Question

find each missing value.

1. $p = \\$1,000$, $r = 5\\%$, $t = 2$ years

$i = \underline{\quad\quad} \cdot \underline{\quad\quad} \cdot \underline{\quad\quad}$
$i = \underline{\quad\quad}$

1. $p = \\$600$, $r = 4\\%$, $t = 3$ years

$i = \underline{\quad\quad} \cdot \underline{\quad\quad} \cdot \underline{\quad\quad}$
$i = \underline{4.5}$

1. $i = \\$330$, $r = 3\\%$, $t = 1$ year

$\underline{\quad\quad} = p \cdot \underline{\quad\quad} \cdot \underline{\quad\quad}$
$p = \underline{1.1}$

1. $i = \\$270$, $r = 5\\%$, $t = 3$ years

$\underline{\quad\quad} = p \cdot \underline{\quad\quad} \cdot \underline{\quad\quad}$
$p = \underline{16.2}$

1. $i = \\$600$, $p = \\$2,500$, $t = 4$ years

$\underline{\quad\quad} = \underline{\quad\quad} \cdot r \cdot \underline{\quad\quad}$
$r = \underline{\quad\quad}$

1. $i = \\$108$, $p = \\$900$, $t = 3$ years

$\underline{\quad\quad} = \underline{\quad\quad} \cdot r \cdot \underline{\quad\quad}$
$r = \underline{\quad\quad}$

1. $p = \\$250$, $r = 6\\%$, $t = 5$ years

$i = \underline{\quad\quad}$

1. $p = \\$3,000$, $r = 7\\%$, $t = 4$ years

$i = \underline{\quad\quad}$

1. $i = \\$750$, $r = 4\\%$, $t = 5$ years

$p = \underline{\quad\quad}$

1. $i = \\$696$, $r = 3\\%$, $t = 4$ years

$p = \underline{\quad\quad}$

1. $i = \\$425$, $p = \\$1,700$, $t = 5$ years

$r = \underline{\quad\quad}$

1. $i = \\$1,680$, $p = \\$12,000$, $t = 2$ years

$r = \underline{\quad\quad}$

1. you deposit $\\$5,000$ in an account that earns $5\\%$ simple interest.

how long will it be before the total amount is $\\$6,000$?
\underline{4 years}

1. after 6 years, an account that earns $4\\%$ simple interest has

earned $\\$480$ in interest. how much was the initial deposit?
\underline{\\$2,000}

1. a deposit of $\\$7,500$ earns $\\$3,900$ over a period of 8 years.

what is the simple interest rate?
\underline{6.5\\%}

1. you deposit $\\$4,500$ in an account that earns $6\\%$ simple interest.

how much will be in your account after 5 years?
\underline{\\$1,350}$

Explanation:

Step1: Recall Simple Interest Formula

The formula for simple interest is \( I = p \cdot r \cdot t \), where \( I \) is the interest, \( p \) is the principal amount, \( r \) is the annual interest rate (in decimal), and \( t \) is the time in years.

Step2: Solve Problem 1

Given \( p = \$1000 \), \( r = 5\% = 0.05 \), \( t = 2 \) years.
Substitute into the formula: \( I = 1000 \cdot 0.05 \cdot 2 \).
Calculate: \( 1000 \cdot 0.05 = 50 \), \( 50 \cdot 2 = 100 \). So \( I = \$100 \).

Step3: Solve Problem 2

Given \( p = \$600 \), \( r = 4\% = 0.04 \), \( t = 3 \) years.
Use \( I = p \cdot r \cdot t \): \( I = 600 \cdot 0.04 \cdot 3 \).
Calculate: \( 600 \cdot 0.04 = 24 \), \( 24 \cdot 3 = 72 \). So \( I = \$72 \) (note: the original handwritten 4.5 is incorrect).

Step4: Solve Problem 3

Given \( I = \$330 \), \( r = 3\% = 0.03 \), \( t = 1 \) year.
From \( I = p \cdot r \cdot t \), solve for \( p \): \( p = \frac{I}{r \cdot t} \).
Substitute: \( p = \frac{330}{0.03 \cdot 1} = \frac{330}{0.03} = 11000 \). So \( p = \$11000 \) (note: the original handwritten 1.1 is incorrect).

