QuanticaAlgebra IOpen in the courseWhat an amount is compared against is called the base, the number a percent is taken of. Twelve percent of one base and twelve percent of another are different amounts, so a percent alone leaves the amount unknown. Prealgebra chapter 9 did the arithmetic. New here is naming an unknown with a letter, which turns a percent sentence into an equation.
Problem
A quality check clears \(64\%\) of a shipment. Percent means per hundred, so \(64\%\) is the ratio \(\frac{64}{100}\). Reduce it to lowest terms and enter the fraction.
Show a hint
- Both parts of a ratio can be divided by the same number without changing it, and \(64\) and \(100\) share a factor larger than \(2\).
- That factor is \(4\), so divide above and below the bar by \(4\), then check whether the result reduces again.
Show the full solution
Dividing both parts by \(4\) gives \(\frac{64}{100}=\boxed{\frac{16}{25}}\). Since \(16\) and \(25\) share no common factor above \(1\), that is lowest terms. As a decimal it is \(64\div 100=0.64\), the same number with the point moved two places left.
Problem
A sensor logs \(\frac{7}{20}\) of its readings as out of range. A percent is a ratio whose second part is \(100\), so rewrite \(\frac{7}{20}\) with \(100\) below the bar. Enter the percent, just the number.
Show a hint
- Multiplying both parts of a ratio by the same number leaves it unchanged, so find the factor that turns the \(20\) below the bar into \(100\).
- \(20\times 5=100\), so multiply above the bar by \(5\) as well.
Show the full solution
Multiplying both parts by \(5\) gives \(\frac{7}{20}=\frac{35}{100}\), so the percent is \(\boxed{35}\). A second route gives the same number, \(\frac{7}{20}\times 100=35\), and it still works when the denominator does not divide \(100\).
Problem
A label states that a bottle of saline is \(6.5\%\) salt by mass, so the ratio of salt to total mass is \(\frac{6.5}{100}\). Lowest terms means the top and the bottom are whole numbers with no common factor other than \(1\). Enter \(6.5\%\) as a fraction in lowest terms.
Show a hint
- Multiplying the top and the bottom by the same number does not change the value, and multiplying by \(10\) removes the decimal point.
- That gives \(\frac{65}{1000}\), and \(65\) and \(1000\) share a factor of \(5\).
Show the full solution
Multiplying the top and the bottom by \(10\) gives \(\frac{6.5}{100}=\frac{65}{1000}\), and dividing both by \(5\) gives \(\boxed{\frac{13}{200}}\). Since \(13\) is prime and does not divide \(200\), that is lowest terms. As a decimal it is \(0.065\).
Problem
A machinist finds that \(\frac{5}{6}\) of a batch passes inspection. That fraction is already in lowest terms and no whole number times \(6\) gives \(100\), so it cannot be rescaled to a whole number over \(100\). Multiply \(\frac{5}{6}\) by \(100\) anyway and enter the exact percent as a fraction in lowest terms.
Show a hint
- Rescaling is blocked here, so the route left is plain arithmetic. Multiplying \(\frac{5}{6}\) by \(100\) multiplies the numerator by \(100\) and leaves the \(6\) underneath alone.
- \(\frac{5}{6}\times 100=\frac{500}{6}\), and \(500\) and \(6\) are both even.
Show the full solution
\(\frac{5}{6}\times 100=\frac{500}{6}\), and dividing the numerator and the denominator by \(2\) gives \(\boxed{\frac{250}{3}}\). The same percent is often written \(83\frac{1}{3}\%\). Long division gives \(250\div 3=83.333\ldots\), which never terminates, so \(83.3\%\) and \(83.33\%\) are roundings while \(\frac{250}{3}\%\) is exact.
A percent question names three numbers, an amount, a percent, and the whole the percent is taken of. The sentence \(a\) is \(p\%\) of \(b\) says \(a\) equals \(\frac{p}{100}\) of \(b\), and of means multiply, so the sentence is \(a=\frac{p}{100}b\). Two of the three are given and one is the letter. The next two problems leave out a different one.
Problem
A vineyard has 875 vines and \(24\%\) of them are a new cultivar. Here \(p=24\) and \(b=875\), and the amount \(a\) is missing. Substitute into \(a=\frac{p}{100}b\) and enter the number of new-cultivar vines.
Show a hint
- \(24\%\) is the number \(\frac{24}{100}\), and of means multiply, so the count is \(\frac{24}{100}\) times \(875\).
- \(\frac{24}{100}\) reduces to \(\frac{6}{25}\), and \(875\div 25=35\), so divide before multiplying.
