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PSLE

PSLE Foundation Mathematics

<p>In the $1^{st}$ month, a farm has a pair of rabbits. Then in the $3^{rd}$ month, the pair of rabbits grow up and reproduce another pair of rabbits. Suppose that: </p>
<p>1. It always takes two months for rabbits to grow up and be able to reproduce baby rabbits. </p>
<p>2. After growing up, every pair of rabbits reproduce one pair of rabbits every month. </p>
<p>3. No rabbit dies on this farm.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the number of pair(s) of rabbits on the farm with respect to time (in month), by completing the following table. (Hint: from the $3^{\text{rd}}$ month onwards, the number of pair(s) of rabbits in each month equals the sum of the number of pair(s) of rabbits in the previous month and the number of pair(s) of new-born rabbits this month.)</div></li></ol>
<div class="content-block content-block--table"><div class="table-container"><table style="table-layout: fixed; width: 400px;"><colgroup><col style="width: 37.5000%;"><col style="width: 62.5000%;"></colgroup><thead><tr><th style="text-align: center; --cell-justify: center">Time (Month)</th><th style="text-align: center; --cell-justify: center">Number of Pair(s) of Rabbits, $P_n$</th></tr></thead><tbody><tr><td style="text-align: center; --cell-justify: center">$1$</td><td style="text-align: center; --cell-justify: center">$1$</td></tr><tr><td style="text-align: center; --cell-justify: center">$2$</td><td style="text-align: center; --cell-justify: center">$1$</td></tr><tr><td style="text-align: center; --cell-justify: center">$3$</td><td style="text-align: center; --cell-justify: center">$2$</td></tr><tr><td style="text-align: center; --cell-justify: center">$4$</td><td style="text-align: center; --cell-justify: center">$3$</td></tr><tr><td style="text-align: center; --cell-justify: center">$5$</td><td style="text-align: center; --cell-justify: center">$5$</td></tr><tr><td style="text-align: center; --cell-justify: center">$6$</td><td style="text-align: center; --cell-justify: center">$8$</td></tr><tr><td style="text-align: center; --cell-justify: center">$7$</td><td style="text-align: center; --cell-justify: center"></td></tr><tr><td style="text-align: center; --cell-justify: center">$8$</td><td style="text-align: center; --cell-justify: center"></td></tr><tr><td style="text-align: center; --cell-justify: center">$9$</td><td style="text-align: center; --cell-justify: center"></td></tr></tbody></table></div></div>
<ol class="i-list" start="2"><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">State a rule of the number sequence formed by the number of pair(s) of rabbits on the farm, $P_n$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Based on your answers in <strong>(i)</strong> and <strong>(ii)</strong>, Write down the next two terms of the following sequences.</div><ol class="abc-list"><li data-list-item-id="item-iii-a" data-list-item-path="item-iii/item-iii-a"><div class="list-item-paragraph">$-3, 2, -1, 1, 0, \dots$</div></li><li data-list-item-id="item-iii-b" data-list-item-path="item-iii/item-iii-b"><div class="list-item-paragraph">$3, 2, 5, 7, 12, \dots$</div></li></ol></li></ol>

In the $1^{st}$ month, a farm has a pair of rabbits. Then in the $3^{rd}$ month, the pair of rabbits grow up and reproduce another pair of rabbits. Suppose that: 

1. It always takes two months for rabbits to grow up and be able to reproduce baby rabbits. 

2. After growing up, every pair of rabbits reproduce one pair of rabbits every month. 

Find the number of pair(s) of rabbits on the farm with respect to time (in month), by completing the following table. (Hint: from the $3^{\text{rd}}$ month onwards, the number of pair(s) of rabbits in each month equals the sum of the number of pair(s) of rabbits in the previous month and the number of pair(s) of new-born rabbits this month.)

State a rule of the number sequence formed by the number of pair(s) of rabbits on the farm, $P_n$.

