Calculate the value of r.

Calculate 14 000 USD in INR.

Calculate the amount of this investment, in INR, in this account after five years.

Harinder chose option B. At the end of five years, Harinder converted this investment back to USD. The exchange rate, at that time, was 1 USD = 67.16 INR.

Calculate how much more money, in USD, Harinder earned by choosing option B instead of option A.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$17500=14000{\left(1+\frac{r}{100}\right)}^5$     (M1)(A1)

Note: Award (M1) for substitution into the compound interest formula, (A1) for correct substitution. Award at most (M1)(A0) if not equated to 17500.

OR

N = 5

PV = ±14000

FV = $\mp$17500

P/Y = 1

C/Y = 1     (A1)(M1)

Note: Award (A1) for C/Y = 1 seen, (M1) for all other correct entries. FV and PV must have opposite signs.

= 4.56 (%)  (4.56395… (%))     (A1) (G3)

[3 marks]

14000 × 66.91     (M1)

Note: Award (M1) for multiplying 14000 by 66.91.

936740 (INR)     (A1) (G2)

Note: Answer must be given to the nearest whole number.

[2 marks]

$936740\times {\left(1+\frac{5.2}{12\times 100}\right)}^{12\times 5}$     (M1)(A1)(ft)

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

OR

N = 60

I% = 5.2

PV = ±936740

P/Y= 12

C/Y= 12    (A1)(M1)

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

OR

N = 5

I% = 5.2

PV = ±936740

P/Y= 1

C/Y= 12    (A1)(M1)

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

= 1214204 (INR)     (A1)(ft) (G3)

Note: Follow through from part (b). Answer must be given to the nearest whole number.

[3 marks]

$\frac{1214204}{67.16}$     (M1)

Note: Award (M1) for dividing their (c) by 67.16.

$\left(\frac{1214204}{67.16}\right)-17500=579$ (USD)     (M1)(A1)(ft) (G3)

Note: Award (M1) for finding the difference between their conversion and 17500. Answer must be given to the nearest whole number. Follow through from part (c).

[3 marks]

Find $h$, the height of the tank.

Show that the volume of the tank is $624 000 {\text{m}}^3$, correct to three significant figures.

Write down the common difference, $d$.

Find the amount of fuel pumped into the tank in the $\text{13th}$ hour.

Find the value of $n$ such that $u_n=0$.

Write down the number of hours that the pump was pumping fuel into the tank.

Find the total amount of fuel pumped into the tank in the first $8$ hours.

Show that the tank will never be completely filled using this pump.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\sin 60°=\frac{h}{40}$  OR  $\tan 60°=\frac{h}{20}$           (M1)

Note: Award (M1) for correct substitutions in trig ratio.

OR

${20}^2+h^2={40}^2 \left(\sqrt{{40}^2-{20}^2}\right)$           (M1)

Note: Award (M1) for correct substitutions in Pythagoras’ theorem.

$\left(h=\right) 34.6 \left(\text{m}\right) \left(\sqrt{1200}, 20\sqrt{3}, 34.6410\dots \right)$       (A1)(G2)

[2 marks]

$\frac{1}{2}\left(70+110\right)\left(34.6410\dots \right)\times 200$           (M1)(M1)

Note: Award (M1) for their correctly substituted area of trapezium formula, provided all substitutions are positive. Award (M1) for multiplying by $200$. Follow through from part (a).

OR

$\left(2\times \frac{1}{2}\times 20\times 34.6410\dots +70\times 34.6410\dots \right)\times 200$           (M1)(M1)

Note: Award (M1) for the addition of correct areas for two triangles and one rectangle. Award (M1) for multiplying by $200$. Follow through from part (a).

OR

$70\times 34.6410\dots \times 200+2\times \frac{1}{2}\times 34.6410\dots \times 20\times 200$           (M1)(M1)

Note: Award (M1) for their correct substitution in volume of cuboid formula. Award (M1) for correctly substituted volume of triangular prism(s). Follow through from part (a).

$623538\dots$         (A1)

$624000 \left({\text{m}}^3\right)$            (AG)

Note: Both an unrounded answer that rounds to the given answer and the rounded value must be seen for the (A1) to be awarded.

[3 marks]

$\left(d=\right) -1800$           (A1)

[1 mark]

$\left(u_{13}=\right) 45000+\left(13-1\right)\left(-1800\right)$           (M1)

Note: Award (M1) for correct substitutions in arithmetic sequence formula.
OR
Award (M1) for a correct ${\text{4}}^{\text{th}}$ term seen as part of list.

