If Hugh chooses Option A, calculate the total value of his allowance at the end of the two year period.

If Hugh chooses Option B, calculate

(i)     the amount of money he will receive in the 17th month;

(ii)     the total value of his allowance at the end of the two year period.

If Hugh chooses Option C, calculate

(i)     the amount of money Hugh would receive in the 13th month;

(ii)     the total value of his allowance at the end of the two year period.

If Hugh chooses Option D, calculate the total value of his allowance at the end of the two year period.

State which of the options, A, B, C or D, Hugh should choose to give him the greatest total value of his allowance at the end of the two year period.

Another bank guarantees Hugh an amount of \(\$1750\) after two years of investment if he invests $1500 at this bank. The interest is compounded annually.

Calculate the interest rate per annum offered by the bank.

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

\(60 \times 24\)     (M1)

Note: Award (M1) for correct product.

\( = 1440\)     (A1)(G2)

[2 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(i)     \(10 + (17 - 1)(5)\)     (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.

\( = 90\)     (A1)(G2)

(ii)     \(\frac{{24}}{2}\left( {2(10) + (24 - 1)(5)} \right)\)     (M1)

OR

\(\frac{{24}}{2}\left( {10 + 125} \right)\)     (M1)

Note: Award (M1) for correct substitution in arithmetic series formula.

\( = 1620\)     (A1)(ft)(G1)

Note: Follow through from part (b)(i).

[5 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

(i)     \(15{(1.1)^{12}}\)     (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.

\( = 47\)     (A1)(G2)

Note: Award (M1)(A1)(A0) for \(47.08\).

     Award (G1) for \(47.08\) if workings are not shown.

(ii)     \(\frac{{15({{1.1}^{24}} - 1)}}{{1.1 - 1}}\)     (M1)

Note: Award (M1) for correct substitution in geometric series formula.

\( = 1327\)     (A1)(ft)(G1)

Note: Follow through from part (c)(i).

[5 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

\(1500{\left( {1 + \frac{6}{{100(12)}}} \right)^{12(2)}}\)     (M1)(A1)

Note: Award (M1) for substituted compound interest formula, (A1) for correct substitutions.

OR

\(N = 2\)

\(I\%  = 6\)

\(PV = 1500\)

\(P/Y = 1\)

\(C/Y = 12\)     (A1)(M1)

Note: Award (A1) for \(C/Y = 12\) seen, (M1) for other correct entries.

OR

\(N = 24\)

\(I\%  = 6\)

\(PV = 1500\)

\(P/Y = 12\)

\(C/Y = 12\)     (A1)(M1)

Note: Award (A1) for \(C/Y = 12\) seen, (M1) for other correct entries.

\( = 1691\)     (A1)(G2)

[3 marks]

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that part is not awarded.

Option D     (A1)(ft)

Note: Follow through from their parts (a), (b), (c) and (d). Award (A1)(ft) only if values for the four options are seen and only if their answer is consistent with their parts (a), (b), (c) and (d).

[1 mark]

\(1750 = 1500{\left( {1 + \frac{r}{{100}}} \right)^2}\)     (M1)(A1)

Note: Award (M1) for substituted compound interest formula equated to \(1750\), (A1) for correct substitutions into formula.

OR

\(N = 2\)

\(PV = 1500\)

\(FV =  - 1750\)

\(P/Y = 1\)

\(C/Y = 1\)     (A1)(M1)

Note: Award (A1) for \(FV = 1750\) seen, (M1) for other correct entries.

\( = 8.01\% {\text{ (8.01234}} \ldots \% ,{\text{ }}0.0801{\text{)}}\)     (A1)(G2)

[3 marks]

Show that A lies on \({L_1}\).

Find the coordinates of M.

Find the length of AC.

Show that the equation of \({L_2}\) is \(2y + x - 19 = 0\).

Find the coordinates of D.

Write down the length of MD correct to five significant figures.

Find the area of ABCD.

\(2 \times 4 - 1 - 7 = 0\) (or equivalent)     (R1)

Note:     For (R1) accept substitution of \(x = 1\) or \(y = 4\) into the equation followed by a confirmation that \(y = 4\) or \(x = 1\).

(since the point satisfies the equation of the line,) A lies on \({L_1}\)     (A1)

Note:     Do not award (A1)(R0).

[2 marks]

\(\frac{{1 + 5}}{2}\) OR \(\frac{{4 + 12}}{2}\) seen     (M1)

Note:     Award (M1) for at least one correct substitution into the midpoint formula.

\((3,{\text{ }}8)\)    (A1)(G2)

Notes:     Accept \(x = 3,{\text{ }}y = 8\).

Award (M1)(A0) for \(\left( {\frac{{1 + 5}}{2},{\text{ }}\frac{{4 + 12}}{2}} \right)\).

Award (G1) for each correct coordinate seen without working.

[2 marks]

\(\sqrt {{{(5 - 1)}^2} + {{(12 - 4)}^2}} \)    (M1)

Note:     Award (M1) for a correct substitution into distance between two points formula.

\( = 8.94{\text{ }}\left( {4\sqrt 5 ,{\text{ }}\sqrt {80} ,{\text{ }}8.94427 \ldots } \right)\)    (A1)(G2)

[2 marks]

gradient of \({\text{AC}} = \frac{{12 - 4}}{{5 - 1}}\)     (M1)

Note:     Award (M1) for correct substitution into gradient formula.

\( = 2\)    (A1)

Note:     Award (M1)(A1) for gradient of \({\text{AC}} = 2\) with or without working

gradient of the normal \( =  - \frac{1}{2}\)     (M1)

Note:     Award (M1) for the negative reciprocal of their gradient of AC.

\(y - 8 =  - \frac{1}{2}(x - 3)\) OR \(8 =  - \frac{1}{2}(3) + c\)     (M1)

Note:     Award (M1) for substitution of their point and gradient into straight line formula. This (M1) can only be awarded where \( - \frac{1}{2}\) (gradient) is correctly determined as the gradient of the normal to AC.

\(2y - 16 =  - (x - 3)\) OR \( - 2y + 16 = x - 3\) OR \(2y =  - x + 19\)     (A1)

Note:     Award (A1) for correctly removing fractions, but only if their equation is equivalent to the given equation.

\(2y + x - 19 = 0\)    (AG)

Note:     The conclusion \(2y + x - 19 = 0\) must be seen for the (A1) to be awarded.

Where the candidate has shown the gradient of the normal to \({\text{AC}} =  - 0.5\), award (M1) for \(2(8) + 3 - 19 = 0\) and (A1) for (therefore) \(2y + x - 19 = 0\).

Simply substituting \((3,{\text{ }}8)\) into the equation of \({L_2}\) with no other prior working, earns no marks.

[5 marks]

\((6,{\text{ }}6.5)\)    (A1)(A1)(G2)

Note:     Award (A1) for 6, (A1) for 6.5. Award a maximum of (A1)(A0) if answers are not given as a coordinate pair. Accept \(x = 6,{\text{ }}y = 6.5\).

Award (M1)(A0) for an attempt to solve the two simultaneous equations \(2y - x - 7 = 0\) and \(2y + x - 19 = 0\) algebraically, leading to at least one incorrect or missing coordinate.

[2 marks]

3.3541     (A1)

Note:     Answer must be to 5 significant figures.

[1 mark]

\(2 \times \frac{1}{2} \times \sqrt {80}  \times \frac{{\sqrt {45} }}{2}\)     (M1)(M1)

Notes:     Award (M1) for correct substitution into area of triangle formula.

If their triangle is a quarter of the rhombus then award (M1) for multiplying their triangle by 4.

If their triangle is a half of the rhombus then award (M1) for multiplying their triangle by 2.

OR

\(\frac{1}{2} \times \sqrt {80}  \times \sqrt {45} \)    (M1)(M1)

Notes:     Award (M1) for doubling MD to get the diagonal BD, (M1) for correct substitution into the area of a rhombus formula.

Award (M1)(M1) for \(\sqrt {80}  \times \) their (f).

\( = 30\)    (A1)(ft)(G3)

Notes:     Follow through from parts (c) and (f).

\(8.94 \times 3.3541 = 29.9856 \ldots \)

[3 marks]

State the values of u₁ and d for this sequence.

Use an appropriate formula to show that the sum of the natural numbers from 1 to n is given by \(\frac{1}{2}n (n +1)\).

Calculate the sum of the natural numbers from 1 to 200.

The sum of the first n terms of G₁ is 29 524. Find n.

A second geometric progression G₂ has the form \(1,\frac{1}{3},\frac{1}{9},\frac{1}{{27}}...\)

Calculate the sum of the first 10 terms of G₂.

Explain why the sum of the first 1000 terms of G₂ will give the same answer as the sum of the first 10 terms, when corrected to three significant figures.

Using your results from parts (a) to (c), or otherwise, calculate the sum of the first 10 terms of the sequence \(2,3\frac{1}{3},9\frac{1}{9},27\frac{1}{{27}}...\)

Give your answer correct to one decimal place.

\(u_1 = d = 1\).     (A1)(A1)

[2 marks]

Sum is \(\frac{1}{2}n(2{u_1} + d(n - 1))\) or \(\frac{1}{2}n({u_1} + {u_n})\)     (M1)

Award (M1) for either sum formula seen, even without substitution.

So sum is \(\frac{1}{2}n(2 + (n - 1)) = \frac{1}{2}n(n + 1)\)     (A1)(AG)

Award (A1) for substitution of \({u_1} = 1 = d\) or \({u_1} = 1\) and \({u_n} = n\) with simplification where appropriate. \(\frac{1}{2}n(n + 1)\) must be seen to award this (A1).

[2 marks]

\(\frac{1}{2}(200)(201) = 20 100\)     (M1)(A1)(G2)

(M1) is for correct formula with correct numerical input. Original sum formula with u, d and n can be used.

[2 marks]

\(\frac{{1 - {3^n}}}{{1 - 3}} = 29524\)     (M1)(A1)

(M1) for correctly substituted formula on one side, (A1) for = 29524 on the other side.

n = 10.     (A1)(G2)

Trial and error is a valid method. Award (M1) for at least \(\frac{{1 - {3^{10}}}}{{1 - 3}}\) seen and then (A1) for = 29524, (A1) for \(n = 10\). For only unproductive trials with \(n \ne 10\), award (M1) and then (A1) if the evaluation is correct.

[3 marks]

Common ratio is \(\frac{1}{3}\), (0.333 (3sf) or 0.3)     (A1)

Accept ‘divide by 3’.

[1 mark]

\(\frac{{1 - {{\left( {\frac{1}{3}} \right)}^{10}}}}{{1 - \frac{1}{3}}}\)     (M1)

= 1.50 (3sf)     (A1)(ft)(G1)

1.5 and \(\frac{3}{2}\) receive (A0)(AP) if AP not yet used Incorrect formula seen in (a) or incorrect value in (b) can follow through to (c). Can award (M1) for \(1 + \left( {\frac{1}{3}} \right) + \left( {\frac{1}{9}} \right) + ......\)

[2 marks]

Both \({\left( {\frac{1}{3}} \right)^{10}}\) and \({\left( {\frac{1}{3}} \right)^{1000}}\) (or those numbers divided by 2/3) are 0 when corrected to 3sf, so they make no difference to the final answer.     (R1)

Accept any valid explanation but please note: statements which only convey the idea of convergence are not enough for (R1). The reason must show recognition that the convergence is adequately fast (though this might be expressed in a much less technical manner).

[1 mark]

The sequence given is \(G_1 + G_2\)     (M1)

The sum is 29 524 + 1.50     (A1)(ft)

= 29 525.5     (A1)(ft)(G2)

The (M1) is implied if the sum of the two numbers is seen. Award (G1) for 29 500 with no working. (M1) can be awarded for
\(2 + 3\frac{1}{3} + ...\) Award final (A1) only for answer given correct to 1dp.

[3 marks]

(i) Identification of u₁ and d was fine. In (b) many candidates failed to recognise the need for a general proof and simply gave an example substitution. Part (c) was well done.

(i) Identification of u₁ and d was fine. In (b) many candidates failed to recognise the need for a general proof and simply gave an example substitution. Part (c) was well done.

(i) Identification of u₁ and d was fine. In (b) many candidates failed to recognise the need for a general proof and simply gave an example substitution. Part (c) was well done.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

(ii) Too many candidates here failed to swap to the GP formulae. Those who did know what was happening here often performed the calculations well and got decent marks. The explanations in (d) were often unsatisfactory but some allowance was made for the language difficulties encountered by candidates writing in a 2^nd or higher language. The last part (e) of the question, intended as a high-grade discriminator performed that task very well.