Show the full solution
\(a=\frac{24}{100}(875)=\frac{6}{25}(875)=6(35)=\boxed{210}\). Reducing \(\frac{24}{100}\) first turns the work into \(875\div 25\) and \(6\times 35\), shorter than multiplying \(875\) by \(24\) and then dividing by \(100\).
Problem
A hall has 650 seats and 273 of them are booked. This time \(a=273\) and \(b=650\) are known and \(p\) is missing. Solve \(a=\frac{p}{100}b\) for \(p\) and enter the percent of the seats booked, just the number.
Show a hint
- Substituting gives \(273=\frac{p}{100}(650)\), and multiplying both sides by \(100\) clears the fraction.
- That leaves \(650p=27300\), a one-letter equation of the kind 4.2 solves in one division.
Show the full solution
From \(273=\frac{p}{100}(650)\), multiplying both sides by \(100\) gives \(650p=27300\), so \(p=\boxed{42}\). Dividing part by whole gives the same number, since \(\frac{273}{650}=0.42\) and \(0.42\times 100=42\).
The third case goes wrong most often. In \(a=\frac{p}{100}b\) the base is the quantity being multiplied, so when the base is what is missing the equation has to be undone rather than run forward. The next two problems are both of that kind, and each one names the number the wrong route produces.
Problem
At a theatre, 350 seats are sold, and that is \(28\%\) of the seats in the house. A student computes \(28\%\) of \(350\) and reports a house of \(98\) seats. Enter the correct number of seats in the house.
Show a hint
- The \(28\%\) is taken of the whole house, so the house is the whole \(b\) and the \(350\) is the amount \(a\).
- \(350=\frac{28}{100}b\), and multiplying both sides by \(100\) gives \(35000=28b\).
Show the full solution
From \(350=\frac{28}{100}b\), multiplying both sides by \(100\) gives \(28b=35000\), so \(b=\boxed{1250}\). Since \(350\) is the part rather than the whole, the correct move is to divide by \(\frac{28}{100}\) instead of multiplying, and \(98\) is smaller than the \(350\) seats already sold.
Problem
An inspector passes \(62\%\) of a shipment and sets the other 209 items aside. The 209 is not the \(62\%\) part, so work out what percent of the shipment it is before writing the equation. Enter the number of items in the shipment.
Show a hint
- The whole shipment is \(100\%\) of itself, and the set-aside items are what is left once the passing \(62\%\) is taken out.
- The set-aside items are \(38\%\), so the equation is \(209=\frac{38}{100}b\).
Show the full solution
The set-aside items are \(100-62=38\) percent of the shipment, so \(209=\frac{38}{100}b\) gives \(38b=20900\) and \(b=\boxed{550}\). Dividing \(209\) by \(0.62\) instead pairs the count with the wrong percent and gives about \(337\). The check is \(\frac{62}{100}(550)=341\), the number that passed, and \(341+209=550\).
The same amount is a different percent of different bases, and one percent taken against different bases gives different amounts. Fifteen percent of \(200\) is \(30\), and fifteen percent of \(800\) is \(120\). When a story states two totals, each percent belongs to the total its own sentence names, and adding the two gives a percent of neither.
Problem
A cannery runs two filling lines. The morning line fills 720 tins and \(45\%\) of them hold soup. The afternoon line fills 450 tins and \(16\%\) of them hold soup. Each percent is taken of its own line's run. Enter the total number of tins of soup.
Show a hint
- The \(45\%\) is taken of the morning run only and the \(16\%\) of the afternoon run only, so the two bases are different and the counts have to be worked out separately.
- The morning count is \(\frac{45}{100}(720)\), and the afternoon count is the same form with \(16\) as the percent and \(450\) as the base.
Show the full solution
The morning count is \(\frac{45}{100}(720)=324\) and the afternoon count is \(\frac{16}{100}(450)=72\), so the total is \(324+72=\boxed{396}\). Adding the percents and taking \(61\%\) of all \(1170\) tins gives \(713.7\), which is wrong because neither percent was measured against \(1170\).
Problem
A car park holds only vans and cars, 96 vans and 384 cars. Enter the percent of all the vehicles in the car park that are vans, just the number.
Show a hint
- The comparison is against every vehicle in the car park, and that total is not printed in the problem.
- The whole is \(96+384=480\), so the equation is \(96=\frac{p}{100}(480)\).
Show the full solution
The car park holds \(96+384=480\) vehicles, so \(96=\frac{p}{100}(480)\) gives \(480p=9600\) and \(p=\boxed{20}\). Taking \(384\) as the whole instead gives \(25\), which is 6.1's part-to-part comparison rather than a share of every vehicle.
