QuanticaAlgebra IOpen in the courseLesson 6.3 turned every percent question into \(a=\frac{p}{100}b\), with one of the three quantities unknown. Here the amount itself changes. A rise or a fall of \(k\) percent is a change of \(\frac{k}{100}\) of the amount before it, so that earlier amount is the base. The first two problems take a rise of the same percent by two separate routes.
Problem
A quarry shipped 3400 tonnes of gravel in June, and July's shipment was 15 percent higher. Work it in two steps. Find 15 percent of 3400 first, then add that onto 3400, and enter July's shipment in tonnes.
Show a hint
- The 15 percent is measured against June's 3400 tonnes, so \(\frac{15}{100}(3400)\) is the size of the rise and not July's shipment.
- Dividing 3400 by 100 leaves 34, so \(\frac{15}{100}(3400)\) is just \(15\times 34\). That result is what gets added onto 3400.
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\(\frac{15}{100}(3400)=510\), and \(3400+510=\boxed{3910}\). The percent is taken of June's shipment, so 3400 is the base and the 510 is the rise rather than the new total.
Problem
A second pit shipped 2600 tonnes in June, and its shipments rose 15 percent in July. July's shipment is \(2600+\frac{15}{100}(2600)\), and 2600 is a factor of both terms. Factor it out, do the one multiplication that remains, and enter July's shipment in tonnes.
Show a hint
- Both terms contain 2600, since \(2600=2600(1)\). Pulling a shared factor out front is the factoring of 2.3, the distributive property read right to left.
- Factoring gives \(2600\left(1+\frac{15}{100}\right)\), and \(1+\frac{15}{100}\) is a single number.
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Factoring gives \(2600\left(1+\frac{15}{100}\right)=2600(1.15)=\boxed{2990}\). Going the long way, \(\frac{15}{100}(2600)=390\) and \(2600+390=2990\), so a 15 percent rise is one multiplication rather than a percent step followed by an addition.
Problem
A pressroom used 5400 litres of ink last quarter and cut its use by 28 percent this quarter. Use one multiplication to find this quarter's ink use, and enter it in litres.
Show a hint
- A cut leaves part of the original, so the multiplier is below 1. Take 28 percent away from the whole 100 percent to see what percent of the ink is left.
- The multiplier is \(1-\frac{28}{100}=0.72\), so the work is \(5400(0.72)\).
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\(5400\left(1-\frac{28}{100}\right)=5400(0.72)=\boxed{3888}\). Subtracting instead gives \(\frac{28}{100}(5400)=1512\) and \(5400-1512=3888\), the same number, because multiplying by \(0.72\) does that subtraction in one step.
Problem
A regional grid operator reports that this winter's peak demand was exactly 0.925 times last winter's peak. Enter the percent by which peak demand fell, a number only.
Show a hint
- Every fall multiplier has the form \(1-\frac{k}{100}\), so compare \(0.925\) with that form and read off what \(\frac{k}{100}\) has to be.
- \(1-0.925=0.075\), so \(\frac{k}{100}=0.075\) and \(k=100(0.075)\).
Show the full solution
A fall multiplier is \(1-\frac{k}{100}\), and \(1-0.925=0.075\), so \(k=100(0.075)=\boxed{7.5}\). Answering 92.5 gives the percent of last winter's peak that is left rather than the percent lost, and those two always total 100.
For two changes in a row, one multiplier is applied and then the other, so the amount ends at \(xm_1m_2\). Each change is a percent of the amount just before it, and after the first change that amount is no longer \(x\). The next two problems check whether the percents combine as simply as the multipliers do.
Problem
A workshop counted 6400 orders in June. The count rose 25 percent in July, then fell 25 percent in August, with the August fall taken off the July count. Enter the August order count.
Show a hint
- The July rise and the August fall are the same percent but not the same number of orders, since the August fall is 25 percent of the July count.
- The July count is \(6400(1.25)\). Take 25 percent off that count, not off \(6400\).
Show the full solution
\(6400(1.25)=8000\), and \(8000(0.75)=\boxed{6000}\). The rise is \(1600\) orders and the fall is \(2000\) orders, since 25 percent of \(8000\) is more than 25 percent of \(6400\), so the August count is \(400\) below the June count.
Problem
A library's holdings grew 30 percent in one decade and 40 percent in the next, each rise measured against the holdings at the start of that decade. A clerk adds the percents and reports that the holdings are now 1.7 times the original. Enter the number the original holdings are actually multiplied by.
Show a hint
- The second rise is 40 percent of the holdings at the start of the second decade, not 40 percent of the original. Call the original holdings \(x\) and write what the holdings are after each decade.
