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In how many years will a sum of money double itself with the rate of 10% per annum simple interest?Answer Verified
Hint: To solve the problem, we should know the definition of annual simple interest. We have, Complete step-by-step answer: Note: While solving questions related to principal interest, it is important to keep in mind that simple interest calculated from the formula, Simple Interest (I) = $\dfrac{P\times R\times t}{100}$ , doesn’t represent the total amount of money. In fact, the total amount is the sum of Principal amount (P) and simple interest. Thus, in this case, when money was doubled, the total amount was 2P and simple interest was P. Compound and Continuous Interest FormulasRecall that compound interest occurs when interest accumulated for one period is added to the principal investment before calculating interest for the next period. The amount A accrued in this manner over time t is modeled by the compound interest formula: A(t)=P(1+rn)nt Here the initial principal P is accumulating compound interest at an annual rate r where the value n represents the number of times the interest is compounded in a year. Example 1Susan invested $500 in an account earning 412% annual interest that is compounded monthly.
Solution: In this example, the principal P=$500, the interest rate r=412%=0.045, and because the interest is compounded monthly, n=12. The investment can be modeled by the following function: A(t)=500(1+ 0.04512)12tA(t)=500(1.00375)12t
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The period of time it takes a quantity to double is called the doubling timeThe period of time it takes a quantity to double.. We next outline a technique for calculating the time it takes to double an initial investment earning compound interest. Example 2Mario invested $1,000 in an account earning 6.3% annual interest that is compounded semi-annually. How long will it take the investment to double? Solution: Here the principal P=$1,000, the interest rate r=6.3%=0.063, and because the interest is compounded semi-annually n=2. This investment can be modeled as follows: A(t)=1,000(1+0.0632)2tA(t)=1,000(1.0315)2t Since we are looking for the time it takes to double $1,000, substitute $2,000 for the resulting amount A (t) and then solve for t. 2,000=1,000(1.0315)2t2,0001,000=(1.0315)2t2=(1.0315)2t At this point we take the common logarithm of both sides. 2=(1.0315)2tlog2 =log(1.0315)2tlog2=2tlog(1.0315)log 22log(1.0315)=t Using a calculator we can approximate the time it takes: t=log( 2)/(2*log(1.0315))≈11.17years Answer: Approximately 11.17 years to double at 6.3%. If the investment in the previous example was one million dollars, how long would it take to double? To answer this we would use P=$1,000,000 and A(t)=$2,000,000: A(t)=1,000(1.0315)2t2,000,000=1,000,000(1.0315)2t Dividing both sides by 1,000,000 we obtain the same exponential function as before. 2=(1.0315)2t Hence, the result will be the same, about 11.17 years. In fact, doubling time is independent of the initial investment P. Interest is typically compounded semi-annually (n = 2), quarterly (n = 4), monthly (n = 12), or daily (n = 365). However if interest is compounded every instant we obtain a formula for continuously compounding interest: A(t)=Pert Here P represents the initial principal amount invested, r represents the annual interest rate, and t represents the time in years the investment is allowed to accrue continuously compounded interest. Example 3Mary invested $200 in an account earning 534% annual interest that is compounded continuously. How long will it take the investment to grow to $350? Solution: Here the principal P=$200 and the interest rate r=534%=5.75%=0.0575. Since the interest is compounded continuously, use the formula A(t)=Pe rt. Hence, the investment can be modeled by the following, A(t)=200e0.0575t To calculate the time it takes to accumulate to $350, set A(t)=350 and solve for t. A(t)=200e0.0575t 350=200e0.0575t Begin by isolating the exponential expression. 350200=e0.0575t74=e0.0575t1.75= e0.0575t Because this exponential has base e, we choose to take the natural logarithm of both sides and then solve for t. ln(1.75) =lne0.0575tApplythepowerrulefor logarithms.ln(1.75)=0.0575tlneRecal lthatlne=1.ln(1.75)=0.0575t⋅1ln(1.75 )0.0575=t Using a calculator we can approximate the time it takes: t=ln(1.75)/0.0575≈9.73 years Answer: It will take approximately 9.73 years. When solving applications involving compound interest, look for the keyword “continuous,” or the keywords that indicate the number of annual compoundings. It is these keywords that determine which formula to choose. Try this! Mario invested $1,000 in an account earning 6.3% annual interest that is compounded continuously. How long will it take the investment to double? Answer: Approximately 11 years. Modeling Exponential Growth and DecayIn the sciences, when a quantity is said to grow or decay exponentially, it is specifically meant to be modeled using the exponential growth/decay formulaA formula that models exponential growth or decay: P(t)=P0ekt.: P(t)=P0ekt Here P0, read “P naught,” or “P zero,” represents the initial amount, k represents the growth rate, and t represents the time the initial amount grows or decays exponentially. If k is negative, then the function models exponential decay. Notice that the function looks very similar to that of continuously compounding interest formula. We can use this formula to model population growth when conditions are optimal. Example 4It is estimated that the population of a certain small town is 93,000 people with an annual growth rate of 2.6%. If the population continues to increase exponentially at this rate:
Solution: We begin by constructing a mathematical model based on the given information. Here the initial population P0=93,000 people and the growth rate r=2.6%=0.026. The following model gives population in terms of time measured in years: P(t)=93,000e0.026t
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Often the growth rate k is not given. In this case, we look for some other information so that we can determine it and then construct a mathematical model. The general steps are outlined in the following example. Example 5Under optimal conditions Escherichia coli (E. coli) bacteria will grow exponentially with a doubling time of 20 minutes. If 1,000 E. coli cells are placed in a Petri dish and maintained under optimal conditions, how many E. coli cells will be present in 2 hours? Figure 7.3 Escherichia coli (E. coli) (Wikipedia) Solution: The goal is to use the given information to construct a mathematical model based on the formula P(t)=P0e kt.