Calculate the value of his investment at the end of the third year for each investment option, correct to two decimal places.

Determine Daniel’s best investment option.

Calculate u₁, the first term in the sequence.

Show that the common difference is 7.

S_n is the sum of the first n terms of the sequence.

Find an expression for S_n. Give your answer in the form S_n = An² + Bn, where A and B are constants.

The first term, v₁, of a geometric sequence is 20 and its fourth term v₄ is 67.5.

Show that the common ratio, r, of the geometric sequence is 1.5.

T_n is the sum of the first n terms of the geometric sequence.

Calculate T₇, the sum of the first seven terms of the geometric sequence.

T_n is the sum of the first n terms of the geometric sequence.

Use your graphic display calculator to find the smallest value of n for which T_n > S_n.

Option 1:     Amount    \( = 25\,000{\left( {1 + \frac{5}{{100}}} \right)^3}\)     (M1)(A1)

= \(28\,940.63\)     (A1)(G2)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitution. Give full credit for use of lists.

Option 2:     Amount     \( = 25\,000{\left( {1 + \frac{{4.8}}{{12(100)}}} \right)^{3 \times 12}}\)     (M1)

= \(28\,863.81\)     (A1)(G2)

Note: Award (M1) for correct substitution in the compound interest formula. Give full credit for use of lists.

[8 marks]

Option 1 is the best investment option.     (A1)(ft)

[1 mark]

u₁ = 135 + 7(1)     (M1)

= 142     (A1)(G2)

[2 marks]

u₂ = 135 + 7(2) = 149     (M1)

d = 149 – 142     OR alternatives     (M1)(ft)

d = 7     (AG)

[2 marks]

\({S_n} = \frac{{n[2(142) + 7(n - 1)]}}{2}\)     (M1)(ft)

Note: Award (M1) for correct substitution in correct formula.

\( = \frac{{n[277 + 7n]}}{2}\)     OR equivalent     (A1)

\( = \frac{{7{n^2}}}{2} + \frac{{277n}}{2}\)     (= 3.5n² + 138.5n)     (A1)(G3)

[3 marks]

20r³ = 67.5     (M1)

r³ = 3.375     OR \(r = \sqrt[3]{{3.375}}\)     (A1)

r = 1.5     (AG)

[2 marks]

\({T_7} = \frac{{20({{1.5}^7} - 1)}}{{(1.5 - 1)}}\)     (M1)

Note: Award (M1) for correct substitution in correct formula.

= 643 (accept 643.4375)     (A1)(G2)

[2 marks]

\(\frac{{20({{1.5}^n} - 1)}}{{(1.5 - 1)}} > \frac{{7{n^2}}}{2} + \frac{{277n}}{2}\)     (M1)

Note: Award (M1) for an attempt using lists or for relevant graph.

n = 10     (A1)(ft)(G2)

Note: Follow through from their (c).

[2 marks]

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common error was to make all the comparisons using interest alone; though much credit was given for doing this, candidates should be aware of what is being asked for in the question.

Many did not understand the notion of monthly compounding periods.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common error was to make all the comparisons using interest alone; though much credit was given for doing this, candidates should be aware of what is being asked for in the question.

Many did not understand the notion of monthly compounding periods.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

For many, this question came as a welcome relief following the previous two questions. For those with a sound grasp of the topic, there were many very successful attempts.

A common weakness was seen in the “show that” parts of the question where, despite a lenient approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for S_n in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only the highly competent reached the correct answer.

Find, in BRL, the amount of money Estela has after commission.

Find, in GBP, the amount of money Estela receives.

After her trip to the United Kingdom Estela has 400 GBP left. At the airport she changes this money back into BRL. The exchange rate is now 1.00 BRL = 0.3125 GBP.

Find, in BRL, the amount of money that Estela should receive.

Estela actually receives 1216.80 BRL after commission.

Find, in BRL, the commission charged to Estela.

The commission rate is t % . Find the value of t.

Show that after three years Daniel will have $1109.70 in his account, correct to two decimal places.

Write down the interest Daniel receives after three years.

4000 × 0.97 = 3880.00 (3880)     (M1)(A1)(G2)

Note: Award (M1) for multiplication of correct numbers.

OR

3 % of 4000 = 120     (A1)

4000 – 120 = 3880.00 (3880)     (A1)(G2)

[2 marks]

3880 × 0.3071 = 1191.55     (M1)(A1)(ft)(G2)

Note: Award (M1) for multiplication of correct numbers. Follow through from their answer to part (a).

[2 marks]

\(\frac{{400}}{{0.3125}}\)     (M1)

= 1280.00 (1280)     (A1)(G2)

Note: Award (M1) for division of correct numbers.

[2 marks]

63.20 (A1)(ft)

Note: Follow through (their (c) –1216.80).

[1 mark]

\(t = \frac{{63.20 \times 100}}{{1280}}\)     (M1)

t = 4.94     (A1)(ft)(G2)

Note: Follow through from their answers to parts (c) and (d).

[2 marks]

\({\text{A}} = 1000{\left( {1 + \frac{{3.5}}{{2 \times 100}}} \right)^6} = 1109.7023...\)     (M1)(A1)(A1)

= 1109.70     (AG)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for unrounded answer. If 1109.70 not seen award at most (M1)(A1)(A0).

OR

\({\text{I}} = 1000{\left( {1 + \frac{{3.5}}{{2 \times 100}}} \right)^6} - 1000 = 109.7023\)     (M1)(A1)

A = 1109.7023...     (A1)

= 1109.70     (AG)

Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for unrounded answer.

[3 marks]

109.70     (A1)

Note: No follow through here.

[1 mark]

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

The first three parts were well done. Some students gave their answer to part (d) in (e) and their answer to (e) in (d). This means that when reading commission they directed their answers to a percentage (commission rate).

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

The first three parts were well done. Some students gave their answer to part (d) in (e) and their answer to (e) in (d). This means that when reading commission they directed their answers to a percentage (commission rate).

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

The first three parts were well done. Some students gave their answer to part (d) in (e) and their answer to (e) in (d). This means that when reading commission they directed their answers to a percentage (commission rate).

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

The first three parts were well done. Some students gave their answer to part (d) in (e) and their answer to (e) in (d). This means that when reading commission they directed their answers to a percentage (commission rate).

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

The first three parts were well done. Some students gave their answer to part (d) in (e) and their answer to (e) in (d). This means that when reading commission they directed their answers to a percentage (commission rate).

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

Most of the students used the correct formula but not all made the correct substitution. From those that made the correct substitution, very few showed the unrounded answer. Part (b) was well done. In part (c) the majority did not put the interest (only) in the formula but the total amount $1109.70.

Most of the students were penalized in this question for not given their money answers correct to the specified accuracy (2 decimal places).

Most of the students used the correct formula but not all made the correct substitution. From those that made the correct substitution, very few showed the unrounded answer. Part (b) was well done. In part (c) the majority did not put the interest (only) in the formula but the total amount $1109.70.

Find the length of the road AB.

Find the size of the angle CAB.

Village D is halfway between A and B. A new road perpendicular to AB and passing through D is built. Let T be the point where this road cuts AC. This information is shown in the diagram below.

Write down the distance from A to D.

Show that the distance from D to T is 2.06 km correct to three significant figures.

A bus starts and ends its journey at A taking the route AD to DT to TA.

Find the total distance for this journey.

The average speed of the bus while it is moving on the road is 70 km h^–1. The bus stops for 5 minutes at each of D and T .

Estimate the time taken by the bus to complete its journey. Give your answer correct to the nearest minute.

AB² = 10² + 8² – 2 × 10 × 8 × cos150°     (M1)(A1)

AB = 17.4 km     (A1)(G2)

Note: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer.

[3 marks]

\(\frac{8}{{\sin {\text{C}}\hat {\rm A}{\text{B}}}} = \frac{{17.4}}{{\sin 150^\circ }}\)     (M1)(A1)

\({\text{C}}\hat {\rm A}{\text{B}} = 13.3^\circ \)     (A1)(ft)(G2)

Notes: Award (M1) for substitution into correct formula, (A1) for correct substitution, (A1) for correct answer. Follow through from their answer to part (a).

[3 marks]

AD = 8.70 km (8.7 km)     (A1)(ft)

Note: Follow through from their answer to part (a).

[1 mark]

DT = tan (13.29...°) × 8.697... = 2.0550...     (M1)(A1)

= 2.06     (AG)

Notes: Award (M1) for correct substitution in the correct formula, award (A1) for the unrounded answer seen. If 2.06 not seen award at most (M1)(AO).

[2 marks]

\(\sqrt {{{8.70}^2} + {{2.06}^2}}  + 8.70 + 2.06\)     (A1)(M1)

= 19.7 km     (A1)(ft)(G2)

Note: Award (A1) for AT, (M1) for adding the three sides of the triangle ADT, (A1)(ft) for answer. Follow through from their answer to part (c).

[3 marks]

\(\frac{{19.7}}{{70}} \times 60 + 10\)     (M1)(M1)

= 26.9     (A1)(ft)

Note: Award (M1) for time on road in minutes, (M1) for adding 10, (A1)(ft) for unrounded answer. Follow through from their answer to (e).

= 27  (nearest minute)     (A1)(ft)(G3)

Note: Award (A1)(ft) for their unrounded answer given to the nearest minute.

[4 marks]

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

The weak students answered parts (a) and (b) using right-angled trigonometry. Different types of mistakes were seen in (a) when applying the cosine rule: some forgot to square root their answer and others calculated each part separately and then missed the 2 minuses. Part (b) was better done than (a). Follow through was applied from (a) to (c). Part (d) was not well done. Most of the students lost one mark in this part question as they did not show the unrounded answer (2.0550...). Part (e) was fairly well done by those who attempted it. In (f) there were very few correct answers. Students found it difficult to find the time when the average speed and distance were given.

Write down an equation showing this information, taking \(b\) to be the cost of one tin of beans and \(c\) to be the cost of one packet of cereal in \({\text{AUD}}\).

Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).

Write down another equation to represent this information.

Stephen thinks that Mal has not bought enough so he buys \(12\) more tins of beans and \(6\) more packets of cereal. He pays \(66{\text{ AUD}}\).

Find the cost of one tin of beans.

(i)     Sketch the graphs of the two equations from parts (a) and (b).

(ii)    Write down the coordinates of the point of intersection of the two graphs.

Using the letters on the diagram draw a triangle showing the position of a \({70^ \circ }\) angle.

Show that the height of the pyramid is \(13.7{\text{ cm}}\), to 3 significant figures.

Calculate

(i)     the length of \({\text{EG}}\);

(ii)    the size of angle \({\text{DEC}}\).

Find the total surface area of the pyramid.

Find the volume of the pyramid.

\(50b + 20c = 260\)     (A1)

[1 mark]

\(12b + 6c = 66\)     (A1)

[1 mark]

Solve to get \(b = 4\)     (M1)(A1)(ft)(G2)

Note: (M1) for attempting to solve the equations simultaneously.

[2 marks]

(i)

     (A1)(A1)(A1)

Notes: Award (A1) for labels and some idea of scale, (A1)(ft)(A1)(ft) for each line.
The axis can be reversed.

(ii)    \((4,3)\) or \((3,4)\)     (A1)(ft)

Note: Accept \(b = 4\), \(c = 3\)

[4 marks]

     (A1)

[1 mark]

\(\tan 70 = \frac{h}{5}\)     (M1)

\(h = 5\tan 70 = 13.74\)     (A1)

\(h = 13.7{\text{ cm}}\)     (AG)

[2 marks]

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

(i)     \({\text{E}}{{\text{G}}^2} = {5^2} + {13.7^2}\) OR \({5^2} + {(5\tan 70)^2}\)     (M1)

(UP)     \({\text{EG}} = 14.6{\text{ cm}}\)     (A1)(G2)

(ii)    \({\text{DEC}} = 2 \times {\tan ^{ - 1}}\left( {\frac{5}{{14.6}}} \right)\)     (M1)

\( = {37.8^ \circ }\)     (A1)(ft)(G2)

[4 marks]

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

\({\text{Area}} = 10 \times 10 + 4 \times 0.5 \times 10 \times 14.619\)     (M1)

(UP)     \( = 392{\text{ c}}{{\text{m}}^2}\)     (A1)(ft)(G2)

[2 marks]

Unit penalty (UP) is applicable in this part of the question where indicated in the left hand column.

