QuanticaAlgebra IOpen in the courseA ratio compares two amounts at one moment. This lesson is about two amounts that both change, and about the combination of them that stays fixed while they do. Sometimes that combination is the quotient of the two amounts, the same number at every moment. Sometimes it is the product. The first two problems have a fixed quotient.
Problem
A press stamps tags out of sheets of metal, and every sheet gives the same number of tags. A run of 12 sheets gives 84 tags. A later run uses 36 sheets. Enter the number of tags the later run gives.
Show a hint
- Every sheet gives the same number of tags, so the two runs differ only in how many sheets went through. Compare 36 sheets with 12 sheets.
- \(36\div 12=3\), so the later run is three copies of the first run.
Show the full solution
The later run uses \(36\div 12=3\) times as many sheets, and every sheet gives the same number of tags, so the tag count is \(3(84)=\boxed{252}\). How many tags one sheet gives was never needed, only the fact that it does not change.
Problem
The same press, the same sheets, so 12 sheets still give 84 tags. A new run uses 55 sheets, and 55 is not a whole number of twelves. Work out how many tags one sheet gives first, then enter the number of tags the 55-sheet run gives.
Show a hint
- The 12-sheet run gives 84 tags and every sheet gives the same number, so one sheet's share is a division.
- \(84\div 12=7\) tags per sheet, and the run has 55 sheets.
Show the full solution
One sheet gives \(84\div 12=7\) tags, so 55 sheets give \(7\times 55=\boxed{385}\). Working out the per-sheet count first handles any run, while multiplying the 12-sheet run by a whole-number factor only works when the new sheet count is a multiple of 12.
Problem
The mass of a steel sheet is directly proportional to its area. A sheet of area 18 square meters has mass 234 kilograms. Enter the mass in kilograms of a sheet of area 31 square meters.
Show a hint
- Here \(k\) is the mass of one square meter of the steel. The 18 square meter sheet is enough to find it.
- \(234\div 18=13\), so every square meter has mass 13 kilograms.
Show the full solution
\(k=234\div 18=13\) kilograms per square meter, so the larger sheet has mass \(13(31)=\boxed{403}\). Going from 18 straight to 31 without \(k\) means multiplying 234 by \(\frac{31}{18}\), which is not a whole number, so finding \(k\) first keeps every step in whole numbers.
Problem
Dyeing fabric uses an amount of dye directly proportional to the mass of the fabric. Dyeing 14 kilograms of fabric uses 6 grams of dye. Enter the number of grams of dye needed for 35 kilograms of the same fabric.
Show a hint
- \(k\) is the dye used per kilogram of fabric, and it is not a whole number here.
- \(\frac{6}{14}=\frac{3}{7}\) grams per kilogram, and 35 is a multiple of 7.
Show the full solution
\(k=\frac{6}{14}=\frac{3}{7}\) grams of dye per kilogram of fabric, so 35 kilograms need \(\frac{3}{7}(35)=\boxed{15}\). A constant of proportionality is a rate, and there is no reason a rate should come out whole. Scaling gives the same value, since \(35\div 14=\frac{5}{2}\) and \(\frac{5}{2}(6)=15\).
A fixed quotient is one way for two amounts to vary together. A fixed product is the other. Cut a strip into equal pieces and the number of pieces times the length of one piece is the length of the whole strip every time, so more pieces means shorter pieces. Two amounts whose product is the same at every pair are inversely proportional.
Problem
All the chairs in a hall are set out in rows, with the same number of chairs in every row, and every arrangement uses all of the chairs. One arrangement is 18 rows of 42 chairs. Another arrangement has 27 rows. Enter the number of chairs in each row of the 27-row arrangement.
Show a hint
- Both arrangements use the same chairs, so the number of rows times the number of chairs in each row comes out the same both times.
- 27 rows is \(\frac{3}{2}\) times 18 rows, so each row of the 27-row arrangement holds \(\frac{2}{3}\) as many chairs.
Show the full solution
From 18 rows to 27 rows the row count is multiplied by \(\frac{27}{18}=\frac{3}{2}\), so the number in each row is divided by that same factor, giving \(42\div\frac{3}{2}=\boxed{28}\). The product of the two counts is fixed at \(18(42)=756\), which is what inverse proportion means here, and \(756\div 27=28\) confirms it.