$23400 \left({\text{m}}^3\right)$        (A1)(ft)(G2)

Note: Follow through from part (c) for their value of $d$.

[2 marks]

$0=45000+\left(n-1\right)\left(-1800\right)$           (M1)

Note: Award (M1) for their correct substitution into arithmetic sequence formula, equated to zero.

$\left(n=\right) 26$        (A1)(ft)(G2)

Note: Follow through from part (c). Award at most (M1)(A0) if their $n$ is not a positive integer.

[2 marks]

$25$           (A1)(ft)

Note: Follow through from part (e)(i), but only if their final answer in (e)(i) is positive. If their $n$ in part (e)(i) is not an integer, award  (A1)(ft) for the nearest lower integer.

[1 mark]

$\left(S_8=\right) \frac{8}{2}\left(2\times 45000+\left(8-1\right)\times \left(-1800\right)\right)$           (M1)

Note: Award (M1) for their correct substitutions in arithmetic series formula. If a list method is used, award (M1) for the addition of their $8$ correct terms.

$310 000 \left({\text{m}}^3\right) \left(309 600\right)$       (A1)(ft)(G2)

Note: Follow through from part (c). Award at most (M1)(A0) if their final answer is greater than $624 000$.

[2 marks]

$\left(S_{25}=\right) \frac{25}{2}\left(2\times 45000+\left(25-1\right)\times \left(-1800\right)\right) , \left(S_{25}=\right) \frac{25}{2}\left(45000+1800\right)$           (M1)

Note: Award (M1) for their correct substitutions into arithmetic series formula.

$S_{25}=585000 \left({\text{m}}^3\right)$       (A1)(ft)(G1)

Note: Award (M1)(A1) for correctly finding $S_{26}=585000 \left({\text{m}}^3\right)$, provided working is shown e.g. $\left(S_{26}=\right) \frac{26}{2}\left(2\times 45000+\left(26-1\right)\times \left(-1800\right)\right)$ , $\left(S_{26}=\right) \frac{26}{2}\left(45000+0\right)$. Follow through from part (c) and either their (e)(i) or (e)(ii). If $d<0$ and their final answer is greater than $624 000$, award at most (M1)(A1)(ft)(R0). If $d>0$, there is no maximum, award at most (M1)(A0)(R0). Award no marks if their number of terms is not a positive integer.

$585000 \left({\text{m}}^3\right)<624000 \left({\text{m}}^3\right)$        (R1)

Hence it will never be filled        (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.
For unsupported $\left(S_{25}\right)=585000$ seen, award at most (G1)(R1)(AG). Working must be seen to follow through from parts (c) and (e)(i) or (e)(ii).

OR

$\left(S_n=\right) \frac{n}{2}\left(2\times 45000+\left(n-1\right)\times \left(-1800\right)\right)$           (M1)

Note: Award (M1) for their correct substitution into arithmetic series formula, with $n$.

Maximum of this function $585225 \left({\text{m}}^3\right)$       (A1)

Note: Follow through from part (c). Award at most (M1)(A1)(ft)(R0) if their final answer is greater than $624 000$. Award at most (M1)(A0)(R0) if their common difference is not $–1800$. Award at most (M1)(A0)(R0) if $585 225$ is not explicitly identified as the maximum of the function.

$585225 \left({\text{m}}^3\right)<624000 \left({\text{m}}^3\right)$        (R1)

Hence it will never be filled        (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.

OR

sketch with concave down curve and labelled $624000$ horizontal line           (M1)

Note: Accept a label of “tank volume” instead of a numerical value. Award (M0) if the line and the curve intersect.

curve explicitly labelled as $\left(S_n=\right) \frac{n}{2}\left(2\times 45000+\left(n-1\right)\times \left(-1800\right)\right)$ or equivalent       (A1)

Note: Award (A1) for a written explanation interpreting the sketch. Accept a comparison of values, e.g $585225 \left({\text{m}}^3\right)<624000 \left({\text{m}}^3\right)$, where $585225$ is the graphical maximum. Award at most (M1)(A0)(R0) if their common difference is not $–1800$.

the line and the curve do not intersect        (R1)

hence it will never be filled        (AG)

Note: The (AG) line must be seen. If it is omitted do not award the final (R1). Do not follow through within the part.

OR

$624000=\frac{n}{2}\left(2\times 45000+\left(n-1\right)\times \left(-1800\right)\right)$           (M1)

Note: Award (M1) for their correctly substituted arithmetic series formula equated to $624000 \left(623538\right)$.