\({\text{Volume}} = \frac{1}{3} \times 10 \times 10 \times 13.7\)     (M1)

(UP)     \( = 457{\text{ c}}{{\text{m}}^3}\) (\(458{\text{ c}}{{\text{m}}^3}\))     (A1)(G2)

[2 marks]

Most candidates managed to write down the equation.

Most candidates managed to write down the equation.

Many managed to find the correct answer and the others tried their best but made some mistake in the process.

(i)     Few candidates sketched the graphs well. Few used a ruler.

(ii)    Many candidates could not be awarded ft from their graph because the answer they gave was not possible.

Very few correct drawings.

Some managed to show this more by good fortune and ignoring their original triangle than by good reasoning.

(i)     Many found this as ft from the previous part. Some lost a UP here.

(ii)    This was not well done. The most common answer was \({40^ \circ }\).

Many managed this or were awarded ft points.

This was well done and most candidates also remembered their units on this part.

Write down the distance Rosa runs in the third training session;

Write down the distance Rosa runs in the \(n\)th training session.

Find the value of \(k\).

Calculate the total distance, in kilometres, Rosa runs in the first 50 training sessions.

Find the distance Carlos runs in the fifth month of training.

Calculate the total distance Carlos runs in the first year.

3800 m     (A1)

[1 mark]

\(3000 + (n - 1)400{\text{ m}}\)\(\,\,\,\)OR\(\,\,\,\)\(2600 + 400n{\text{ m}}\)     (M1)(A1)

Note:     Award (M1) for substitution into arithmetic sequence formula, (A1) for correct substitution.

[2 marks]

\(3000 + (k - 1)400 > 42195\)     (M1)

Notes:     Award (M1) for their correct inequality. Accept \(3 + (k - 1)0.4 > 42.195\).

Accept \( = \) OR \( \geqslant \). Award (M0) for \(3000 + (k - 1)400 > 42.195\).

\((k = ){\text{ }}99\)     (A1)(ft)(G2)

Note:     Follow through from part (a)(ii), but only if \(k\) is a positive integer.

[2 marks]

\(\frac{{50}}{2}\left( {2 \times 3000 + (50 - 1)(400)} \right)\)     (M1)(A1)(ft)

Note:     Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution.

\(640\,000{\text{ m}}\)     (A1)

Note:     Award (A1) for their \(640\,000\) seen.

\( = 640{\text{ km}}\)     (A1)(ft)(G3)

Note:     Award (A1)(ft) for correctly converting their answer in metres to km; this can be awarded independently from previous marks.

OR

\(\frac{{50}}{2}\left( {2 \times 3 + (50 - 1)(0.4)} \right)\)     (M1)(A1)(ft)(A1)

Note:     Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct substitution, (A1) for correctly converting 3000 m and 400 m into km.

\( = 640{\text{ km}}\)     (A1)(G3)

[4 marks]

\(7500 \times {1.2^{5 - 1}}\)     (M1)(A1)

Note:     Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

\( = 15\,600{\text{ m }}(15\,552{\text{ m}})\)     (A1)(G3)

OR

\(7.5 \times {1.2^{5 - 1}}\)     (M1)(A1)

Note:     Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

\( = 15.6{\text{ km}}\)     (A1)(G3)

[3 marks]

\(\frac{{7500({{1.2}^{12}} - 1)}}{{1.2 - 1}}\)     (M1)(A1)

Notes:     Award (M1) for substitution into sum of a geometric series formula, (A1) for correct substitutions. Follow through from their ratio (\(r\)) in part (d). If \(r < 1\) (distance does not increase) or the final answer is unrealistic (eg \(r = 20\)), do not award the final (A1).

\( = 297\,000{\text{ m }}(296\,853 \ldots {\text{ m}},{\text{ }}297{\text{ km}})\)     (A1)(G2)

[3 marks]

Show that the width of the plot, in metres, is given by \(100 - x\).

Write down the area of the plot in terms of \(x\).

Find the value of \(x\) that maximizes the area of the plot.

Show that Violeta earns 5000 BGN from selling the flowers grown on the plot.

Find the amount of money that Violeta would have after 6 years. Give your answer correct to two decimal places.

Find how long it would take for the interest earned to be 2000 BGN.

Find the lowest possible value for \(r\).

\(\frac{{200 - 2x}}{2}\) (or equivalent)     (M1)

OR

\(2x + 2y = 200\) (or equivalent)     (M1)

Note:     Award (M1) for a correct expression leading to \(100 - x\) (the \(100 - x\) does not need to be seen). The 200 must be seen for the (M1) to be awarded. Do not accept \(100 - x\) substituted in the perimeter of the rectangle formula.

\(100 - x\)     (AG)

[1 mark]

\(({\text{area}} = ){\text{ }}x(100 - x)\)\(\,\,\,\)OR\(\,\,\,\)\( - {x^2} + 100x\) (or equivalent)     (A1)

[1 mark]

\(x = \frac{{ - 100}}{{ - 2}}\)\(\,\,\,\)OR\(\,\,\,\)\( - 2x + 100 = 0\)\(\,\,\,\)OR\(\,\,\,\)graphical method     (M1)

Note:     Award (M1) for use of axis of symmetry formula or first derivative equated to zero or a sketch graph.

\(x = 50\)     (A1)(ft)(G2)

Note:     Follow through from part (b), provided x is positive and less than 100.

[2 marks]

\(50(100 - 50) \times 2\)     (M1)(M1)

Note:     Award (M1) for substituting their \(x\) into their formula for area (accept “\(50 \times 50\)” for the substituted formula), and (M1) for multiplying by 2. Award at most (M0)(M1) if their calculation does not lead to 5000 (BGN), although the 5000 (BGN) does not need to be seen explicitly.

Substitution of 50 into area formula may be seen in part (c).

5000 (BGN)     (AG)

[2 marks]

\(5000{\left( {1 + \frac{4}{{2 \times 100}}} \right)^{2 \times 6}}\)     (M1)(A1)

Note:     Award (M1) for substitution into compound interest formula, (A1) for correct substitution.

OR

\({\text{N}} = 6\)

\({\text{I}}\%  = 4\)

\({\text{PV}} =  - 5000\)

\({\text{P/Y}} = 1\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for \({\text{C/Y}} = 2\)  seen, (M1) for other correct entries.

OR

\({\text{N}} = 12\)

\({\text{I}}\%  = 4\)

\({\text{PV}} =  - 5000\)

\({\text{P/Y}} = 2\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for \({\text{C/Y}} = 2\) seen, (M1) for other correct entries.

6341.21 (BGN)     (A1)(G3)

Note:     Award (A1) for correct answer, to two decimal places only.

Award (G2) for 6341.20 or a correct, unrounded final answer if no working is seen (6341.2089…).

[3 marks]

\(5000{\left( {1 + \frac{4}{{2 \times 100}}} \right)^{2 \times t}} = 7000\)     (M1)(A1)(ft)

Note:     Award (M1) for using the compound interest formula with a variable for time, (A1)(ft) for substituting the correct values and equating to 7000. Follow through for their “2” from part (e)(i).

OR

\({\text{I% }} = 4\)

\({\text{PV}} = ( \pm )5000\)

\({\text{FV}} =  \mp 7000\)

\({\text{P/Y}} = 1\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for 7000 seen, (M1) for the other correct entries.

Award (M1) for their C/Y from part (e)(i).

OR

\({\text{I% }} = 4\)

\({\text{PV}} = ( \pm )5000\)

\({\text{FV}} =  \mp 7000\)

\({\text{P/Y}} = 2\)

\({\text{C/Y}} = 2\)     (M1)(A1)

Note:     Award (A1) for 7000 seen, (M1) for the other correct entries.

Award (M1) for their C/Y from part (e)(i).

OR

     (M1)(A1)(ft)

Note:     Award (M1) for a sketch with a straight line intercepted by appropriate curve, (A1)(ft) for numerical answer in the range of 8.4 and 8.5.

Follow through from their part (e)(i).

\(t = 8.50{\text{ (years) }}(8.49564 \ldots )\)     (A1)(ft)(G3)

Note:     Award only (A1) if 16.9912… is seen without working. If working is seen, award at most (M1)(A1)(A0).

[3 marks]

\(5000{\left( {1 + \frac{r}{{100}}} \right)^{12}} = 10000\)     (M1)

Note: Award (M1) for correct substitution into compound interest formula with 10 000 seen. 

OR

\(2 = {\left( {1 + \frac{r}{{100}}} \right)^{12}}\)     (M1)

Note:     Award (M1) for correct substitution and simplification of compound interest formula, equating to 2.

\(r = 5.95{\text{ (% ) }}(5.94630 \ldots )\)     (A1)(G2)

[2 marks]

Write down the distance, \({a_1}\), in metres that she has to run in order to collect the first pumpkin.

The distances she runs to collect each pumpkin form a sequence \({a_1},{\text{ }}{a_2},{\text{ }}{a_3}, \ldots \) .

(i)     Find \({a_2}\).

(ii)     Find \({a_3}\).

Write down the common difference, \(d\), of the sequence.

The final pumpkin Sirma collected was 24 metres from the start.

(i)     Find the total number of pumpkins that Sirma collected.

(ii)     Find the total distance that Sirma ran to collect these pumpkins.

Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.

Calculate the total number of pumpkins that Peter collected.

Peter also plays the game. When the signal is given for the end of the game he has run 940 metres.

Calculate Peter’s distance from the start when the signal is given.

\(6{\text{ (m)}}\)    (A1)(G1)

(i)     \(8\)     (A1)(ft)

(ii)     \(10\)     (A1)(ft)(G2)

Note: Follow through from part (a).

\(2{\text{ (m)}}\)    (A1)(ft)

Note: Follow through from parts (a) and (b).

(i)     \(2 \times 24 = 6 + 2(n - 1)\;\;\;\)OR\(\;\;\;24 = 3 + (n - 1)\)     (M1)

Note: Award (M1) for correct substitution in arithmetic sequence formula.

\(n = 22\)     (A1)(ft)(G1)

Note: Follow through from parts (a) and (c).

(ii)     \(\frac{{(6 + 48)}}{2} \times 22\)     (M1)(A1)(ft)

Note: Award (M1) for substitution in arithmetic series formula, (A1)(ft) for correct substitution.

\( = 594\)     (A1)(ft)(G2)

Note: Follow through from parts (a) and (d)(i).

\(\frac{{\left[ {2 \times 6 + 2(n - 1)} \right] \times n}}{2} = 940\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution in arithmetic series formula, (A1) for their correct substituted formula equated to \(940\). Follow through from parts (a) and (c).

\({n^2} + 5n - 940 = 0\)

\(n = 28.2611 \ldots \)

\(n = 28\)     (A1)(ft)(G2)

\(\frac{{\left[ {2 \times 6 + 2(28 - 1)} \right] \times 28}}{2}\)     (M1)

Notes: Award (M1) for substituting their \(28\) into the arithmetic series formula.

\( = 16{\text{ (m)}}\)     (A1)(ft)(G2)

(i) Write an equation for u₆ in terms of d and u₁.

(ii) Write an equation for u₁₀ in terms of d and u₁.

Write down the value of

(i) d ;

(ii) u₁ .

Find the total number of seats in the amphitheatre.

A few years later, a second level was added to increase the amphitheatre’s capacity by another 1600 seats. Each row has four more seats than the previous row. The first row on this level has 70 seats.

Find the number of rows on the second level of the amphitheatre.

Write down the next two terms of this geometric sequence.

Write down the common ratio of this geometric sequence.

Calculate the number of people who will receive the text message at 12:30.

Calculate the total number of people who will have received the text message by 12:30.

Calculate the exact time at which a total of 29 524 people will have received the text message.

(i) u₁ + 5d = 100     (A1)

(ii) u₁ + 9d = 124     (A1)

[2 marks]

(i) 6     (G1)(ft)

(ii) 70     (G1)(ft)

Notes: Follow through from their equations in parts (a) and (b) even if working not seen. Their answers must be integers. Award (M1)(A0) for an attempt to solve two equations analytically.

[2 marks]

\(S_{20} = \frac{20}{2}(2 \times 70 + (20 - 1) \times 6)\)     (M1)(A1)(ft)

Note: Award (M1) for substituted sum of AP formula, (A1)(ft) for their correct substituted values.

= 2540     (A1)(ft)(G2)

Note: Follow through from their part (b).