Problem
On a level balance beam, the mass hung on one side times its distance from the pivot is the same number for every load that balances the beam. A 27 kilogram mass balances it at 16 centimeters from the pivot. Enter the distance in centimeters at which a 12 kilogram mass balances the same beam.
Show a hint
- The fixed number here is a mass times its distance from the pivot, and the 27 kilogram load gives it.
- \(27(16)=432\), so mass times distance is 432 for any load that balances.
Show the full solution
\(k=27(16)=432\), so the 12 kilogram mass balances at \(432\div 12=\boxed{36}\). The lighter mass sits further from the pivot because the product has to stay 432. Scaling agrees, since the mass is multiplied by \(\frac{12}{27}=\frac{4}{9}\) and \(16\div\frac{4}{9}=36\).
A description or a table of pairs does not say which relationship holds, so test it. Divide \(y\) by \(x\) at each pair and see whether one number comes out every time. Multiply \(x\) by \(y\) at each pair and do the same. At most one of the two tests passes, and for some amounts that vary together neither one does.
Problem
A table lists three pairs from one relationship. $$\begin{array}{c|ccc} x & 8 & 12 & 20 \\ \hline y & 45 & 30 & 18 \end{array}$$ Decide whether \(y\) is directly proportional to \(x\) or inversely proportional to \(x\), then enter the value of \(y\) when \(x=40\).
Show a hint
- Run both tests on the first two pairs. Only one of \(\frac{y}{x}\) and \(xy\) repeats.
- \(8(45)=360\) and \(12(30)=360\), so check the third pair against 360 as well.
Show the full solution
The quotients \(\frac{45}{8}\), \(\frac{30}{12}\) and \(\frac{18}{20}\) are three different numbers, while \(8(45)\), \(12(30)\) and \(20(18)\) are all 360, so \(xy=360\) and \(y=360\div 40=\boxed{9}\). One pair can never settle which relationship holds, since a single pair has both a quotient and a product. Two pairs are the least that can rule one out.
Problem
A taxi company charges the same way for every trip. A 5 kilometre trip costs 17 dollars and a 12 kilometre trip costs 31 dollars. Test whether the fare is directly proportional to the distance, or inversely proportional, before assuming either. Enter the cost in dollars of a 20 kilometre trip.
Show a hint
- Test the two given trips. \(\frac{17}{5}\) and \(\frac{31}{12}\) are different, and \(17(5)\) and \(31(12)\) are different, so neither the quotient nor the product is fixed.
- The extra 7 kilometres between the two trips cost \(31-17=14\) dollars, and the pickup charge is the same on both trips.
Show the full solution
The extra 7 kilometres cost \(31-17=14\) dollars, so each kilometre costs \(14\div 7=2\) dollars, the pickup charge is \(17-5(2)=7\) dollars, and 20 kilometres costs \(20(2)+7=\boxed{47}\). Neither \(\frac{y}{x}\) nor \(xy\) is fixed here, since \(\frac{17}{5}\) and \(\frac{31}{12}\) differ and so do \(17(5)\) and \(31(12)\), so the fare is proportional to neither. Assuming direct proportion and multiplying the 5 kilometre fare by 4 gives 68, which charges the 7 dollar pickup four times over instead of once.
The comparison can be with a power of \(x\). Saying \(y\) is directly proportional to \(x^2\) means \(\frac{y}{x^2}\) is the same number at every pair, so \(y=kx^2\). Saying \(y\) is inversely proportional to \(x^2\) means \(yx^2\) is the same number at every pair, so \(y=\frac{k}{x^2}\). One known pair still gives \(k\).
Problem
The area of a badge is directly proportional to the square of its height. A badge 6 centimeters tall has area 27 square centimeters. A second badge of the same shape is 14 centimeters tall. Enter its area in square centimeters.
Show a hint
- The constant here is the area divided by the square of the height, not the area divided by the height, so divide \(27\) by \(6^2\) rather than by \(6\).