Demonstrates there is no solution       (A1)

Note: Award (A1) for a correct working that the discriminant is less than zero OR correct working indicating there is no real solution in the quadratic formula.

There is no (real) solution (to this equation)       (R1)

hence it will never be filled        (AG)

Note: At most (M1)(A0)(R0) for their correctly substituted arithmetic series formula $=624000, 623538$ or $622800$ with a statement "no solution". Follow through from their part (b).

[3 marks]

Find the range of the average body weights for these seven species of mammal.

For the data from these seven species calculate $r$, the Pearson’s product–moment correlation coefficient;

For the data from these seven species describe the correlation between the average body weight and the average weight of the brain.

Write down the equation of the regression line $y$ on $x$, in the form $y=mx+c$.

Use your regression line to estimate the average weight of the brain of grey wolves.

Find the percentage error in your estimate in part (d).

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$529-3$     (M1)

$=526\text{ (kg)}$     (A1)(G2)

[2 marks]

$0.922(0.921857\dots )$     (G2)

[2 marks]

(very) strong, positive     (A1)(ft)(A1)(ft)

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

[2 marks]

$y=0.000986x+0.0923(y=0.000985837\dots x+0.0923391\dots )$     (A1)(A1)

Note:     Award (A1) for $0.000986x$, (A1) for 0.0923.

Award a maximum of (A1)(A0) if the answer is not an equation in the form $y=mx+c$.

[2 marks]

$0.000985837\dots (36)+0.0923391\dots$     (M1)

Note:     Award (M1) for substituting 36 into their equation.

$0.128\text{ (kg) }\left(0.127829\dots \text{ (kg)}\right)$     (A1)(ft)(G2)

Note:     Follow through from part (c). The final (A1) is awarded only if their answer is positive.

[2 marks]

$\left|\frac{0.127829\dots -0.120}{0.120}\right|\times 100$     (M1)

Note:     Award (M1) for their correct substitution into percentage error formula.

$6.52(\mathrm{\%})\left(\mathrm{6.52442...}(\mathrm{\%})\right)$     (A1)(ft)(G2)

Note: Follow through from part (d). Do not accept a negative answer.

[2 marks]

Use your graphic display calculator to write down $\overline{y}$, the mean examination score.

Use your graphic display calculator to write down r , Pearson’s product–moment correlation coefficient.

Find the exact value of m and of c for these data.

Use the regression line y on x to estimate Jerome’s examination score.

Justify whether it is valid to use the regression line y on x to estimate Jerome’s examination score.

54     (G1)

[1 mark]

0.5     (G2)

[2 marks]

m = 0.875, c = 41.75  $\left(m=\frac{7}{8}\text{,}c=\frac{167}{4}\right)$        (A1)(A1)

Note: Award (A1) for 0.875 seen. Award (A1) for 41.75 seen. If 41.75 is rounded to 41.8 do not award (A1).

[2 marks]

y = 0.875(17) + 41.75      (M1)

Note: Award (M1) for correct substitution into their regression line.

= 56.6   (56.625)      (A1)(ft)(G2)

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

[2 marks]

the estimate is valid      (A1)

since this is interpolation and the correlation coefficient is large enough      (R1)

OR

the estimate is not valid      (A1)

since the correlation coefficient is not large enough      (R1)

Note: Do not award (A1)(R0). The (R1) may be awarded for reasoning based on strength of correlation, but do not accept “correlation coefficient is not strong enough” or “correlation is not large enough”.

Award (A0)(R0) for this method if no numerical answer to part (a)(iii) is seen.

[2 marks]

Calculate, in CAD, the total amount John pays for the bicycle.

Find the value of the bicycle during the 5th year. Give your answer to two decimal places.

Calculate, in years, when the bicycle value will be less than 50 USD.

Find the total amount John has paid to insure his bicycle for the first 5 years.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

1.042 × 880 × 1.25  OR  (880 + 0.042 × 880) × 1.25      (M1)(M1)

Note: Award (M1) for multiplying 880 by 1.042 and (M1) for multiplying 880 by 1.25.

1150 (CAD)  (1146.20 (CAD))      (A1)(G2)

Note: Accept 1146.2 (CAD)

[3 marks]

$\frac{704}{880}$  OR  $\frac{563.20}{704}$      (M1)

Note: Award (M1) for correctly dividing sequential terms to find the common ratio, or 0.8 seen.

880(0.8)⁵⁻¹      (M1)

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

360.45 (USD)      (A1)(G3)

Note: Do not award the final (A1) if the answer is not correct to 2 decimal places. Award at most (M0)(M1)(A0) if $r=1.25$.

