chapter 7 lesson 8 -10 simple interest\nlesson 9 (sect 7.3) future value s = p(1+rt)\nproblem a\nexercise…

Problem A

Explanation:

Step1: Identify values

$P = 800$, $r=0.0275$, $t=\frac{210}{365}$

Step2: Use future - value formula

$S = P(1 + rt)=800\times(1 + 0.0275\times\frac{210}{365})$ $S = 800\times(1+0.0158219)$ $S = 800\times1.0158219$ $S\approx812.66$

Answer:

The future value of the deposit is approximately $$812.66$

Problem B

Explanation:

Step1: Calculate number of days

From March 1, 2019 to October 20, 2019: March has 31 days, April has 30 days, May has 31 days, June has 30 days, July has 31 days, August has 31 days, September has 30 days, and from October 1 - 20 there are 20 days. Total days $d=31 + 30+31 + 30+31 + 31+30 + 20=234$ days. So $t=\frac{234}{365}$ $P = 7750$, $r = 0.0625$

Step2: Calculate interest

$I=Prt=7750\times0.0625\times\frac{234}{365}$ $I = 7750\times0.0625\times0.6411$ $I=7750\times0.04007$ $I\approx310.54$

Answer:

The interest included is approximately $$310.54$

Problem C

Explanation:

Step1: Identify values

$P = 1500$, $S = 1562.04$, time from June 1, 2022 to December 1, 2022 is 6 months or $t=\frac{6}{12}=0.5$

Step2: Rearrange future - value formula for $r$

$S = P(1+rt)$ implies $1+rt=\frac{S}{P}$ and $r=\frac{\frac{S}{P}-1}{t}$ $r=\frac{\frac{1562.04}{1500}-1}{0.5}$ $r=\frac{1.04136 - 1}{0.5}$ $r=\frac{0.04136}{0.5}$ $r = 0.08272$ or $8.272%$

Answer:

The rate of interest is $8.272%$

Problem D

Explanation:

Step1: Identify values

$P = 3100$, $S = 3195.72$, $r=0.0575$

Step2: Rearrange future - value formula for $t$

$S = P(1+rt)$ implies $1+rt=\frac{S}{P}$ and $t=\frac{\frac{S}{P}-1}{r}$ $t=\frac{\frac{3195.72}{3100}-1}{0.0575}$ $t=\frac{1.02765 - 1}{0.0575}$ $t=\frac{0.02765}{0.0575}$ $t = 0.481$ years

Step3: Convert years to days

$t$ in days $=0.481\times365\approx175$ days

Answer:

It will take approximately 175 days.