[3 marks]

\(\frac{n}{2}(2 \times 70 + (n - 1) \times 4) = 1600\)     (M1)(A1)

Note: Award (M1) for substituted sum of AP formula, (A1) for their correct substituted values.

4n² +136n – 3200 = 0     (M1)

Note: Award (M1) for this equation (or other equivalent expanded quadratic) seen, may be implied if correct final answer seen.

n = 16     (A1)(G3)

Note: Do not award the final (A1) for n = 16, – 50 given as final answer, award (G2) if n = 16, – 50 given as final answer without working.

[4 marks]

9, 27     (A1)

[1 mark]

3     (A1)

[1 mark]

1 × 3⁶     (M1)

= 729     (A1)(ft)(G2)

Note: Award (M1) for correctly substituted GP formula. Follow through from their answer to part (b).

[2 marks]

\(\frac{{1(3^7 - 1)}}{(3 - 1)}\)     (M1)

Note: Award (M1) for correctly substituted GP formula. Accept sum 1+ 3 + 9 + 27 + ... + 729. If lists are used, award (M1) for correct list that includes 1093. (1, 4, 13, 40, 121, 364, 1093, 3280…)

= 1093     (A1)(ft)(G2)

Note: Follow through from their answer to part (b). For consistent use of n = 6 from part (c) (243) to part (d) leading
to an answer of 364, treat as double penalty and award (M1)(A1)(ft) if working is shown.

[2 marks]

\(\frac{{1(3^n - 1)}}{(3 - 1)} = 29524\)     (M1)

Note: Award (M1) for correctly substituted GP formula. If lists are used, award (M1) for correct list that includes 29524. (1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573...). Accept alternative methods, for example continuation of sum in part (d).

n = 10     (A1)(ft)

Note: Follow through from their answer to part (b).

Exact time = 12:45     (A1)(ft)(G2)

[3 marks]

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part A: Arithmetic

The contextual nature of this question posed problems for many, though there were many fine attempts. Failure to discriminate between the sequence and series formulas was the cause of the most errors. The final part saw many able to substitute into the formula for the series, but then unable to continue. The use of the GDC in such situations is encouraged; either by graphing each side of the equation and drawing the resultant sketch or by the solver function.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with many choosing the incorrect value of n when using the formulas. Weaker candidates were more successful via counting. The context again proved challenging in the final part, with the incorrect time being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each side of the equation is encouraged; however, if teachers feel that such questions require the use (and teaching) of logarithms, such an approach is, of course, given full credit.

(i)     Show that the vertical height of the cone is \(2.65\) cm correct to three significant figures.

(ii)     Calculate the volume of the perfume bottle.

The bottle contains \({\text{125 c}}{{\text{m}}^{\text{3}}}\) of perfume. The bottle is not full and all of the perfume is in the cylinder part.

Find the height of the perfume in the bottle.

Temi makes some crafts with perfume bottles, like the one above, once they are empty. Temi wants to know the surface area of one perfume bottle.

Find the total surface area of the perfume bottle.

Temi covers the perfume bottles with a paint that costs 3 South African rand (ZAR) per millilitre. One millilitre of this paint covers an area of \({\text{7 c}}{{\text{m}}^{\text{2}}}\).

Calculate the cost, in ZAR, of painting the perfume bottle. Give your answer correct to two decimal places.

Temi sells her perfume bottles in a craft fair for 325 ZAR each. Dominique from France buys one and wants to know how much she has spent, in euros (EUR). The exchange rate is 1 EUR = 13.03 ZAR.

Find the price, in EUR, that Dominique paid for the perfume bottle. Give your answer correct to two decimal places.

(i)     \({x^2} + {3^2} = {4^2}\)     (M1)

Note: Award (M1) for correct substitution into Pythagoras’ formula.

Accept correct alternative method using trigonometric ratios.

\(x = 2.64575 \ldots \)     (A1)

\(x = 2.65{\text{ }}({\text{cm}})\)     (AG)

Note: The unrounded and rounded answer must be seen for the (A1) to be awarded.

OR

\(\sqrt {{4^2} - {3^2}} \)     (M1)

Note: Award (M1) for correct substitution into Pythagoras’ formula.

\( = \sqrt 7 \)     (A1)

\( = 2.65{\text{ (cm)}}\)     (AG)

Note: The exact answer must be seen for the final (A1) to be awarded.

(ii)     \(\pi  \times {3^2} \times 4.5 + \frac{1}{3}\pi  \times {3^2} \times 2.65\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into the volume of a cylinder formula, (M1) for correct substitution into the volume of a cone formula, (M1) for adding both of their volumes.

\( = 152{\text{ c}}{{\text{m}}^3}\;\;\;(152.210 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}48.45\pi {\text{ c}}{{\text{m}}^3})\)     (A1)(G3)

\(\pi {3^2}h = 125\)     (M1)

Note: Award (M1) for correct substitution into the volume of a cylinder formula.

Accept alternative methods. Accept \(4.43\) (\(4.42913 \ldots \)) from using rounded answers in \(h = \frac{{125 \times 4.5}}{{127}}\).

\(h = 4.42{\text{ (cm)}}\;\;\;\left( {4.42097 \ldots {\text{ (cm)}}} \right)\)     (A1)(G2)

\(2\pi  \times 3 \times 4.5 + \pi  \times 3 \times 4 + \pi  \times {3^2}\)     (M1)(M1)(M1)

Note: Award (M1) for correct substitution into curved surface area of a cylinder formula, (M1) for correct substitution into the curved surface area of a cone formula, (M1) for adding the area of the base of the cylinder to the other two areas.

\( = 151{\text{ c}}{{\text{m}}^2}\;\;\;(150.796 \ldots {\text{ c}}{{\text{m}}^2},{\text{ }}48\pi {\text{ c}}{{\text{m}}^2})\)     (A1)(G3)

\(\frac{{150.796 \ldots }}{7} \times 3\)     (M1)(M1)

Notes: Award (M1) for dividing their answer to (c) by \(7\), (M1) for multiplying by \(3\). Accept equivalent methods.

\( = 64.63{\text{ (ZAR)}}\)     (A1)(ft)(G2)

Notes: The (A1) is awarded for their correct answer, correctly rounded to 2 decimal places. Follow through from their answer to part (c). If rounded answer to part (c) is used the answer is \(64.71\) (ZAR).

\(\frac{{325}}{{13.03}}\)     (M1)

Note: Award (M1) for dividing \(325\) by \(13.03\).

\( = 24.94{\text{ (EUR)}}\)     (A1)(G2)

Note: The (A1) is awarded for the correct answer rounded to 2 decimal places, unless already penalized in part (d).

Calculate the value of the common ratio.

Calculate the 10^th term of this sequence.

The k^th term is the first term which is greater than 2000. Find the value of k.

Write in words \((q \wedge \neg r) \Rightarrow \neg p\).

Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write this statement in symbolic form.

Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write the contrapositive of this statement in symbolic form.

u₁r⁴ = 324     (A1)

u₁r = 12     (A1)

r³ = 27     (M1)

r = 3     (A1)(G3)

Note: Award at most (G3) for trial and error.

[4 marks]

4 × 3⁹ = 78732 or 12 × 3⁸ = 78732     (A1)(M1)(A1)(ft)(G3)

Note: Award (A1) for u₁ = 4 if n = 9 , or u₁ = 12 if n = 8, (M1) for correctly substituted formula.

(ft) from their (a).

[3 marks]

4 × 3^k−1 > 2000     (M1)

Note: Award (M1) for correct substitution in correct formula. Accept an equation.

k > 6     (A1)

k = 7     (A1)(ft)(G2)

Notes: If second line not seen award (A2) for correct answer. (ft) from their (a).

Accept a list, must see at least 3 terms including the 6^th and 7^th.

Note: If arithmetic sequence formula is used consistently in parts (a), (b) and (c), award (A0)(A0)(M0)(A0) for (a) and (ft) for parts (b) and (c).

[3 marks]

If the number is even and the number does not end in zero, (then) the number is not a multiple of five.     (A1)(A1)(A1)

Note: Award (A1) for “if…(then)”, (A1) for “the number is even and the number does not end in zero”, (A1) for the number is not a multiple of 5.

[3 marks]

\((p \wedge \neg q) \Rightarrow \neg r\)     (A1)(A1)(A1)(A1)

(A1) for \(\Rightarrow\), (A1) for \(\wedge\), (A1) for p and \(\neg q\), (A1) for \(\neg r\)

Note: If parentheses not present award at most (A1)(A1)(A1)(A0).

[4 marks]

\(r \Rightarrow (\neg p \vee q)\)   OR   \(r \Rightarrow \neg (p \wedge \neg q)\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for reversing the order, (A1) for negating the statements on both sides.

If parentheses not present award at most (A1)(ft)(A0).

Do not penalise twice for missing parentheses in (i) and (ii).

[2 marks]

An easy ratio to find and the majority of candidates found r = 3, though many had trouble showing the appropriate method, thus losing marks.

A fairly straightforward part for most candidates.

The majority found k − 7; many without supporting work which lost them a mark. Where candidates had difficulty in this part, it was generally a case of poor algebraic skills.

This question on logic was straightforward for most candidates who scored full marks for parts (a) and (b) (i). A few omitted the brackets in part (b).

This question on logic was straightforward for most candidates who scored full marks for parts (a) and (b) (i). A few omitted the brackets in part (b).

Very poorly answered with many candidates scoring just one mark. The main error was to open the bracket and not use the “or”.

Write down the size of angle AQB.

Calculate the length of AQ.

Calculate the length of AC.

Show that the length of CQ is 50.37 m, correct to 4 significant figures.

Find the size of the angle AQC.

Calculate the total area of the glass needed to construct

(i) the two rectangular faces of the greenhouse;

(ii) the two triangular faces of the greenhouse.

The cost of one square metre of glass used to construct the greenhouse is 4.80 USD.

Calculate the cost of glass to make the greenhouse. Give your answer correct to the nearest 100 USD.

110°     (A1)

\(\frac{{AQ}}{{\sin 35^\circ }} = \frac{{10}}{{\sin 110^\circ }}\)     (M1)(A1)

Note: Award (M1) for substituted sine rule formula, (A1) for their correct substitutions.

OR

\(AQ = \frac{5}{{\cos 35^\circ }}\)     (A1)(M1)

Note: Award (A1) for 5 seen, (M1) for correctly substituted trigonometric ratio.

\(AQ = 6.10\) (6.10387...)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (a).

\(AC^2 = 10^2 + 50^2\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\(AC = 51.0 (\sqrt{2600}, 50.9901...)\)     (A1)(G2)

\(QC^2 = (6.10387...)^2 + (50)^2\)     (M1)

Note: Award (M1) for correctly substituted Pythagoras formula.

\(QC = 50.3711...\)     (A1)

\(= 50.37\)     (AG)

Note: Both the unrounded and rounded answers must be seen to award (A1).

   If 6.10 is used then 50.3707... is the unrounded answer.

   For an incorrect follow through from part (b) award a maximum of (M1)(A0) – the given answer must be reached to award the final (A1)(AG).

\(\cos AQC = \frac{{{{(6.10387...)}^2} + {{(50.3711...)}^2} - {{(50.9901...)}^2}}}{{2(6.10387...)(50.3711...)}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted cosine rule formula, (A1)(ft) for their correct substitutions.

= 92.4°   (\({92.3753...^\circ }\))     (A1)(ft)(G2)

Notes: Follow through from their answers to parts (b), (c) and (d). Accept 92.2 if the 3 sf answers to parts (b), (c) and (d) are used.

     Accept 92.5° (\({92.4858...^\circ }\)) if the 3 sf answers to parts (b), (c) and 4 sf answers to part (d) used.

(i) \(2(50 \times 6.10387...)\)     (M1)

Note: Award (M1) for their correctly substituted rectangular area formula, the area of one rectangle is not sufficient.

= 610 m² (610.387...)     (A1)(ft)(G2)

Notes: Follow through from their answer to part (b).

    The answer is 610 m². The units are required.

(ii) Area of triangular face \( = \frac{1}{2} \times 10 \times 6.10387... \times \sin 35^\circ \)     (M1)(A1)(ft)

OR

Area of triangular face \( = \frac{1}{2} \times 6.10387... \times 6.10387... \times \sin 110^\circ \)     (M1)(A1)(ft)

\(= 17.5051...\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correct substitutions.

