Calculate the common ratio of the sequence;

Calculate the eleventh term of the sequence;

Calculate the sum of the first 23 terms of the sequence.

2r² = 2.205     (M1)

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

r = 1.05     (A1)     (C2)

[2 marks]

2(1.05)¹⁰     (M1)

Note: Award (M1) for the correct substitution, using their answer to part (a), in geometric sequence formula.

= 3.26     (3.25778…)     (A1)(ft)     (C2)

Note: Follow through from their part (a).

[2 marks]

\(\frac{{2({{1.05}^{23}} - 1)}}{{(1.05 - 1)}}\)     (M1)

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

= 82.9     (82.8609…)     (A1)(ft)     (C2)

Notes: Accept an answer of 3.97221...if r = −1.05 is found in part (a) and used again in part (c). Follow through from their part (a).

[2 marks]

In part (a), 1.1025 proved to be a popular, but erroneous, answer. Similarly to question 4, such candidates failed to find a square root. Whilst this accuracy mark was lost for such candidates, much good work was seen in this question reflecting how well drilled the majority of candidates were in both arithmetic and geometric sequence techniques.

In part (a), 1.1025 proved to be a popular, but erroneous, answer. Similarly to question 4, such candidates failed to find a square root. Whilst this accuracy mark was lost for such candidates, much good work was seen in this question reflecting how well drilled the majority of candidates were in both arithmetic and geometric sequence techniques.

In part (a), 1.1025 proved to be a popular, but erroneous, answer. Similarly to question 4, such candidates failed to find a square root. Whilst this accuracy mark was lost for such candidates, much good work was seen in this question reflecting how well drilled the majority of candidates were in both arithmetic and geometric sequence techniques.

In this question give all answers correct to two decimal places.

Diogo deposited \(8000\) Argentine pesos, \({\text{ARS}}\), in a bank account which pays a nominal annual interest rate of \(15\% \), compounded monthly.

Find how much interest Diogo has earned after \(2\) years.

Carmen also deposited \({\text{ARS}}\) in a bank account. Her account pays a nominal annual interest rate of \(17\% \), compounded yearly. After three years, the total amount in Carmen’s account is \({\text{10}}\,{\text{000}}\,{\text{ARS}}\).

Find the amount that Carmen deposited in the bank account.

The first time an answer is not given to two decimal places the final (A1) is not awarded.

\(FV = 8000 \times {\left( {1 + \frac{{15}}{{100 \times 12}}} \right)^{2 \times 12}}\,\,\,\left( {FV = 8000 \times {{(1.0125)}^{2 \times 12}}} \right)\)       (M1)(A1)

OR

\(I = 8000 \times {\left( {1 + \frac{{15}}{{100 \times 12}}} \right)^{2 \times 12}} - 8000\)       (M1)(A1)

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

OR

\(N = 24\)

\(I\%  = 15\)

\(PV =  - 8000\)

\(PMT = 0\)

\((FV = 10778.8084)\)

\(P/Y = 12\)

\(C/Y = 12\)            (A1)(M1)

Note: Award (A1) for \(C/Y = 12\) seen, (M1) for all other correct entries. \(FV = 10778.8084\) need not be seen.

OR

\(N = 2\)

\(I\%  = 15\)

\(PV =  - 8000\)

\(PMT = 0\)

\((FV = 10778.8084)\)

\(P/Y = 1\)

\(C/Y = 12\)            (A1)(M1)

Note: Award (A1) for \(C/Y = 12\) seen, (M1) for all other correct entries. \(FV = 10778.8084\) need not be seen.

interest \( = 2778.81\)            (A1)    (C3)

Note: Final answer must be to two decimal places for the (A1) to be awarded.

\(N = 3\)

\(I\%  = 17\)

\((PV =  \pm 6243.705564)\)

\(PMT = 0\)

\((FV =  \mp 10\,000)\)

\(P/Y = 1\)

\(C/Y = 1\)          (A1)(M1)

Note: Award (A1) for \(FV =  \mp 10\,000\), (M1) for all other correct entries. \(PV =  \pm 6243.705564\) need not be seen.

OR

\(10\,000 = PV \times {\left( {1 + \frac{{17}}{{100}}} \right)^3}\)          (M1)(A1)

Note: Award (M1) for substituting into compound interest formula, (A1) for equating \(10\,000\) to the correctly substituted compounded interest formula.

\((PV = )\,\,6243.71\)                 (A1)         (C3)

Note: Answer must be to two decimal places.

Question 12: Compound Interest
The use of the TVM solver, with lack of working, remains a source of concern, though many more are writing down the calculator display; candidates are advised still to write down substituted formulas prior to using the TVM solver. Compounding periods remain a source of confusion
The use of the 0.15 and 0.17 was a common error, as was the computation of the amount in part (a).

Question 12: Compound Interest
The use of the TVM solver, with lack of working, remains a source of concern, though many more are writing down the calculator display; candidates are advised still to write down substituted formulas prior to using the TVM solver. Compounding periods remain a source of confusion
The use of the 0.15 and 0.17 was a common error, as was the computation of the amount in part (a).

Kwai Fan changes \({\text{S\$ }}500\) to Indian rupees.

Calculate the number of Indian rupees she will receive using this exchange rate. Give your answer correct to the nearest rupee.

On her return to Singapore, Kwai Fan has \(2500\) Indian rupees left from her trip. She wishes to exchange these rupees back to Singapore dollars. There is a \(3\% \) commission charge for this transaction and the exchange rate is \(100{\text{ INR}} = {\text{S\$}}3.672\).

Calculate the commission in Indian rupees that she is charged for this exchange.

On her return to Singapore, Kwai Fan has \(2500\) Indian rupees left from her trip. She wishes to exchange these rupees back to Singapore dollars. There is a \(3\% \) commission charge for this transaction and the exchange rate is \(100{\text{ INR}} = {\text{S\$}}3.672\).

Calculate the amount of money she receives in Singapore dollars, correct to two decimal places.

Financial penalty (FP) applies in this question.

\(500 \times \frac{{100}}{{3.684}}\)     (M1)

FP     \( = 13572\)     (A1)     (C2)

Note: (M1) for multiplication by \(\frac{{100}}{{3.684}}\)

[2 marks]

\(2500 \times 0.03\)     (M1)

\( = 75{\text{ }}(75.0{\text{, }}75.00)\)     (A1)     (C2)

If \(2500 \times 0.03 \times \frac{{3.672}}{{100}}\)

\( = 2.75\)

Award (M1)(A0)

[2 marks]

Financial penalty (FP) applies in this question.

\(2425 \times \frac{{3.672}}{{100}}\)     (M1)(ft)

FP     \( = 89.05\)     (A1)(ft)

OR

\(\frac{{3.672}}{{100}} \times 0.97 \times 2500\)     (M1)(ft)

FP     \( = 89.05\)     (A1)(ft)

OR

\(3\% {\text{ of }}91.8 = 2.754\)

\(91.8 - 2.754\)     (M1)(ft)

FP     \( = 89.05\)     (A1)(ft)     (C2)

Note: (ft) in (c) if the conversion process is reversed consistently through the question, i.e. multiplication in (a) followed by division in (c).

[2 marks]

This caused problems for many candidates. The form of the exchange rate proved difficult.

Most candidates managed to answer this correctly.

This part also proved problematic for many candidates.

Using simplest terms, write an equation in b and m for Thursday’s sales.

Find b and m.

Draw a sketch, in the space provided, to show how the prices can be found graphically.

Thursday's sales, \(6b + 9m = 23.40\)     (A1)

\(2b + 3m = 7.80\)     (A1)     (C2)

[2 marks]

\(m = 1.40\)     (accept 1.4)     (A1)(ft)

\(b = 1.80\)     (accept 1.8)     (A1)(ft)

Award (A1)(d) for a reasonable attempt to solve by hand and answer incorrect.     (C2)

[2 marks]

     (A1)(A1)(ft)

(A1) each for two reasonable straight lines. The intersection point must be approximately correct to earn both marks, otherwise penalise at least one line.

Note: The follow through mark is for candidate’s line from (a).     (C2)

[2 marks]

a) Nearly all the candidates were able to write the equation but very few simplified it.

b) A majority of candidates were able to find the values of b and m. Some used the right method but made arithmetical errors, many of which were due to them using the method of substitution which involved fractions. GDC use was expected.

c) A majority of candidates did not attempt this part. For those who did, very few were able to sketch the graph correctly. Common errors were to plot the point (1.4, 1.8) or draw a straight line through that point and the origin.

Calculate the common ratio for the increasing sequence of fees.

Give your answer correct to 2 decimal places.

The fees continue to increase in the same ratio.

Find the fees paid for 2006.

Give your answer correct to 2 decimal places.

The fees continue to increase in the same ratio.

A student attends the school for eight years, starting in 2000.

Find the total fees paid for these eight years.

\(r = \frac{{8320}}{{8000}}\) (or equivalent)     (M1)

Note: Award (M1) for dividing correct terms.

r = 1.04     (A1)     (C2)

Notes: In (b) and (c) (ft) from candidate’s r.

Allow lists, graphs etc. as working in (b) and (c).

[2 marks]

Financial penalty (FP) applies in this part

Fees = 8000 (1.04)⁶     (M1)

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

(FP)     Fees = 10122.55 USD (USD not required)     (A1)(ft)     (C2)

Note: Special exception to the note above.

Award maximum of (M1)(A0) if 5 is used as the power.

[2 marks]

Financial penalty (FP) applies in this part

\({\text{Total}} = \frac{{8000({{1.04}^8} - 1)}}{{1.04 - 1}}\)     (M1)

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

Give full credit for solution by lists.

(FP)     Total = 73713.81 USD (USD not required)     (A1)(ft)     (C2)

[2 marks]

Many marks were lost through incorrect rounding or premature rounding (if a year by year approach was used).

This part was well attempted, errors being the use of 4% as the common ratio.

Many marks were lost through incorrect rounding or premature rounding (if a year by year approach was used).

The common error here was the use of the incorrect index in the formula.

Many marks were lost through incorrect rounding or premature rounding (if a year by year approach was used).

Attempts at calculation without use of the formula were largely unsuccessful.

5000 AUD is invested at a nominal annual interest rate of 2.5 % compounded half yearly.

Calculate the length of time in years for the interest on this investment to exceed 446.25 AUD.