- A 30 percent rise multiplies by \(1.3\) and a 40 percent rise multiplies by \(1.4\), so the holdings come to \(x(1.3)(1.4)\), and a product can be regrouped.
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Writing the original holdings as \(x\), the first rise brings them to \(1.3x\) and the second brings them to \(1.4(1.3x)\), so the multiplier is \((1.3)(1.4)=\boxed{1.82}\). Adding the percents gives the clerk's 1.7 and leaves out the 40 percent taken on the first decade's growth, worth \(0.4(0.3x)=0.12x\), and \(1.7+0.12=1.82\).
Sometimes the amount after the change is given and the amount before it is wanted. The change is still one multiplication, so \(N=Om\), one equation with \(O\) unknown, and chapter 4 divides both sides by \(m\). Taking the same percent off \(N\) is a different calculation, since that percent was measured against \(O\) and \(N\) is not \(O\).
Problem
A season pass costs 336 dollars after a 12 percent rise. A student takes 12 percent off 336 and reports 295.68 dollars as the old price. Check that against the stated rise, then enter the correct old price in dollars.
Show a hint
- The 12 percent was a percent of the old price, not of 336, so name the old price \(p\) and write the rise as an equation in \(p\).
- The stated fact is \(1.12p=336\), so divide both sides by 1.12.
Show the full solution
Naming the old price \(p\), the stated rise gives \(1.12p=336\), so \(p=\frac{336}{1.12}=\boxed{300}\). Raising the student's 295.68 by 12 percent gives 331.1616, not 336. The 12 percent belongs to the old price, so undoing the rise is a division by 1.12, and 0.88 is not \(\frac{1}{1.12}\).
Problem
A factory's output fell 20 percent in the spring and then rose 5 percent in the summer, with each change measured against the output just before that change. Summer output was 8400 units a day. Enter the daily output, in units a day, before the 20 percent drop.
Show a hint
- Two changes in a row are one multiplication by the product of the two multipliers, so collapse the pair into a single multiplier before working backwards.
- Write \(O\) for the daily output before the drop. Since \(0.8(1.05)=0.84\), the equation is \(0.84O=8400\).
Show the full solution
Writing \(O\) for the daily output before the drop, the two multipliers give \(0.8(1.05)=0.84\), so \(0.84O=8400\) and \(O=\frac{8400}{0.84}=\boxed{10000}\). Check forward, \(10000(0.8)=8000\) and \(8000(1.05)=8400\). A 20 percent drop then a 5 percent rise is not a 15 percent drop, since 0.84 is not 0.85.
A percent change is the change measured against the amount before it, so from \(O\) to \(N\) the percent change is \(\frac{N-O}{O}\times 100\). The denominator is \(O\) every time. Swapping which of the two amounts comes first swaps the denominator, so the two directions between one fixed pair of amounts are different percents.
Problem
A workshop finished 450 units in March and 720 units in April. Find the percent increase from March to April, then find the percent decrease from April back down to March. Enter the percent increase minus the percent decrease.
Show a hint
- Both percents come from the same gap of 270 units, and they differ only in what that gap is divided by.
- The increase is measured against March's 450, and the decrease is measured against April's 720.
Show the full solution
Up is \(\frac{270}{450}=0.6\), so 60 percent, and down is \(\frac{270}{720}=0.375\), so 37.5 percent, giving \(60-37.5=\boxed{22.5}\). The gap of 270 units is the same both ways and only the base changes, so the percent up and the percent back down are not equal.
Problem
A bakery cuts the price of a loaf by 10 percent and finds it then sells 30 percent more loaves. Revenue is price times number sold. No price and no number of loaves is stated anywhere, so write the old price as \(p\) and the old number sold as \(n\). Enter the percent by which revenue rises, a number only.
Show a hint
- Nothing pins down \(p\) or \(n\), so carry both letters and write the new revenue as the product of the two new quantities.
- New revenue is \((0.9p)(1.3n)\). Compare that with the old revenue \(pn\), and see what is left once the letters cancel.
Show the full solution
New revenue is \((0.9p)(1.3n)=1.17pn\) against the old \(pn\), so \(\frac{1.17pn-pn}{pn}=0.17\) and the percent rise is \(\boxed{17}\). Both letters cancel, so the answer holds for every price and every number of loaves. Adding the percents to get 20 leaves out the 30 percent gain being counted on the reduced price.
A mixture problem gives a percent for each ingredient and asks for a percent of the blend. Two percents cannot be averaged here, since each one is a percent of a different amount. The amounts of the substance itself do add, so count the litres of it on each side, total those, and divide by the total volume of the blend.