OR

(Height of triangle) \( = {(6.10387...)^2} - {5^2}\)

\(= 3.50103...\)

Area of triangular face \( = \frac{1}{2} \times 10 \times their{\text{ }} height\)

\(= 17.5051...\)

Note: Award (M1) for substituted triangle area formula, (A1)(ft) for correctly substituted area formula. If 6.1 is used, the height is 3.49428... and the area of both triangular faces 34.9 m²

Area of both triangular faces = 35.0 m² (35.0103...)     (A1)(ft)(G2)

Notes: The answer is 35.0 m². The units are required. Do not penalize if already penalized in part (f)(i). Follow through from their part (b).

(610.387... + 35.0103...) × 4.80     (M1)

= 3097.90...     (A1)(ft)

Notes: Follow through from their answers to parts (f)(i) and (f)(ii).

    Accept 3096 if the 3 sf answers to part (f) are used.

= 3100     (A1)(ft)(G2)

Notes: Follow through from their unrounded answer, irrespective of whether it is correct. Award (M1)(A2) if working is shown and 3100 seen without the unrounded answer being given.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result. The use of Pythagoras’ theorem was also successful, the major source of error being the lack of unrounded and rounded answers in part (d).

Again, most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates were able to recognize sine rule, substitute correctly and reach the required result. Part (e) was less well answered, due in part to the triangle being in three dimensions. However, all three sides had either been asked for in previous parts or given and all that was required was a sketch of a triangle with the vertices labelled; such a diagram was never on any script and this technique should be encouraged.

Again, most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Most candidates used the appropriate area formula – however, some did not read the question with the attention it required and found the area of three rectangles – one of which being the stated “concrete base”.

Write down the total number of fish in the pond.

Leanne catches a fish.

Find the probability that she

(i) catches an undersized bream;

(ii) catches either a flathead or an undersized fish or both;

(iii) does not catch an undersized whiting;

(iv) catches a whiting given that the fish was normal.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Copy and complete the probability tree diagram below.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Calculate the probability that it is windy and Leanne catches a fish on a particular day.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Calculate the probability that Leanne catches a fish on a particular day.

Use your answer to part (e) to calculate the probability that Leanne catches a fish on two consecutive days.

Leanne notices that on windy days, the probability she catches a fish is 0.1 while on non-windy days the probability she catches a fish is 0.65. The probability that it will be windy on a particular day is 0.3.

Given that Leanne catches a fish on a particular day, calculate the probability that the day was windy.

90     (A1)

[1 mark]

(i) \(\frac{3}{{90}}(0.0\bar 3,{\text{ }}0.0333,{\text{ }}0.0333...,{\text{ }}3.\bar 3\% ,{\text{ }}3.33\% )\)     (A1)(ft)

Note: For the denominator follow through from their answer in part (a).

(ii) \(\frac{{53}}{{90}}(0.5\bar 8,{\text{ }}0.588...,{\text{ }}0.589,{\text{ }}58.\bar 8\% ,{\text{ }}58.9\% )\)     (A1)(A1)(ft)(G2)

Notes: Award (A1) for the numerator. (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iii) \(\frac{{72}}{{90}}{\text{(0.8, 80}}\%)\)     (A1)(ft)(A1)(ft)(G2)

Notes: Award (A1)(ft) for the numerator, (their part (a) –18) (A1)(ft) for denominator. For the denominator follow through from their answer in part (a).

(iv) \(\frac{{24}}{{48}}(0.5,{\text{ 50}}\% )\)     (A1)(A1)(G2)

Note: Award (A1) for numerator, (A1) for denominator.

[7 marks]

     (A1)(A1)(A1)

Notes: Award (A1) for each correct entry. Tree diagram must be seen for marks to be awarded.

[3 marks]

\(0.3 \times 0.1 = 0.03\left( {\frac{3}{{100}}} \right)\)     (M1)(A1)(G2)

Note: Award (M1) for correct product seen.

[2 marks]

\(0.3 \times 0.1+ 0.7\times0.65\)     (M1)(M1)

Notes: Award (M1) for \(0.7\times0.65\) (or 0.455) seen, (M1) for adding their 0.03. Follow through from their answers to parts (c) and (d).

\( = 0.485\left( {\frac{{485}}{{1000}},\frac{{97}}{{200}}} \right)\)     (A1)(ft)(G2)

Note: Follow through from their tree diagram and their answer to part (d).

[3 marks]

\(0.485 \times 0.485\)     (M1)

\(0.235\left( {\frac{{9409}}{{40000}}{\text{, }}0.235225} \right)\)     (A1)(ft)(G2)

Note: Follow through from their answer to part (e).

[2 marks]

\(\frac{{0.03}}{{0.485}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substituted conditional probability formula, (A1)(ft) for their (d) as numerator and their (e) as denominator.

\(0.0619\left( {\frac{{6}}{{97}}}\text{, 0.0618556...} \right) \)     (A1)(ft)(G2)

Note: Follow through from their parts (d) and (e).

[3 marks]

(a) Most candidates found this correctly although a few wrote 180 instead of 90.

(b) This was also answered well. The main errors were putting 65/90 in part (ii) and 24/90 in part (iv).

(c) The tree diagram was completed correctly in most scripts. It appears that some candidates may have answered this on their question paper and this was not sent to the scanning centre with the answer papers.

(d) Many answered this correctly. Some added instead of multiplying.

(e) Surprisingly well answered. Again some added and multiplied in the wrong place.

(f) Most candidates added here and then divided by 2 rather than multiplying.

(g) This was badly done with very few correct answers seen.

Factorise \(3{x^2} + 13x - 10\).

Solve the equation \(3{x^2} + 13x - 10 = 0\).

Consider a function \(f(x) = 3{x^2} + 13x - 10\) .

Find the equation of the axis of symmetry on the graph of this function.

Consider a function \(f(x) = 3{x^2} + 13x - 10\) .

Calculate the minimum value of this function.

Show that \(4{x^2} + 6xy = 300\).

Find an expression for \(y\) in terms of \(x\).

Hence show that the volume \(V\) of the box is given by \(V = 100x - \frac{4}{3}{x^3}\).

Find \(\frac{{{\text{d}}V}}{{{\text{d}}x}}\).

(i)     Hence find the value of \(x\) and of \(y\) required to make the volume of the box a maximum.

(ii)    Calculate the maximum volume.

\((3x - 2)(x + 5)\)     (A1)(A1)

[2 marks]

\((3x - 2)(x + 5) = 0\)

\(x = \frac{2}{3}\) or \(x = - 5\)     (A1)(ft)(A1)(ft)(G2)

[2 marks]

\(x = \frac{{ - 13}}{6}{\text{ }}( - 2.17)\)     (A1)(A1)(ft)(G2)

Note: (A1) is for \(x = \), (A1) for value. (ft) if value is half way between roots in (b).

[2 marks]

Minimum \(y = 3{\left( {\frac{{ - 13}}{6}} \right)^2} + 13\left( {\frac{{ - 13}}{6}} \right) - 10\)     (M1)

Note: (M1) for substituting their value of \(x\) from (c) into \(f(x)\) .

\( = - 24.1\)     (A1)(ft)(G2)

[2 marks]

\({\text{Area}} = 2(2x)x + 2xy + 2(2x)y\)     (M1)(A1)

Note: (M1) for using the correct surface area formula (which can be implied if numbers in the correct place). (A1) for using correct numbers.

\(300 = 4{x^2} + 6xy\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.

[2 marks]

\(6xy = 300 - 4{x^2}\)     (M1)

\(y = \frac{{300 - 4{x^2}}}{{6x}}\) or \(\frac{{150 - 2{x^2}}}{{3x}}\)     (A1)

[2 marks]

\({\text{Volume}} = x(2x)y\)     (M1)

\(V = 2{x^2}\left( {\frac{{300 - 4{x^2}}}{{6x}}} \right)\)     (A1)(ft)

\( = 100x - \frac{4}{3}{x^3}\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.

[2 marks]

\(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 100 - \frac{{12{x^2}}}{3}\)  or  \(100 - 4{x^2}\)     (A1)(A1)

 Note: (A1) for each term.

[2 marks]

Unit penalty (UP) is applicable where indicated in the left hand column

(i)     For maximum \(\frac{{{\text{d}}V}}{{{\text{d}}x}} = 0\)  or  \(100 - 4{x^2} = 0\)     (M1)

\(x = 5\)     (A1)(ft)

\(y = \frac{{300 - 4{{(5)}^2}}}{{6(5)}}\)  or  \(\left( {\frac{{150 - 2{{(5)}^2}}}{{3(5)}}} \right)\)     (M1)

\( = \frac{{20}}{3}\)     (A1)(ft)

(UP)     (ii)    \(333\frac{1}{3}{\text{ c}}{{\text{m}}^3}{\text{ }}(333{\text{ c}}{{\text{m}}^3})\)

Note: (ft) from their (e)(i) if working for volume is seen.

[5 marks]

Most candidates made a good attempt to factorise the expression.

Many gained both marks here from a correct answer or ft from the previous part.

Many used the formula correctly. Some forgot to put \(x = \) .

Most candidates found this value from their GDC.

A good attempt was made to show the correct surface area.

Many could rearrange the equation correctly.

Although this was not a difficult question it probably looked complicated for the candidates and it was often left out.

Those who reached this length could usually manage the differentiation.

(i)     Many found the correct value of \(x\) but not of \(y\).

(ii)    This was well done and again the units were included in most scripts.

Find the amount of money that Ruby will have in the bank after 3 years.

Show that Ying will have 7545 USD in the bank at the end of 10 years.

Find the number of complete years it will take for Ying’s investment to first exceed 6500 USD.

Find the number of complete years it will take for Ying’s investment to exceed Ruby’s investment.

Ruby moves from the USA to Italy. She transfers 6610 USD into an Italian bank which has an exchange rate of 1 USD = 0.735 Euros. The bank charges 1.8 % commission.

Calculate the amount of money Ruby will invest in the Italian bank after commission.

Ruby returns to the USA for a short holiday. She converts 800 Euros at a bank in Chicago and receives 1006.20 USD. The bank advertises an exchange rate of 1 Euro = 1.29 USD.

Calculate the percentage commission Ruby is charged by the bank.

5000 + 3 × 230 = 5690     (M1)(A1)(G2)

Note: Accept alternative method.

[2 marks]

\({\text{A}} = 5000{\left( {1 + \frac{{4.2}}{{100}}} \right)^{10}}\) or equivalent     (M1)(A1)

\( = 7544.79 \ldots \)     (A1)

\( = 7545{\text{ USD}}\)     (AG)

Note: Award (M1) for correct substituted compound interest formula, (A1) for correct substitutions, (A1) for unrounded answer seen.

If final line not seen award at most (M1)(A1)(A0).

[3 marks]

5000(1.042)ⁿ > 6500     (M1)(A1)

Notes: Award (M1) for setting up correct equation/inequality, (A1) for correct values.

Follow through from their formula in part (b).

OR

List of values seen with at least 2 terms     (M1)

Lists of values including at least the terms with n = 6 and n = 7     (A1)

Note: Follow through from their formula in part (b).

OR

Sketch showing 2 graphs, one exponential, the other a horizontal line     (M1)

Point of intersection identified or vertical line     (M1)

Note: Follow through from their formula in part (b).

n = 7     (A1)(ft)(G2)

[3 marks]

5000(1.042)ⁿ > 5000 + 230n     (M1)(A1)

Note: Award (M1) for setting up correct equation/inequality, (A1) for correct values.

OR

2 lists of values seen (at least 2 terms per list)     (M1)

Lists of values including at least the terms with n = 5 and n = 6     (A1)

Note: One of the lists may be written under (c).

OR

Sketch showing 2 graphs of correct shape     (M1)

Point of intersection identified or vertical line     (M1)

n = 6     (A1)(ft)(G2)

Note: Follow through from their formulae used in parts (a) and (b).

[3 marks]

6610 × 0.735     (M1)

= 4858.35     (A1)

4858.35 × 0.982(= 4770.8997...)     (M1)

= 4771 Euros     (A1)(ft)(G3)

Note: Accept alternative method.

[4 marks]

800 × 1.29 (= 1032 USD)     (M1)(A1)

Note: Award (M1) for multiplying by 1.29, (A1) for 1032. Award (G2) for 1032 if product not seen.

(1032 – 1006.20 = 25.8)

\(25.8 \times \frac{100}{1032} \%\)     (A1)(M1)

Note: Award (A1) for 25.8 seen, (M1) for multiplying by \(\frac{100}{1032}\).