\(446.25 = 5000{\left( {1 + \frac{{2.5}}{{2(100)}}} \right)^{2n}} - 5000\)     (M1)(A1)

Notes: Award (M1) for substitution into compound interest formula. Award (A1) for correct values.

\(5446.25 = 5000{\left( {1 + \frac{{2.5}}{{2(100)}}} \right)^{2n}}\)     (A1)

n = 3.44

n = 3.5     (A1)

OR

5446.25 = 5000(1.0125)²ⁿ     (A1)(M1)(A1)

Notes: Award (A1) for 5446.25 seen.

Award (M1) for substitution into compound interest formula.

Award (A1) for correct values.

n = 3.44 years

3.5 years required     (A1)     (C4)

Notes: For incorrect substitution into compound interest formula award at most (M1)(A0)(A1)(A0).

Award (A3) for 3.44 seen without working.

Allow solution by lists. In this case

Award (A1) for half year rate 1.25 % seen.

(A1) for 5446.25 seen.

(M1) for at least 2 correct uses of multiplication by 1.0125

5000 × 1.0125 = 5062.5 and 5062.5 × 1.0125 = 5125.78125

(A1) n = 3.5

If yearly rate used then award (A0)(A1)(M1)(A0)

[4 marks]

This question was poorly answered by many of the candidates. Candidates confuse interest with principal in the formulas.

The idea of compounding periods and the implication for determining the level of interest is poorly understood. The correct answer is 3.5 years. Interpretation of compounding periods is expected.

Calculate the amount of commission charged in SEK.

Write down the amount of money she receives from the bank after commission.

Chiara returns to Italy with 296 SEK. She changes this money back into euros at a bank and receives 32€. The bank does not charge commission.

Calculate the value in SEK of 1€.

350 × 10.275 × 0.02     (M1)(M1)

Note: Award (M1) for ×10.275, (M1) for ×0.02.

71.93 (SEK)     (A1)     (C3)

[3 marks]

3524.33(SEK)     (A1)(ft)     (C1)

Note: Accept 3524.32. Follow through from their answer to part (a).

[1 mark]

\(\frac{{296}}{{32}}\)     (M1)

9.25     (A1)     (C2)

[2 marks]

A common error in Question 9 was finding the amount of money received for part (a) rather than just the commission. Some candidates had difficulties giving the appropriate number of decimal places.

A common error in Question 9 was finding the amount of money received for part (a) rather than just the commission. Some candidates had difficulties giving the appropriate number of decimal places.

A common error in Question 9 was finding the amount of money received for part (a) rather than just the commission. Some candidates had difficulties giving the appropriate number of decimal places.

Find \(f ' (x) \).

Write down the value of \(f ′(0)\).

The function has a local maximum at x = −2.

Calculate the value of a.

\( f '(x) = 3ax^2 - 3\)     (A1)(A1)     (C2)

Note: Award a maximum of (A1)(A0) if any extra terms are seen.

−3     (A1)(ft)     (C1)

Note: Follow through from their part (a).

\(f '(x) = 0\)     (M1)

Note: This may be implied from line below.

\(3a(-2)^2 - 3 = 0\)     (M1)

\((a =) \frac{1}{4}\)     (A1)(ft)     (C3)

Note: Follow through from their part (a).

Many candidates could find the derivative of the cubic function and find the value of the derivative at \(x = 0\). For part (c) many candidates calculated the value of the function rather than the derivative at \(x = - 2\).

Many candidates could find the derivative of the cubic function and find the value of the derivative at \(x = 0\).

Many candidates could find the derivative of the cubic function and find the value of the derivative at \(x = 0\). For part (c) many candidates calculated the value of the function rather than the derivative at \(x = - 2\). However only the best realized that the derivative is zero at the maximum and so calculated the value of \(a\).

Find the exact value of \(z\).

Write your answer to part (a) correct to 2 decimal places.

Write your answer to part (a) correct to three significant figures.

Write your answer to part (a) in the form \(a \times {10^k}\), where 1 ≤ a < 10, \(k \in {\mathbb{Z}}\) .

\(z = \frac{{12\cos ({{60}^ \circ })}}{{(4(8) + 32)}}\)     (M1)

Note: Award (M1) for correct substituted formula seen.

\( = 0.09375\left( {\frac{3}{{32}}} \right)\)     (A1)(C2)

[2 marks]

\(0.09\)     (A1)(ft)     (C1)

[1 mark]

\(0.0938\)     (A1)(ft)     (C1)

[1 mark]

\(9.375 \times {10^{ - 2}}\) (\(9.38 \times {10^{ - 2}}\))     (A1)(ft)(A1)(ft)     (C2)

Note: Award (A1)(ft) for \(9.375\), (A1)(ft) for \( \times {10^{ - 2}}\). Follow through from their part (a).

[2 marks]

Although the use of radians leading to an incorrect answer of \( - 0.1785774388\) was seen on a minority of scripts, many candidates produced correct answers for parts (a) and (b)(i). The requirement for an answer to 3 significant figures led many to count the first zero after the decimal point and as a consequence gave an incorrect answer of \(0.094\). Despite any previous incorrect working, it was pleasing to see that most candidates were able to express their answer to part (a) in standard form.

Although the use of radians leading to an incorrect answer of \( - 0.1785774388\) was seen on a minority of scripts, many candidates produced correct answers for parts (a) and (b)(i). The requirement for an answer to 3 significant figures led many to count the first zero after the decimal point and as a consequence gave an incorrect answer of \(0.094\). Despite any previous incorrect working, it was pleasing to see that most candidates were able to express their answer to part (a) in standard form

Although the use of radians leading to an incorrect answer of \( - 0.1785774388\) was seen on a minority of scripts, many candidates produced correct answers for parts (a) and (b)(i). The requirement for an answer to 3 significant figures led many to count the first zero after the decimal point and as a consequence gave an incorrect answer of \(0.094\). Despite any previous incorrect working, it was pleasing to see that most candidates were able to express their answer to part (a) in standard form.

Although the use of radians leading to an incorrect answer of \( - 0.1785774388\) was seen on a minority of scripts, many candidates produced correct answers for parts (a) and (b)(i). The requirement for an answer to 3 significant figures led many to count the first zero after the decimal point and as a consequence gave an incorrect answer of \(0.094\). Despite any previous incorrect working, it was pleasing to see that most candidates were able to express their answer to part (a) in standard form.

When Bermuda \({\text{(B)}}\), Puerto Rico \({\text{(P)}}\), and Miami \({\text{(M)}}\) are joined on a map using straight lines, a triangle is formed. This triangle is known as the Bermuda triangle.

According to the map, the distance \({\text{MB}}\) is \(1650\,{\text{km}}\), the distance \({\text{MP}}\) is \(1500\,{\text{km}}\) and angle \({\text{BMP}}\) is \(57^\circ \).

Calculate the distance from Bermuda to Puerto Rico, \({\text{BP}}\).

Calculate the area of the Bermuda triangle.

\({\text{B}}{{\text{P}}^2} = {1650^2} + {1500^2} - 2 \times 1650 \times 1500\,\cos \,(57^\circ )\)                  (M1)(A1)

\(1510\,({\text{km}})\,\,\,\left( {1508.81...\,({\text{km}})} \right)\)       (A1)     (C3)

Notes: Award (M1) for substitution in the cosine rule formula, (A1) for correct substitution.

\(\frac{1}{2} \times 1650 \times 1500 \times \sin \,57^\circ \)        (M1)(A1)

\( = 1\,040\,000\,({\text{k}}{{\text{m}}^2})\,\,\,\left( {1\,037\,854.82...\,({\text{k}}{{\text{m}}^2})} \right)\)       (A1)    (C3)

Note: Award (M1) for substitution in the area of triangle formula, (A1) for correct substitution.

Question 6: Non-right angle trigonometry.
Instead of using the law of cosines weaker candidates substituted into Pythagoras’ theorem and likewise used \(A = \frac{1}{2}bh\) instead of \(A = \frac{1}{2}ab\sin C\). Those that did select the correct formula almost always made correct substitutions but were not always able to calculate the correct answer.

Question 6: Non-right angle trigonometry. Instead of using the law of cosines weaker candidates substituted into Pythagoras’ theorem and likewise used \(A = \frac{1}{2}bh\) instead of \(A = \frac{1}{2}ab\sin C\). Those that did select the correct formula almost always made correct substitutions but were not always able to calculate the correct answer.

(i) Find the common difference.

(ii) Find the first term of the sequence.

Calculate the eighty-fourth term.

Calculate the sum of the first 200 terms.

(i) \(u_5 = u_1 + 4d = 20\)

\(u_{12} = u_1 + 11d = 41\)     (M1)

(M1) for both equations correct (or (M1) for \(20 + 7d = 41\))

\(7d = 21\)

\(d = 3\)     (A1)     (C2)

(ii) \(u_1 + 12 = 20\)

\(u_1 = 8\)     (A1)(ft)     (C1)

[3 marks]

\(u_{84} = 8 + (84 - 1)3\)

\(= 257\)     (A1)(ft)     (C1)

[1 mark]

\(S_{200} = 100(16 + 199 \times 3)\)     (M1)

\( = 61300\)     (A1)(ft)     (C2)

[2 marks]

This question was generally answered well. Most of the candidates had a good understanding of how to use the formulae for an arithmetic sequence.

This question was generally answered well. Most of the candidates had a good understanding of how to use the formulae for an arithmetic sequence.

This question was generally answered well. Most of the candidates had a good understanding of how to use the formulae for an arithmetic sequence.

Calculate

(i)     the common difference of the sequence;

(ii)     the first term of the sequence.

The first, second and fifth terms of this arithmetic sequence are the first three terms of a geometric sequence.

Calculate the seventh term of the geometric sequence.

(i)     \({u_1} + d = 30,{\text{ }}{u_1} + 4d = 90,{\text{ }}3d = 90 - 30\;\;\;\)(or equivalent)     (M1)

Note: Award (M1) for one correct equation. Accept a list of at least 5 correct terms.