Step5: Solve Problem 4

Given \( I = \$270 \), \( r = 5\% = 0.05 \), \( t = 3 \) years.
From \( I = p \cdot r \cdot t \), solve for \( p \): \( p = \frac{I}{r \cdot t} \).
Substitute: \( p = \frac{270}{0.05 \cdot 3} = \frac{270}{0.15} = 1800 \). So \( p = \$1800 \) (note: the original handwritten 16.2 is incorrect).

Step6: Solve Problem 5

Given \( I = \$600 \), \( p = \$2500 \), \( t = 4 \) years.
From \( I = p \cdot r \cdot t \), solve for \( r \): \( r = \frac{I}{p \cdot t} \).
Substitute: \( r = \frac{600}{2500 \cdot 4} = \frac{600}{10000} = 0.06 = 6\% \).

Step7: Solve Problem 6

Given \( I = \$108 \), \( p = \$900 \), \( t = 3 \) years.
From \( I = p \cdot r \cdot t \), solve for \( r \): \( r = \frac{I}{p \cdot t} \).
Substitute: \( r = \frac{108}{900 \cdot 3} = \frac{108}{2700} = 0.04 = 4\% \).

Step8: Solve Problem 7

Given \( p = \$250 \), \( r = 6\% = 0.06 \), \( t = 5 \) years.
Use \( I = p \cdot r \cdot t \): \( I = 250 \cdot 0.06 \cdot 5 \).
Calculate: \( 250 \cdot 0.06 = 15 \), \( 15 \cdot 5 = 75 \). So \( I = \$75 \).

Step9: Solve Problem 8

Given \( p = \$3000 \), \( r = 7\% = 0.07 \), \( t = 4 \) years.
Use \( I = p \cdot r \cdot t \): \( I = 3000 \cdot 0.07 \cdot 4 \).
Calculate: \( 3000 \cdot 0.07 = 210 \), \( 210 \cdot 4 = 840 \). So \( I = \$840 \).

Step10: Solve Problem 9

Given \( I = \$750 \), \( r = 4\% = 0.04 \), \( t = 5 \) years.
From \( I = p \cdot r \cdot t \), solve for \( p \): \( p = \frac{I}{r \cdot t} \).
Substitute: \( p = \frac{750}{0.04 \cdot 5} = \frac{750}{0.2} = 3750 \). So \( p = \$3750 \).

Step11: Solve Problem 10

Given \( I = \$696 \), \( r = 3\% = 0.03 \), \( t = 4 \) years.
From \( I = p \cdot r \cdot t \), solve for \( p \): \( p = \frac{I}{r \cdot t} \).
Substitute: \( p = \frac{696}{0.03 \cdot 4} = \frac{696}{0.12} = 5800 \). So \( p = \$5800 \).

Step12: Solve Problem 11

Given \( I = \$425 \), \( p = \$1700 \), \( t = 5 \) years.
From \( I = p \cdot r \cdot t \), solve for \( r \): \( r = \frac{I}{p \cdot t} \).
Substitute: \( r = \frac{425}{1700 \cdot 5} = \frac{425}{8500} = 0.05 = 5\% \).

Step13: Solve Problem 12

Given \( I = \$1680 \), \( p = \$12000 \), \( t = 2 \) years.
From \( I = p \cdot r \cdot t \), solve for \( r \): \( r = \frac{I}{p \cdot t} \).
Substitute: \( r = \frac{1680}{12000 \cdot 2} = \frac{1680}{24000} = 0.07 = 7\% \).

Step14: Solve Problem 13

Total amount \( A = p + I = \$6000 \), \( p = \$5000 \), so \( I = 6000 - 5000 = \$1000 \).
\( r = 5\% = 0.05 \), \( p = 5000 \). From \( I = p \cdot r \cdot t \), solve for \( t \): \( t = \frac{I}{p \cdot r} \).
Substitute: \( t = \frac{1000}{5000 \cdot 0.05} = \frac{1000}{250} = 4 \) years. So \( t = 4 \) years (matches the handwritten answer).

Step15: Solve Problem 14

Given \( I = \$480 \), \( r = 4\% = 0.04 \), \( t = 6 \) years.
From \( I = p \cdot r \cdot t \), solve for \( p \): \( p = \frac{I}{r \cdot t} \).
Substitute: \( p = \frac{480}{0.04 \cdot 6} = \frac{480}{0.24} = 2000 \). So \( p = \$2000 \) (matches the handwritten answer).