Generating Terms from Recurrence Relations

Modelling with Recurrence Relations

 Number Patterns

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data-height="143" style="width:550px;height:143px;max-width:100%;display:block;margin:0 auto;"></div><div class="list-item-paragraph"><p>Snowflakes can be constructed by starting with an equilateral triangle (Fig. $1$), then repeatedly altering each line segment of the resulting polygon as follows:</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$1.$ Divide each outer line segments into three segments of equal length.</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$2.$ Add an equilateral triangle that has the middle segment from step $1$ as its base.</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$3.$ Remove the line segment that is the base of the triangle from step $2$.</p></div><div class="list-item-paragraph" style="margin-left: 3rem"><p>$4.$ Repeat the above steps for a number of iterations, $n$.</p></div><div class="list-item-paragraph"><p>The $1$st, $2$nd and $3$rd iteration produces the snowflakes in Fig. $2$, Fig. $3$ and Fig. $4$ respectively.</p></div><ol class="i-list"><li data-list-item-id="subitem-1760173916843-lcp2lyh2g" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173916843-lcp2lyh2g"><div class="list-item-paragraph">If&nbsp;${{a}_{0}}$ denotes the area of the original triangle and the area of each new triangle added in the ${{n}^{th}}$ iteration is denoted by ${{a}_{n}}$, show that</div><div class="list-item-paragraph" style="text-align: center">${{a}_{n}}=\frac{1}{9}{{a}_{n-1}}$ for all positive integers $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760173934499-q0bb8y1qp" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173934499-q0bb8y1qp"><div class="list-item-paragraph"><p>Write down the number of sides in the polygons in Fig.$1$, Fig. $2$ and Fig. $3$, and deduce, with clear explanations, that the number of new triangles added in the ${{n}^{th}}$ iteration is ${{T}_{n}}=k\left( {{4}^{n}} \right)$ where&nbsp;$k$ is a constant to be determined.</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760173941871-845h0bf3u" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173941871-845h0bf3u"><div class="list-item-paragraph">Find the total area of triangles added in the ${{n}^{th}}$ iteration, ${{A}_{n}}$in terms of ${{a}_{0}}$ and $n$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="subitem-1760173953489-ffk11k6yl" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760173953489-ffk11k6yl"><div class="list-item-paragraph">Show that the total area of the snowflake produced after the ${{n}^{th}}$ iteration is ${{a}_{0}}\left[ \frac{8}{5}-\frac{3}{5}{{\left( \frac{4}{9} \right)}^{n}} \right]$ $\mathrm{unit}{{\mathrm{s}}^{2}}$.<span class="inline-marks">[3]</span></div><div class="list-item-paragraph">The snowflake formed when the above steps are followed indefinitely is called the Koch snowflake.</div></li><li data-list-item-id="subitem-1760174001548-7ootpzwbg" data-list-item-path="item-1760173813868-4qjn9vwrk/subitem-1760174001548-7ootpzwbg"><div class="list-item-paragraph"><p>Determine the least number of iterations needed for the area of the snowflake to exceed $99\%$ of the area of a Koch snowflake.</p><span class="inline-marks">[3]</span></div></li></ol></li></ol>

Problems involving inequalities

Geometric Progression

Sum to infinity

2019 ASRJC P1 Q11

Sequence and Series

<ol class="abc-list"><li><div class="list-item-paragraph">The volume of water, in litres, in a storage tank decreases by $10\%$ each day. However, $90$ litres of water is also pumped into the tank at the end of each day. The volume of water in the tank at the end of $n$ days is denoted by ${{x}_{n}}$ and ${{x}_{0}}$ is the initial volume of water in the tank.</div><ol class="i-list"><li><div class="list-item-paragraph">Write down a recurrence relation to represent the above situation.</div><p style="text-align: right;">[1]</p><div class="list-item-paragraph">Given that $a{{r}^{n-1}}+...+a{{r}^{2}}+ar+a=a\left( \frac{1-r{{\,}^{n}}}{1-r} \right)$,</div></li><li><div class="list-item-paragraph">Show that ${{x}_{n}}={{0.9}^{n}}\left( {{x}_{0}}-900 \right)+900$.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">Deduce the limit of ${{x}_{n}}$ as $n$ becomes very large.</div><p style="text-align: right;">[1]</p></li></ol></li></ol>

Limit of a Sequence and Behaviour of a Sequence

Finding Explicit Terms of Linear Recurrences

2008 MJC P1 Q8

<p>The number of bacteria (in millions) in Pond ${A}$ at the start of the ${n}$th week, before any chemical treatment, is given by ${u_n}$. Pond ${A}$ is treated at the start of each week with Chemical ${A}$, which kills $70\%$ of all bacteria instantly. At the end of each week, $6$ million new bacteria is reproduced.</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Write down a recurrence relation of the form ${u_{n+1} = au_n + b}$, where ${a}$ and ${b}$ are constants to be determined.</span></div><p style="text-align: right;">[1]</p></li><li><div class="list-item-content"><span class="list-item-text">Show that ${{u}_{n}}={{0.3}^{n-1}}{{u}_{1}}+\frac{60}{7}\left( 1-{{0.3}^{n-1}} \right)$.</span></div><p style="text-align: right;">[2]</p></li></ol>
<p>The number of bacteria (in millions) in Pond ${B}$ at the start of the ${n}$th week, before any chemical treatment, is given by ${v_n}$.</p>
<p>Pond ${B}$ is treated at the start of each week with Chemical ${B}$. It is known that ${v_n}$ follows the recurrence relation</p>
<p style="text-align: center">${{v}_{n+1}}=0.01{{v}_{n}}^{2}+6$</p>
<p>It is given that if the sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$, … converges to a limit, it converges to either ${\alpha}$ or ${\beta}$, where ${\alpha < \beta}$.</p>
<ol class="i-list" start="3"><li><div class="list-item-content"><span class="list-item-text">Find ${\alpha}$ and ${\beta}$. Explain whether ${v_n}$ necessarily converges to ${\alpha}$ or ${\beta}$.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">If ${u_1 = v_1 = 30}$, determine which chemical would be more effective in killing the bacteria in the long run.</span></div><p style="text-align: right;">[2]</p></li></ol>
<p>Pond ${C}$ is treated with Chemical ${C}$. To account for the bacteria’s increasing resistance to the chemical, the dosage of Chemical ${C}$ is increased by $5$ ml each week. The first dose is $20$ ml.</p>
<ol class="i-list" start="5"><li><div class="list-item-content"><span class="list-item-text">How many weeks does it take to finish the first $3$ litres of chemical in the treatment of Pond ${C}$?</span></div><p style="text-align: right;">[3]</p></li></ol>