\((d = ){\text{ }}20\)     (A1)

(ii)     \(({u_1} = ){\text{ }}10\)     (A1)(ft)     (C3)

Note: Follow through from (a)(i), irrespective of working shown if \({u_1} = 30 - {\text{ (their }}d)\;\;\;\)OR\(\;\;\;{u_1} = 90 - 4 \times {\text{ (their }}d{\text{)}}\)

\(({u_7} = ){\text{ }}10({3^{(7 - 1)}}\;\;\;\)OR\(\;\;\;({u_7} = ){\text{ 10}} \times {3^6}\)     (M1)(A1)(ft)

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

OR

\(10;{\text{ }}30;{\text{ }}90;{\text{ }}270;{\text{ }}810;{\text{ }}2430;{\text{ }}7290\)     (M1)(A1)(ft)

Note: Award (M1) for a list of at least 5 consecutive terms of a geometric sequence, (A1)(ft) for terms corresponding to their answers in part (a).

\( = 7290\)     (A1)(ft)     (C3)

Note: Follow through from part (a).

Part (a) was answered correctly by many candidates, but working using equations was rarely seen. A “trial and error” method, based upon a list of terms was the most seen method.

In part (b) many found the correct answer, but many others did not. Some gave the seventh term of the arithmetic sequence, some gave a term of an incorrect order and some a completely incorrect answer. Finding the correct ratio was the most common problem. Often repeated multiplication was used to find the answer, but also the formula for the nth term of a geometric sequence was used. Several did not use the correct three terms from the question.

Write down an expression for A in terms of x.

Find the value of x given that A = 109.25 m².

The owner of the garden puts a fence around the shaded region. Find the length of this fence.

(x + 1)² – 1  or  x² + 2x     (A1)     (C1)

[1 mark]

(x + 1)² – 1 = 109.25     (M1)

x² + 2x – 109.25 = 0     (M1)

Notes: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for correctly removing the brackets.

OR

(x + 1)² – 1 = 109.25     (M1)

x + 1 = \(\sqrt {110.25} \)     (M1)

Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for taking the square root of both sides.

OR

(x + 1)² – 10.5² = 0     (M1)

(x – 9.5) (x + 11.5) = 0     (M1)

Note: Award (M1) for writing an equation consistent with their expression in (a) (accept equivalent forms), (M1) for factorised left side of the equation.

x = 9.5     (A1)(ft)     (C3)

Note: Follow through from their expression in part (a).

The last mark is lost if x is non positive.

If the follow through equation is not quadratic award at most (M1)(M0)(A1)(ft).

[3 marks]

4 × (9.5 + 1) = 42 m     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for correct method for finding the length of the fence. Accept equivalent methods.

[2 marks]

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

Some candidates were able to answer this question correctly, but the majority experienced difficulty in finding the correct expression for the area of the shaded region. Those who showed working could then be awarded follow through marks for correctly equating their expressions to the given area and for their found value of x. Many candidates also could not find the perimeter of the shaded region in part c) even though they had found the value of x correctly.

Factorise the expression \({x^2} - kx\) .

Hence solve the equation \({x^2} - kx = 0\) .

The diagram below shows the graph of the function \(f(x) = {x^2} - kx\) for a particular value of \(k\).

Write down the value of \(k\) for this function.

The diagram below shows the graph of the function \(f(x) = {x^2} - kx\) for a particular value of \(k\).

Find the minimum value of the function \(y = f(x)\) .

\(x(x - k)\)     (A1)     (C1)

[1 mark]

\(x = 0\) or \(x = k\)     (A1)     (C1)

Note: Both correct answers only.

[1 mark]

\(k = 3\)     (A1)     (C1)

[1 mark]

\({\text{Vertex at }}x = \frac{{ - ( - 3)}}{{2(1)}}\)     (M1)

Note: (M1) for correct substitution in formula.

\(x = 1.5\)     (A1)(ft)

\(y = - 2.25\)     (A1)(ft)

OR

\(f'(x) = 2x - 3\)     (M1)

Note: (M1) for correct differentiation.

\(x = 1.5\)     (A1)(ft)
\(y = - 2.25\)     (A1)(ft)

OR

for finding the midpoint of their 0 and 3     (M1)
\(x = 1.5\)     (A1)(ft)
\(y = - 2.25\)     (A1)(ft)

Note: If final answer is given as \((1.5{\text{, }}{- 2.25})\) award a maximum of (M1)(A1)(A0)

[3 marks]

This question was poorly answered by all but the best candidates. The links between the parts were not made. The idea of the line of symmetry for the graph was seldom investigated. The “minimum value of the function” was often incorrectly given as a coordinate pair.

This question was poorly answered by all but the best candidates. The links between the parts were not made. The idea of the line of symmetry for the graph was seldom investigated. The “minimum value of the function” was often incorrectly given as a coordinate pair.

This question was poorly answered by all but the best candidates. The links between the parts were not made. The idea of the line of symmetry for the graph was seldom investigated. The “minimum value of the function” was often incorrectly given as a coordinate pair.

This question was poorly answered by all but the best candidates. The links between the parts were not made. The idea of the line of symmetry for the graph was seldom investigated. The “minimum value of the function” was often incorrectly given as a coordinate pair.

Find the amount of money that Minta will have in the bank after 3 years. Give your answer correct to two decimal places.

Minta will withdraw the money from her bank account when the interest earned is 300 euros.

Find the time, in years, until Minta withdraws the money from her bank account.

\(1000{\left( {1 + \frac{5}{{4 \times 100}}} \right)^{4 \times 3}}\)     (M1)(A1)

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

OR

\({\text{N}} = 3\)

\({\text{I}}\%  = 5\)

\({\text{PV}} =  - 1000\)

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

\({\text{C/Y}} = 4\)     (A1)(M1)

Note: Award (A1) for \({\text{C/Y}} = 4\) seen, (M1) for other correct entries.

OR

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

\({\text{I}}\%  = 5\)

\({\text{PV}} =  - 1000\)

\({\text{P/Y}} = 4\)

\({\text{C/Y}} = 4\)     (A1)(M1)

Note: Award (A1) for \({\text{C/Y}} = 4\) seen, (M1) for other correct entries.

\( = 1160.75\) (€)    (A1)     (C3)

\(1000{\left( {1 + \frac{5}{{4 \times 100}}} \right)^{4 \times t}} = 1300\)     (M1)(A1)

Note: Award (M1) for using the compound interest formula with a variable for time, (A1) for substituting correct values and equating to \(1300\).

OR

\({\text{I}}\%  = 5\)

\({\text{PV}} =  \pm 1000\)

\({\text{FV}} =  \mp 1300\)

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

\({\text{C/Y}} = 4\)     (A1)(M1)

Note: Award (A1) for 1300 seen, (M1) for the other correct entries.

OR

\({\text{I}}\%  = 5\)

\({\text{PV}} =  \pm 1000\)

\({\text{FV}} =  \mp 1300\)

\({\text{P/Y}} = 4\)

\({\text{C/Y}} = 4\)     (A1)(M1)

Note: Award (A1) for 1300 seen, (M1) for the other correct entries.

OR

Sketch drawn of two appropriate lines which intersect at a point

Note: Award (M1) for a sketch with a straight line intercepted by appropriate curve, (A1) for a numerical answer in the range \(5.2-5.6\).

\(t = 5.28{\text{ (years)}}\;\;\;(5.28001 \ldots )\)     (A1) (C3)

Find the first term of the sequence.

Find the 21^st term of the sequence.

Find the sum of the first 30 terms of the sequence.

32 = u₁ + (10 − 1) × (−6)     (M1)

Notes: Award (M1) for correct substitution in correct formula. Accept correct alternative methods.

u₁ = 86     (A1)     (C2)

[2 marks]

u₂₁ = 86 + (21 − 1) × (−6)     (M1)

u₂₁ = −34     (A1)(ft)

Notes: Award (M1) for correct substitution in correct formula. Accept correct alternative methods. Award (M1) for a list of at least 5 correct terms seen. Follow through from their answer to part (a).

OR

u₂₁ = 32 + 11 × (−6)     (M1)

u₂₁ = −34     (A1)     (C2)

[2 marks]

\({S_{30}} = \frac{{30}}{2}(2 \times 86 + (30 - 1) \times ( - 6))\)     (M1)

Notes: Award (M1) for their correct substitution in correct formula. Accept correct alternative methods. For a list award (M1) for the correct addition of at least 10 terms.

\({S_{30}} = -30\)     (A1)(ft)     (C2)

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

[2 marks]

This question was very well answered by most candidates. Correct working was clearly shown. Many candidates used 32 as their first term and many others subtracted 6 rather than multiplied by -6, indicating a lack of attention in their algebraic notation and manipulation.

This question was very well answered by most candidates. Correct working was clearly shown. Many candidates used 32 as their first term and many others subtracted 6 rather than multiplied by -6, indicating a lack of attention in their algebraic notation and manipulation.

This question was very well answered by most candidates. Correct working was clearly shown. Many candidates used 32 as their first term and many others subtracted 6 rather than multiplied by -6, indicating a lack of attention in their algebraic notation and manipulation.

Find the value of \(r\), the common ratio of the sequence.

Find the value of \(n\) for which \({u_n} = 2\).

Find the sum of the first 30 terms of the sequence.

\(\frac{{162}}{{486}}\)\(\,\,\,\)OR\(\,\,\,\)\(\frac{{54}}{{162}}\)     (M1)

Note:     Award (M1) for dividing any \({u_{n + 1}}\) by \({u_n}\).

\( = \frac{1}{3}{\text{ }}(0.333,{\text{ }}0.333333 \ldots )\)     (A1)     (C2)

[2 marks]

\(486{\left( {\frac{1}{3}} \right)^{n - 1}} = 2\)     (M1)

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

\(n = 6\)     (A1)(ft)     (C2)

Note:     Follow through from part (a).

Award (A1)(A0) for \({u_6} = 2\) or \({u_6}\) with or without working.

[2 marks]

\({S_{30}} = \frac{{486\left( {1 - {{\frac{1}{3}}^{30}}} \right)}}{{1 - \frac{1}{3}}}\)     (M1)

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

\( = 729\)     (A1)(ft)     (C2)

[2 marks]

Write down the distance he runs on the second day of training.

Calculate the total distance Javier runs in the first seven days of training.

Javier stops training on the day his total distance exceeds 100 km.

Calculate the number of days Javier has trained for the running race.

1.65 (km) or 1650 (m)     (A1)     (C1)

[1 mark]

\(\frac{{1.5({{1.1}^7} - 1)}}{{1.1 - 1}}\)     (M1)

Notes: Award (M1) for correct substitution of candidate’s 10 % into the correct formula. Accept a list.

