1. Problem: Find the interest rate per annum if the third to the last payment is 1040.40 from an initial payment of 1000. 2. Formula: For compound interest, the payment grows as $P(1+r)^n$ where $P$ is principal, $r$ is rate per period, and $n$ is number of periods. 3. Here, $1040.40 = 1000(1+r)^n$ where $n$ is the number of periods before the third to last payment. 4. Simplify: $\frac{1040.40}{1000} = (1+r)^n \Rightarrow 1.0404 = (1+r)^n$. 5. Taking the $n$th root: $1+r = (1.0404)^{1/n}$. 6. Without $n$, assume $n=1$ for annual rate: $r = 1.0404 - 1 = 0.0404$ or 4.04% per annum. --- 1. Problem: Find unknowns in simple interest table. 2. Formula: Simple Interest $I = P \times r \times t$. 3. Calculate missing values: - For $P=10000$, $r=8\% = 0.08$, $t=15$, $I = 10000 \times 0.08 \times 15 = 12000$. - For $r=2\% = 0.02$, $t=5$, $P=10000$, $I = 10000 \times 0.02 \times 5 = 1000$. - For $I=360000$, $r=2\% = 0.02$, $t=3$, $P = \frac{I}{r \times t} = \frac{360000}{0.02 \times 3} = 6000000$. - For $P=500000$, $r=10.5\% = 0.105$, $I=175000$, $t = \frac{I}{P \times r} = \frac{175000}{500000 \times 0.105} = 3.333$ years. - For $P=880000$, $r=9.25\% = 0.0925$, $t=2.5$, $I = P \times r \times t = 880000 \times 0.0925 \times 2.5 = 203500$. --- 1. Problem: Evaluate $g(x) = 2x$ for given values. 2. Calculate: - $g(2) = 2 \times 2 = 4$ - $g(4) = 2 \times 4 = 8$ - $g(0) = 2 \times 0 = 0$ - $g(-2) = 2 \times (-2) = -4$ - $g(-4) = 2 \times (-4) = -8$ --- 1. Problem: Find present value of 50000 due in 4 years at 12% compounded semi-annually. 2. Formula: Present Value $PV = \frac{FV}{(1 + \frac{r}{m})^{mt}}$ where $FV=50000$, $r=0.12$, $m=2$, $t=4$. 3. Calculate: $PV = \frac{50000}{(1 + \frac{0.12}{2})^{2 \times 4}} = \frac{50000}{(1.06)^8} = \frac{50000}{1.59385} = 31362.04$. --- 1. Problem: Find initial investment given simple interest 21000 at 7% for 3 years. 2. Formula: $I = P \times r \times t$. 3. Rearrange: $P = \frac{I}{r \times t} = \frac{21000}{0.07 \times 3} = 100000$. --- 1. Problem: Number of months of artificial payments for appliance costing 2000 for 1 year starting from 7th month. 2. Since payments start at month 7 and last 1 year, payments are from month 7 to month 18. 3. Number of months = $18 - 7 + 1 = 12$ months. --- 1. Problem: Interest rate per annum if second to last payment is 918 from initial 900. 2. Using compound interest: $918 = 900(1+r)^n$. 3. Simplify: $1.02 = (1+r)^n$. 4. Assuming $n=1$, $r = 0.02$ or 2% per annum. --- 1. Problem: Solve $10x + 3 = 3x - 4$. 2. Rearrange: $10x - 3x = -4 - 3$. 3. Simplify: $7x = -7$. 4. Solve: $x = -1$. --- 1. Problem: Given function points {(5,10), (6,12), (7,14)}, find domain and range. 2. Domain: $5 6 7$. 3. Range: $10 12 14$. --- 1. Problem: Number of periods of artificial payments for quarterly payments of 8000 for 8 years starting 2 years from now. 2. Total periods = $8$ years $\times 4$ quarters/year = 32 quarters. 3. Since payments start 2 years later, payments are for $8 - 2 = 6$ years. 4. Number of payments = $6 \times 4 = 24$ periods. Final answers summarized in content above.