[3 marks]

     (M1)

Note: Award (M1) for correct substitution into geometric sequence formula and (in)equating to 50. Accept weak or strict inequalities. Accept an equation. Follow through from their common ratio in part (b). Accept a sketch of their GP with $y=50$ as a valid method.

OR

$u_{13}=60.473$  AND  $u_{14}=48.379$      (M1)

Note: Award (M1) for their $u_{13}$ and $u_{14}$ both seen. If the student states , without $u_{13}=60.473$ seen, this is not sufficient to award (M1).

14 or “14th year” or “after the 13th year”     (A1)(ft)(G2)

Note: The context of the question requires the final answer to be an integer. Award at most (M1)(A0) for a final answer of 13.9 years. Follow through from their 0.8 in part (b).

[2 marks]

$\frac{5}{2}\left(\left(2\times 120\right)+\left(-3.5\left(5-1\right)\right)\right)$    (M1)(A1)

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

565 (USD)    (A1)(G2)

[3 marks]

Calculate the volume of this pan.

Find the radius of the sphere in cm, correct to one decimal place.

Find the value of $a$.

Find the temperature that the pizza will be 5 minutes after it is taken out of the oven.

Calculate, to the nearest second, the time since the pizza was taken out of the oven until it can be eaten.

In the context of this model, state what the value of 19 represents.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(V=)\pi \times {\text{(17.5)}}^2\times 0.5$     (A1)(M1)

Notes:     Award (A1) for 17.5 (or equivalent) seen.

Award (M1) for correct substitutions into volume of a cylinder formula.

$=481\text{ c}{\text{m}}^3(481.056\dots \text{ c}{\text{m}}^3,153.125\pi \text{ c}{\text{m}}^3)$     (A1)(G2)

[3 marks]

$\frac{4}{3}\times \pi \times r^3=481.056\dots$     (M1)

Note:     Award (M1) for equating their answer to part (a) to the volume of sphere.

$r^3=\frac{3\times 481.056\dots }{4\pi }(=114.843\dots )$     (M1)

Note:     Award (M1) for correctly rearranging so $r^3$ is the subject.

$r=4.86074\dots \text{ (cm)}$     (A1)(ft)(G2)

Note:     Award (A1) for correct unrounded answer seen. Follow through from part (a).

$=4.9\text{ (cm)}$     (A1)(ft)(G3)

Note:     The final (A1)(ft) is awarded for rounding their unrounded answer to one decimal place.

[4 marks]

$230=a(2.06{)}^0+19$     (M1)

Note:     Award (M1) for correct substitution.

$a=211$     (A1)(G2)

[2 marks]

$(P=)211\times (2.06{)}^{-5}+19$      (M1)

Note:     Award (M1) for correct substitution into the function, $P(t)$. Follow through from part (c). The negative sign in the exponent is required for correct substitution.

$=24.7$ (°C) $(24.6878\dots$ (°C))     (A1)(ft)(G2)

[2 marks]

$45=211\times (2.06{)}^{-t}+19$     (M1)

Note:     Award (M1) for equating 45 to the exponential equation and for correct substitution (follow through for their $a$ in part (c)).

$(t=)2.89711\dots$     (A1)(ft)(G1)

$174\text{ (seconds) }\left(173.826\dots \text{ (seconds)}\right)$     (A1)(ft)(G2)

Note:     Award final (A1)(ft) for converting their $\text{2.89711}\dots$ minutes into seconds.

[3 marks]

the temperature of the (dining) room     (A1)

OR

the lowest final temperature to which the pizza will cool     (A1)

[1 mark]

Write down a formula for $A$, the surface area to be coated.

Express this volume in $\text{c}{\text{m}}^3$.

Write down, in terms of $r$ and $h$, an equation for the volume of this water container.

Show that $A=\pi r^2+\frac{1000000}{r}$.

Find $\frac{\text{d}A}{\text{d}r}$.

Using your answer to part (e), find the value of $r$ which minimizes $A$.

Find the value of this minimum area.

Find the least number of cans of water-resistant material that will coat the area in part (g).

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$(A=)\pi r^2+2\pi rh$    (A1)(A1)

Note:     Award (A1) for either $\pi r^2$ OR $2\pi rh$ seen. Award (A1) for two correct terms added together.

[2 marks]

$500000$    (A1)

Notes:     Units not required.

[1 mark]

$500000=\pi r^{2}h$    (A1)(ft)

Notes:     Award (A1)(ft) for $\pi r^{2}h$ equating to their part (b).

