booklets, stapled to sewn in the ratio \(7:5\). Doubling both parts of \(7:5\) gives \(14:10\), which totals \(24\), and tripling gives \(21:15\), which totals \(36\). Keep scaling both parts by the same number until the two counts total \(552\), then enter the number of stapled booklets.
Show a hint
- The scaled totals \(24\) and \(36\) are both multiples of \(12\), and \(12\) is \(7+5\). Scaling both parts by the same number multiplies the total by that number as well.
- \(552\div 12=46\), so both parts of \(7:5\) get multiplied by \(46\).
Show the full solution
Since \(7+5=12\) and \(552\div 12=46\), both parts get multiplied by \(46\), and the stapled count is \(7(46)=\boxed{322}\). The sewn count is \(5(46)=230\), and \(322+230=552\). Every scaled total is a multiple of \(12\), so dividing \(552\) by \(12\) gives the multiplier in one step instead of testing \(2\), \(3\), \(4\) and on up.
Problem
A stadium has 493 seats, covered seats to open seats in the ratio \(9:8\). Write the two counts as \(9k\) and \(8k\) for a single number \(k\), form one equation from the total, and enter the number of covered seats.
Show a hint
- Let \(k\) be the size of one part, so the covered seats number \(9k\), the open seats number \(8k\), and the two counts together make \(493\).
- \(9k+8k\) collects to \(17k\), so the equation reads \(17k=493\), a one-letter equation of the kind 4.2 solves.
Show the full solution
\(9k+8k=493\) collects to \(17k=493\), so \(k=29\) and the covered seats number \(9(29)=\boxed{261}\). The open seats number \(8(29)=232\), and \(261+232=493\). Naming the part size \(k\) makes the whole problem the single division \(493\div 17\).
Problem
A workshop owns 378 hand tools, all of them hammers or chisels, with hammers to chisels in the ratio \(5:9\). A student computes \(\frac{5}{9}\) of \(378\) and reports \(210\) hammers. Enter the correct number of hammers.
Show a hint
- The ratio \(5:9\) compares hammers with chisels, while \(378\) is the total of both kinds. A fraction of that total is taken over both kinds together, not over the chisels alone.
- The tools split into \(5+9=14\) equal parts, and the hammers are \(5\) of those parts.
Show the full solution
The tools split into \(5+9=14\) equal parts, so one part is \(378\div 14=27\) tools and the number of hammers is \(5(27)=\boxed{135}\). There are \(9(27)=243\) chisels, and \(135+243=378\). Hammers are \(\frac{5}{14}\) of the tools, not \(\frac{5}{9}\), and with \(210\) hammers only \(168\) chisels remain, more hammers than chisels, the reverse of \(5:9\).
Problem
A nursery grows 984 saplings, oak, maple and pine, in the ratio \(6:7:11\). Write the three counts as \(6k\), \(7k\) and \(11k\), and enter the number of pine saplings.
Show a hint
- All three counts are built from the same \(k\), so they add to \(6k+7k+11k\), and that sum is \(984\).
- \(6k+7k+11k\) collects to \(24k\), so \(24k=984\).
Show the full solution
\(6k+7k+11k=984\) collects to \(24k=984\), so \(k=41\) and the pine count is \(11(41)=\boxed{451}\). The others are \(6(41)=246\) oak and \(7(41)=287\) maple, and \(246+287+451=984\). With three parts the sum has one more term than with two, and the rest of the method is the same.
Writing \(x:y=a:b\) is writing \(\frac{x}{y}=\frac{a}{b}\), and that is an equation in two letters. Multiply both sides by \(y\) and then by \(b\), which is 4.4's move done twice, and what is left is \(bx=ay\). Each numerator is multiplied whole, parentheses and all. Neither side is a fraction any more, so \(bx=ay\) is linear in \(x\) and \(y\).
Problem
The crates and pallets in a depot are in the ratio \(12:7\), and the depot has 224 pallets. Let \(c\) be the number of crates, so that statement is the equation \(\frac{c}{224}=\frac{12}{7}\). Cross-multiply, solve, and enter the number of crates.
Show a hint
- Both sides are single fractions, so clear the denominators. Multiply both sides by \(224\) and then by \(7\), and see what is left.
- Clearing gives \(7c=224\times 12\). Since \(224=7\times 32\), divide by \(7\) before multiplying by \(12\), which keeps the numbers small.