Step16: Solve Problem 15

Given \( p = \$7500 \), \( I = \$3900 \), \( t = 8 \) years.
From \( I = p \cdot r \cdot t \), solve for \( r \): \( r = \frac{I}{p \cdot t} \).
Substitute: \( r = \frac{3900}{7500 \cdot 8} = \frac{3900}{60000} = 0.065 = 6.5\% \) (matches the handwritten answer).

Step17: Solve Problem 16

First, find interest \( I \): \( p = \$4500 \), \( r = 6\% = 0.06 \), \( t = 5 \) years.
\( I = 4500 \cdot 0.06 \cdot 5 = 4500 \cdot 0.3 = 1350 \).
Total amount \( A = p + I = 4500 + 1350 = 5850 \). So total is \$5850 (note: the original handwritten 1350 is only interest, not total).

Answer (Correcting Errors):

1. \( I = 1000 \cdot 0.05 \cdot 2 \); \( I = \boldsymbol{\$100} \)
2. \( I = 600 \cdot 0.04 \cdot 3 \); \( I = \boldsymbol{\$72} \)
3. \( 330 = p \cdot 0.03 \cdot 1 \); \( p = \boldsymbol{\$11000} \)
4. \( 270 = p \cdot 0.05 \cdot 3 \); \( p = \boldsymbol{\$1800} \)
5. \( 600 = 2500 \cdot r \cdot 4 \); \( r = \boldsymbol{6\%} \)
6. \( 108 = 900 \cdot r \cdot 3 \); \( r = \boldsymbol{4\%} \)
7. \( I = 250 \cdot 0.06 \cdot 5 \); \( I = \boldsymbol{\$75} \)
8. \( I = 3000 \cdot 0.07 \cdot 4 \); \( I = \boldsymbol{\$840} \)
9. \( 750 = p \cdot 0.04 \cdot 5 \); \( p = \boldsymbol{\$3750} \)
10. \( 696 = p \cdot 0.03 \cdot 4 \); \( p = \boldsymbol{\$5800} \)
11. \( 425 = 1700 \cdot r \cdot 5 \); \( r = \boldsymbol{5\%} \)
12. \( 1680 = 12000 \cdot r \cdot 2 \); \( r = \boldsymbol{7\%} \)
13. Time: \(\boldsymbol{4}\) years
14. Principal: \(\boldsymbol{\$2000}\)
15. Rate: \(\boldsymbol{6.5\%}\)
16. Total Amount: \(\boldsymbol{\$5850}\) (Interest: \(\$1350\))

Answer:

(Correcting Errors):

1. \( I = 1000 \cdot 0.05 \cdot 2 \); \( I = \boldsymbol{\$100} \)
2. \( I = 600 \cdot 0.04 \cdot 3 \); \( I = \boldsymbol{\$72} \)
3. \( 330 = p \cdot 0.03 \cdot 1 \); \( p = \boldsymbol{\$11000} \)
4. \( 270 = p \cdot 0.05 \cdot 3 \); \( p = \boldsymbol{\$1800} \)
5. \( 600 = 2500 \cdot r \cdot 4 \); \( r = \boldsymbol{6\%} \)
6. \( 108 = 900 \cdot r \cdot 3 \); \( r = \boldsymbol{4\%} \)
7. \( I = 250 \cdot 0.06 \cdot 5 \); \( I = \boldsymbol{\$75} \)
8. \( I = 3000 \cdot 0.07 \cdot 4 \); \( I = \boldsymbol{\$840} \)
9. \( 750 = p \cdot 0.04 \cdot 5 \); \( p = \boldsymbol{\$3750} \)
10. \( 696 = p \cdot 0.03 \cdot 4 \); \( p = \boldsymbol{\$5800} \)
11. \( 425 = 1700 \cdot r \cdot 5 \); \( r = \boldsymbol{5\%} \)
12. \( 1680 = 12000 \cdot r \cdot 2 \); \( r = \boldsymbol{7\%} \)
13. Time: \(\boldsymbol{4}\) years
14. Principal: \(\boldsymbol{\$2000}\)
15. Rate: \(\boldsymbol{6.5\%}\)
16. Total Amount: \(\boldsymbol{\$5850}\) (Interest: \(\$1350\))