PERCENTAGE

$$I = \frac{2500\times4.5\times6}{100} = \$675$$

$$2500+675=\$3175$$

$$I=\frac{8000\times7\times3}{100}=\$1680$$

Total owed $=8000+1680=\$9680$. Monthly instalment $=9680\div36\approx\$268.89$.

Year 1: $1000+60=1060$. Year 2: $1000+120=1120$. Year 3: $1000+180=1180$. Year 4: $1000+240=1240$.

This is an arithmetic sequence — the total increases by exactly \$60 every year (the same fixed amount).

Simple interest is always calculated on the ORIGINAL principal only, never on previously-earned interest — so the same fixed dollar amount ($6\%$ of the original \$1000, i.e. \$60) is added every single year, creating a straight arithmetic (linear) pattern rather than a growing (compounding) one.

$$I=\frac{5000\times8\times2.5}{100}=\$1000$$ — matches the student's answer.

Because simple interest adds the SAME fixed amount each year (unlike compound interest), it's mathematically valid to just add up whole and partial years directly — \$400 for each full year, plus half of \$400 for the half year, giving the same total as the formula.

$$I=\frac{12000\times3.5\times4}{100}=1680. \quad \text{Total}=12000+1680=\$13680$$

Yes, \$13,680 is enough. They will have $13680-13500=\$180$ left over after the purchase.

$$I=\frac{3400\times4.5\times7}{100}=\$1071$$

$$3400+1071=\$4471$$

$I=\frac{15000\times3.8\times5}{100}=\$2850$. Total: $15000+2850=\$17850$.

$\frac{5T}{100}=1 \Rightarrow T=20$ years — this works for ANY principal $P$, since $P$ cancels out of the equation entirely.

$\frac{8T}{100}=1 \Rightarrow T=12.5$ years. Pattern: doubling time $T=\frac{100}{\text{rate}}$ — as the rate increases, the doubling time decreases proportionally (specifically, it's inversely proportional to the rate).

$\frac{5000\times4\times T}{100}=800 \Rightarrow 200T=800 \Rightarrow T=4$ years.

$\frac{5000\times8\times T}{100}=800 \Rightarrow 400T=800 \Rightarrow T=2$ years — exactly half of 4 years, confirming that (for a FIXED target interest amount and fixed principal) doubling the rate does exactly halve the required time, since time and rate are inversely proportional in the simple interest formula when $I$ and $P$ are held constant.

The student forgot the $\div100$ in the formula $I=\frac{PRT}{100}$ — the rate 5% must be converted to a decimal-equivalent proportion of the principal, which requires dividing by 100. Correct: $I=\frac{6000\times5\times3}{100}=\$900$.

\$90,000 interest on a \$6,000 investment would mean the investment grew to 16 TIMES its original value in just 3 years — an absurdly high return rate for a simple 5% interest rate; real-world simple interest at modest rates like 5% should yield interest that's a small FRACTION of the principal over a few years, not many times larger than it.

Suppose \$2,400 (the principal, $P$) is invested at an annual simple interest rate of 3.5% ($R$) for 6 years ($T$). Using the formula $I=\frac{PRT}{100}$: $$I=\frac{2400\times3.5\times6}{100}=\$504$$ The total amount in the account after 6 years is therefore $2400+504=\$2904$.

$I=\frac{8000\times6\times4}{100}=\$1920$. Total repayment: $8000+1920=\$9920$.

4 years of profit: $2600\times4=\$10{,}400$. Since $10400>9920$ (the repayment amount), the business CAN repay the loan using 4 years of profit, with a surplus of $10400-9920=\$480$.

$$I=\frac{45000\times4.2\times1}{100}=\$1890 \text{ per year}$$

\$1,890 is far short of the \$12,000 needed — INSUFFICIENT. Required total principal for \$12,000 annual interest: $\frac{P\times4.2\times1}{100}=12000 \Rightarrow P=\frac{1200000}{4.2}\approx\$285{,}714$. Additional principal needed: $285714-45000\approx\$240{,}714$.

$I=\frac{18000\times7.5\times5}{100}=\$6750$. Total: $18000+6750=\$24750$.

Monthly instalment: $24750\div60=\$412.50\approx\$413$ (rounding to nearest dollar). Whether this is 'reasonable' depends heavily on the borrower's income and other expenses — for many households, a \$413 monthly car payment represents a significant but often manageable portion of a typical budget, though it would need to be weighed against other essential costs (housing, food, utilities) to determine true affordability for a specific borrower.

$$I=\frac{120000\times3.2\times1}{100}=\$3840 \text{ per year}$$

Total withdrawn over 15 years: $3840\times15=\$57{,}600$. This simplified model assumes the \$120,000 stays completely untouched and keeps earning interest forever at a fixed rate — in reality, banks might not offer guaranteed fixed simple interest indefinitely, inflation could erode the real value of a fixed \$3,840 annual withdrawal over 15 years, and if the retiree ever needed to withdraw MORE than just the interest (e.g. for an emergency), the principal would shrink, reducing future interest earned — making this a simplified, optimistic model rather than a fully realistic retirement income plan.

Standard form: Let $P=5500$, $R=5.5$ (as a percentage), $T=3$ years. $$I=\frac{PRT}{100}=\frac{5500\times5.5\times3}{100}=\$907.50$$ The student's version used $R=0.055$ (a decimal) with no division by 100 — this happens to give the same numeric answer, but deviates from the conventional formula structure, which could cause confusion or errors if applied inconsistently in other problems.

$I=\frac{25000\times5.8\times3}{100}=\$4350$. Total: $25000+4350=\$29350$.

3 years of extra revenue: $8500\times3=\$25{,}500$. Since $25500<29350$, the extra revenue is NOT enough to fully cover the loan repayment. Shortfall: $29350-25500=\$3850$ — the farmer would need additional funds beyond the projected equipment-driven revenue increase to fully repay the loan within 3 years.