- \(\frac{27}{36}=\frac{3}{4}\), and \(14^2=196\).
Show the full solution
The constant is \(k=\frac{27}{6^2}=\frac{27}{36}=\frac{3}{4}\), so the taller badge has area \(\frac{3}{4}(14^2)=\frac{3}{4}(196)=\boxed{147}\). Dividing \(27\) by \(6\) instead gives \(4.5\) and then \(63\), which would be the answer if the area were proportional to the height rather than to its square.
Problem
A meter reads the light falling on a screen, and the reading is inversely proportional to the square of the screen's distance from the lamp. At a distance of 3 metres the meter reads 400 units. Enter the reading at a distance of 15 metres.
Show a hint
- Inversely proportional to the square of the distance means the reading times the square of the distance is the same number at every distance, not the reading times the distance.
- The distance is multiplied by \(15\div 3=5\), so the square of the distance is multiplied by \(25\).
Show the full solution
The distance is multiplied by \(15\div 3=5\), so the reading is divided by \(5^2=25\), giving \(400\div 25=\boxed{16}\). Finding \(k=400(3^2)=3600\) and computing \(\frac{3600}{15^2}\) gives 16 as well, with larger numbers. Dividing by 5 rather than 25 gives 80, which is what scaling by the distance instead of by its square produces.
When only a factor is wanted, \(k\) never has to be found. Under direct proportion, multiplying \(x\) by a number multiplies \(y\) by that same number. Under inverse proportion it divides \(y\) by that number. With a power of \(x\), the factor is raised to that power before it is used. The last two problems are much shorter this way.
Problem
A rectangle is reshaped with its area kept the same, so its length and its width are inversely proportional. The length is changed to \(\frac{5}{8}\) of what it was. Enter the number the width is multiplied by, as a fraction in lowest terms.
Show a hint
- No length, width or area is given, so the constant \(k\) cannot be found. Work from the fact that the product of length and width is the same before and after.
- Let the old length and width be \(L\) and \(W\), and let the width be multiplied by \(c\). The old area is \(LW\) and the new area is \(\frac{5}{8}L \cdot cW\), and those two are equal.
Show the full solution
The product of length and width is the same before and after, so if the length is multiplied by \(\frac{5}{8}\), the width must be divided by \(\frac{5}{8}\), and dividing by \(\frac{5}{8}\) is the same as multiplying by \(\boxed{\frac{8}{5}}\). No length, width or area was given, so \(k\) is unknown here, and the factor was found without it.
Problem
Storage bins in one product line all have the same shape. For that shape the outside surface area is directly proportional to the square of the height, and the volume is directly proportional to the cube of the height. A bin 10 centimetres tall has surface area 280 square centimetres and volume 300 cubic centimetres. A larger bin of the same shape has volume 2400 cubic centimetres. Enter its surface area in square centimetres.
Show a hint
- The height of the larger bin is not given, so work in factors instead. Both volumes are known, so start by dividing one by the other.
- The volume is multiplied by \(2400\div 300=8\), and the volume is proportional to the cube of the height.
Show the full solution
The volume is multiplied by \(2400\div 300=8\), and volume is proportional to the cube of the height, so the height is multiplied by 2 because \(2^3=8\). Surface area is proportional to the square of the height, so it is multiplied by \(2^2=4\), giving \(280(4)=\boxed{1120}\). Multiplying 280 by 8 instead gives 2240, but 8 is the volume factor and a surface area is two lengths multiplied together, so the height factor of 2 belongs in the product twice and not three times. A 6 by 5 by 10 box has volume 300 and surface area 280, and doubling every edge gives 2400 and 1120.
Most of these problems came down to one fixed number, found from one pair and then used on another, or skipped when only a factor was wanted. The taxi fare needed two, which put it outside both kinds of proportion. Lesson 6.6, Joint Proportion and Rates, has one amount varying with two others at once, so the statement is \(x=kyz\).
Practice these ideas
Practice
The cost of nylon rope is directly proportional to its length. A 7 meter piece costs 42 dollars. Enter the cost of a 35 meter piece of the same rope, in dollars.