Show the full solution
Cross-multiplying \(\frac{c}{224}=\frac{12}{7}\) gives \(7c=2688\), so \(c=\boxed{384}\). Check, \(\frac{384}{224}\) reduces to \(\frac{12}{7}\). In the ratio, crates come first, so \(12\) is the numerator on the same side as \(c\). With the \(12\) and \(7\) swapped the result would be \(\frac{224\times 7}{12}\), not a whole number of crates.
Problem
The ratio of \(5(x-4)+3x\) to \(2(x+21)\) is \(3:1\). Collect like terms on each side first, then cross-multiply and enter the value of \(x\).
Show a hint
- A ratio of \(3:1\) says the first quantity is \(3\) times the second. Distribute the 2.1 way and combine like terms the 2.4 way on each side, then write that sentence as an equation.
- The sides collect to \(8x-20\) and \(2x+42\), so the statement reads \(8x-20=3(2x+42)\).
Show the full solution
Distributing gives \(5x-20+3x=8x-20\) on the left and \(2x+42\) on the right, so \(\frac{8x-20}{2x+42}=\frac{3}{1}\), and cross-multiplying gives \(8x-20=6x+126\), then \(2x=146\) and \(x=\boxed{73}\). Check, \(564\) to \(188\) is \(3\) to \(1\). Writing \(5(x-4)\) as \(5x-4\) instead gives \(x=65\), where the sides are \(500\) and \(172\), not a \(3:1\) pair.
An amount can be added to one part, taken from one part, or moved between the two, and the problem states the new ratio. Write the starting amounts as \(ak\) and \(bk\), adjust each by what moved, and that second statement is one equation in \(k\). The two ratios have separate multipliers, so \(13k\) and \(4k\) cannot also be \(10k\) and \(3k\).
Problem
A club's juniors and seniors are in the ratio \(11:8\). Then 42 juniors join, no one leaves, and the ratio of juniors to seniors is now \(3:2\). Enter how many juniors the club had before those 42 joined.
Show a hint
- The senior count never changes, so write both starting counts from the first ratio and adjust only the junior count.
- With \(11k\) juniors and \(8k\) seniors, the counts afterwards are \(11k+42\) and \(8k\), so the second statement is \(\frac{11k+42}{8k}=\frac{3}{2}\).
Show the full solution
With \(11k\) juniors and \(8k\) seniors, the ratio after the 42 join is \(\frac{11k+42}{8k}=\frac{3}{2}\). Cross-multiplying gives \(22k+84=24k\), so \(2k=84\) and \(k=42\), and the juniors before were \(11(42)=\boxed{462}\). Check, the \(336\) seniors are unchanged and \(504:336\) reduces to \(3:2\), so the \(504\) is the count after, not the one asked for. Naming both counts instead gives the system \(8j=11s\) and \(2j+84=3s\), and elimination reaches the same \(j=462\).
The same quantity can appear in two ratios with a different number in each. In \(x:y=7:6\) the quantity \(y\) is \(6\) parts, and in \(y:z=9:2\) it is \(9\), and those parts are not the same size. Scale each ratio, both parts by the same factor, until the two numbers for \(y\) agree. The three amounts are then in one ratio \(x:y:z\).
Problem
In a stockroom, the ratio of blue crates to green crates is \(15:8\), and the ratio of green crates to red crates is \(8:11\). There are \(165\) blue crates. Enter the number of red crates.
Show a hint
- Both ratios use the same number, \(8\), for the green crates, so blue, green, and red fit into one three-part ratio without rescaling either ratio.
- Write that three-part ratio, then compare its blue part with the \(165\) actual blue crates to get the multiplier that turns parts into counts.
Show the full solution
Both ratios use \(8\) for green, so blue to green to red is \(15:8:11\). The blue part gives \(15k=165\), so \(k=11\), and the red count is \(11k=\boxed{121}\). Green is \(8k=88\), and the checks pass, \(165:88=15:8\) and \(88:121=8:11\). The chain is direct only because the shared quantity has the same number in both ratios.
Problem
For three quantities \(x\), \(y\), and \(z\), \(x:y=16:9\) and \(y:z=6:13\). Enter \(\frac{z}{x}\) as a fraction in lowest terms.
Show a hint
- In one statement \(y\) is \(9\) parts and in the other it is \(6\), and those parts are not the same size. Scale each ratio, multiplying both of its parts by the same factor, until the two numbers for \(y\) match.
- The least common multiple of \(9\) and \(6\) is \(18\), so double both parts of \(16:9\) and triple both parts of \(6:13\).
Show the full solution
Doubling \(16:9\) gives \(32:18\) and tripling \(6:13\) gives \(18:39\), so with \(y\) at \(18\) parts in both, \(x:y:z=32:18:39\) and \(\frac{z}{x}=\boxed{\frac{39}{32}}\). Stacking the printed numbers gives \(\frac{13}{16}\) instead, which is the trap, since the \(9\) and the \(6\) stand for different sizes of part.