14.2 (km)     (A1)(ft)     (C2)

[2 marks]

\(\frac{{1.5({{1.1}^n} - 1)}}{{1.1 - 1}} > 100\)     (M1)

Note: Award (M1) for setting up their inequality/equation. Accept a list.

n = 21.371...     (A1)(ft)

n = 22     (A1)(ft)     (C3)

Notes: Follow through from their values of 1.1 and 1.5 in part (b). The final (A1)(ft) is for rounding up their answer for n to a whole number of days.

[3 marks]

Most candidates could answer the first part of this question, although a number found it difficult to find the total distance run after 7 days. Many gave the correct answer of 1.65 km or 1650 m for part (a).

In part (b), stronger candidates answered correctly, however many used a list or the incorrect arithmetic formula.

In part (c), the most common mistake was to use the arithmetic formula. Many candidates rounded their answer down rather than up.

Find the amount of Euros that Claudia receives. Give your answer correct to two decimal places.

Find the value of \(r\).

\(8000 \times 0.09819 \times 0.98\)     (M1)(M1)

Note:     Award (M1) for multiplying 8000 by 0.09819, (M1) for multiplying by 0.98 (or equivalent).

769.81 (EUR)     (A1)     (C3)

[3 marks]

\(r\%  \times \frac{{85}}{{0.08753}} = 14.57\)     (M1)(M1)

Note:     Award (M1) for dividing 85 by 0.08753, and (M1) for multiplying their \(\frac{{85}}{{0.08753}}\) by \(r\% \) and equating to 14.57.

OR

\(\frac{{85}}{{0.08753}} = {\text{971.095}} \ldots \)     (M1)

Note:     Award (M1) for dividing 85 by 0.08753.

\(\frac{{14.57}}{{9.71095 \ldots }}\)\(\,\,\,\)OR\(\,\,\,\)\(\frac{{14.57}}{{971.095 \ldots }} \times 100\)     (M1)

Note:     Award (M1) for dividing 14.57 by 9.71095… or equivalent.

\(r = 1.50{\text{ }}(1.50036 \ldots )\)     (A1)     (C3)

[3 marks]

Write down the second term of this sequence.

Write down the common difference of this sequence.

Write down the fourth term of this sequence.

The n^th term is the first term in this sequence which is greater than 1000. Find the value of n.

8     (A1)     (C1)

[1 mark]

5     (A1)(ft)     (C1)

[1 mark]

18     (A1)(ft)     (C1)

[1 mark]

3 + 5 × (n – 1) > 1000     (M1)

Note: Allow equality sign and equal to 1001

n > 200.4     (A1)

Note: Accept n = 200.4 or 5n =1002

OR

(M1) for attempt at listing, (A1) for 998 and 1003 seen.     (M1)(A1)

n = 201     (A1)(ft)     (C3)

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

[3 marks]

Most candidates achieved at least the first three marks on this question and a significant number gained full marks. Up to five marks could be awarded even to students who did not find correctly the value of the nth term, but showed correct method and attempted to find the value of n. Some candidates lost the last mark for giving a non integer value.

Most candidates achieved at least the first three marks on this question and a significant number gained full marks. Up to five marks could be awarded even to students who did not find correctly the value of the nth term, but showed correct method and attempted to find the value of n. Some candidates lost the last mark for giving a non integer value.

Most candidates achieved at least the first three marks on this question and a significant number gained full marks. Up to five marks could be awarded even to students who did not find correctly the value of the nth term, but showed correct method and attempted to find the value of n. Some candidates lost the last mark for giving a non integer value.

Most candidates achieved at least the first three marks on this question and a significant number gained full marks. Up to five marks could be awarded even to students who did not find correctly the value of the nth term, but showed correct method and attempted to find the value of n. Some candidates lost the last mark for giving a non integer value.

Find the volume of the box.

Joaquin estimates the volume of one cube to be 500 cm³. He uses this value to estimate the number of cubes in the box.

Find Joaquin’s estimated number of cubes in the box.

The actual number of cubes in the box is 350.

Find the percentage error in Joaquin’s estimated number of cubes in the box.

\(50 \times 100 \times 40 = 200\,000{\text{ c}}{{\text{m}}^3}\)     (M1)(A1)     (C2)

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

[2 marks]

\(\frac{{200\,000}}{{500}} = 400\)     (M1)(A1)(ft)     (C2)

Note: Award (M1) for dividing their answer to part (a) by 500.

[2 marks]

\(\frac{{400 - 350}}{{350}} \times 100 = 14.3{\text{ }}\% \)     (M1)(A1)(ft)     (C2)

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

Award (A1) for answer, follow through from part (b).

Accept –14.3 %.

% sign not necessary.

[2 marks]

This question proved to be the one that most candidates answered correctly. Many received full marks and the only error seen was incorrect substitution in the percentage error formula.

This question proved to be the one that most candidates answered correctly. Many received full marks and the only error seen was incorrect substitution in the percentage error formula.

This question proved to be the one that most candidates answered correctly. Many received full marks and the only error seen was incorrect substitution in the percentage error formula.

Calculate the number of SGD Pietro receives.

Pietro also has 100 GBP that he wishes to change to USD for a trip to Cambodia.

To perform this transaction, the GBP must first be converted to SGD and then to USD.

Calculate the number of USD Pietro receives.

\(800 \times 1.55\)     (M1)

Note: Award (M1) for multiplication by \(1.55\).

\( = 1240\)     (A1)     (C2)

\(\frac{{100 \times 1.92}}{{1.28}}\)     (A1)(M1)(M1)

Notes: Award (M1) for multiplication by a GBP rate (\(1.92\) or \(2.05\)), (M1) for division by a USD rate (\(1.28\) or \(1.15\)), (A1) for two correct rates used.

\( = 150\)     (A1) (C4)

Note: Award a maximum of (A1)(ft)(M1)(M1)(A1)(ft) for \(\frac{{100 \times 2.05}}{{1.15}}\), if in part (a) a rate of \(1.75\) is used.

Award a maximum of (A1)(ft)(M1)(M1)(A1)(ft) if division by an EUR rate is seen in part (a) and multiplication by \(1.28\) and division by \(1.92\) is seen in (b).

Many candidates lost at least two marks on this question for using an incorrect rate. The difference between “Bank buys” and “Bank sells” was not understood by many candidates. Their use of the table was often not consistent, leading to the candidates losing 4 marks, 2 in part (a) and 2 in part (b). Only very few candidates were confused on when to multiply and when to divide by a conversion rate. It was disappointing to see that so many candidates were not able to apply their knowledge of currency conversion in the real world context where both rates are given and the candidate had to decide which one to use. Methods marks were given out frequently, showing candidates were confident to calculate currency conversion with given rates.

Many candidates lost at least two marks on this question for using an incorrect rate. The difference between “Bank buys” and “Bank sells” was not understood by many candidates. Their use of the table was often not consistent, leading to the candidates losing 4 marks, 2 in part (a) and 2 in part (b). Only very few candidates were confused on when to multiply and when to divide by a conversion rate. It was disappointing to see that so many candidates were not able to apply their knowledge of currency conversion in the real world context where both rates are given and the candidate had to decide which one to use. Methods marks were given out frequently, showing candidates were confident to calculate currency conversion with given rates.

Find the value of \(m\).

State what \(m\) represents in this context.

Find the value of \(n\).

State what \(n\) represents in this context.

\(\frac{{600 - 150}}{{6 - 1}}\)     (M1)

OR

\(600 = 150 + (6 - 1)m\)     (M1)

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

\( = 90\)     (A1)     (C2)

the annual rate of growth of the number of apartments     (A1)     (C1)

Note: Do not accept common difference.

\(150 = 90 \times (1) + n\)     (M1)

Note: Award (M1) for correct substitution of their gradient and one of the given points into the equation of a straight line.

\(n = 60\)     (A1)(ft)     (C2)

Note: Follow through from part (a).

the initial number of apartments     (A1)     (C1)

Note: Do not accept “first number in the sequence”.

Calculate exactly \(\frac{{{{(3 \times 2.1)}^3}}}{{7 \times 1.2}}\).

Write the answer to part (a) correct to 2 significant figures.

Calculate the percentage error when the answer to part (a) is written correct to 2 significant figures.

Write your answer to part (c) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10{\text{ and }}k \in \mathbb{Z}\).

\(29.7675\)     (A1)     (C1)

Note: Accept exact answer only.

[1 mark]

\(30\)     (A1)(ft)     (C1)

[1 mark]

\(\frac{{30 - 29.7675}}{{29.7675}} \times 100\% \)     (M1)

For correct formula with correct substitution.

\( = 0.781\% \)     accept \(0.78\% \) only if formula seen with \(29.7675\) as denominator     (A1)(ft)     (C2)

[2 marks]

\(7.81 \times {10^{ - 1}}\% \) (\(7.81 \times {10^{ - 3}}\) with no percentage sign)     (A1)(ft)(A1)(ft)     (C2)

[2 marks]

Many candidates gained maximum marks. The common errors were in the initial calculation due to forgetting to use brackets when entering the denominator into the GDC; using 2 decimal places instead of 2 significant figures in part (b); and using the wrong value as the denominator in part (c). Some candidates were still using the old Information booklet with the wrong formula for percentage error. Because examiners believed that the wording was ambiguous the correct answer given to 2 significant figures was awarded full marks for this part.

Many candidates gained maximum marks. The common errors were in the initial calculation due to forgetting to use brackets when entering the denominator into the GDC; using 2 decimal places instead of 2 significant figures in part (b); and using the wrong value as the denominator in part (c). Some candidates were still using the old Information booklet with the wrong formula for percentage error. Because examiners believed that the wording was ambiguous the correct answer given to 2 significant figures was awarded full marks for this part.

Many candidates gained maximum marks. The common errors were in the initial calculation due to forgetting to use brackets when entering the denominator into the GDC; using 2 decimal places instead of 2 significant figures in part (b); and using the wrong value as the denominator in part (c). Some candidates were still using the old Information booklet with the wrong formula for percentage error. Because examiners believed that the wording was ambiguous the correct answer given to 2 significant figures was awarded full marks for this part.

Many candidates gained maximum marks. The common errors were in the initial calculation due to forgetting to use brackets when entering the denominator into the GDC; using 2 decimal places instead of 2 significant figures in part (b); and using the wrong value as the denominator in part (c). Some candidates were still using the old Information booklet with the wrong formula for percentage error. Because examiners believed that the wording was ambiguous the correct answer given to 2 significant figures was awarded full marks for this part.

Write down the median length of these leaves.