Show the solution
\(k=42\div 7=6\) dollars per meter, so 35 meters cost \(6(35)=\boxed{210}\). As a check, 35 meters is five 7 meter pieces, and \(5(42)=210\).
Practice
\(y\) is directly proportional to \(x\), and \(y=51\) when \(x=17\). Enter the value of \(y\) when \(x=23\).
Show the solution
\(k=\frac{51}{17}=3\), so \(y=3x\) and \(y=3(23)=\boxed{69}\). Since 23 is not a multiple of 17, there is no whole-number scaling between the two pairs, and \(y=3x\) is true at every \(x\), including this one.
Practice
\(x\) and \(y\) are inversely proportional, and \(y=24\) when \(x=11\). Enter the value of \(y\) when \(x=33\).
Show the solution
\(k=24(11)=264\), so \(y=264\div 33=\boxed{8}\). Tripling \(x\) from 11 to 33 makes \(y\) a third of what it was, so \(24\div 3=8\) gives the same answer without finding \(k\).
Practice
A fixed volume of soup is ladled out equally into bowls, with none left over. Enter the word direct or the word inverse for the relationship between the number of bowls and the amount of soup in each bowl.
Show the solution
The number of bowls times the amount in each bowl equals the fixed total, so the amount in each bowl is the total divided by the number of bowls, which makes the relationship \(\boxed{\text{inverse}}\). Doubling the number of bowls halves the amount in each, while in a direct proportion doubling one would double the other.
Practice
\(y\) is directly proportional to \(x\), and \(y=21\) when \(x=6\). Enter the constant of proportionality \(k\) as a fraction in lowest terms.
Show the solution
\(k=\frac{21}{6}=\boxed{\frac{7}{2}}\). A constant of proportionality is a rate, so a fraction is a perfectly ordinary value for it.
Practice
The mass of a bar of alloy is directly proportional to its length. A bar 26 centimeters long has mass 65 grams. Enter the length in centimeters of a bar of the same alloy with mass 100 grams.
Show the solution
The constant of proportionality is \(\frac{65}{26}=\frac{5}{2}\) grams per centimeter, so the length is \(100\div\frac{5}{2}=\boxed{40}\). Scaling works too, since the mass is multiplied by \(\frac{100}{65}=\frac{20}{13}\) and \(26\cdot\frac{20}{13}=40\).
Practice
\(y\) is inversely proportional to \(x\), and \(y=14\) when \(x=6\). Enter the value of \(x\) when \(y=4\).
Show the solution
\(xy=14(6)=84\) for every pair, so \(4x=84\) and \(x=\boxed{21}\). Dividing \(y\) by \(\frac{7}{2}\) to get from \(14\) to \(4\) multiplies \(x\) by \(\frac{7}{2}\), and \(6\cdot\frac{7}{2}=21\) as well.
Practice
A fixed load of gravel fills 24 barrows when each barrow carries 35 kilograms. Enter the number of kilograms in each barrow when the same load is spread evenly over 42 barrows.
Show the solution
The whole load is \(24(35)=840\) kilograms, and spread over 42 barrows that gives \(840\div 42=\boxed{20}\). Check, \(42(20)=840\) kilograms, the same load. Barrow count times kilograms per barrow stays fixed, which is what inverse proportion means.
Practice
A table lists three pairs from one relationship. $$\begin{array}{c|ccc} x & 3 & 7 & 12 \\ \hline y & 12 & 28 & 48 \end{array}$$ Decide whether \(y\) is directly proportional to \(x\) or inversely proportional to \(x\), then enter the value of \(y\) when \(x=20\).
Show the solution
All three quotients equal 4 while the products 36, 196 and 576 differ, so \(y=4x\) and \(y=4(20)=\boxed{80}\). Checking only one pair proves nothing, since a single pair has both a quotient and a product.
Practice
A fixed number of tiles is laid in rows of equal length, so the number of rows and the number of tiles in each row are inversely proportional. The number of rows is tripled. Enter the number that the tile count in each row is multiplied by, as a fraction in lowest terms.