A ratio is one equation in two letters, so neither amount follows from it alone. Adding a second fact about the same two amounts makes a system, and 5.1 counted two equations in two letters as usually enough. It can be a total, a difference, one amount, or the ratio after a change. Cross-multiply the ratio, then use substitution or elimination.
Problem
A caterer orders bread rolls and pastries, rolls to pastries in the ratio \(11:6\). Rolls cost 2 dollars each, pastries cost 5 dollars each, and the order comes to 468 dollars. Enter the number of pastries.
Show a hint
- The ratio is one fact about the two counts and the total cost is another, so there are two equations in two letters here.
- With \(r\) rolls and \(p\) pastries, cross-multiplying the ratio gives \(6r=11p\), and the cost gives \(2r+5p=468\). Tripling the cost equation makes its \(r\)-term \(6r\), which matches the ratio equation.
Show the full solution
Cross-multiplying the ratio gives \(6r=11p\), and the cost gives \(2r+5p=468\). Tripling the cost equation makes it \(6r+15p=1404\), so replacing \(6r\) with \(11p\) leaves \(26p=1404\) and \(p=\boxed{54}\). Then \(r=99\), and \(2(99)+5(54)=468\) checks, though the \(99\) is the roll count and not the ask. Writing the counts as \(11k\) and \(6k\) instead gives \(52k=468\) and the same pair.
Problem
A library shelves fiction and nonfiction in the ratio \(11:7\). Then 45 books move from fiction to nonfiction, nothing is added or discarded, and the ratio of fiction to nonfiction reads \(4:3\). Enter how many fiction books the library had before the move.
Show a hint
- After the move there are \(45\) fewer fiction books and \(45\) more nonfiction books. Write both starting counts with one letter, using the first ratio.
- With \(11k\) fiction and \(7k\) nonfiction, the second ratio gives \(3(11k-45)=4(7k+45)\).
Show the full solution
With \(11k\) fiction and \(7k\) nonfiction the new ratio gives \(\frac{11k-45}{7k+45}=\frac{4}{3}\), so \(33k-135=28k+180\), then \(5k=315\) and \(k=63\). The fiction count before the move was \(11(63)=\boxed{693}\). The nonfiction count was \(441\), and after the move \(648:486\) reduces to \(4:3\). The usual slip is adding the \(45\) to fiction and taking it off nonfiction, which gives \(5k=-315\) and a negative count, so it rules itself out.
Problem
A festival sells full-price, student, and child tickets. Of the tickets sold, full-price to student is \(7:4\), and student to child is \(6:11\). The festival then sells another 75 child tickets, with no change to the other two counts, and full-price to child is now \(7:9\). Enter the number of student tickets sold.
Show a hint
- In one ratio the student tickets are \(4\) parts and in the other they are \(6\), and those parts are not the same size. Combine the two statements into a single three-part ratio before naming any letter.
- Matching the student number at \(12\) makes the three-part ratio \(21:12:22\), so the counts are \(21k\), \(12k\) and \(22k\), and the last fact cross-multiplies to \(9(21k)=7(22k+75)\).
Show the full solution
Matching the student number at \(12\) turns \(7:4\) into \(21:12\) and \(6:11\) into \(12:22\), so the counts are \(21k\), \(12k\) and \(22k\). The last fact gives \(\frac{21k}{22k+75}=\frac{7}{9}\), so \(189k=154k+525\), then \(35k=525\) and \(k=15\). The student count is \(12(15)=\boxed{180}\). Check, full-price \(315\) against child \(405\) is \(7:9\). The letter solved for is \(k\), so \(15\) is a part size and not a ticket count.
Every problem here put one letter on the size of a part or cross-multiplied a ratio into \(bx=ay\), and no count came without a second fact. Lesson 6.2, Conversion Factors, uses a ratio of another kind, where the two parts are one amount written in two units. The question there is what such a ratio is worth and what multiplying by it does.
Practice these ideas
Practice
A crate holds 340 tiles, each one either plain or patterned. The ratio of plain tiles to patterned tiles is \(9:11\). Enter the number of plain tiles.
Show the solution
\(9k+11k=340\) gives \(20k=340\), so \(k=17\) and the plain tiles number \(9(17)=\boxed{153}\). The patterned count is \(11(17)=187\), and \(153+187=340\).
Practice
A crate holds pears and quinces and nothing else, in the ratio \(6:11\). Enter the fraction of the fruit that is quinces, in lowest terms.