Write down the number of leaves with a length less than or equal to 8 cm.

Use the graph to find the value of \(k\).

Before measuring, the researcher estimated \(k\) to be approximately 9.5 cm. Find the percentage error in her estimate.

9 (cm)     (A1)     (C1)

[1 mark]

40 (leaves)     (A1)     (C1)

[1 mark]

\((200 \times 0.90 = ){\text{ }}180\) or equivalent     (M1)

Note:     Award (M1) for a horizontal line drawn through the cumulative frequency value of 180 and meeting the curve (or the corresponding vertical line from 10.5 cm).

\((k = ){\text{ }}10.5{\text{ (cm)}}\)     (A1)     (C2)

Note:     Accept an error of ±0.1.

[2 marks]

\(\left| {\frac{{9.5 - 10.5}}{{10.5}}} \right| \times 100\% \)     (M1)

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

\({\text{9.52 (% ) }}\left( {{\text{9.52380}} \ldots {\text{ (% )}}} \right)\)     (A1)(ft)     (C2)

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

Award (A1)(A0) for an answer of \( - 9.52\) with or without working.

[2 marks]

Write down the value of the car when Shiyun bought it.

Calculate the value of the car three years after Shiyun bought it. Give your answer correct to two decimal places.

Calculate the time for the car to depreciate to half of its value since Shiyun bought it.

\(25000{\text{ USD}}\)     (A1)     (C1)

[1 mark]

\(25000 \times {1.5^{ - 0.6}}\)     (M1)
\(19601.32{\text{ USD}}\)     (A1)     (C2)

[2 marks]

\(12500 = 25000 \times {1.5^{ - 0.2t}}\)     (A1)(ft)(M1)

Notes: Award (A1)(ft) for \(12 500\) seen. Follow through from their answer to part (a). Award (M1) for equating their half value to \(25000 \times {1.5^{ - 0.2t}}\) . Allow the use of an inequality.

OR

Graphical method (sketch):

(A1)(ft) for \(y =12 500\) seen on the sketch. Follow through from their answer to part (a).     (A1)(ft)

(M1) for the exponent model and an indication of their intersection with their horizontal line.     (M1)

\(8.55\)     (A1)(ft)     (C3)

[3 marks]

A substituted value of \(t = 1\) in part (a) saw many incorrect answers of \(23052.70\) for this part of the question. Part (b) was better attempted with many correct answers seen. Many candidates picked up the first two marks of part (c) equating a correct expression to half their answer found in part (a). Many though did not seem to know the correct process of using their GDC to find the required answer. Much trial and improvement was seen here with varying degrees of success.

A substituted value of \(t = 1\) in part (a) saw many incorrect answers of \(23052.70\) for this part of the question. Part (b) was better attempted with many correct answers seen. Many candidates picked up the first two marks of part (c) equating a correct expression to half their answer found in part (a). Many though did not seem to know the correct process of using their GDC to find the required answer. Much trial and improvement was seen here with varying degrees of success.

A substituted value of \(t = 1\) in part (a) saw many incorrect answers of \(23052.70\) for this part of the question. Part (b) was better attempted with many correct answers seen. Many candidates picked up the first two marks of part (c) equating a correct expression to half their answer found in part (a). Many though did not seem to know the correct process of using their GDC to find the required answer. Much trial and improvement was seen here with varying degrees of success.

Find the common ratio of the sequence.

Find u₁, the first term of the sequence.

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

\(\frac{{101.25}}{{135}}\)     (M1)

\( = \frac{3}{4}(0.75)\)     (A1)     (C2)

\({u_1}{\left( {\frac{3}{4}} \right)^4} = 101.25\)     (M1)

OR

\({u_1}{\left( {\frac{3}{4}} \right)^3} = 135\)     (M1)

OR

(by list)

\({u_3} = 180,{\text{ }}{u_2} = 240\)     (M1)

Notes: Award (M1) for their correct substitution in geometric sequence formula, or stating explicitly \({u_3}\) and \({u_2}\).

\(({u_1} = )320\)     (A1)(ft)     (C2)

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

\({S_{10}} = \frac{{320\left( {1 - {{\left( {\frac{3}{4}} \right)}^{10}}} \right)}}{{1 - \left( {\frac{3}{4}} \right)}}\)     (M1)

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

    Accept a list of all their ten geometric terms.

= 1210 (1207.918...)     (A1)(ft)     (C2)

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

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as specified in the question.

Using the mapping diagram, write down two equations in terms of \(a\) and \(b\).

Solve the equations to find the value of

(i)     \(a\);

(ii)     \(b\).

Find the value of \(c\).

\(a{(1)^2} + b =  - 9\)     (A1)

\(a{(3)^2} + b = 119\)     (A1)     (C2)

Note: Accept equivalent forms of the equations.

[2 marks]

(i)     \(a = 16\)     (A1)(ft)

(ii)     \(b =  - 25\)     (A1)(ft)     (C2)

Note: Follow through from part (a) irrespective of whether working is seen.

     If working is seen follow through from part (i) to part (ii).

[2 marks]

\(16{c^2} - 25 = 171\)     (M1)

Note: Award (M1) for correct quadratic with their \(a\) and \(b\) substituted.

\(c = 3.5\)     (A1)(ft)     (C2)

Note: Accept \(x\) instead of \(c\).

     Follow through from part (b).

     Award (A1) only, for an answer of \( \pm 3.5\) with or without working.

[2 marks]

This question was answered reasonably well with many candidates able to write down the two equations and solve them for a and b. Errors such as mistaking the equation given for \(3{a^2} + b = 119\) meant that marks were lost even though the candidates appeared to know what they needed to do. Most candidates who were able to set up the equation in part (c) solved it correctly. Follow through marks were awarded to many candidates for correct working with their substituted values from part (b).

This question was answered reasonably well with many candidates able to write down the two equations and solve them for a and b. Errors such as mistaking the equation given for \(3{a^2} + b = 119\) meant that marks were lost even though the candidates appeared to know what they needed to do. Most candidates who were able to set up the equation in part (c) solved it correctly. Follow through marks were awarded to many candidates for correct working with their substituted values from part (b).

This question was answered reasonably well with many candidates able to write down the two equations and solve them for a and b. Errors such as mistaking the equation given for \(3{a^2} + b = 119\) meant that marks were lost even though the candidates appeared to know what they needed to do. Most candidates who were able to set up the equation in part (c) solved it correctly. Follow through marks were awarded to many candidates for correct working with their substituted values from part (b).

Calculate the distance travelled by the Earth in one day.

Give your answer to part (a) in the form \(a \times {10^k}\) where \(1 \leqslant a \leqslant 10\) and \(k \in \mathbb{Z}\) .

\(2\pi \frac{{150000000}}{{365}}\)     (M1)(A1)(M1)

Notes: Award (M1) for substitution in correct formula for circumference of circle.

Award (A1) for correct substitution.

Award (M1) for dividing their perimeter by \(365\).

Award (M0)(A0)(M1) for \(\frac{{150000000}}{{365}}\) .

\(2580000{\text{ km}}\)     (A1)     (C4)

[4 marks]

\(2.58 \times {10^6}\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for \(2.58\), (A1)(ft) for \({10^6}\) . Follow through from their answer to part (a). The follow through for the index should be dependent not only on the answer to part (a), but also on the value of their mantissa. No (AP) penalty for first (A1) provided their value is to 3 sf or is all their digits from part (a).

[2 marks]

A significant number of candidates simply divided \(150 000 000\) by \(365\) and consequently lost all but one method mark in part (a). Presumably these candidates assumed that the given value was the circumference rather than the radius.

A significant number of candidates simply divided \(150 000 000\) by \(365\) and consequently lost all but one method mark in part (a). Presumably these candidates assumed that the given value was the circumference rather than the radius. Recovery in part (b) did, however, result in many getting both marks here. It was noted on some answers to part (b) that the index power was negative rather than positive suggesting a misunderstanding by candidates of standard form.

Calculate the amount that Albena receives in EUR.

At the airport shop, Albena buys chocolates that cost 34.50 BGN. She uses 20 EUR to pay for the chocolates but receives her change in BGN. The shop’s exchange rate is 1 EUR = 1.90 BGN.

Find how many BGN she receives as change.

\(\frac{{0.97 \times 3550}}{{1.95}}\)     (M1)(M1)(M1)

Note: Award (M1) for \(0.97\) seen, (M1) for \(0.97 \times 3550\), (M1) for division by \(1.95\).

OR

\((3550 - 0.03 \times 3550) \times \frac{1}{{1.95}}\)     (M1)(M1)(M1)

Note: Award (M1) for \(0.03 \times 3550\) seen, (M1) for subtracting \(0.03 \times 3550\) from \(3550\), (M1) for division by \(1.95\).

\( = 1765.90{\text{ (EUR)}}\)     (A1)     (C4)

\(20 \times 1.90 - 34.50\)     (M1)

Note: Award (M1) for subtraction of \(34.50\) from their product of \(20 \times 1.90\).

\( = 3.50\,\,\,{\text{(BGN)}}\)     (A1)     (C2)

Notes: Award at most (M1)(A0) for an answer of \(4\), but only if working seen.

Find the common difference of this arithmetic sequence.

There are 10 rows in the choir and 11 singers in the first row.

Find the total number of singers in the choir.

20 = u₁ + 3d     (A1)

32 = u₁ + 7d     (A1)

Note: Award (A1) for each equation, (A1) for correct answer.

OR

\(d = \frac{{32 - 20}}{4}\)     (A1)(A1)

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

d = 3     (A1)     (C3)

[3 marks]

\(\frac{{10}}{2}(2 \times 11 + 9 \times 3)\) or \(\frac{{10}}{2}(11 + 38)\)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1) for correct substitution, follow through from their answer to part (a).

OR

11 + 14 + ... + 38     (M1)(A1)(ft)

Note: Award (M1) for attempt at the sum of a list, (A1)(ft) for all correct numbers, follow through from their answer to part (a).

= 245     (A1)(ft)     (C3)

[3 marks]

This question was very well answered with most candidates finding the common difference and the total number of singers. Most candidates used the given formulae, rather than making lists. A common mistake was to find the number of singers in the back row, rather than find the total number of singers in the choir.

This question was very well answered with most candidates finding the common difference and the total number of singers. Most candidates used the given formulae, rather than making lists. A common mistake was to find the number of singers in the back row, rather than find the total number of singers in the choir.