The number of bacteria (in millions) in Pond ${A}$ at the start of the ${n}$th week, before any chemical treatment, is given by ${u_n}$. Pond ${A}$ is treated at the start of each week with Chemical ${A}$, which kills $70\%$ of all bacteria instantly. At the end of each week, $6$ million new bacteria is reproduced.

Write down a recurrence relation of the form ${u_{n+1} = au_n + b}$, where ${a}$ and ${b}$ are constants to be determined.

Show that ${{u}_{n}}={{0.3}^{n-1}}{{u}_{1}}+\frac{60}{7}\left( 1-{{0.3}^{n-1}} \right)$.

The number of bacteria (in millions) in Pond ${B}$ at the start of the ${n}$th week, before any chemical treatment, is given by ${v_n}$.

Pond ${B}$ is treated at the start of each week with Chemical ${B}$. It is known that ${v_n}$ follows the recurrence relation

It is given that if the sequence ${{v}_{1}}$, ${{v}_{2}}$, ${{v}_{3}}$, … converges to a limit, it converges to either ${\alpha}$ or ${\beta}$, where ${\alpha < \beta}$.

Find ${\alpha}$ and ${\beta}$. Explain whether ${v_n}$ necessarily converges to ${\alpha}$ or ${\beta}$.

If ${u_1 = v_1 = 30}$, determine which chemical would be more effective in killing the bacteria in the long run.

Pond ${C}$ is treated with Chemical ${C}$. To account for the bacteria’s increasing resistance to the chemical, the dosage of Chemical ${C}$ is increased by $5$ ml each week. The first dose is $20$ ml.

How many weeks does it take to finish the first $3$ litres of chemical in the treatment of Pond ${C}$?

Arithmetic Progression

2016 DHS P2 Q3

<p>Joe puts $\$10$ on $1$ January $2025$ into a bank account which pays compound interest at a rate of ${2\%}$ per month on the last day of each month. He puts a further $\$10$ into the account on the first day of each subsequent month.</p>
<ol class="i-list"><li data-list-item-id="item-1758616974246-ehmh00du4" data-list-item-path="item-1758616974246-ehmh00du4"><div class="list-item-paragraph">How much compound interest has his original $\$10$ earned at the end of ${2}$ years?</div></li><li data-list-item-id="item-1758616975056-xqpcg9qyf" data-list-item-path="item-1758616975056-xqpcg9qyf"><div class="list-item-paragraph">Content for Let&nbsp;${a_n}$ denote the total amount in the account at the end of ${n}$ months. Find the value of ${a_1}$ and the recurrence relation in the form ${a_{n+1} = f(a_n)}$.</div></li><li data-list-item-id="item-1758616975918-g09py0m4v" data-list-item-path="item-1758616975918-g09py0m4v"><div class="list-item-paragraph">With the aid of a graphing calculator, determine</div><ol class="abc-list"><li data-list-item-id="subitem-1758616977178-8ags06khk" data-list-item-path="item-1758616975918-g09py0m4v/subitem-1758616977178-8ags06khk"><div class="list-item-paragraph">the total amount in the account at the end of ${2}$ years.</div></li><li data-list-item-id="subitem-1758616978806-0kmx0yygg" data-list-item-path="item-1758616975918-g09py0m4v/subitem-1758616978806-0kmx0yygg"><div class="list-item-paragraph">the number of complete months for the total in the account to first exceed $\$2000$.</div></li></ol></li></ol>

Joe puts $\$10$ on $1$ January $2025$ into a bank account which pays compound interest at a rate of ${2\%}$ per month on the last day of each month. He puts a further $\$10$ into the account on the first day of each subsequent month.

How much compound interest has his original $\$10$ earned at the end of ${2}$ years?

Content for Let ${a_n}$ denote the total amount in the account at the end of ${n}$ months. Find the value of ${a_1}$ and the recurrence relation in the form ${a_{n+1} = f(a_n)}$.

With the aid of a graphing calculator, determine

Math Question Bank