1. the principal “p” is borrowed at simple interest rate r for a period of time t. find the simple interest…
1. the principal “p” is borrowed at simple interest rate r for a period of time t. find the simple interest owed for the use of the money.\n a. p = $260, r = 4%, t = 3 years.\n b. p = $18,000, r = 7.5%, t = 18 months.\n2. the principal “p” is borrowed at simple interest rate “r” for a period of time t”. find the loan’s future value, “a”, or the total amount due at time “t”.\n a. p = $3000, r = 7%, t = 2 years.\n b. p = $9000, r = 6.5%, t = 8 months.\n3. the principal “p” is borrowed and the loan’s future value, “a”, at time “t” is given. determine the loan’s simple interest rate, “r” to the nearest tenth a percent.\n a. p = $5000, a = $5900, t = 2 years\n b. p = $2300, a = $2840, t = 9 months.\n4. determine the present value, “p” you must invest to have the future value “a” at simple interest rate “r” after time “t”. round answers to the nearest cent.\n a. a = $14,000, r = 9.5%, t = 6 years.\n b. a = $5000, r = 14.5%, t = 9 months.
Answer
Explanation:
Step1: Recall simple - interest formulas
The simple - interest formula is $I = Prt$, where $I$ is the interest, $P$ is the principal, $r$ is the annual interest rate (in decimal form), and $t$ is the time in years. The future - value formula is $A=P + I=P(1 + rt)$. To find $r$ from $A = P(1+rt)$, we can solve for $r$: $A=P + Prt$, $A - P=Prt$, $r=\frac{A - P}{Pt}$. To find $P$ from $A = P(1+rt)$, we have $P=\frac{A}{1 + rt}$.
Step2: Solve 1a
Given $P = 260$, $r=0.04$, $t = 3$. Using $I = Prt$, we substitute the values: $I=260\times0.04\times3=31.2$.
Step3: Solve 1b
First, convert 18 months to years. Since 1 year = 12 months, $t=\frac{18}{12}=1.5$ years. Given $P = 18000$, $r = 0.075$, using $I=Prt$, we have $I=18000\times0.075\times1.5 = 2025$.
Step4: Solve 2a
Given $P = 3000$, $r=0.07$, $t = 2$. Using $A=P(1+rt)$, we substitute the values: $A=3000\times(1 + 0.07\times2)=3000\times(1 + 0.14)=3000\times1.14 = 3420$.
Step5: Solve 2b
First, convert 8 months to years. $t=\frac{8}{12}=\frac{2}{3}$ years. Given $P = 9000$, $r=0.065$, using $A=P(1+rt)$, we have $A=9000\times(1+0.065\times\frac{2}{3})=9000\times(1+\frac{0.13}{3})\approx9000\times(1 + 0.0433)=9000\times1.0433 = 9389.7$.
Step6: Solve 3a
Given $P = 5000$, $A = 5900$, $t = 2$. Using $r=\frac{A - P}{Pt}$, we substitute the values: $r=\frac{5900 - 5000}{5000\times2}=\frac{900}{10000}=0.09 = 9.0%$.
Step7: Solve 3b
First, convert 9 months to years. $t=\frac{9}{12}=0.75$ years. Given $P = 2300$, $A = 2840$, using $r=\frac{A - P}{Pt}$, we have $r=\frac{2840 - 2300}{2300\times0.75}=\frac{540}{1725}\approx0.313 = 31.3%$.
Step8: Solve 4a
Given $A = 14000$, $r=0.095$, $t = 6$. Using $P=\frac{A}{1+rt}$, we substitute the values: $P=\frac{14000}{1+0.095\times6}=\frac{14000}{1 + 0.57}=\frac{14000}{1.57}\approx8917.20$.
Step9: Solve 4b
First, convert 9 months to years. $t=\frac{9}{12}=0.75$ years. Given $A = 5000$, $r=0.145$, using $P=\frac{A}{1+rt}$, we have $P=\frac{5000}{1+0.145\times0.75}=\frac{5000}{1 + 0.10875}=\frac{5000}{1.10875}\approx4509.58$.
Answer:
1a. $I = 31.2$ 1b. $I = 2025$ 2a. $A = 3420$ 2b. $A\approx9389.7$ 3a. $r = 9.0%$ 3b. $r\approx31.3%$ 4a. $P\approx8917.20$ 4b. $P\approx4509.58$