C1.13; E1.13 - Percentage change and interest

C1.13; E1.13 - Percentage change and interest

Percentages can change a price, measure a rise in earnings or describe interest on savings. You will use multipliers for increases and decreases, calculate percentage change, and distinguish simple interest from compound interest.

Percentage multipliers

Two savings accounts start with \1000andpayand pay10%$ per year. One calculates each year's interest from the original deposit; the other includes earlier interest in the next calculation. Will their values still agree after two years? First learn how to change an amount once.

An increase of 15%15\% keeps the original 100%100\% and adds 15%15\%, leaving 115%115\%. A decrease keeps 100%−15%=85%100\%-15\%=85\%. This turns a two-step calculation into one multiplication.

Percentage multipliers

increase by p%:multiply by 1+p100decrease by p%:multiply by 1−p100\begin{aligned} \text{increase by }p\% &: \quad\text{multiply by }1+\frac{p}{100}\\ \text{decrease by }p\% &: \quad\text{multiply by }1-\frac{p}{100} \end{aligned}

For an ordinary decrease of a non-negative price, 0≤p≤1000\le p\le100. A 100%100\% decrease leaves zero; a 0%0\% change leaves the amount unchanged.

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For a positive percentage increase, the multiplier is greater than 11; a discount multiplier lies between 00 and 11, including the endpoints. A large increase is possible: an increase of 150%150\% leaves 250%250\%, so its multiplier is 2.52.5, not 1.51.5.

Worked example 1: an increase

Question. A bicycle costs \200.Itspriceincreasesby. Its price increases by 15%$. Calculate the new price.

Plan. Retain 115%115\% of the original price.

100%+15%=115%original plus increase200×1.15=230multiply by 115/100\begin{aligned} 100\%+15\%&=115\% &&\text{original plus increase}\\ 200\times1.15&=230 &&\text{multiply by }115/100 \end{aligned}

Answer. \230$.

Check. The increase is 230-200=\30;; 30/200=0.15$, as required.

Change only the direction: use a 15%15\% discount rather than a 15%15\% increase.

Worked example 2: a discount

Question. A bicycle costs \200.Itspricedecreasesby. Its price decreases by 15%$. Calculate the sale price.

Plan. Keep 85%85\% of the original price.

100%−15%=85%percentage left to pay200×0.85=170multiply by 85/100\begin{aligned} 100\%-15\%&=85\% &&\text{percentage left to pay}\\ 200\times0.85&=170 &&\text{multiply by }85/100 \end{aligned}

Answer. \170$.

Check. The discount is \30,and, and 170+30=200.Thesavingis. The saving is 15%$ of the listed price.

A deposit uses a related distinction. For a 20%20\% deposit, multiply the full price by 0.200.20 to find the amount paid now. Multiply by 0.800.80 to find the balance still due, assuming no additional charges. These are different requested amounts.

Watch out

Multiplying by 0.150.15 finds the change amount, not the final amount after a 15%15\% change. Decide whether the question asks for the saving, increase, deposit, balance or new total.

Measuring percentage change

An increase of \10meansdifferentthingstosomeoneearningmeans different things to someone earning$50andsomeoneearningand someone earning$500$. Percentage change compares the change amount with the original value. This makes changes to different starting amounts comparable.

Percentage change

percentage change=new value−original valueoriginal value×100%\text{percentage change}=\frac{\text{new value}-\text{original value}}{\text{original value}}\times100\%

For a positive original value, a positive result means an increase and a negative result means a decrease. You may instead report the positive size of a decrease and label it clearly. An original value of zero cannot be used as a denominator.

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Worked example 3: a rise in earnings

Question. Weekly earnings rise from \320toto$368$. Calculate the percentage increase.

Plan. Find the extra earnings and compare them with the original \320$.

368−320=48increase in dollars48320×100%=15%compare with original earnings\begin{aligned} 368-320&=48 &&\text{increase in dollars}\\ \frac{48}{320}\times100\%&=15\% &&\text{compare with original earnings} \end{aligned}

Answer. A 15%15\% increase.

Check. 320×1.15=368320\times1.15=368, so the multiplier recovers the new earnings.

Keep the original earnings and change only the new value so that the change is a fall of the same size.

Worked example 4: a fall in earnings

Question. Weekly earnings fall from \320toto$272$. Calculate the percentage decrease.

Plan. Find the loss of earnings and compare it with the original amount.

320−272=48decrease in dollars48320×100%=15%use the same original base\begin{aligned} 320-272&=48 &&\text{decrease in dollars}\\ \frac{48}{320}\times100\%&=15\% &&\text{use the same original base} \end{aligned}

Answer. A 15%15\% decrease.

Check. 320×0.85=272320\times0.85=272, returning to the new earnings.

For a trader, cost price plays the role of the original value and selling price the new value. If cost is \125andsellingpriceisand selling price is$150,theprofitis, the profit is $25andthepercentageprofitisand the percentage profit is25/125\times100%=20%.Ifsoldfor. If sold for $100instead,thelossisinstead, the loss is$25,or, or 20%$ of the cost. Label profit or loss rather than leaving the direction unclear.

Watch out

Using 48/36848/368 in the rise example compares the increase with the new earnings. Percentage increase uses the starting amount, so the denominator is 320320.