Site S has a width of \(20\) m. Write down the area of S.

Site T has the same area as site S, but a different width. Find the width of T.

When the width of the construction site is \(b\) metres, the site has a maximum area.

(i)     Write down the value of \(b\).

(ii)     Write down the maximum area.

The range of \(A(x)\) is \(m \leqslant A(x) \leqslant n\).

Hence write down the value of \(m\) and of \(n\).

\(3600{\text{ (}}{{\text{m}}^2})\)     (A1)(C1)

\(x(200 - x) = 3600\)     (M1)

Note: Award (M1) for setting up an equation, equating to their \(3600\).

\(180{\text{ (m)}}\)     (A1)(ft)     (C2)

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

(i)     \(100{\text{ (m)}}\)     (A1)     (C1)

(ii)     \(10\,000{\text{ (}}{{\text{m}}^2})\)     (A1)(ft)(C1)

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

\(m = 3600\;\;\;\)and\(\;\;\;n = 10\,000\)     (A1)(ft)     (C1)

Notes: Follow through from part (a) and part (c)(ii), but only if their \(m\) is less than their \(n\). Accept the answer \(3600 \leqslant A \leqslant 10\,000\).

Find the exchange rate between GBP and EUR in the form 1 GBP = k EUR, where k is a constant. Give your answer correct to two decimal places.

Isabella changes 400 USD into Euros and is charged 2 % commission.

Calculate how many Euros she receives. Give your answer correct to two decimal places.

k = 2.034 × 0.632     (M1)

= 1.29 (1 GBP = 1.29 EUR)     (A1)     (C2)

Note: Accept 1.29 only

[2 marks]

Financial penalty (FP) applies in part (b).

400 × 0.632     (M1)

= 252.80 EUR     (A1)

2 % of 252.80 = 5.06 EUR     (A1)

(FP)     She receives 247.74 EUR     (A1)

OR

(FP)     0.98 × 252.80 = 247.74 EUR     (A1)(A1)     (C4)

Note: Accept (A1) for 0.98 seen.

[4 marks]

Most students gained full marks on this question. However, some students found the required format of the answer in part (a) confusing.

Most students gained full marks on this question. However, some students found the required format of the answer in part (a) confusing.

Write down the value of r = c × d.

Write down your value of r in the form a × 10^k, where 1 ≤ a < 10 and \(k \in \mathbb{Z}\).

Consider the following statements about c, d and r. Only three of these statements are true.

Circle the true statements.

r = 0.01924     (A1)     (C1)

Note: Accept 0.0192 and 0.019

[1 mark]

r = 1.924 × 10⁻²     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1) for 1.924, (A1) for 10⁻². Accept 1.92 and 1.9. Follow through from their part (a).

[2 marks]

     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for each true statement circled. Do not follow through from part (a). Award (A1)(A1)(A0) if 1 extra term seen. Award (A1)(A0)(A0) if 2 extra terms seen. Award (A0)(A0)(A0) if all terms circled. Accept other indications of the correct statements i.e. highlighted or ticks.

[3 marks]

Most candidates could find the value of r and give it in standard form, although some did not give it to the correct degree of accuracy. Some candidates gave a positive index and others used calculator notation rather than standard form.

Most candidates could find the value of r and give it in standard form, although some did not give it to the correct degree of accuracy. Some candidates gave a positive index and others used calculator notation rather than standard form.

There were a number of candidates who were unable to find the three true statements about set notation.

Write down this radius, in kilometres, in the form \(a \times {10^k}\), where \(1 \leqslant a < 10,{\text{ }}k \in \mathbb{Z}\).

The Earth travels around the Sun in one orbit. It takes one year for the Earth to complete one orbit.

Calculate the distance, in kilometres, the Earth travels around the Sun in one orbit, assuming that the orbit is a circle.

Today is Anna’s 17th birthday.

Calculate the total distance that Anna has travelled around the Sun, since she was born.

\(1.5 \times {10^8}{\text{ (km)}}\)     (A2)     (C2)

Notes: Award (A2) for the correct answer.

     Award (A1)(A0) for 1.5 and an incorrect index.

     Award (A0)(A0) for answers of the form \(15 \times {10^7}\).

[2 marks]

\(2\pi 1.5 \times {10^8}\)     (M1)

\( = 942\,000\,000{\text{ (km)   (942}}\,{\text{477}}\,{\text{796.1}} \ldots {\text{, }}3 \times {10^8}\pi ,{\text{ }}9.42 \times {10^8})\)     (A1)(ft)     (C2)

Notes: Award (M1) for correct substitution into correct formula. Follow through from part (a).

     Do not accept calculator notation \(9.42{\text{E}}8\).

     Do not accept use of \(\frac{{22}}{7}\) or\( 3.14\) for \(\pi\).

[2 marks]

\(17 \times 942\,000\,000\)     (M1)

\( = 1.60 \times {10^{10}}{\text{ (km) }}\left( {{\text{1.60221}} \ldots  \times {{10}^{10}}{\text{, 1.6014}} \times {{10}^{10}},{\text{ 16}}\,{\text{022}}\,{\text{122}}\,{\text{530, }}(5.1 \times {{10}^9})\pi } \right)\)     (A1)(ft)     (C2)

Note: Follow through from part (b).

[2 marks]

Find the value of

(i) p ;

(ii) q .

Write down the value of r .

Find the value of s .

(i) 2p + q = 11 and 4p + q = 17     (M1)

Note: Award (M1) for either two correct equations or a correct equation in one unknown equivalent to 2p = 6 .

p = 3     (A1)

(ii) q = 5     (A1)     (C3)

Notes: If only one value of p and q is correct and no working shown, award (M0)(A1)(A0).

[3 marks]

r = 8     (A1)(ft)     (C1)

Note: Follow through from their answers for p and q irrespective of whether working is seen.

[1 mark]

3 × 2^s + 5 = 197     (M1)

Note: Award (M1) for setting the correct equation.

s = 6     (A1)(ft)     (C2)

Note: Follow through from their values of p and q.

[2 marks]

Candidates both understood how to interpret a mapping diagram correctly and did very well on this question or the question was very poorly answered or not answered at all. Writing down two simultaneous equations in part (a) proved to be elusive to many and this prevented further work on this question. Candidates who were able to find values of p and q (correct or otherwise) invariably made a good attempt at finding the value of s in part (c).

Candidates both understood how to interpret a mapping diagram correctly and did very well on this question or the question was very poorly answered or not answered at all. Writing down two simultaneous equations in part (a) proved to be elusive to many and this prevented further work on this question. Candidates who were able to find values of p and q (correct or otherwise) invariably made a good attempt at finding the value of s in part (c).

Candidates both understood how to interpret a mapping diagram correctly and did very well on this question or the question was very poorly answered or not answered at all. Writing down two simultaneous equations in part (a) proved to be elusive to many and this prevented further work on this question. Candidates who were able to find values of p and q (correct or otherwise) invariably made a good attempt at finding the value of s in part (c).

Write down the common difference of the sequence.

Calculate the sum of the first \(50\) terms of the sequence.

\({u_k}\) is the first term in the sequence which is negative.

Find the value of \(k\).

\(d = - 7\)     (A1)     (C1)

[1 mark]

\({S_{50}} = \frac{{50}}{2}(2(124) + 49( - 7))\)     (M1)

Note: (M1) for correct substitution.

\( = - 2375\)     (A1)(ft)     (C2)

[2 marks]

\(124 - 7(k - 1) < 0\)     (M1)

\(k > 18.7\) or \(18.7\) seen     (A1)(ft)

\(k = 19\)     (A1)(ft)     (C3)

Note: (M1) for correct inequality or equation seen or for list of values seen or for use of trial and error.

[3 marks]

(a) Again, the omission of the negative sign was a too common fault.

This was generally well attempted.

The common misconception was confusion between \(k\) and the value of the \(k\)^th term. Close reading of this part was required from the candidates.

A ladder is standing on horizontal ground and leaning against a vertical wall. The length of the ladder is \(4.5\) metres. The distance between the bottom of the ladder and the base of the wall is \(2.2\) metres.

Use the above information to sketch a labelled diagram showing the ground, the ladder and the wall.

Calculate the distance between the top of the ladder and the base of the wall.

Calculate the obtuse angle made by the ladder with the ground.

(A1)   (C1)

Notes: Award (A1) for drawing an approximately right angled triangle, with correct labelling of the distances \(4.5\,({\text{m}})\) and \(2.2\,({\text{m}})\).

\(\sqrt {\,{{4.5}^2} - {{2.2}^2}} \,\,\,({\text{accept eqivalent }}eg\,\,{d^2} + {2.2^2} = {4.5^2})\)         (M1)

\( = 3.93\,({\text{m}})\,\,\,\left( {\sqrt {\,15.41} \,({\text{m}}),\,\,3.92555...\,({\text{m}})} \right)\)          (A1)    (C2)

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

\(180^\circ  - {\cos ^{ - 1}}\left( {\frac{{2.2}}{{4.5}}} \right)\)        (M1)(M1)

OR

\(180^\circ  - {\tan ^{ - 1}}\left( {\frac{{3.92555...}}{{2.2}}} \right)\)        (M1)(M1)

OR

\(180^\circ  - {\sin ^{ - 1}}\left( {\frac{{3.92555...}}{{4.5}}} \right)\)        (M1)(M1)

Note: Award (M1) for a correct substitution in the correct trigonometric ratio.
Award (M1) for subtraction from \(180^\circ \) (this may be implied if the sum of their inverse of the trigonometric ratio and their final answer equals \(180\)).

\( = 119^\circ \,\,\,(119.267...^\circ )\)        (A1)(ft)    (C3)

Note: Follow through from their part (b) if cosine is not used. Accept \(119.239...\) or \(119.151...\) from use of \(3\) sf values.

Question 3: Right angle trigonometry.
Candidates sketched the ladder leaning against the wall and recognized that Pythagoras’ theorem was needed to find the distance between the top of the ladder and the base of the wall (but not always correctly). Although it was a right triangle a number of the candidates used the law of sines (instead of Pythagoras’ theorem) and law of cosines (instead of a trigonometry ratio). Many candidates failed to find the obtuse angle made by the ladder with the ground even though the word obtuse was in bold type in the question.