The equation \(a=\frac{p}{100}b\) holds for any \(p\), not only for \(p\) between \(1\) and \(100\). With \(p\) above \(100\), \(\frac{p}{100}\) is greater than \(1\), so the amount comes out larger than the base. With \(p\) below \(1\), \(\frac{p}{100}\) is smaller than \(\frac{1}{100}\), so the base is more than a hundred times the amount.
Problem
Depot A holds 2380 tonnes of grain, which is \(175\%\) of what Depot B holds. Here \(\frac{175}{100}\) is greater than \(1\). Enter the number of tonnes Depot B holds.
Show a hint
- The 2380 tonnes is compared against what Depot B holds, so Depot B is the base. Call that base \(b\).
- \(2380=\frac{175}{100}b\), and \(\frac{175}{100}\) reduces to \(\frac{7}{4}\).
Show the full solution
Depot B is the base, so \(2380=\frac{175}{100}b=\frac{7}{4}b\). Multiplying both sides by \(4\) gives \(7b=9520\), so \(b=\boxed{1360}\). With a percent above \(100\) the amount is larger than the base, so any answer above 2380 would be wrong.
Problem
A quarry's tailings are \(0.15\%\) recoverable metal by mass, and that metal weighs 42 tonnes. Of those same tailings, 5040 tonnes have already been shipped to a second site. Enter what percent of the total tailings mass has been shipped, just the number.
Show a hint
- \(0.15\%\) is \(\frac{0.15}{100}\), which is \(0.0015\), and the metal is that fraction of all the tailings, so the tailings mass is a missing base.
- Multiplying \(42=\frac{0.15}{100}b\) by \(100\) gives \(0.15b=4200\), so \(b\) comes out by dividing by \(0.15\). With \(b\) known, solve \(5040=\frac{p}{100}b\) for \(p\).
Show the full solution
From \(42=\frac{0.15}{100}b\), multiplying both sides by \(100\) gives \(0.15b=4200\), so \(b=28000\) tonnes, and then \(5040=\frac{p}{100}(28000)\) gives \(28000p=504000\) and \(p=\boxed{18}\). Since the percent is below \(1\), the base is more than a hundred times the amount, which is why 42 tonnes of metal comes from 28000 tonnes of tailings.
Every question here was \(a=\frac{p}{100}b\) with a different one of the three letters unknown, and the work each time was deciding which number is the base. Lesson 6.4, Percent Problems, keeps that equation and takes up the case where a quantity rises or falls by a percent, and the base there is what the quantity was before the change.
Practice these ideas
Practice
Write \(85\%\) as a fraction in lowest terms.
Show the solution
Dividing numerator and denominator of \(\frac{85}{100}\) by \(5\) gives \(\boxed{\frac{17}{20}}\). Since \(17\) is prime and \(20\) is not a multiple of \(17\), no further reducing is possible.
Practice
Write \(9\%\) as a decimal.
Show the solution
\(9\%\) is \(\frac{9}{100}\), so move the decimal point in \(9.0\) two places left to divide by \(100\), which gives \(9\%=\boxed{0.09}\). One place left would give \(0.9\), and that is \(90\%\).
Practice
A pressure gauge reads \(0.58\) of full scale. Enter that reading as a percent, just the number.
Show the solution
Multiplying by \(100\) gives \(0.58\times 100=\boxed{58}\). Two decimal places means hundredths, so \(0.58=\frac{58}{100}\), which is \(58\) per hundred.
Practice
A crate is \(\frac{13}{25}\) full. Rewrite that ratio with \(100\) below the bar and enter the percent, just the number.
Show the solution
Multiplying both parts by \(4\) gives \(\frac{13}{25}=\frac{52}{100}\), so the percent is \(\boxed{52}\).
Practice
A battery loses \(2.4\%\) of its charge overnight. Write \(2.4\%\) as a decimal.
Show the solution
Dividing by \(100\) slides the decimal point two places to the left, so \(2.4\%=\frac{2.4}{100}=\boxed{0.024}\). One hop left of \(2.4\) gives \(0.24\), and \(0.24\) is \(24\%\), so the second hop is the one that matters.
Practice
A service fee is \(7.5\%\) of an order. Write \(7.5\%\) as a fraction in lowest terms, in the form \(a/b\).
Show the solution
Multiplying top and bottom of \(\frac{7.5}{100}\) by \(10\) gives \(\frac{75}{1000}\), and dividing both by \(25\) gives \(\boxed{\frac{3}{40}}\). Multiplying top and bottom by the same number never changes the value, so \(\frac{7.5}{100}\) and \(\frac{3}{40}\) are the same number.