OR

\(\frac{1006.20}{1032} = 0.975\)     (M1)(A1)

OR

\(\frac{1006.20}{1032} \times 100 = 97.5\)     (M1)(A1)

\( = 2.5{\text{ }}\% \)     (A1)(G3)

Notes: If working not shown award (G3) for 2.5.

Accept alternative method.

[5 marks]

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

It was nice to see many students recovering after part (d) and to gain full marks in the last two parts of the question.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

It was nice to see many students recovering after part (d) and to gain full marks in the last two parts of the question.

Most of the students read carefully the instruction written in the heading of the question and therefore gave their answers with the accuracy stated but some did not.

Simple interest was well done as well as compound interest with only a small minority of candidates making no progress. A number of students lost the answer mark in (b) for not showing the unrounded answer before writing the answer given. It is also important to mention that calculator commands are not accepted as correct working and therefore full marks are not awarded. Also, some candidates wrote their answers without showing any working leading to a number of marks being lost.

It was nice to see many students recovering after part (d) and to gain full marks in the last two parts of the question.

If Clara chooses option 1,

(i) write down the values of the second and third installments;

(ii) calculate the value of the final installment;

(iii) show that the total amount that Clara would pay for the land is \(\$ 88000\).

If Clara chooses option 2,

(i) find the value of the second installment;

(ii) show that the value of the fifth installment is \(\$ 2721\).

The price of the land is \(\$ 80000\). In option 1 her total repayments are \(\$ 88000\) over the 20 months. Find the annual rate of simple interest which gives this total.

Clara knows that the total amount she would pay for the land is not the same for both options. She wants to spend the least amount of money. Find how much she will save by choosing the cheaper option.

(i) Second installment \( = \$ 2700\)     (A1)

Third installment \( = \$ 2900\)     (A1)

(ii) Final installment \( = 2500 + 200 \times 19\)     (M1)(A1)

Note: (M1) for substituting in correct formula or listing, (A1) for correct substitutions.

\( = \$ 6300\)     (A1)(G2)

(iii) Total amount \( = \frac{{20}}{2}(2500 + 6300)\)

OR

\( \frac{{20}}{2}(5000 + 19 \times 200)\)     (M1)(A1)

Note: (M1) for substituting in correct formula or listing, (A1) for correct substitution.

\( = \$ 88000\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.

[7 marks]

(i) Second installment \(2000 \times 1.08 = \$ 2160\)     (M1)(A1)(G2)

Note: (M1) for multiplying by \(1.08\) or equivalent, (A1) for correct answer.

(ii) Fifth installment \( = 2000 \times {1.08^4} = 2720.98 = \$ 2721\)     (M1)(A1)(AG)

Notes: (M1) for correct formula used with numbers from the problem. (A1) for correct substitution. The \(2720.9 \ldots \) must be seen for the (A1) mark to be awarded. Accept list of 5 correct values. If values are rounded prematurely award (M1)(A0)(AG).

[4 marks]

Interest is \( = \$ 8000\)     (A1)

\(80000 \times \frac{r}{{100}} \times \frac{{20}}{{12}} = 8000\)     (M1)(A1)

Note: (M1) for attempting to substitute in simple interest formula, (A1) for correct substitution.

Simple Interest Rate \( = 6\% \)     (A1)(G3)

Note: Award (G3) for answer of \(6\% \) with no working present if interest is also seen award (A1) for interest and (G2) for correct answer.

[4 marks]

Financial accuracy penalty (FP) is applicable where indicated in the left hand column.

(FP)     Total amount for option 2 \( = 2000\frac{{(1 - {{1.08}^{20}})}}{{(1 - 1.08)}}\)     (M1)(A1)

Note: (M1) for substituting in correct formula, (A1) for correct substitution.

\( = \$ 91523.93\) (\( = \$ 91524\))     (A1)
\(91523.93 - 88000 = \$ 3523.93 = \$ 3524\) to the nearest dollar     (A1)(ft)(G3)

Note: Award (G3) for an answer of \(\$ 3524\) with no working. The difference follows through from the sum, if reasonable. Award a maximum of (M1)(A0)(A0)(A1)(ft) if candidate has treated option 2 as an arithmetic sequence and has followed through into their common difference. Award a maximum of (M1)(A1)(A0)(ft)(A0) if candidate has consistently used \(0.08\) in (b) and (d).

[4 marks]

This question was answered correctly by many. Candidates were able to restart if they failed to complete a particular part. Many candidates wasted much time because their understanding was limited to a recursive method and hence wrote out all the terms rather than using the formula for the nth term or sum. A surprising number of students were not able to use the simple interest formula for a period which was not a whole number of years. Also hardly anyone knew to calculate interest first before substituting into the formula. Many students who attempted part (d) lost a point due to FP. A number of students rounded their answers prematurely to the nearest dollar.

This question was answered correctly by many. Candidates were able to restart if they failed to complete a particular part. Many candidates wasted much time because their understanding was limited to a recursive method and hence wrote out all the terms rather than using the formula for the nth term or sum. A surprising number of students were not able to use the simple interest formula for a period which was not a whole number of years. Also hardly anyone knew to calculate interest first before substituting into the formula. Many students who attempted part (d) lost a point due to FP. A number of students rounded their answers prematurely to the nearest dollar.

This question was answered correctly by many. Candidates were able to restart if they failed to complete a particular part. Many candidates wasted much time because their understanding was limited to a recursive method and hence wrote out all the terms rather than using the formula for the nth term or sum. A surprising number of students were not able to use the simple interest formula for a period which was not a whole number of years. Also hardly anyone knew to calculate interest first before substituting into the formula. Many students who attempted part (d) lost a point due to FP. A number of students rounded their answers prematurely to the nearest dollar.

This question was answered correctly by many. Candidates were able to restart if they failed to complete a particular part. Many candidates wasted much time because their understanding was limited to a recursive method and hence wrote out all the terms rather than using the formula for the nth term or sum. A surprising number of students were not able to use the simple interest formula for a period which was not a whole number of years. Also hardly anyone knew to calculate interest first before substituting into the formula. Many students who attempted part (d) lost a point due to FP. A number of students rounded their answers prematurely to the nearest dollar.

Show that the common ratio is \(\frac{1}{2}\) .

Find the value of the eleventh term.

Find the sum of the first eight terms.

Find the number of terms in the sequence for which the sum first exceeds \(2047.968\).

Find the value of the eleventh term.

The sum of the first \(n\) terms of this sequence is \(\frac{n}{2}(3n - 1)\).

(i)     Find the sum of the first 100 terms in this arithmetic sequence.

(ii)    The sum of the first \(n\) terms is \(477\).

     (a)     Show that \(3{n^2} - n - 954 = 0\) .

     (b)     Using your graphic display calculator or otherwise, find the number of terms, \(n\) .

\(1024{r^3} = 128\)     (M1)

\({r^3} = \frac{1}{8}\) or \(r = \sqrt[3]{{\frac{1}{8}}}\)     (M1)

\(r = \frac{1}{2}{\text{ }}(0.5)\)     (AG)

Notes: Award at most (M1)(M0) if last line not seen. Award (M1)(M0) if \(128\) is found by repeated multiplication (division) of \(1024\) by \(0.5\) \((2)\).

[2 marks]

\(1024 \times {0.5^{10}}\)     (M1)

Notes: Award (M1) for correct substitution into correct formula. Accept an equivalent method.

1     (A1)(G2)

[2 marks]

\({S_8} = \frac{{1024\left( {1 - {{\left( {\frac{1}{2}} \right)}^8}} \right)}}{{1 - \frac{1}{2}}}\)     (M1)(A1)

Note: Award (M1) for substitution into the correct formula, (A1) for correct substitution.

OR

(A1) for complete and correct list of eight terms     (A1)

(M1) for their eight terms added     (M1)

\({S_8} = 2040\)     (A1)(G2)

[3 marks]

\(\frac{{1024\left( {1 - {{\left( {\frac{1}{2}} \right)}^n}} \right)}}{{1 - \frac{1}{2}}} > 2047.968\)     (M1)(M1)(ft)

Notes: Award (M1) for correct substitution into the correct formula for the sum, (M1) for comparing to \(2047.968\) . Accept equation. Follow through from their expression for the sum used in part (c).

OR

If a list is used: \({S_{15}} = 2047.9375\)     (M1)

\({S_{16}} = 2047.96875\)     (M1)

\(n = 16\)     (A1)(ft)(G2)

Note: Follow through from their expression for the sum used in part (c).

[3 marks]

\({\text{common difference}} = 3\) (may be implied)     (A1)

\({u_{11}} = 31\)     (A1)(G2)

[2 marks]

(i)     \(\frac{{100}}{2}(3 \times 100 - 1)\)     OR     \(\frac{{100(2 + 99 \times 3)}}{2}\)     (M1)

         \(14 950\)     (A1)(G2)

(ii)     (a)     \(\frac{n}{2}(3n - 1) = 477\)     OR     \(\frac{n}{2}(2 + 3(n - 1)) = 477\)     (M1)

                   \(3{n^2} - n = 954\)     (M1)

                   \(3{n^2} - n - 954 = 0\)     (AG)

Notes: Award second (M1) for correct removal of denominator or brackets and no further incorrect working seen. Award at most

(M1)(M0) if last line not seen.

        (b)     \(18\)     (G2)

Note: If both solutions to the quadratic equation are seen and the correct value is not identified as the required answer, award (G1)(G0).

[6 marks]

Part A: Geometric sequences/series

The majority of the candidates were not able to offer a satisfactory justification in a) and only scored 1 mark.

Part A: Geometric sequences/series

Parts b) and c) were mostly well answered.

Part A: Geometric sequences/series

Parts b) and c) were mostly well answered.

Part A: Geometric sequences/series

The responses to part d) were often weak. Those candidates who set up the equation scored two marks but very few of them were able to reach the correct final answer.

Part B: Arithmetic sequences/series

Parts a), and b)(i) were mostly answered correctly.

Part B: Arithmetic sequences/series

Parts a), and b)(i) were mostly answered correctly. Parts b)(ii)a) and b)(ii)b) were poorly answered. Many candidates did not know how to approach the “show that” question. A few were able to solve the quadratic equation using the GDC. Those who attempted to solve it without the GDC generally failed to find the correct answer.

Calculate the volume of the cylinder in cm³. Give your answer correct to two decimal places.

The cylinder is used for storing tennis balls. Each ball has a radius of 3.25 cm.

Calculate how many balls Jenny can fit in the cylinder if it is filled to the top.

(i) Jenny fills the cylinder with the number of balls found in part (b) and puts the lid on. Calculate the volume of air inside the cylinder in the spaces between the tennis balls.

(ii) Convert your answer to (c) (i) into cubic metres.

(i) Find the value of angle \({\text{B}}\hat {\rm T}{\text{L}}\).

(ii) Use triangle BTL to calculate the sloping distance BT from the base, B to the top, T of the tower.

Calculate the vertical height TG of the top of the tower.

Leonardo now walks to point M, a distance 200 m from B on the opposite side of the tower. Calculate the distance from M to the top of the tower at T.

\(\pi \times 3.25^2 \times 39\)     (M1)(A1)

(= 1294.1398)

Answer 1294.14 (cm³)(2dp)     (A1)(ft)(G2)

(UP) not applicable in this part due to wording of question. (M1) is for substituting appropriate numbers from the problem into the correct formula, even if the units are mixed up. (A1) is for correct substitutions or correct answer with more than 2dp in cubic centimetres seen. Award (G1) for answer to > 2dp with no working and no attempt to correct to 2dp. Award (M1)(A0)(A1)(ft) for \(\pi  \times {32.5^2} \times 39{\text{ c}}{{\text{m}}^3}\) (= 129413.9824) = 129413.98

Use of \(\pi = \frac{22}{7}\) or 3.142 etc is premature rounding and is awarded at most (M1)(A1)(A0) or (M1)(A0)(A1)(ft) depending on whether the intermediate value is seen or not. For all other incorrect substitutions, award (M1)(A0) and only follow through the 2 dp correction if the intermediate answer to more decimal places is seen. Answer given as a multiple of \(\pi\) is awarded at most (M1)(A1)(A0). As usual, an unsubstituted formula followed by correct answer only receives the G marks.

[3 marks]

39/6.5 = 6     (A1)

[1 mark]

Unit penalty (UP) is applicable where indicated in the left hand column.