Question 3: Right angle trigonometry.
Candidates sketched the ladder leaning against the wall and recognized that Pythagoras’ theorem was needed to find the distance between the top of the ladder and the base of the wall (but not always correctly). Although it was a right triangle a number of the candidates used the law of sines (instead of Pythagoras’ theorem) and law of cosines (instead of a trigonometry ratio). Many candidates failed to find the obtuse angle made by the ladder with the ground even though the word obtuse was in bold type in the question.

Question 3: Right angle trigonometry.
Candidates sketched the ladder leaning against the wall and recognized that Pythagoras’ theorem was needed to find the distance between the top of the ladder and the base of the wall (but not always correctly). Although it was a right triangle a number of the candidates used the law of sines (instead of Pythagoras’ theorem) and law of cosines (instead of a trigonometry ratio). Many candidates failed to find the obtuse angle made by the ladder with the ground even though the word obtuse was in bold type in the question.

Write down, and simplify, an expression for the car’s value when Gabriella purchased it.

Find the value of \(k\).

Find the value of \(n\).

\(12870 - k{(1.1)^0}\)    (M1)

Note:     Award (M1) for correct substitution into \(V(t)\).

\( = 12870 - k\)    (A1)     (C2)

Note:     Accept \(12870 - 3080\) OR 9790 for a final answer.

[2 marks]

\(9143.20 = 12870 - k{(1.1)^2}\)    (M1)

Note:     Award (M1) for correct substitution into \(V(t)\).

\((k = ){\text{ }}3080\)    (A1)     (C2)

[2 marks]

\(12870 - 3080{(1.1)^n} = 0\)    (M1)

Note:     Award (M1) for correct substitution into \(V(t)\).

OR

     (M1)

Note:     Award (M1) for a correctly shaped curve with some indication of scale on the vertical axis.

\((n = ){\text{ }}15.0{\text{ }}(15.0033 \ldots )\)    (A1)(ft)     (C2)

Note:     Follow through from part (b).

[2 marks]

4

\(\frac{1}{3}\)

\(\pi \)

\(0.38\)

\(\sqrt 5 \)

\(-0.25\)

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note: Award (A1) for each number correctly placed.

     Award (A0) for any entry in more than one region.

[1 mark]

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note: Award (A1) for each number correctly placed.

     Award (A0) for any entry in more than one region.

[1 mark]

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note: Award (A1) for each number correctly placed.

     Award (A0) for any entry in more than one region.

[1 mark]

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note: Award (A1) for each number correctly placed.

     Award (A0) for any entry in more than one region.

[1 mark]

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note: Award (A1) for each number correctly placed.

     Award (A0) for any entry in more than one region.

[1 mark]

     (A1)(A1)(A1)(A1)(A1)(A1)     (C6)

Note: Award (A1) for each number correctly placed.

     Award (A0) for any entry in more than one region.

[1 mark]

Write down the radius of the satellite’s orbit.

Calculate the distance travelled by the satellite in one orbit of the Earth. Give your answer correct to the nearest km.

Write down your answer to (b) in the form \(a \times {10^k}\) , where \(1 \leqslant a < 10{\text{, }}k \in \mathbb{Z}\) .

\(6900\) km     (A1)     (C1)

[1 mark]

\(2\pi (6900)\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into circumference formula, (A1)(ft) for correct substitution. Follow through from part (a).

\( = 43354\)     (A1)(ft)     (C3)

Notes: Follow through from part (a). The final (A1) is awarded for rounding their answer correct to the nearest km. Award (A2) for \(43 400\) shown with no working.

[3 marks]

\(4.3354 \times {10^4}\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for \(4.3354\), (A1)(ft) for \( \times {10^4}\) . Follow through from part (b). Accept \(4.34 \times {10^4}\) .

[2 marks]

Candidates appeared to be confused by the context in this question. They had difficulty identifying the radius and many used the formula for the area of a circle, rather than the circumference.

Candidates appeared to be confused by the context in this question. They had difficulty identifying the radius and many used the formula for the area of a circle, rather than the circumference. A large number of candidates misread the final sentence in part b and did not write their answer to the nearest kilometre.

Candidates appeared to be confused by the context in this question. They had difficulty identifying the radius and many used the formula for the area of a circle, rather than the circumference.

List the elements of \(A\).

List the elements of \(B\).

Complete the Venn diagram with all the elements of \(U\).

\(2, 4, 6, 8, 10\)     (A1)     (C1)

Note: Do not penalize the use of \(\left\{ {{\text{   }}} \right\}\).

[1 mark]

\(3, 6, 9\)     (A1)     (C1)

Note: Do not penalize the use of \(\left\{ {{\text{   }}} \right\}\).

     Follow through from part (a) only if their \({\text{U}}\) is listed.

[1 mark]

   
 (A1)(ft)(A1)(ft)(A1)(ft)(A1)(ft)     (C4)

Notes: Award (A1)(ft) for the correct placement of \(6\).

     Award (A1)(ft) for the correct placement of \(8\) and \(9\) and the empty region.

     Award (A1)(ft) for the correct placement of \(2\), \(4\), \(3\), \(7\), and \(10\).

     Award (A1)(ft) for the correct placement of \(1\) and \(5\).

     If an element is in more than one region, award (A0) for that element.

     Follow through from their answers to parts (a) and (b).

[4 marks]

This question was done well by most candidates. The most frequent error was to omit the placement of 1 and 5 or to include 0 in the set of even integers.

This question was done well by most candidates. The most frequent error was to omit the placement of 1 and 5 or to include 0 in the set of even integers.

This question was done well by most candidates. The most frequent error was to omit the placement of 1 and 5 or to include 0 in the set of even integers.

List the elements of the following set.

A

List the elements of the following set.

\(A \cap B'\)

Write down one element of \((A \cup B)'\).

One of the statements in the table below is false. Indicate with an X which statement is false. Give a reason for your answer.

6, 9, 12     (A1)     (C1)

[1 mark]

9     (A1)(ft)     (C1)

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

[1 mark]

any element from {5, 7, 8, 10, 11}     (A1)(A1)(ft)     (C2)

Note: Award (A1)(ft) for finding \((A \cup B)\), follow through from their A.

Award full marks if all correct elements of \((A \cup B)'\) are listed.

[2 marks]

\(15 \notin U\)     (R1)(A1)     (C2)

Notes: Accept correct reason in words.

If the reason is incorrect, both marks are lost.

Do not award (R0)(A1).

[2 marks]

The question was not well answered by the majority of the candidates. Many did not identify the universal set correctly and so took 3 to be a member of this set. This affected their answers in a)(i) and a)(ii).

The question was not well answered by the majority of the candidates. Many did not identify the universal set correctly and so took 3 to be a member of this set. This affected their answers in a)(i) and a)(ii).

Not many students answered (b) correctly. Some listed all correct elements of the given set instead of just one, which shows that they did not read the question carefully.

Although many candidates could indicate which statement in the table in c) was false, often they were unable either to identify or articulate a correct reason for it.

Find the number of statues that are non-defective.

The manufacturer sells each non-defective statue for \(5.25\) British pounds (GBP) to an American company. The exchange rate from GBP to US dollars (USD) is \(1{\text{ GBP}} = 1.6407{\text{ USD}}\).

Calculate the amount in USD paid by the American company for all the non-defective statues. Give your answer correct to two decimal places.

The American company sells one of the statues to an Australian customer for \(12{\text{ USD}}\). The exchange rate from Australian dollars (AUD) to USD is \(1{\text{ AUD}} = 0.8739{\text{ USD}}\) .

Calculate the amount that the Australian customer pays, in AUD, for this statue. Give your answer correct to two decimal places.

\(0.88 \times 16000\)     OR     \(16000 - 0.12 \times 16000\)     (M1)

\(14080\)     (A1)     (C2)

[2 marks]

\(1.6407 \times 5.25 \times 14080\)     (M1)

\(121 280.54{\text{ USD}}\)     (A1)(ft)     (C2)

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

[2 marks]

\(12 \times \frac{1}{{0.8739}}\)     (M1)

\(13.73{\text{ AUD}}\)     (A1)     (C2)

Note: If division used in part (b) and multiplication used in part (c), award (M0)(A0) for part (b) and (M1)(A1)(ft) for part (c).

[2 marks]

This question was generally well answered with much correct working seen in parts (a) and (b). The most popular incorrect answer in part (a) was \(1920\) – candidates simply stating the number of defective items rather than the number of non-defective items. Unfortunately in part (c) many candidates multiplied by \(0.8739\) rather than divided and \(10.49\) proved a popular, but erroneous, answer.

This question was generally well answered with much correct working seen in parts (a) and (b). The most popular incorrect answer in part (a) was \(1920\) – candidates simply stating the number of defective items rather than the number of non-defective items. Unfortunately in part (c) many candidates multiplied by \(0.8739\) rather than divided and \(10.49\) proved a popular, but erroneous, answer.

This question was generally well answered with much correct working seen in parts (a) and (b). The most popular incorrect answer in part (a) was \(1920\) – candidates simply stating the number of defective items rather than the number of non-defective items. Unfortunately in part (c) many candidates multiplied by \(0.8739\) rather than divided and \(10.49\) proved a popular, but erroneous, answer.

Calculate the size of angle ACB.

Nadia makes an accurate drawing of triangle ABC. She measures angle BAC and finds it to be 29°.

Calculate the percentage error in Nadia’s measurement of angle BAC.

\(\frac{{\sin {\text{A}}{\operatorname{\hat B}}{\text{C}}}}{{13.4}} = \frac{{\sin 30^\circ }}{{6.7}}\)     (M1)(A1)

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

\({\text{A}}{\operatorname{\hat B}}{\text{C}}\) = 90°     (A1)

\({\text{A}}{\operatorname{\hat C}}{\text{B}}\) = 60°     (A1)(ft)     (C4)

Note: Radians give no solution, award maximum (M1)(A1)(A0).

[4 marks]

\(\frac{{29 - 30}}{{30}} \times 100\)     (M1)

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

% error = −33.3 %     (A1)     (C2)

Notes: Percentage symbol not required. Accept positive answer.

[2 marks]

Use of cosine rule was common. The assumption of a right angle in the given diagram was minimal.

The incorrect denominator was often seen in the error formula.

Use the above information to write down an equation in \(x\) and \(y\).

Use the information about the cost of tickets to write down a second equation in \(x\) and \(y\).

Find the value of \(x\) and the value of \(y\).

Find the percentage error.