Do not accept unless $V=\pi r^{2}h$ is explicitly defined as their part (b).

[1 mark]

$A=\pi r^2+2\pi r\left(\frac{500000}{\pi r^2}\right)$    (A1)(ft)(M1)

Note:     Award (A1)(ft) for their $\frac{500000}{\pi r^2}$ seen.

Award (M1) for correctly substituting only $\frac{500000}{\pi r^2}$ into a correct part (a).

Award (A1)(ft)(M1) for rearranging part (c) to $\pi rh=\frac{500000}{r}$ and substituting for $\pi rh$ in expression for $A$.

$A=\pi r^2+\frac{1000000}{r}$    (AG)

Notes:     The conclusion, $A=\pi r^2+\frac{1000000}{r}$, must be consistent with their working seen for the (A1) to be awarded.

Accept ${10}^6$ as equivalent to $1000000$.

[2 marks]

$2\pi r-\frac{\text{1}\text{000}\text{000}}{r^2}$    (A1)(A1)(A1)

Note:     Award (A1) for $2\pi r$, (A1) for $\frac{1}{r^2}$ or $r^{-2}$, (A1) for $-\text{1}\text{000}\text{000}$.

[3 marks]

$2\pi r-\frac{1000000}{r^2}=0$    (M1)

Note:     Award (M1) for equating their part (e) to zero.

$r^3=\frac{1000000}{2\pi }$ OR $r=\sqrt[3]{\frac{1000000}{2\pi }}$     (M1)

Note:     Award (M1) for isolating $r$.

OR

sketch of derivative function     (M1)

with its zero indicated     (M1)

$(r=)54.2(\text{cm})(54.1926\dots )$    (A1)(ft)(G2)

[3 marks]

$\pi (54.1926\dots {)}^2+\frac{1000000}{(54.1926\dots )}$    (M1)

Note:     Award (M1) for correct substitution of their part (f) into the given equation.

$=27700(\text{c}{\text{m}}^2)(27679.0\dots )$    (A1)(ft)(G2)

[2 marks]

$\frac{27679.0\dots }{2000}$    (M1)

Note:     Award (M1) for dividing their part (g) by 2000.

$=13.8395\dots$    (A1)(ft)

Notes:     Follow through from part (g).

14 (cans)     (A1)(ft)(G3)

Notes:     Final (A1) awarded for rounding up their $13.8395\dots$ to the next integer.

[3 marks]

Use Giovanni’s diagram to show that angle ABC, the angle at which the Tower is leaning relative to the
horizontal, is 85° to the nearest degree.

Use Giovanni's diagram to calculate the length of AX.

Use Giovanni's diagram to find the length of BX, the horizontal displacement of the Tower.

Find the percentage error on Giovanni’s diagram.

Giovanni adds a point D to his diagram, such that BD = 45 m, and another triangle is formed.

Find the angle of elevation of A from D.

$\frac{\text{sin BAC}}{37}=\frac{\text{sin 60}}{56}$    (M1)(A1)

Note: Award (M1) for substituting the sine rule formula, (A1) for correct substitution.

angle $\text{B}\stackrel{\wedge }{\text{A}}\text{C}$ = 34.9034…°    (A1)

Note: Award (A0) if unrounded answer does not round to 35. Award (G2) if 34.9034… seen without working.

angle $\text{A}\stackrel{\wedge }{\text{B}}\text{C}$ = 180 − (34.9034… + 60)     (M1)

Note: Award (M1) for subtracting their angle BAC + 60 from 180.

85.0965…°    (A1)

85°     (AG)

Note: Both the unrounded and rounded value must be seen for the final (A1) to be awarded. If the candidate rounds 34.9034...° to 35° while substituting to find angle $\text{A}\stackrel{\wedge }{\text{B}}\text{C}$, the final (A1) can be awarded but only if both 34.9034...° and 35° are seen.
If 85 is used as part of the workings, award at most (M1)(A0)(A0)(M0)(A0)(AG). This is the reverse process and not accepted.

sin 85… × 56     (M1)

= 55.8 (55.7869…) (m)     (A1)(G2)

Note: Award (M1) for correct substitution in trigonometric ratio.

$\sqrt{{56}^2-55.7869{\dots }^2}$     (M1)

Note: Award (M1) for correct substitution in the Pythagoras theorem formula. Follow through from part (a)(ii).

OR

cos(85) × 56     (M1)

Note: Award (M1) for correct substitution in trigonometric ratio.