Show the solution
The counts are \(6k\) and \(11k\), so the crate holds \(17k\) pieces and the quinces are \(\frac{11k}{17k}\) of it, which is \(\boxed{\frac{11}{17}}\). A common wrong answer is \(\frac{11}{6}\), the quince count over the pear count rather than over the whole crate, and a share of the crate cannot be more than \(1\).
Practice
In a choir, the ratio of altos to sopranos is \(8:13\), and there are 296 altos. Enter the number of sopranos.
Show the solution
The altos are \(8\) parts, so \(8k=296\) and \(k=37\), which makes the sopranos \(13(37)=\boxed{481}\). Check, \(296:481\) reduces to \(8:13\). No total is stated here and none is needed, since one amount fixes \(k\) on its own.
Practice
A jar holds only amber, jade and onyx marbles, in the ratio \(9:16:5\). Enter the fraction of the marbles in the jar that are jade, in lowest terms.
Show the solution
The counts are \(9k\), \(16k\) and \(5k\), so the jar holds \(30k\) marbles and the jade share is \(\frac{16k}{30k}=\boxed{\frac{8}{15}}\). With three parts the share of the whole is still one part over the sum of all the parts, and \(\frac{16}{9}\) or \(\frac{16}{5}\) would each compare jade with one other colour instead.
Practice
A silo holds 494 sacks of wheat, rye and oats in the ratio \(5:9:12\). Enter the number of sacks of oats.
Show the solution
Write the counts as \(5k\), \(9k\), and \(12k\). Then \(26k=494\), so \(k=19\) and the oat count is \(12(19)=\boxed{228}\). The other two are \(5(19)=95\) wheat and \(9(19)=171\) rye, and \(95+171+228=494\).
Practice
A workshop stocks screws, bolts and rivets in the ratio \(11:6:4\), and there are 156 bolts. Enter the total number of screws, bolts and rivets.
Show the solution
Each part is \(k\) pieces, and the bolts are \(6\) parts, so \(6k=156\) and \(k=26\). The three counts together are \(11+6+4=21\) parts, so the total is \(21\times 26=\boxed{546}\). Counting by kind, that is \(286\) screws, \(156\) bolts and \(104\) rivets, and \(286+156+104=546\).
Practice
The equation \(\frac{x}{84}=\frac{17}{12}\) holds. Cross-multiply and enter \(x\).
Show the solution
Cross-multiplying gives \(12x=1428\), so \(x=\boxed{119}\). Cancelling the \(12\) into the \(84\) first turns the whole thing into \(17\times 7\), which is faster than multiplying and then dividing.
Practice
Cross-multiply \(\frac{7x-12}{2x+50}=\frac{5}{3}\), collect the \(x\)-terms on one side and the numbers on the other, and enter \(x\).
Show the solution
Cross-multiply, \(3(7x-12)=5(2x+50)\), so \(21x-36=10x+250\), then \(11x=286\) and \(x=\boxed{26}\). Check, at \(x=26\) the fraction is \(\frac{170}{102}\), which reduces to \(\frac{5}{3}\).
Practice
Two accounts hold money in the ratio \(17:8\), and the first holds 288 dollars more than the second. Enter the balance of the smaller account, in dollars.
Show the solution
With balances \(17k\) and \(8k\), the gap is \(17k-8k=9k\), so \(9k=288\) and \(k=32\), and the smaller balance is \(8(32)=\boxed{256}\). The other balance is \(544\), and \(544-256=288\), with \(544:256\) equal to \(17:8\) in lowest terms. A difference is enough to find \(k\), the same as a total would be.
Practice
A stall sells notebooks and folders in the ratio \(9:7\). Notebooks cost 6 dollars each, folders cost 11 dollars each, and the stall takes 1572 dollars for them. Enter the number of folders sold.
Show the solution
With \(9k\) notebooks and \(7k\) folders, the money gives \(54k+77k=1572\), so \(131k=1572\) and \(k=12\), which makes the folder count \(7(12)=\boxed{84}\). There are \(108\) notebooks, and \(648+924=1572\) checks. The ratio alone fixes neither count, and the takings are the second fact that does.
Practice
Two positive whole numbers are in the ratio \(13:6\), and both are less than \(200\). Enter how many such pairs there are.
Show the solution
Every such pair is \(13k\) and \(6k\) for a positive whole number \(k\), and \(6k\) is the smaller of the two, so the only condition is that \(13k\) stays under \(200\). Since \(13\times 15=195\) and \(13\times 16=208\), \(k\) can be \(1\) through \(15\), a count of \(\boxed{15}\). Without the limit there would be one such pair for every \(k\), endlessly many.