(UP) (i) Volume of one ball is \(\frac{4}{3} \pi \times 3.25^3 {\text{ cm}}^3\)     (M1)

\({\text{Volume of air}} = \pi  \times {3.25^2} \times 39 - 6 \times \frac{4}{3}\pi  \times {3.25^3} = 431{\text{ c}}{{\text{m}}^3}\)     (M1)(A1)(ft)(G2)

Award first (M1) for substituted volume of sphere formula or for numerical value of sphere volume seen (143.79… or 45.77… \( \times \pi\)). Award second (M1) for subtracting candidate’s sphere volume multiplied by their answer to (b). Follow through from parts (a) and (b) only, but negative or zero answer is always awarded (A0)(ft)

(UP) (ii) 0.000431m³ or  4.31×10⁻⁴ m³     (A1)(ft)

[4 marks]

Unit penalty (UP) is applicable where indicated in the left hand column.

(i) \({\text{Angle B}}\widehat {\text{T}}{\text{L}} = 180 - 80 - 26.5\) or \(180 - 90 - 26.5 - 10\)     (M1)

\(= 73.5^\circ\)     (A1)(G2)

(ii) \(\frac{{BT}}{{\sin (26.5^\circ )}} = \frac{{120}}{{\sin (73.5^\circ )}}\)     (M1)(A1)(ft)

(UP) BT = 55.8 m (3sf)     (A1)(ft)

[5 marks]

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but
negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

Unit penalty (UP) is applicable where indicated in the left hand column.

TG = 55.8sin(80°) or 55.8cos(10°)     (M1)

(UP) = 55.0 m (3sf)     (A1)(ft)(G2)

Apply (AP) if 0 missing

[2 marks]

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but
negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

Unit penalty (UP) is applicable where indicated in the left hand column.

\({\text{MT}}^2 = 200^2 + 55.8^2 - 2 \times 200 \times 55.8 \times \cos(100^\circ)\)     (M1)(A1)(ft)

(UP) MT = 217 m  (3sf)     (A1)(ft)

Follow through only from part (ii)(a)(ii). Award marks at discretion for any valid alternative method.

[3 marks]

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but
negative lengths are always awarded (A0)(ft).

The answers are (all 3sf)

(ii)(a)     – 124 m (A0)(ft)

(ii)(b)     123 m (A0)

(ii)(c)     313 m (A0)

If radian mode has been used throughout the question, award (A0) to the first incorrect answer then follow through, but negative lengths are always awarded (A0)(ft)

(i) Many candidates incurred the new one-off unit penalty here. Too many ignored the call for two decimal places and some extrapolated that instruction to later parts (which was clearly not intended). There was the predictable confusion of using radius instead of diameter. Another common error was to divide the cylinder volume by that of the ball, to decide how many would fit. Some follow-through was allowed later from this error, however, this led to zero or negligible air volume, which was clearly ridiculous.

Choice and use of the formulae for volumes was often competent but the conversion to cubic metres was very badly done. Almost no correct answers were seen at all.

(i) Many candidates incurred the new one-off unit penalty here. Too many ignored the call for two decimal places and some extrapolated that instruction to later parts (which was clearly not intended). There was the predictable confusion of using radius instead of diameter. Another common error was to divide the cylinder volume by that of the ball, to decide how many would fit. Some follow-through was allowed later from this error, however, this led to zero or negligible air volume, which was clearly ridiculous.

Choice and use of the formulae for volumes was often competent but the conversion to cubic metres was very badly done. Almost no correct answers were seen at all.

(i) Many candidates incurred the new one-off unit penalty here. Too many ignored the call for two decimal places and some extrapolated that instruction to later parts (which was clearly not intended). There was the predictable confusion of using radius instead of diameter. Another common error was to divide the cylinder volume by that of the ball, to decide how many would fit. Some follow-through was allowed later from this error, however, this led to zero or negligible air volume, which was clearly ridiculous.

Choice and use of the formulae for volumes was often competent but the conversion to cubic metres was very badly done. Almost no correct answers were seen at all.

(ii) Candidates were often sloppy in reading the information. In particular, despite the statement BL = 120 clearly written, many took GL as 120. Triangle TBL was often taken as right-angled. Angle BTL presented few problems, though sometimes the method was very long-winded. Candidates often managed part (a) then went awry in later parts. Many unit penalties were applied, if not already used in questions 1 or 2.

(ii) Candidates were often sloppy in reading the information. In particular, despite the statement BL = 120 clearly written, many took GL as 120. Triangle TBL was often taken as right-angled. Angle BTL presented few problems, though sometimes the method was very long-winded. Candidates often managed part (a) then went awry in later parts. Many unit penalties were applied, if not already used in questions 1 or 2.

(ii) Candidates were often sloppy in reading the information. In particular, despite the statement BL = 120 clearly written, many took GL as 120. Triangle TBL was often taken as right-angled. Angle BTL presented few problems, though sometimes the method was very long-winded. Candidates often managed part (a) then went awry in later parts. Many unit penalties were applied, if not already used in questions 1 or 2.

Find the value of \({u_{20}}\).

Find the sum of the first 10 terms of the sequence.

Now consider the sequence \({v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots \) where

\[{v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24\]

This sequence continues in the same manner.

Find the exact value of \({v_{10}}\).

Now consider the sequence \({v_1},{\text{ }}{v_2},{\text{ }}{v_3},{\text{ }} \ldots ,{\text{ }}{v_n},{\text{ }} \ldots \) where

\[{v_1} = 3,{\text{ }}{v_2} = 6,{\text{ }}{v_3} = 12,{\text{ }}{v_4} = 24\]

This sequence continues in the same manner.

Find the sum of the first 8 terms of this sequence.

\(k\) is the smallest value of \(n\) for which \({v_n}\) is greater than \({u_n}\).

Calculate the value of \(k\).

\(600 + (20 - 1) \times 17\)     (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 20 correct terms seen.

\( = 923\)     (A1)(G3)

[3 marks]

\(\frac{{10}}{2}\left[ {2 \times 600 + (10 - 1) \times 17} \right]\)     (M1)(A1)

Note: Award (M1) for substituted arithmetic series formula, (A1) for their correct substitutions. Follow through from part (a). For consistent use of geometric series formula in part (b) with the geometric sequence formula in part (a) award a maximum of (M1)(A1)(A0) since their final answer cannot be an integer.

OR

\({u_{10}} = 600 + (10 - 1)17 = 753\)     (M1)

\({S_{10}} = \frac{{10}}{2}\left( {600 + {\text{their }}{u_{10}}} \right)\)     (M1)

Note: Award (M1) for their correctly substituted arithmetic sequence formula, (M1) for their correctly substituted arithmetic series formula. Follow through from part (a) and within part (b).

Note: If a list is used, award (M1) for at least 10 correct terms seen, award (A1) for these terms being added.

\( = 6765\)   (accept \(6770\))     (A1)(ft)(G2)

[3 marks]

\(3 \times {2^9}\)     (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions. If a list is used, award (M1) for at least 6 correct terms seen, award (A1) for at least 8 correct terms seen.

\( = 1536\)     (A1)(G3)

Note: Exact answer only. If both exact and rounded answer seen, award the final (A1).

[3 marks]

\(\frac{{3 \times \left( {{2^8} - 1} \right)}}{{2 - 1}}\)     (M1)(A1)(ft)

Note: Award (M1) for substituted geometric series formula, (A1) for their correct substitutions. Follow through from part (c). If a list is used, award (M1) for at least 8 correct terms seen, award (A1) for these 8 correct terms being added. For consistent use of arithmetic series formula in part (d) with the arithmetic sequence formula in part (c) award a maximum of (M1)(A1)(A1).

\( = 765\)     (A1)(ft)(G2)

[3 marks]

\(3 \times {2^{k - 1}} > 600 + (k - 1)(17)\)     (M1)

Note: Award (M1) for their correct inequality; allow equation. 

Follow through from parts (a) and (c). Accept sketches of the two functions as a valid method.

\(k > 8.93648 \ldots \)   (may be implied)     (A1)(ft)

Note: Award (A1) for \(8.93648…\) seen. The GDC gives answers of \(-34.3\) and \(8.936\) to the inequality; award (M1)(A1) if these are seen with working shown.

OR

\({v_8} = 384\)     \({u_8} = 719\)     (M1)

\({v_9} = 768\)     \({u_9} = 736\)     (M1)

Note: Award (M1) for \({v_8}\) and \({u_8}\) both seen, (M1) for \({v_9}\) and \({u_9}\) both seen.

\(k = 9\)     (A1)(ft)(G2)

Note: Award (G1) for \(8.93648…\) and \(-34.3\) seen as final answer without working. Accept use of \(n\).

[3 marks]

Find the size of BĈA.

Calculate the length of AD.

The farmer wants to build a fence around ABD.

Calculate the total length of the fence.

The farmer wants to build a fence around ABD.

The farmer pays 802.50 USD for the fence. Find the cost per metre.

Calculate the area of the triangle ABD.

A layer of earth 3 cm thick is removed from ABD. Find the volume removed in cubic metres.

\(\frac{{\sin {\text{BCA}}}}{{35}} = \frac{{\sin 105^\circ }}{{80}}\)     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

\({\text{B}}{\operatorname{\hat C}}{\text{A}} = 25.0^{\circ}\)     (A1)(G2)

[3 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Length BD = 40 m     (A1)

Angle ABC = 180° − 105° − 25° = 50°     (A1)(ft)

Note: (ft) from their answer to (a).

AD² = 35² + 40² − (2 × 35 × 40 × cos 50°)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitutions.

(UP)     AD = 32.0 m     (A1)(ft)(G3)

Notes: If 80 is used for BD award at most (A0)(A1)(ft)(M1)(A1)(ft)(A1)(ft) for an answer of 63.4 m.

If the angle ABC is incorrectly calculated in this part award at most (A1)(A0)(M1)(A1)(ft)(A1)(ft).

If angle BCA is used award at most (A1)(A0)(M1)(A0)(A0).

[5 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

length of fence = 35 + 40 + 32     (M1)

(UP)     = 107 m     (A1)(ft)(G2)

Note: (M1) for adding 35 + 40 + their (b).

[2 marks]

cost per metre \( = \frac{802.50}{107}\)     (M1)

Note: Award (M1) for dividing 802.50 by their (c).

cost per metre = 7.50 USD (7.5 USD) (USD not required)     (A1)(ft)(G2)

[2 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Area of ABD \( = \frac{1}{2} \times 35 \times 40 \times \sin 50^\circ \)     (M1)

= 536.2311102     (A1)(ft)

(UP)     = 536 m²     (A1)(ft)(G2)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitution, (ft) from their value of BD and their angle ABC in (b).

[3 marks]

Volume = 0.03 × 536     (A1)(M1)

= 16.08

= 16.1     (A1)(ft)(G2)

Note: Award (A1) for 0.03, (M1) for correct formula. (ft) from their (e).

If 3 is used award at most (A0)(M1)(A0).

[3 marks]

This was a simple application of non-right angled trigonometry and most candidates answered it well. Some candidates lost marks in both parts due to the incorrect setting of the calculators. Those that did not score well overall primarily used Pythagoras.

This was a simple application of non-right angled trigonometry and most candidates answered it well. Some candidates lost marks in both parts due to the incorrect setting of the calculators. Those that did not score well overall primarily used Pythagoras.

Most candidates scored full marks, many by follow through from an incorrect part (b). The main error was using the value for BC and not BD.

Most candidates scored full marks, many by follow through from an incorrect part (b). The main error was using the value for BC and not BD.

Done well; again some candidates used the right-angled formula.

This part was poorly done; many candidates unable to convert 3 cm to 0.03 m. A significant number used the wrong formula, multiplying their answer by 1/3.

Find the initial temperature of the coffee.

Write down the equation of the horizontal asymptote.

Find the room temperature.

Find the temperature of the coffee after 10 minutes.

Find the temperature of Robert’s coffee after being heated in the microwave for 30 seconds after it has reached the temperature in part (d).

Calculate the length of time it would take a similar cup of coffee, initially at 20°C, to be heated in the microwave to reach 100°C.

Calculate correct to 2 decimal places the amount of euros he receives.

He buys 1 kilogram of Belgian chocolates at 1.35 euros per 100 g.

Calculate the cost of his chocolates in GBP correct to 2 decimal places.