= 4.88 (4.88072…) (m)     (A1)(ft)(G2)

Note: Accept 4.73 (4.72863…) (m) from using their 3 s.f answer. Accept equivalent methods.

[2 marks]

$\left|\frac{4.88-3.9}{3.9}\right|\times 100$     (M1)

Note: Award (M1) for correct substitution into the percentage error formula.

= 25.1  (25.1282) (%)     (A1)(ft)(G2)

Note: Follow through from part (a)(iii).

[2 marks]

$\text{ta}{\text{n}}^{-1}\left(\frac{55.7869\dots }{40.11927\dots }\right)$     (A1)(ft)(M1)

Note: Award (A1)(ft) for their 40.11927… seen. Award (M1) for correct substitution into trigonometric ratio.

OR

(37 − 4.88072…)² + 55.7869…²

(AC =) 64.3725…

64.3726…² + 8² − 2 × 8 × 64.3726… × cos120

(AD =) 68.7226…

$\frac{\text{sin 120}}{68.7226\dots }=\frac{\text{sin A}\stackrel{\wedge }{\text{D}}\text{C}}{64.3725\dots }$    (A1)(ft)(M1)

Note: Award (A1)(ft) for their correct values seen, (M1) for correct substitution into the sine formula.

= 54.3°  (54.2781…°)     (A1)(ft)(G2)

Note: Follow through from part (a). Accept equivalent methods.

[3 marks]

Find the slant height of the cone-shaped container.

Find the slant height of the cone-shaped container.

Show that the total surface area of the cone-shaped container is 314 cm², correct to three significant figures.

Find the height, $h$, of this cylinder-shaped container.

The factory director wants to increase the volume of coconut water sold per container.

State whether or not they should replace the cone-shaped containers with cylinder‑shaped containers. Justify your conclusion.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\frac{\pi {\left(5.2\right)}^2\times 13}{3}$    (M1)

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

368  (368.110…) cm³     (A1)(G2)

Note: Accept 117.173…$\pi$ cm³ or $\frac{8788}{75}\pi$ cm³.

[2 marks]

(slant height²) = (5.2)² + 13²   (M1)

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

14.0  (14.0014…) (cm)     (A1)(G2)

[2 marks]

14.0014… × (5.2) × $\pi$ + (5.2)² × $\pi$     (M1)(M1)

Note: Award (M1) for their correct substitution in the curved surface area formula for cone; (M1) for adding the correct area of the base. The addition must be explicitly seen for the second (M1) to be awarded. Do not accept rounded values here as may come from working backwards.

313.679… (cm²)     (A1)

Note: Use of 3 sf value 14.0 gives an unrounded answer of 313.656….

314 (cm²)     (AG)

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

[3 marks]

2 × $\pi$ × (5.2) × $h$ + 2 × $\pi$ × (5.2)² = 314     (M1)(M1)(M1)

Note: Award (M1) for correct substitution in the curved surface area formula for cylinder; (M1) for adding two correct base areas of the cylinder; (M1) for equating their total cylinder surface area to 314 (313.679…). For this mark to be awarded the areas of the two bases must be added to the cylinder curved surface area and equated to 314. Award at most (M1)(M0)(M0) for cylinder curved surface area equated to 314.

($h$ =) 4.41 (4.41051…) (cm)     (A1)(G3)

[4 marks]

$\pi$ × (5.2)² × 4.41051…     (M1)

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

375  (374.666…) (cm³)     (A1)(ft)(G2)

Note: Follow through from part (d).

375 (cm³) > 368 (cm³)      (R1)(ft)

OR

“volume of cylinder is larger than volume of cone” or similar    (R1)(ft)

Note: Follow through from their answer to part (a). The verbal statement should be consistent with their answers from parts (e) and (a) for the (R1) to be awarded.

replace with the cylinder containers     (A1)(ft)

Note: Do not award (A1)(ft)(R0). Follow through from their incorrect volume for the cylinder in this question part but only if substitution in the volume formula shown.

[4 marks]

Find the size of angle $\text{C}\stackrel{\wedge }{\text{A}}\text{D}$.

Find the size of angle $\text{A}\stackrel{\wedge }{\text{C}}\text{D}$.

(CAD =) 53.1°  (53.0521…°)       (A1)(ft)

Note: Follow through from their part (b)(i) only if working seen.

[1 mark]

(ACD = ) 70° − (180° − 125° − 31.9478°…)      (M1)

Note: Award (M1) for subtracting their angle $\text{A}\stackrel{\wedge }{\text{C}}\text{B}$ from 70°.

