chapter 7 lesson 8 -10 simple interest\nlesson 8 (sections 7.1 & 7.2) simple interest i = prt\nproblem…
chapter 7 lesson 8 -10 simple interest\nlesson 8 (sections 7.1 & 7.2) simple interest i = prt\nproblem a\nexercise 7.1 page 273\n#4 lin yan borrowed $1800 from her parents to finance a vacation. if interest was charged on the loan at 5.2%, how much interest will she have to pay in 220 days?\nproblem b\n#2 on july 17, 2021, leah deposited $1500 into a savings account that earned simple interest at 1.05%. how much interest was earned and paid into leah’s account on december 1, 2021?\nproblem c\nexercise 7.2 page 278 p, r, t\ni finding p - principal\n#2 determine the deposit that must be made to earn $39.27 in 225 days at 2.75%.\nproblem d\nii finding r - the annual rate of interest\n#12 what rate of interest is required for $740.48 to earn $42.49 interest from september 9, 2022 to march 4, 2023?\nproblem e\niii finding t - time expressed in terms of a year\nexample 1\nhow long will it take $2500 to earn $62.50 at 5% simple interest?\nproblem f\nexample 2\nhow many months will it take $3000 to earn $175 at 7% simple interest?\nproblem g\nexercise 7.2 page 278\n#10 how many days will it take $1200 to earn $12.22 interest at 16.9%?
Answer
Problem A
Explanation:
Step1: Identify values
$P = 1800$, $r=0.052$, $t=\frac{220}{365}$
Step2: Use simple - interest formula
$I = Prt=1800\times0.052\times\frac{220}{365}$ $I=\frac{1800\times0.052\times220}{365}\approx 55.48$
Answer:
$55.48$
Problem B
Explanation:
Step1: Calculate number of days
July has 31 days, August has 31 days, September has 30 days, October has 31 days, November has 30 days. From July 17 to July 31 is $31 - 17=14$ days. Total days $t=14 + 31+30 + 31+30+1=137$ days, so $t=\frac{137}{365}$ $P = 1500$, $r = 0.0105$
Step2: Use simple - interest formula
$I=Prt=1500\times0.0105\times\frac{137}{365}$ $I=\frac{1500\times0.0105\times137}{365}\approx 5.79$
Answer:
$5.79$
Problem C
Explanation:
Step1: Rearrange simple - interest formula for $P$
$I = Prt$, so $P=\frac{I}{rt}$. $I = 39.27$, $r = 0.0275$, $t=\frac{225}{365}$
Step2: Calculate $P$
$P=\frac{39.27}{0.0275\times\frac{225}{365}}=\frac{39.27\times365}{0.0275\times225}\approx 2300$
Answer:
$2300$
Problem D
Explanation:
Step1: Calculate number of days
September has 30 days, October has 31 days, November has 30 days, December has 31 days, January has 31 days, February (2023 is not a leap - year) has 28 days, March 1 to March 4 is 4 days. From September 9 to September 30 is $30 - 9 = 21$ days. Total days $t=21+31+30+31+31+28+4 = 176$ days, so $t=\frac{176}{365}$ $P = 740.48$, $I = 42.49$
Step2: Rearrange simple - interest formula for $r$
$I = Prt$, so $r=\frac{I}{Pt}$ $r=\frac{42.49}{740.48\times\frac{176}{365}}=\frac{42.49\times365}{740.48\times176}\approx 0.12$ or $12%$
Answer:
$12%$
Problem E
Explanation:
Step1: Rearrange simple - interest formula for $t$
$I = Prt$, so $t=\frac{I}{Pr}$. $I = 62.50$, $P = 2500$, $r = 0.05$
Step2: Calculate $t$
$t=\frac{62.50}{2500\times0.05}=\frac{62.50}{125}=0.5$ years
Answer:
$0.5$ years
Problem F
Explanation:
Step1: Rearrange simple - interest formula for $t$
$I = Prt$, so $t=\frac{I}{Pr}$. $I = 175$, $P = 3000$, $r = 0.07$ $t=\frac{175}{3000\times0.07}=\frac{175}{210}=\frac{5}{6}$ years
Step2: Convert years to months
Since 1 year has 12 months, number of months $= \frac{5}{6}\times12 = 10$ months
Answer:
$10$ months
Problem G
Explanation:
Step1: Rearrange simple - interest formula for $t$
$I = Prt$, so $t=\frac{I}{Pr}$. $I = 12.22$, $P = 1200$, $r = 0.169$
Step2: Calculate $t$ in years and convert to days
$t=\frac{12.22}{1200\times0.169}=\frac{12.22}{202.8}\approx 0.06$ years Number of days $t\times365=0.06\times365 = 22$ days
Answer:
$22$ days