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

(UP) 90°C     (A1)

[1 mark]

y = 16     (A1)

[1 mark]

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

(UP) 16°C (ft) from answer to part (b)     (A1)(ft)

[1 mark]

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

(UP) 25.4°C     (A1)

[1 mark]  

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

for seeing 2^0.75 or equivalent     (A1)

for multiplying their (d) by their 2^0.75     (M1)

(UP) 42.8°C     (A1)(ft)(G2)

[3 marks]

Unit penalty (UP) is applicable in part (i)(a)(c)(d)(e) and (f)

for seeing  \(20 \times 2^{1.5t} = 100\)     (A1)

for seeing a value of t between 1.54 and 1.56 inclusive     (M1)(A1)

(UP) 1.55 minutes or 92.9 seconds     (A1)(G3)

[4 marks]

Financial accuracy penalty (FP) is applicable in part (ii) only.

\(120 - 3 = 117\)

(FP) \(117 \times 1.37\)     (A1)

= 160.29 euros (correct answer only)     (M1)

first (A1) for 117 seen, (M1) for multiplying by 1.37     (A1)(G2)

[3 marks]

Financial accuracy penalty (FP) is applicable in part (ii) only.

(FP) \(\frac{{13.5}}{{1.37}}\)     (A1)(M1)

9.85 GBP (answer correct to 2dp only)

first (A1) is for 13.5 seen, (M1) for dividing by 1.37     (A1)(ft)(G3)

[3 marks]

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

Many candidates who had not lost a UP in question 2 lost one here. Parts (a), (c) and (d) were reasonably well tackled. Almost everybody had difficulty with the equation of the horizontal asymptote, a common answer being y = 20. Most of the candidates realised that 30 seconds was 0.5 minutes and calculated part (e) correctly. Part (f), solving an exponential equation, was a good discriminator. Trial and error was expected but many students did not think of doing this.

The financial part was the best done question in the paper and a large majority of candidates gained full marks here.

The financial part was the best done question in the paper and a large majority of candidates gained full marks here.

Calculate
(i)     the area of the base of the wastepaper bin;
(ii)    the height, \(h\) , of Nadia’s wastepaper bin;
(iii)   the total external surface area of the wastepaper bin.

State whether Nadia’s design is practical. Give a reason.

Write down an equation in \(h\) and \(r\) , using the given volume of the bin.

Show that the total external surface area, \(A\) , of the bin is \(A = \pi {r^2} + \frac{{16000}}{r}\) .

Write down \(\frac{{{\text{d}}A}}{{{\text{d}}r}}\).

(i)     Find the value of \(r\) that minimizes the total external surface area of the wastepaper bin.
(ii)    Calculate the value of \(h\) corresponding to this value of \(r\) .

Determine whether Merryn’s design is an improvement upon Nadia’s. Give a reason.

(i)     \({\text{Area}} = \pi {(5)^2}\)     (M1)
\( = 78.5{\text{ (c}}{{\text{m}}^2}{\text{)}}\) (\(78.5398 \ldots \))     (A1)(G2)

Note: Accept \(25\pi \) .

(ii)    \(8000 = 78.5398 \ldots  \times h\)     (M1)
\(h = 102{\text{ (cm)}}\) (\(101.859 \ldots \))     (A1)(ft)(G2)

Note: Follow through from their answer to part (a)(i).

(iii)   \({\text{Area}} = \pi {(5)^2} + 2\pi (5)(101.859 \ldots )\)     (M1)(M1)

Note: Award (M1) for their substitution in curved surface area formula, (M1) for addition of their two areas.

\( = 3280{\text{ (c}}{{\text{m}}^2}{\text{)}}\) (\(3278.53 \ldots \))     (A1)(ft)(G2)

Note: Follow through from their answers to parts (a)(i) and (ii).

No, it is too tall/narrow.     (A1)(ft)(R1)

Note: Follow through from their value for \(h\).

\(8000 = \pi {r^2}h\)     (A1)

\(A = \pi {r^2} + 2\pi r\left( {\frac{{8000}}{{\pi {r^2}}}} \right)\)     (A1)(M1)

Note: Award (A1) for correct rearrangement of their part (c), (M1) for substitution of their rearrangement into area formula.

\( = \pi {r^2} + \frac{{16000}}{r}\)     (AG)

\(\frac{{{\text{d}}A}}{{{\text{d}}r}} = 2\pi r - 16000{r^{ - 2}}\)     (A1)(A1)(A1)

Note: Award (A1) for \(2\pi r\) , (A1) for \( - 16000\) (A1) for \({r^{ - 2}}\) . If an extra term is present award at most (A1)(A1)(A0).

(i)     \(\frac{{{\text{d}}A}}{{{\text{d}}r}} = 0\)     (M1)
\(2\pi {r^3} - 16000 = 0\)    (M1)
\(r = 13.7{\text{ cm}}\) (\(13.6556 \ldots \))     (A1)(ft)

Note: Follow through from their part (e).

(ii)    \(h = \frac{{8000}}{{\pi {{(13.65 \ldots )}^2}}}\)     (M1)
\( = 13.7{\text{ cm}}\) (\(13.6556 \ldots \))     (A1)(ft)

Note: Accept \(13.6\) if \(13.7\) used.

Yes or No, accompanied by a consistent and sensible reason.     (A1)(R1)

Note: Award (A0)(R0) if no reason is given.

Write down the mode.

Using your graphic display calculator, write down the value of
(i)     the mean;
(ii)    the standard deviation;
(iii)   the median.

Find the interquartile range.

Draw a box and whisker diagram for this data, on graph paper, using a scale of \(1{\text{ cm}}\) to represent \(5{\text{ cm}}\).

Sam estimated the value of the mean of the measured lengths to be \(43{\text{ cm}}\).

Find the percentage error of Sam’s estimated mean.

\(47.5{\text{ (cm)}}\)     (A1)

(i)     \(45.85{\text{ (cm)}}\)     (G2)

Note: Accept \(45.9\) .

(ii)    \(17.1{\text{ }}(17.0888 \ldots )\)     (G1)
(iii)   \(47.5{\text{ (cm)}}\)     (G1)

\(62.5 - 32.5 = 30\)     (M1)(A1)(G2)

Note: Award (M1) for correct quartiles seen.

(A1) for correct label and scale
(A1)(ft) for correct median
(A1)(ft) for correct quartiles and box
(A1) for endpoints at \(17.5\) and \(77.5\) joined to box by straight lines     (A1)(A1)(ft)(A1)(ft)(A1)

Notes: The final (A1) is lost if the lines go through the box. Follow through from their parts (b) and (c).

\(\varepsilon  = \left| {\frac{{43 - 45.85}}{{45.85}}} \right| \times 100\% \)     (M1)

Note: Award (M1) for their correct substitution in \(\% \) error formula.

\( = 6.22\% \) (\(6.21592 \ldots \))     (A1)(ft)(G2)

Notes: Follow through from their answer to part (b)(i). Accept \(6.32\% \) with use of \(45.9\) .

Write down the value of

(i)     \({S_1}\);

(ii)     \({S_2}\).

The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).

Show that \({u_2} = 9\).

The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).

Find the common difference of the sequence.

The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).

Find \({u_{10}}\).

The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).

Find the lowest value of \(n\) for which \({u_n}\) is greater than \(1000\).

The \({n^{{\text{th}}}}\) term of the arithmetic sequence is given by \({u_n}\).

There is a value of \(n\) for which

\[{u_1} + {u_2} +  \ldots  + {u_n} = 1512.\]

Find the value of \(n\).

(i)     \({S_1} = 7\)     (A1)

(ii)     \({S_2} = 16\)     (A1)

\(({u_2} = ){\text{ }}16 - 7 = 9\)     (M1)(AG)

Note: Award (M1) for subtracting 7 from 16. The 9 must be seen.

OR

\(16 - 7 - 7 = 2\)

\(({u_2} = ){\text{ }}7 + (2 - 1)(2) = 9\)     (M1)(AG)

Note: Award (M1) for subtracting twice \(7\) from \(16\) and for correct substitution in correct arithmetic sequence formula.

The \(9\) must be seen.

Do not accept: \(9 - 7 = 2,{\text{ }}{u_2} = 7 + (2 - 1)(2) = 9\).

\({u_1} = 7\)     (A1)(ft)

\(d = 2{\text{ }}( = 9 - 7)\)     (A1)(ft)(G2)

Notes: Follow through from their \({S_1}\) in part (a)(i).

\(7 + 2 \times (10 - 1)\)     (M1)

Note: Award (M1) for correct substitution in the correct arithmetic sequence formula. Follow through from their parts (a)(i) and (c).

\( = 25\)     (A1)(ft)(G2)

Note: Award (A1)(ft) for their correct tenth term.

\(7 + 2 \times (n - 1) > 1000\)     (A1)(ft)(M1)

Note: Award (A1)(ft) for their correct expression for the \({n^{{\text{th}}}}\) term, (M1) for comparing their expression to \(1000\). Accept an equation. Follow through from their parts (a)(i) and (c).

\(n = 498\)     (A1)(ft)(G2)

Notes: Answer must be a natural number.

\(6n + {n^2} = 1512\;\;\;\)OR\(\;\;\;\frac{n}{2}\left( {14 + 2(n - 1)} \right) = 1512\;\;\;\)OR

\({S_n} = 1512\;\;\;\)OR\(\;\;\;7 + 9 +  \ldots  + {u_n} = 1512\)     (M1)

Notes: Award (M1) for equating the sum of the first \(n\) terms to \(1512\). Accept a sum of at least the first 7 correct terms.

\(n = 36\)     (A1)(G2)

Note: If \(n = 36\) is seen without working, award (G2). Award a maximum of (M1)(A0) if \( - 42\) is also given as a solution.

Show that \({\text{BD}} = 93{\text{ m}}\) correct to the nearest metre.

Calculate angle \({\rm{B\hat CD}}\).

Find the area of ABCD.

Calculate Abdallah’s estimate for the area.

Find the percentage error in Abdallah’s estimate.

\({\text{B}}{{\text{D}}^2} = {40^2} + {84^2}\)     (M1)

Note:     Award (M1) for correct substitution into Pythagoras.

Accept correct substitution into cosine rule.

\({\text{BD}} = 93.0376 \ldots \)     (A1)

\( = 93\)     (AG)

Note:     Both the rounded and unrounded value must be seen for the (A1) to be awarded.

[2 marks]

\(\cos C = \frac{{{{115}^2} + {{60}^2} - {{93}^2}}}{{2 \times 115 \times 60}}{\text{ }}({93^2} = {115^2} + {60^2} - 2 \times 115 \times 60 \times \cos C)\)     (M1)(A1)

Note:     Award (M1) for substitution into cosine formula, (A1) for correct substitutions.

\( = 53.7^\circ {\text{ }}(53.6679 \ldots ^\circ )\)     (A1)(G2)

[3 marks]

\(\frac{1}{2}(40)(84) + \frac{1}{2}(115)(60)\sin (53.6679 \ldots )\)     (M1)(M1)(A1)(ft)

Note:     Award (M1) for correct substitution into right-angle triangle area. Award (M1) for substitution into area of triangle formula and (A1)(ft) for correct substitution.

\( = 4460{\text{ }}{{\text{m}}^2}{\text{ }}(4459.30 \ldots {\text{ }}{{\text{m}}^2})\)     (A1)(ft)(G3)

Notes:     Follow through from part (b).

[4 marks]

\(\frac{{(40 + 60)(84 + 115)}}{4}\)     (M1)

Note:     Award (M1) for correct substitution in the area formula used by ‘Ancient Egyptians’.

\( = 4980{\text{ }}{{\text{m}}^2}{\text{ }}(4975{\text{ }}{{\text{m}}^2})\)     (A1)(G2)

[2 marks]

\(\left| {\frac{{4975 - 4459.30 \ldots }}{{4459.30 \ldots }}} \right| \times 100\)     (M1)

Notes:     Award (M1) for correct substitution into percentage error formula.

\( = 11.6{\text{ }}(\% ){\text{ }}(11.5645 \ldots )\)     (A1)(ft)(G2)

Notes:    Follow through from parts (c) and (d)(i).

[2 marks]

On 1 January 2005, Rebecca invested \(30000{\text{ AUD}}\) in a Supersaver account at a nominal annual rate of \(2.5\% \) compounded annually. Calculate the amount in the Supersaver account after two years.

On 1 January 2005, Rebecca invested \(30000{\text{ AUD}}\) in a Supersaver account at a nominal annual rate of \(2.5\% \) compounded annually.

Find the number of complete years since 1 January 2005 it would take for the amount in Rebecca’s account to exceed the amount in Daniel’s account.