Calculating the amount of money in a savings account after a given amount of time, and calculating how long it will take for the amount of savings to exceed a given value.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "A savings account gives $\\var{int}\\%$ interest per year. The amount of money in the account is modelled as $S=£\\var{start} \\times \\var{1+int/100}^t$, where $t$ is the number of years. Interest is paid into the account monthly.
", "advice": "Part a:
\nTo calculate how much money will be in the savings account after a certain amount of time, we substitute the amount of time for $t$ in the equation \\[ S =£\\var{start} \\times \\var{1+int/100}^t .\\]
\nFor 6 months, $t=0.5$, therefore \\[ \\begin{split} S &\\,= £\\var{start} \\times \\var{1+int/100}^{0.5} \\\\ &\\,= £\\var{sol1} \\,.\\end{split} \\]
\nFor 1 year:
\n\\[ \\begin{split} S &\\,= £\\var{start} \\times \\var{1+int/100} \\\\ &\\,= £\\var{sol2} \\,.\\end{split} \\]
\nFor $\\var{n}$ years:
\n\\[ \\begin{split} S &\\,= £\\var{start} \\times \\var{1+int/100}^\\var{n} \\\\ &\\,= £\\var{sol3} \\,. \\end{split} \\]
\n\nPart b:
\nTo find out how long it will take for the amount of savings to exceed $£\\var{target}$, we need to solve the inequality \\[ \\var{start} \\times \\var{1+int/100}^t > \\var{target} \\] for $t$:
\n\\[ \\begin{split} \\var{start} \\times \\var{1+int/100}^t\\, &\\,> \\var{target} \\\\ \\var{1+int/100}^t &\\,> \\var{target/start} \\\\ t \\log(\\var{1+int/100}) &\\,> \\log(\\var{target/start}) \\\\t &\\,> \\frac{\\log(\\var{target/start})}{\\log(\\var{1+int/100})} \\qquad \\text{(Be careful when rearranging inequalities. Since $\\log(\\var{1+int/100})>0$, this will not change the inequality in this case.)} \\\\ t &\\,> \\var{tt2} \\text{ years (2 d.p.)}\\,.\\end{split} \\]
\nSince $\\var{tt2}$ years is equivalent to $\\var{yt}$ years and $\\var{mt}$ months, it will take $\\var{answer[0]}$ years and $\\var{answer[1]}$ {grammar} for the savings to exceed $£\\var{target}$.
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\n6 months: £[[0]]
\n1 year: £[[1]]
\n$\\var{n}$ years: £[[2]]
\nGive your answers to the nearest penny.
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\nYears:[[0]]
\nMonths:[[1]]
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