Answer: In two hours 64,000 cells will be present. When the growth rate is negative the function models exponential decay. We can describe decreasing quantities using a half-lifeThe period of time it takes a quantity to decay to one-half of the initial amount., or the time it takes to decay to one-half of a given quantity. Example 6Due to radioactive decay, caesium-137 has a half-life of 30 years. How long will it take a 50-milligram sample to decay to 10 milligrams? Solution: Use the half-life information to determine the rate of decay k. In t=30 years the initial amount P0=50 milligrams will decay to half P(30)=25 milligrams. P(t)=P0ek t25=50ek30 Solve for the only variable, k. 25= 50ek302550=e30kln(12)=lne30kln(12)=30klneln1−ln2 30=kRecallthatln1=0.−ln230 =k Note that k=−ln230≈−0.0231 is negative. However, we will use the exact value to construct a model that gives the amount of cesium-137 with respect to time in years. P(t)=50e(−ln2/30)t Use this model to find t when P(t)=10 milligrams. 10=50e(−ln2/30)t 1050=e(−ln2/30)tln(15)=lne(−ln2/30)tln1−ln5=(−ln230 )tlneRecallthatlne=1.−30(ln1− ln5)ln2=t−30(0−ln5)ln2=t 30ln5ln2=t Answer: Using a calculator, it will take t≈69.66 years to decay to 10 milligrams. Radiocarbon dating is a method used to estimate the age of artifacts based on the relative amount of carbon-14 present in it. When an organism dies, it stops absorbing this naturally occurring radioactive isotope, and the carbon-14 begins to decay at a known rate. Therefore, the amount of carbon-14 present in an artifact can be used to estimate the age of the artifact. Example 7An ancient bone tool is found to contain 25% of the carbon-14 normally found in bone. Given that carbon-14 has a half-life of 5,730 years, estimate the age of the tool. Solution: Begin by using the half-life information to find k. Here the initial amount P0 of carbon-14 is not given, however, we know that in t=5,730 years, this amount decays to half, 12P0. P(t )=P0ekt12P0=P0ek5,730 Dividing both sides by P0 leaves us with an exponential equation in terms of k. This shows that half-life is independent of the initial amount. 12=ek5,730 Solve for k. ln(12)=lnek5,730 ln1−ln2=5,730klne0−ln25,730=k−ln 25,730=k Therefore we have the model, P(t)=P0e(−ln 2/5,730)t Next we wish to the find time it takes the carbon-14 to decay to 25% of the initial amount, or P(t)=0.25P0. 0.25P0=P0e(−ln2/5,730)t Divide both sides by P0 and solve for t. 0.25=e(−ln2/5,730)tln (0.25)=lne(−ln2/5,730)tln(0.25)=(− ln25,730)tlne−5,730ln(0.25)ln2=t11,460≈t Answer: The tool is approximately 11,460 years old. Try this! The half-life of strontium-90 is about 28 years. How long will it take a 36 milligram sample of strontium-90 to decay to 30 milligrams? Answer: 7.4 years Key Takeaways
Topic Exercises
Part A: Compound and Continuous Interest
Part B: Modeling Exponential Growth and DecaySolve for the given variable:
Part C: Discussion BoardAnswers
How long will it take the money to double itself at 5% compounded annually?Thus, it will take 14.20 year.
How long does it take a sum of money to double its value at 5% per annum simple interest?Thus, it will take 8 years.
How long will it take money to double itself if invested at 5% compounded annually a 13.7 years b 14.7 years C 14.2 years'd 15.3 years?If an amount X is invested at an interest rate of 5% compounded annually in how many years does the investment amount double itself A 14.2 years, B 13.7 years, C 14.7 years, or d. years. The correct answer is a 14.2 years.
How long will it take an investment to double at 5% compounded monthly?How does the rule of 72 work? Using the rule of 72, you would estimate that an investment with a 5% compound interest rate would double in 14 years (72/5).
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