Simple interest

Interest is money paid on an investment or charged on a loan. The principal is the original amount. With simple interest, each period's interest is calculated from the same original principal, rather than an increasing balance.

Suppose \1000earnsearns10%peryearsimpleinterest.Everyyearearnsper year simple interest. Every year earns$100,soaftertwoyearstheinterestis, so after two years the interest is $200andthetotalisand the total is$1200$. The initial deposit and the interest are separate amounts.

Simple interest

I=P×r100×t,A=P+II=P\times\frac{r}{100}\times t,\qquad A=P+I

Here PP is the principal, rr is the percentage rate per year, tt is time in years, II is interest and AA is the final amount. The rate and time units must match, and the model assumes a constant principal and rate.

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Worked example 5: whole years of simple interest

Question. \800isinvestedatis invested at3%$ per year simple interest for two years. Calculate the interest and the final value.

Plan. Find one year's interest, multiply by the time, then add the principal.

800×0.03=24one year’s interest24×2=48two years’ interest800+48=848principal plus interest\begin{aligned} 800\times0.03&=24 &&\text{one year's interest}\\ 24\times2&=48 &&\text{two years' interest}\\ 800+48&=848 &&\text{principal plus interest} \end{aligned}

Answer. Interest \48;finalvalue; final value $848$.

Check. Two years earn 6%6\% of \800;; 0.06\times800=48$.

Change only the time to eighteen months. A rate per year needs time in years.

Worked example 6: a fraction of a year

Question. \800earnsearns3%$ per year simple interest for eighteen months. Calculate the interest and final value.

Plan. Convert the time to years before multiplying.

18÷12=1.5time in years800×0.03×1.5=36simple interest800+36=836add the original deposit\begin{aligned} 18\div12&=1.5 &&\text{time in years}\\ 800\times0.03\times1.5&=36 &&\text{simple interest}\\ 800+36&=836 &&\text{add the original deposit} \end{aligned}

Answer. Interest \36;finalvalue; final value $836$.

Check. One year earns \24andhalfayearearnsand half a year earns$12,totalling, totalling $36$.

Watch out

800×0.03×18800\times0.03\times18 treats eighteen months as eighteen years. Also, an answer of \36$ is the interest only; include the principal when asked for the final investment value.

Compound interest

With compound interest, interest is added to the balance at the end of each stated period. The next period earns interest on that new balance, including earlier interest. Apply the same percentage multiplier each period, rather than adding the same cash amount.

Worked example 7: following two annual balances

Question. \1000isinvestedatis invested at10%$ per year compound interest. Calculate its value after two years.

Plan. Multiply the balance by 1.101.10 once for each year.

1000×1.10=1100balance after year one1100×1.10=1210year two uses the new balance\begin{aligned} 1000\times1.10&=1100 &&\text{balance after year one}\\ 1100\times1.10&=1210 &&\text{year two uses the new balance} \end{aligned}

Answer. \1210$.

Check. The two interest payments are \100andand$110.Addingthemto. Adding them to $1000givesgives$1210$.

Change only the duration to three years. The third multiplier acts on the balance after year two.

Worked example 8: one more compounding period

Question. Find the value of the same \1000investmentafterthreeyearsatinvestment after three years at10%$ per year compound interest.

Plan. Multiply by 1.101.10 three times.

1000×1.103=1331one factor for each year1331−1000=331total interest earned\begin{aligned} 1000\times1.10^3&=1331 &&\text{one factor for each year}\\ 1331-1000&=331 &&\text{total interest earned} \end{aligned}

Answer. The investment is worth \1331,including, including $331$ interest.

Check. Year three earns 10%10\% of \1210,whichis, which is $121.Then. Then 1210+121=1331$.

Repeated multiplication explains the power in the general rule. The number of factors is the number of compounding periods, not the percentage rate.

Compound interest

A=P(1+r100)nA=P\left(1+\frac{r}{100}\right)^n

Here PP is the principal, rr is the percentage interest rate per period, and nn is the number of complete periods. This rule assumes a constant rate, interest added each period, and no deposits, withdrawals or fees. Total interest is A−PA-P.

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Annual compounding uses whole years as periods. Do not substitute a fractional number of years into this model unless the question specifies how interest is credited for part of a period. On calculator questions, retain unrounded working and round money as requested; a final payment is normally written to two decimal places.

The opening accounts agree after one year, but not after two: simple interest gives \1200,compoundinterest, compound interest $1210.Theextra. The extra $10isinterestonthefirstyear′sis interest on the first year's$100$ interest.

Watch out

1000×(1+2×0.10)=12001000\times(1+2\times0.10)=1200 calculates two years of simple interest. Compound interest needs 1000×1.1021000\times1.10^2, because the second year's base has changed.

Quick Recap

  • Increase and decrease multipliers retain the original whole as well as accounting for the change.
  • Percentage change divides the change amount by the original non-zero value.
  • Simple interest always uses the original principal; match time units to the rate.
  • Compound interest uses a new balance each period; match the exponent to the number of periods.
  • Distinguish interest alone from the final amount including the original deposit.

The opening accounts differ after two years: \1200withsimpleinterestandwith simple interest and$1210$ with compound interest.