Problem
A bottler mixes 45 litres of a syrup that is 60 percent sugar with 75 litres of a syrup that is 20 percent sugar. Averaging the two percents gives 40, which is not the answer. Enter the percent sugar in the blend, a number only.
Show a hint
- The two percents are percents of different volumes, so they do not combine directly. Work out the litres of sugar in each syrup first.
- The blend holds \(45(0.6)+75(0.2)\) litres of sugar in \(45+75\) litres of syrup.
Show the full solution
The sugar totals \(45(0.6)+75(0.2)=27+15=42\) litres in 120 litres of blend, and \(\frac{42}{120}=0.35\), so the percent sugar is \(\boxed{35}\). Averaging to get 40 would be right only for equal volumes, and here the weaker syrup is the larger batch, so the blend lands below the halfway mark.
Problem
A vat holds 280 litres of glaze that is 5 percent pigment by volume. Pure pigment is stirred in until the glaze is 30 percent pigment. One student pours in 70 litres, which is 25 percent of 280, reasoning that the percent has to go up by 25 points, and gets 24 percent instead. Enter the number of litres of pure pigment stirred in.
Show a hint
- Call the litres poured in \(a\). Every litre poured in is a litre of pigment and also a litre of glaze, so the pigment and the total both go up by \(a\).
- The vat starts with \(0.05(280)=14\) litres of pigment, so the equation is \(\frac{14+a}{280+a}=0.3\).
Show the full solution
With \(a\) litres poured in, the pigment is \(0.05(280)+a=14+a\) and the total is \(280+a\), so \(\frac{14+a}{280+a}=0.3\). Cross-multiplying gives \(14+a=84+0.3a\), then \(0.7a=70\) and \(a=\boxed{100}\). Check, 114 litres of pigment in 380 litres of glaze is 30 percent. The 70-litre route gives \(\frac{84}{350}\), or 24 percent, because those 70 litres are part of the new total as well.
Every change here was one multiplication, two changes in a row were one multiplication by the product of the two multipliers, and going backwards was a division. Lesson 6.5, Direct and Inverse Proportion, looks at two quantities that change together. The question there is what stays fixed, a ratio in one case and a product in the other.
Practice these ideas
Practice
A walking trail is 1800 meters long and is lengthened by 10 percent. Enter the new length in meters.
Show the solution
\(1800(1.1)=\boxed{1980}\). The increase alone is \(\frac{10}{100}(1800)=180\), and \(1800+180\) is the same length.
Practice
A farm harvested 5600 kilograms of barley last year, and this year's harvest fell 25 percent. Enter this year's harvest in kilograms.
Show the solution
\(5600(0.75)=\boxed{4200}\). The fall itself is \(0.25(5600)=1400\), and \(5600-1400\) gives the same harvest.
Practice
A distributor multiplies every wholesale price by 1.06 to get the retail price. Enter the percent by which the retail price is above the wholesale price, a number only.
Show the solution
\(1.06=1+0.06=1+\frac{6}{100}\), so the percent rise is \(\boxed{6}\). The amount past 1 is the percent divided by 100, so \(0.06\) means 6 percent, while \(1.06\) itself is the retail price as 106 percent of the wholesale price.
Practice
A sale takes \(18\%\) off every price. Enter the single number that each original price is multiplied by to get the sale price.
Show the solution
After \(18\%\) comes off, \(100\%-18\%=82\%\) of the price remains, so the multiplier is \(1-0.18=\boxed{0.82}\). The value \(0.18\) is the size of the discount, not the share of the price still paid.
Practice
A club grew from 250 members to 315 members. Enter the percent increase, a number only.
Show the solution
The change is \(315-250=65\), and \(\frac{65}{250}=0.26\), so the percent increase is \(\boxed{26}\). Dividing by the new 315 instead gives about 20.6 percent, which compares the change to the final size rather than the starting size.
Practice
A commute that took 105 minutes now takes 84 minutes. Enter the percent decrease, a number only.
Show the solution
The drop is \(105-84=21\) minutes, and \(\frac{21}{105}=0.2\), so the percent decrease is \(\boxed{20}\). Check, \(105(0.8)=84\).
Practice
After an 8 percent rise, the rent on an apartment is 1350 dollars a month. Enter the monthly rent in dollars before the rise.
Show the solution
\(1.08r=1350\) gives \(r=\frac{1350}{1.08}=\boxed{1250}\). Check, \(1250(1.08)=1350\). Taking 8 percent off \(1350\) gives \(1242\), which is not the old rent, since the 8 percent is a percent of the smaller old rent rather than of \(1350\).
Practice
A bus fare is cut by 35 percent, and the new fare is 26 dollars. Enter the fare in dollars before the cut.