Practice
In a model kit, pegs and struts come in the ratio \(10:3\), and struts and panels come in the ratio \(3:4\). A batch of these parts has \(76\) panels. Enter the number of pegs in the batch.
Show the solution
The struts match at \(3\) in both ratios, so pegs to struts to panels is \(10:3:4\). The panels give \(4k=76\), so \(k=19\) and the pegs number \(10(19)=\boxed{190}\). If the two strut counts had differed, both ratios would need scaling to a common strut count first.
Practice
A field holds 682 plants, all of them corn, beans or squash. Corn to beans is \(6:5\) and beans to squash is \(10:9\). Enter the number of squash plants.
Show the solution
Doubling \(6:5\) gives \(12:10\), which matches the beans in \(10:9\), so corn to beans to squash is \(12:10:9\). The three parts add to \(31\), one part is \(\frac{682}{31}=22\) plants, and squash is \(9(22)=\boxed{198}\). Corn and beans come out to \(264\) and \(220\), and \(264+220+198=682\).
Practice
A survey counts \(m\) walkers, \(n\) cyclists and \(p\) drivers, with \(m\) to \(n\) as \(5:12\) and \(n\) to \(p\) as \(8:9\). Scale so the two numbers for \(n\) agree, then enter \(\frac{p}{m}\) as a fraction in lowest terms.
Show the solution
Doubling \(5:12\) gives \(10:24\) and tripling \(8:9\) gives \(24:27\), so \(m:n:p=10:24:27\) and \(\frac{p}{m}=\boxed{\frac{27}{10}}\). Using the printed numbers straight gives \(\frac{9}{5}\), which is wrong because \(n\) is \(12\) parts in the first ratio and \(8\) parts in the second.
Practice
A pantry stocks honey jars and jam jars in the ratio \(10:7\). After 48 more honey jars arrive and no new jam jars, the ratio of honey to jam is \(8:5\). Enter the number of jam jars.
Show the solution
Write the counts as \(10k\) honey jars and \(7k\) jam jars. Then \(\frac{10k+48}{7k}=\frac{8}{5}\), and cross-multiplying gives \(50k+240=56k\), so \(6k=240\), \(k=40\) and the jam count is \(7(40)=\boxed{280}\). Check, the honey counts are \(400\) and \(448\), and \(448:280\) reduces to \(8:5\). The \(400\) is honey, not the count asked for.
Practice
A vault holds silver and copper coins, and the ratio of silver to copper is \(9:4\). The owner removes 32 silver coins and puts in 32 copper coins, after which the ratio of silver to copper is \(5:4\). Enter the number of silver coins the vault held at the start.
Show the solution
With \(9k\) silver and \(4k\) copper at the start, the new ratio gives \(4(9k-32)=5(4k+32)\), so \(36k-128=20k+160\), then \(16k=288\) and \(k=18\). The starting silver count is \(9(18)=\boxed{162}\). The vault began with \(162\) silver and \(72\) copper, and afterwards \(130:104\) reduces to \(5:4\).
Practice
At a gallery, the ratio of prints to paintings is \(16:7\). Then 26 prints are sold and 13 paintings are added, and the ratio of prints to paintings is now \(5:3\). Enter the total number of prints and paintings after the change.
Show the solution
With \(16k\) prints and \(7k\) paintings, the new ratio gives \(3(16k-26)=5(7k+13)\), so \(48k-78=35k+65\), then \(13k=143\) and \(k=11\). The gallery held \(176\) prints and \(77\) paintings, and afterwards \(150\) and \(90\), a total of \(\boxed{240}\). The check is that \(150:90\) reduces to \(5:3\), and the total fell from \(253\) because 26 works left and only 13 came in.
Practice
A cabinet holds oak, ash and birch dowels and nothing else, with oak to ash \(9:5\) and ash to birch \(15:8\). After 105 more birch dowels arrive, the birch count equals the ash count. Enter the total number of dowels the cabinet held before that delivery.
Show the solution
Triple \(9:5\) to get \(27:15\), so both statements now use \(15\) for the ash and oak to ash to birch is \(27:15:8\). With the counts \(27k\), \(15k\) and \(8k\), equal counts after the delivery mean \(8k+105=15k\), so \(7k=105\) and \(k=15\). The parts total \(50\), so the cabinet held \(50(15)=\boxed{750}\). The counts are \(405\) oak, \(225\) ash and \(120\) birch, and \(120+105=225\) as required.