OR

(ADC =) 360 − (85 + 70 + 125) = 80

(ACD =) 180 − 80 − 53.0521...      (M1)

46.9°  (46.9478…°)      (A1)(ft)(G2)

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

[2 marks]

Show that $\text{BD}=93\text{ m}$ correct to the nearest metre.

Calculate angle $\mathrm{B}\hat{\mathrm{C}}\mathrm{D}$.

Find the area of ABCD.

Calculate Abdallah’s estimate for the area.

Find the percentage error in Abdallah’s estimate.

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\text{B}{\text{D}}^2={40}^2+{84}^2$     (M1)

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

Accept correct substitution into cosine rule.

$\text{BD}=93.0376\dots$     (A1)

$=93$     (AG)

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

[2 marks]

$\cos C=\frac{{115}^2+{60}^2-{93}^2}{2\times 115\times 60}({93}^2={115}^2+{60}^2-2\times 115\times 60\times \cos C)$     (M1)(A1)

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

$={53.7}^{\circ }(53.6679{\dots }^{\circ })$     (A1)(G2)

[3 marks]

$\frac{1}{2}(40)(84)+\frac{1}{2}(115)(60)\sin (53.6679\dots )$     (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.

$=4460{\text{m}}^2(4459.30\dots {\text{m}}^2)$     (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’.

$=4980{\text{m}}^2(4975{\text{m}}^2)$     (A1)(G2)

[2 marks]

$\left|\frac{4975-4459.30\dots }{4459.30\dots }\right|\times 100$     (M1)

Notes:     Award (M1) for correct substitution into percentage error formula.

$=11.6(\mathrm{\%})(11.5645\dots )$     (A1)(ft)(G2)

Notes:    Follow through from parts (c) and (d)(i).

[2 marks]

Find the angle $\text{AÔB}$.

Find the area of the shaded segment.

Find the area of a circle with radius $4.5 \text{m}$.

Find the area of the field that can be reached by the goat.

Find the value of $t$ at which the goat is eating grass at the greatest rate.

$\left(\frac{1}{2}\text{AÔB}=\right) \text{arccos}\left(\frac{4}{4.5}\right)=27.266\dots$        (M1)(A1)

$\text{AÔB}=54.532\dots \approx 54.5°$  ($0.951764\dots \approx 0.952$ radians)        A1 

Note: Other methods may be seen; award (M1)(A1) for use of a correct trigonometric method to find an appropriate angle and then A1 for the correct answer.

[3 marks]

finding area of triangle

EITHER

area of triangle $=\frac{1}{2}\times 4.5^2\times \sin \left(54.532\dots \right)$        (M1)

Note: Award M1 for correct substitution into formula.

$=8.24621\dots \approx 8.25 {\text{m}}^2$        (A1)

OR

$\text{AB}=2\times \sqrt{4.5^2-4^2}=4.1231\dots$

area triangle $=\frac{4.1231\dots \times 4}{2}$        (M1)

$=8.24621\dots \approx 8.25 {\text{m}}^2$        (A1)

finding area of sector

EITHER

area of sector $=\frac{54.532\dots }{360}\times \mathrm{\pi }\times 4.5^2$        (M1)

$=9.63661\dots \approx 9.64 {\text{m}}^2$        (A1)

OR

area of sector $=\frac{1}{2}\times 0.9517641\dots \times 4.5^2$        (M1)

$=9.63661\dots \approx 9.64 {\text{m}}^2$        (A1)

THEN

area of segment $=9.63661\dots -8.24621\dots$

$=1.39 {\text{m}}^2 \left(1.39040\dots \right)$        A1 

[5 marks]

$\mathrm{\pi }\times 4.5^2$         (M1)

$63.6 {\text{m}}^2 \left(63.6172\dots {\text{m}}^2\right)$        A1 

[2 marks]

METHOD 1

$4\times 1.39040... (5.56160)$         (A1)

subtraction of four segments from area of circle         (M1)

$=58.1 {\text{m}}^2 \left(58.055\dots \right)$       A1 

METHOD 2

$4\left(0.5\times 4.5^2\times \sin 54.532\dots \right)+4\left(\frac{35.4679}{360}\times \mathrm{\pi }\times 4.5^2\right)$         (M1)

$= 32.9845\dots +25.0707$         (A1)

$=58.1 {\text{m}}^2 \left(58.055\dots \right)$       A1 

[3 marks]

sketch of $\frac{dV}{dt}$   OR   $\frac{dV}{dt}=0.110363\dots$   OR   attempt to find where $\frac{d^{2}V}{d{t}^2}=0$         (M1)