Show the solution
Let \(f\) be the old fare. A 35 percent cut leaves \(1-0.35=0.65\) of it, so \(0.65f=26\) and \(f=\frac{26}{0.65}=\boxed{40}\). Adding 35 percent of \(26\) back on gives \(35.10\), not \(40\), because that 35 percent would be a share of the smaller new fare rather than of the old one.
Practice
An amount \(x\) rises by 50 percent. The new amount then falls by 30 percent of itself. The ending amount equals \(mx\) for a single number \(m\). Enter \(m\) as a decimal.
Show the solution
The two multipliers give \(m=1.5\times 0.7=\boxed{1.05}\). That is a net rise of 5 percent, not the 20 percent that subtracting the percents suggests, since the 30 percent is taken from the raised amount rather than from \(x\).
Practice
A crowd of 5200 grows 45 percent by noon and then shrinks 40 percent by evening. Enter the evening crowd size.
Show the solution
\(5200(1.45)=7540\), and \(7540(0.6)=\boxed{4524}\). Collapsing first works too, since \(1.45(0.6)=0.87\) and \(5200(0.87)=4524\).
Practice
A stock of parts falls 10 percent one month and 10 percent again the next. Enter the overall percent change, a number only, with a minus sign if it is a fall.
Show the solution
Two 10 percent falls multiply the stock by \(0.9(0.9)=0.81\), and \(0.81\) is \(0.19\) less than \(1\), a fall of \(19\) percent, so the overall percent change is \(\boxed{-19}\). The second 10 percent comes off the reduced stock, so two 10 percent falls land short of a 20 percent fall.
Practice
Store B sells a tool for 20 percent less than store A charges for the same tool. Enter the percent by which store A's price is above store B's price, a number only.
Show the solution
Call store A's price \(a\), so store B's price is \(0.8a\) and the gap is \(a - 0.8a = 0.2a\). Measured against store B's price, that gap is \(\frac{0.2a}{0.8a} = 0.25\), so the answer is \(\boxed{25}\). The \(a\) drops out of the ratio, so the same percent holds for any price, and the two directions differ only because one comparison uses \(a\) as its base and the other uses \(0.8a\).
Practice
An amount is raised by 150 percent. Enter the percent of the new amount that must then be taken off to return to the original amount, a number only.
Show the solution
Raising by 150 percent means multiplying by \(1+\frac{150}{100}=2.5\), and undoing that means multiplying by \(\frac{1}{2.5}=0.4\), so the percent taken off is \(\boxed{60}\). Taking off 150 percent would leave a negative amount, so the percent taken off is never equal to the percent added.
Practice
In a survey, 60 percent of the people asked own a bicycle, and 45 percent of those bicycle owners also own a helmet. The number of people asked is not given. Enter the percent of all the people asked who own both a bicycle and a helmet, a number only.
Show the solution
With \(n\) people asked, the number who own both is \(0.45(0.6n)=0.27n\), so the percent is \(\boxed{27}\). The \(n\) cancels, so the answer does not depend on how many people were asked. Adding to get 105 or averaging to get 52.5 both treat the two percents as though they had the same base.
Practice
A blender combines 24 liters of a mix that is 70 percent apple juice with 56 liters of a mix that is 30 percent apple juice. Enter the percent apple juice in the combined mix, a number only.
Show the solution
The apple juice totals \(24(0.7)+56(0.3)=16.8+16.8=33.6\) liters in \(80\) liters of mix, and \(\frac{33.6}{80}=0.42\), so the percent apple juice is \(\boxed{42}\). Averaging 70 and 30 gives 50, which would be right only if the two batches were the same size.
Practice
A tank contains 150 liters of a solution that is 12 percent acid by volume. Pure acid is added until the solution is 20 percent acid. Enter the number of liters of pure acid added.
Show the solution
The tank starts with \(0.12(150)=18\) liters of acid, and adding \(a\) liters of pure acid gives \(\frac{18+a}{150+a}=0.2\), so \(18+a=30+0.2a\), then \(0.8a=12\) and \(a=\boxed{15}\). Check, 33 liters of acid in 165 liters is 20 percent.
Practice
A machine's price rose 20 percent in the spring and fell 15 percent in the autumn, and it now costs 2040 dollars. Enter its price in dollars before the spring rise.
Show the solution
Let \(p\) be the price before the rise. The two changes collapse to one multiplier, \(1.2(0.85)=1.02\), so \(1.02p=2040\) and \(p=\frac{2040}{1.02}=\boxed{2000}\). Check forward, \(2000(1.2)=2400\) and \(2400(0.85)=2040\). A 20 percent rise and a 15 percent fall leave the price 2 percent up, not 5 percent up.