Show the solution
Writing the new counts as \(3r\) rows and \(n\) tiles per row, \(3r\cdot n=rt\) gives \(n=\frac{t}{3}\), so the tile count in each row is multiplied by \(\boxed{\frac{1}{3}}\). The factors \(3\) and \(\frac{1}{3}\) multiply to \(1\), which is what keeps the total the same.
Practice
\(y\) is directly proportional to \(x^2\), and \(y=48\) when \(x=4\). Enter the value of \(y\) when \(x=9\).
Show the solution
\(k=\frac{48}{4^2}=3\), so \(y=3(9^2)=\boxed{243}\). Using \(k=\frac{48}{4}=12\) instead gives \(12(9)=108\), which is the answer only if \(y\) is proportional to \(x\) rather than to \(x^2\).
Practice
\(y\) is inversely proportional to \(x^2\), and \(y=50\) when \(x=3\). Enter the value of \(y\) when \(x=5\).
Show the solution
The constant \(yx^2\) equals \(50(3^2)=450\), so \(y=\frac{450}{25}=\boxed{18}\). Ignoring the square and using \(yx=150\) gives \(30\) instead.
Practice
A courier charges a fixed handling amount plus a further amount for each kilogram. A 4 kilogram parcel costs 29 dollars and a 9 kilogram parcel costs 54 dollars. Enter the cost in dollars of a 15 kilogram parcel.
Show the solution
The extra 5 kilograms cost \(54-29=25\) dollars, so each kilogram costs 5 dollars, the handling amount is \(29-4(5)=9\) dollars, and a 15 kilogram parcel costs \(15(5)+9=\boxed{84}\). Scaling 29 by \(\frac{15}{4}\) gives 108.75, too high because the 9 dollar handling amount is then counted 3.75 times instead of once.
Practice
For solid cubes cut from one material, the mass is directly proportional to the cube of the edge length. Such a cube with edge length 2 centimetres has mass 56 grams. Enter the mass in grams of a cube of the same material with edge length 5 centimetres.
Show the solution
\(k=\frac{56}{2^3}=7\) grams per cubic centimetre, so the cube with edge length 5 centimetres has mass \(7(125)=\boxed{875}\). Scaling the edge by \(\frac{5}{2}\) scales the mass by \(\left(\frac{5}{2}\right)^3=\frac{125}{8}\), and \(56\cdot\frac{125}{8}=875\) as well.
Practice
The reading on a light meter is inversely proportional to the square of the distance from the meter to the lamp. The meter is moved so that this distance becomes \(\frac{2}{5}\) of what it was. Enter the number the reading is multiplied by, as a fraction in lowest terms.
Show the solution
The denominator \(d^2\) is multiplied by \(\left(\frac{2}{5}\right)^2=\frac{4}{25}\), so the reading is divided by \(\frac{4}{25}\), the same as multiplying by \(\boxed{\frac{25}{4}}\). Leaving out the square gives \(\frac{5}{2}\), the common wrong answer.
Practice
Among statues of one shape, mass is directly proportional to the cube of the height. One such statue is 40 centimeters tall and has mass 33 kilograms. A second statue of that shape has mass 891 kilograms. Enter the height of the second statue in centimeters.
Show the solution
The mass is multiplied by \(891\div 33=27\), and mass is proportional to the cube of the height, so the height is multiplied by the number whose cube is 27, namely 3, giving \(40(3)=\boxed{120}\). The check is that \(120^3=1728000\) is 27 times \(40^3=64000\). The common slip is \(40(27)=1080\), but 27 is the factor for the mass and 3 is the factor for the height.
Practice
A table lists three pairs from one relationship. $$\begin{array}{c|ccc} x & 2 & 3 & 6 \\ \hline y & 675 & 300 & 75 \end{array}$$ Neither \(\frac{y}{x}\) nor \(xy\) is the same at every pair, but \(y\) is inversely proportional to a power of \(x\). Enter the value of \(y\) when \(x=5\).
Show the solution
The products \(yx^2\) are \(675(4)\), \(300(9)\) and \(75(36)\), all 2700, so \(yx^2=2700\) and \(y=\frac{2700}{25}=\boxed{108}\). Doubling \(x\) from 3 to 6 leaves \(y\) at a quarter of its value rather than a half, which is what makes the power 2 and not 1.