$t=1$ hour        A1 

[2 marks]

Part (a)(i) proved to be difficult for many candidates. About half of the candidates managed to correctly find the angle $\text{A}\hat{O}B$. A variety of methods were used: cosine to find half of $\text{A}\hat{O}B$ then double it; sine to find angle $\text{A}\hat{O}B$ , then find half of A$\text{A}\hat{O}B$ and double it; Pythagoras to find half of AB and then sine rule to find half of angle $\text{A}\hat{O}B$ then double it; Pythagoras to find half of AB, then double it and use cosine rule to find angle $\text{A}\hat{O}B$ . Many candidates lost a mark here due to premature rounding of an intermediate value and hence the final answer was not correct (to three significant figures).

In part (a)(ii) very few candidates managed to find the correct area of the shaded segment and include the correct units. Some only found the area of the triangle or the area of the sector and then stopped.

In part (b)(i), nearly all candidates managed to find the area of a circle.

In part (b)(ii), finding the area of the field reached by the goat proved troublesome for most of the candidates. It appeared as if the candidates did not fully understand the problem. Very few candidates realized the connection to part (a)(ii).

Part (c) was accessed by only a handful of candidates. The candidates could simply have graphed the function on their GDC to find the greatest value, but most did not realize this.

$a$.

$b$.

$w_n$.

$l_n$.

Show that Eddie needs $144$ tiles.

Find the value of $w_n$ for this path.

Find the total area of the tiles in Eddie’s path. Give your answer in the form $a\times {10}^k$ where $1\le a<10$ and $k$ is an integer.

Find the cost of a single pack of five tiles.

Find the minimum number of packs of tiles Eddie will need to order.

Find the total cost for Eddie’s order.

$30$         A1

[1 mark]

$40$         A1

[1 mark]

arithmetic formula chosen         (M1)

$w_n=20+\left(n-1\right)10 \left(=10+10n\right)$         A1

[2 marks]

arithmetic formula chosen

$l_n=30+\left(n-1\right)10 \left(=20+10n\right)$         A1

[1 mark]

$740=30+\left(n-1\right)10$   OR   $740=20-10n$            M1

$n=72$            A1

$144$ tiles            AG

Note: The AG line must be stated for the final A1 to be awarded.

[2 marks]

$w_{72}=730$         A1

[1 mark]

$\left(10\times 20\right)\times 144$          (M1)

$=28800$          (A1)

$2.88\times {10}^4 {\text{cm}}^2$         A1

Note: Follow through within the question for correctly converting their intermediate value into standard form (but only if the pre-conversion value is seen).

[3 marks]

EITHER

$1$ square metre $=100 \text{cm}\times 100 \text{cm}$          (M1)

(so, $50$ tiles) and hence $10$ packs of tiles in a square metre          (A1)

(so each pack is $\frac{\$24.50}{10 \text{packs}}$)

OR

area covered by one pack of tiles is $\left(0.2 \text{m}\times 0.1 \text{m}\times 5=\right) 0.1 {\text{m}}^2$          (A1)

$24.5\times 0.1$          (M1)

THEN

$\$2.45$ per pack (of $5$ tiles)         A1

[3 marks]

$\frac{1.08\times 144}{5} \left(=31.104\right)$          (M1)(M1)

Note: Award M1 for correct numerator, M1 for correct denominator.

$32$ (packs of tiles)         A1

[3 marks]

$35+\left(32\times 2.45\right)$          (M1)

$\$113 \left(113.4\right)$         A1

[2 marks]

In part (a), most candidates were able to find the correct values for $a$ and $b$.

In part (b), most candidates were able to write down the correct expressions for $w_n$ and $l_n$.

In part (c)(i), candidates continued to struggle with “show that” questions. Some substituted 144 for $n$ and worked backwards, however this is never the intention of the question; candidates should progress towards (not “away from”) the given result. Part (ii) was well answered by many candidates.

Part (d) was poorly answered with many candidates multiplying 720 with 730 instead of 10 with 20. However, the majority managed to convert their answer correctly to standard form which gained them a mark for that particular skill.

Part (e) saw very few candidates find the cost of one packet of tiles. The main reason was the failure to convert cm² to m².

In part (f), about half of the candidates managed to find the correct number of packets. Some gained a mark for finding 8% or dividing by 5.

In part (g), most candidates could use their answers to parts (e) and (f) to score “follow through” marks and find the total cost of their order.