Practice
A kiln fires \(\frac{4}{9}\) of a load at a time. Since \(9\) does not divide \(100\), that share is not a whole number of percent. Enter it as an exact percent, written as a single fraction in lowest terms rather than a mixed number or a rounded decimal.
Show the solution
A percent is a count of hundredths, so the number of percent is \(\frac{4}{9}\times 100=\boxed{\frac{400}{9}}\). The numerator \(400\) and the denominator \(9\) have no common factor other than \(1\), so that is lowest terms, and the decimal form \(44.444\ldots\) does not terminate, which makes \(44.4\%\) a rounding and not the exact share.
Practice
A depot stores 840 tyres and \(65\%\) of them are winter tyres. Enter the number of winter tyres.
Show the solution
\(a=\frac{65}{100}(840)=\frac{13}{20}(840)=13(42)=\boxed{546}\).
Practice
A tank holds 1600 litres of brine, and \(3.5\%\) of it is salt. Enter the number of litres of salt.
Show the solution
\(a=\frac{3.5}{100}(1600)=0.035(1600)=\boxed{56}\).
Practice
A polling office mails out 700 survey forms, and 189 of them come back. Enter what percent of the forms came back, just the number.
Show the solution
Let \(p\) be the percent, so \(189=\frac{p}{100}(700)\). Multiplying both sides by \(100\) gives \(700p=18900\), so \(p=\boxed{27}\). Checking, \(\frac{27}{100}\) of \(700\) is \(27\times 7=189\).
Practice
A shelf holds 640 books, 80 of them reference books. Enter what percent of the books on the shelf are reference books, just the number.
Show the solution
From \(80=\frac{p}{100}(640)\), multiplying both sides by \(100\) gives \(640p=8000\), so \(p=\boxed{12.5}\). A percent need not be a whole number.
Practice
A deposit of 144 dollars is \(32\%\) of a fee. Enter the fee in dollars.
Show the solution
From \(144=\frac{32}{100}b\), multiplying both sides by \(100\) gives \(32b=14400\), so \(b=\boxed{450}\). Checking, \(\frac{32}{100}(450)=144\).
Practice
A club has 78 members who have paid their dues, and that is \(26\%\) of the membership. Multiplying \(78\) by \(26\%\) gives \(20.28\), fewer people than have already paid. Enter the number of members in the club.
Show the solution
With \(b\) the number of members, \(78=\frac{26}{100}b\), and multiplying both sides by \(100\) gives \(26b=7800\), so \(b=\boxed{300}\). The \(20.28\) is \(26\%\) of the part, and a club is never smaller than the number of members who have paid.
Practice
Depot X ships 250 crates and \(24\%\) of them are fragile. Depot Y ships 380 crates and \(15\%\) of them are fragile. Enter the number of fragile crates the two depots ship together.
Show the solution
The counts are \(\frac{24}{100}(250)=60\) and \(\frac{15}{100}(380)=57\), so the total is \(60+57=\boxed{117}\). The two percents cannot be added, because \(24\%\) is taken of \(250\) crates and \(15\%\) of \(380\) crates, and \(39\%\) of all \(630\) crates would be \(245.7\).
Practice
A tank has a capacity of 3060 litres, which is \(120\%\) of a second tank's capacity. Enter the capacity of the second tank in litres.
Show the solution
From \(3060=\frac{120}{100}b=\frac{6}{5}b\), multiplying both sides by \(5\) gives \(6b=15300\), so \(b=\boxed{2550}\). A percent above \(100\) means the first tank is the larger of the two, so the answer had to be less than \(3060\), and \(1.2(2550)=3060\) checks.
Practice
In a batch of 45000 screws, \(0.24\%\) are out of tolerance. Enter the number of screws out of tolerance.
Show the solution
\(0.24\%=\frac{0.24}{100}=0.0024\), so the count is \(0.0024(45000)=\boxed{108}\). Using \(0.24\) instead gives \(10800\), which is \(24\%\) of the batch and a hundred times too many.
Practice
In a depot, \(15\%\) of the vehicles are vans and the rest are trucks, and there are 630 more trucks than vans. Enter the number of vehicles in the depot.
Show the solution
Writing the total as \(b\), the vans number \(\frac{15}{100}b\) and the trucks \(\frac{85}{100}b\), so \(\frac{70}{100}b=630\) and \(b=\boxed{900}\). The check is \(135\) vans and \(765\) trucks, a gap of \(630\).