\(x + y = 154\)    (A1)     (C1)

[1 mark]

\(320x + 85y = 14\,970\)    (A1)     (C1)

[1 mark]

\(x = 8,{\text{ }}y = 146\)    (A1)(ft)(A1)(ft)     (C2)

Note:     Follow through from parts (a) and (b) irrespective of working seen, but only if both values are positive integers.

Award (M1)(A0) for a reasonable attempt to solve simultaneous equations algebraically, leading to at least one incorrect or missing value.

[2 marks]

\(\left| {\frac{{14270 - 14970}}{{14970}}} \right| \times {\text{ }}100\)    (M1)

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

\( = 4.68(\% ){\text{ }}(4.67601 \ldots ){\text{ }}\)    (A1) (C2)

[2 marks]

Factorise \(x^2 - 6x + 8\).

Hence, or otherwise, solve the equation \(x^2 - 6x + 8 = 0\).

Let \(g(x) = x + 3\).

Write down the solutions to the equation \(f (x) = g(x)\).

\( (x - 2)(x - 4)\)     (A1)(A1)     (C2)

[2 marks]

x = 2, x = 4     (A1)(ft)(A1)(ft)     (C2)

[2 marks]

x = 0.807, x = 6.19     (A1)(A1)     (C2)

Note: Award maximum of (A0)(A1) if coordinate pairs given.

OR

(M1) for an attempt to solve \(x^2 - 7x + 5 = 0\) via formula with correct values substituted.     (M1)

\(x = \frac{{7 \pm \sqrt {29} }}{2}\)     (A1)     (C2)

[2 marks]

This was generally well answered, but a number seemed not to know the term “factorise”.

This was generally well answered.

This part proved problematic for many candidates. It was expected that the GDC was used, though many attempted an algebraic solution.

Calculate the exact value of \(h\) when \(\ell  = 0.03625\) and \(d = 0.05\) .

Write down the answer to part (a) correct to three decimal places.

Write down the answer to part (a) correct to three significant figures.

Write down the answer to part (a) in the form \(a \times {10^k}\) , where \(1 \leqslant a < 10{\text{, }}k \in \mathbb{Z}\).

\(h = \sqrt {{{0.03625}^2} - \frac{{{{0.05}^2}}}{4}} \)     (M1)
\( = 0.02625\)     (A1)     (C2)

Note: Award (A1) only for \(0.0263\) seen without working

[2 marks]

\(0.026\)     (A1)(ft)     (C1)

[1 mark]

\(0.0263\)     (A1)(ft)     (C1)

[1 mark]

\(2.625 \times {10^{ - 2}}\)

for \(2.625\) (ft) from unrounded (a) only     (A1)(ft)

for \( \times {10^{ - 2}}\)     (A1)(ft)     (C2)

[2 marks]

This was answered correctly by the majority of the candidates however some candidates entered the numbers incorrectly and arrived at the wrong answer.

Correction to decimal places was less well attempted than to significant figures.

Correction to decimal places was less well attempted than to significant figures.

Most made a successful attempt to change their answer to part (a) into scientific notation. Some were penalised for not using their answer to (a).

Enrico has 475 USD.

How many euros is this worth?

Enrico has 475 USD.

Enrico goes to a bank to exchange his dollars. The bank charges 3 % commission.

How many euros does Enrico receive?

Find the exchange rate from euros to yen.

Note: Financial penalty (FP) applies in this part

(FP)     \(475 \times 0.6337 = 301.01\)     (M1)(A1)

Note: Award (M1) for multiplication by 0.6337.

[2 marks]

Note: Financial penalty (FP) applies in this part\(\frac{3}{{100}} \times 301.01 = 9.03\)     (M1)

Note: Award (M1) for multiplication by 3/100.

(FP)     \(301.01 - 9.03 = 291.98\)     (A1)(ft)

OR

\(0.97 \times 301.01\)     (M1)

(FP)     \(= 291.98\)     (A1)(ft)     (C4)

[2 marks]

Note: Financial penalty (FP) applies in this part

(FP)     \(\frac{{{\rm{99}}{\rm{.7469}}}}{{{\rm{0}}{\rm{.6337}}}}{\rm{ = 157}}{\rm{.40}}\)     (M1)(A1)     (C2)

Notes: Award (M1) for dividing by 0.6337.

[2 marks]

This question was well answered by many candidates, particularly part (a), however a significant number lost the financial penalty mark for not giving an answer correct to two decimal places, as stated in the question.

This question was well answered by many candidates, particularly part (a), however a significant number lost the financial penalty mark for not giving an answer correct to two decimal places, as stated in the question.

This question was well answered by many candidates, particularly part (a), however a significant number lost the financial penalty mark for not giving an answer correct to two decimal places, as stated in the question.

Find the interest Astrid has earned during the five years of her investment. Give your answer correct to two decimal places.

\(I = 1200{\left( {1 + \frac{{7.2}}{{600}}} \right)^{5 \times 12}} - 1200\)     (M1)(A1)
I = 518.15 Euros     (A1)     (C3)

Notes: Award (M1) for substitution in the compound interest formula, (A1) for correct substitutions, (A1) for correct answer.

If final amount found is 1718.15 and working shown award (M1) (A1)(A0).

[3 marks]

Part (a) of this question was incorrectly answered by many candidates not noticing that the rate of interest was compounded monthly rather than annually. A number of candidates gave the final amount as the answer, rather than the interest. 

Calculate the amount of commission on the exchange in SGD.

Calculate the number of American dollars (USD) Neung takes home. Give your answer correct to 2 decimal places.

At the airport in Vietnam, Neung changes 150 USD into Vietnamese dong (VND) to pay for her transport home.

The exchange rate between American dollars (USD) and Vietnamese dong (VND) is

1 USD = 19 495 VND.

There is no commission.

Calculate the number of Vietnamese dong that Neung receives. Give your answer correct to the nearest thousand dong.

5000 × 0.024     (M1)

Note: Award (M1) for multiplication by 0.024.

=120     (A1)     (C2)

\(4880 \times \frac{1}{{1.2945}}\)     (M1)

Note: Award (M1) for multiplication by \(\frac{1}{{1.2945}}\).

= 3769.80     (A1)(ft)     (C2)

Note: Correct answer to 2 dp only. Follow through from their part (a).

\(150 \times 19495\)     (M1)

Note: Award (M1) for \( \times 19495\).

\(= 2924000\)     (A1)     (C2)

Notes: Correct answer to nearest 1000 only. Do not penalize incorrect accuracy in (c) if this has already been penalized in part (b).

Marks were awarded in part (a) for multiplication by 0.024 in part (b) for division by 1.2945 and in part (c) for multiplication by 19495. Candidates did not follow specified levels of accuracy. Candidates were able to answer later parts of the question even if they did not answer the first parts correctly.

Marks were awarded in part (a) for multiplication by 0.024 in part (b) for division by 1.2945 and in part (c) for multiplication by 19495. Candidates did not follow specified levels of accuracy. Candidates were able to answer later parts of the question even if they did not answer the first parts correctly.

Marks were awarded in part (a) for multiplication by 0.024 in part (b) for division by 1.2945 and in part (c) for multiplication by 19495. Candidates did not follow specified levels of accuracy. Candidates were able to answer later parts of the question even if they did not answer the first parts correctly.

Calculate the exact value of the ninth term of the sequence.

Calculate the least number of terms required for the sum of the sequence to be greater than 682.6

\({u_9} = 512{\left( {\frac{1}{4}} \right)^8}\)     (M1)(A1)

Notes: Award (M1) for substituted geometric sequence formula, (A1) for correct substitution.

OR

If a list is used, award (M1) for a list of 9 terms, (A1) for all 9 terms correct.     (M1)(A1)

\( = \frac{1}{{128}}\) (\(0.0078125\))     (A1)     (C3)

Note: Award (A1) for exact answer only.

[3 marks]

\(\frac{{512\left( {1 - {{\left( {\frac{1}{4}} \right)}^n}} \right)}}{{1 - \left( {\frac{1}{4}} \right)}} > 682.6\)     (M1)(A1)(ft)

Notes: Award (M1) for setting substituted geometric sum formula \( > 682.6\) (A1)(ft) for correct substitution into geometric sum formula. Follow through from their common ratio.

OR

If list is used, award (M1) for S(6) and S(7) seen, values don’t have to be correct.

(A1) for correct S(6) and S(7). (S(6) \( = 682.5\) and S(7) \( = 682.625\)).     (M1)(A1)

\(n = 7\)     (A1)(ft)     (C3)

Notes: Follow through from their common ratio. Do not award the final (A1)(ft) if \(n\) is less than \(5\) or if \(n\) is not an integer.

[3 marks]

The upper quartile of candidates scored well on this question with the vast majority scoring more than 4 marks. However, the lower quartile did very badly with the majority scoring less than 2 marks. A fundamental error in part (a) resulted in many less able candidates using a common ratio of \(4\) instead of \({\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}\). Where lists were used in either part of the question, they were either invariably incomplete or contained numerical errors. Indeed, using lists seem to be as problematic in this question as they were in Q6 on arithmetic sequences. Correctly quoted and substituted formula in a correct inequality (= sign was allowed) did earn many candidates two marks here. The required answer of \(7\) however did not always follow.

The upper quartile of candidates scored well on this question with the vast majority scoring more than 4 marks. However, the lower quartile did very badly with the majority scoring less than 2 marks. A fundamental error in part (a) resulted in many less able candidates using a common ratio of \(4\) instead of \({\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}\). Where lists were used in either part of the question, they were either invariably incomplete or contained numerical errors. Indeed, using lists seem to be as problematic in this question as they were in Q6 on arithmetic sequences. Correctly quoted and substituted formula in a correct inequality (= sign was allowed) did earn many candidates two marks here. The required answer of \(7\) however did not always follow.

Calculate the number of leaves that the tree loses on the 21st October.

Find the total number of leaves that the tree loses in the 31 days of the month of October.

\(u_{21} = 24 + (21 - 1)(16)\)     (M1)(A1)

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

\(u_{21} = 344\)     (A1)     (C3)

[3 marks]

\({S_{31}} = \frac{{31}}{2}\left[ {2(24) + (31 - 1)(16)} \right]\)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitutions. (ft) from their value for d.

\(S_{31} = 8184\)     (A1)(ft)     (C3)

[3 marks]

This was generally well attempted, often by listing.

The common misconception was the use of the “term by term” formula. Listing the values in this part was usually not a successful strategy. The incorrect substitution of the answer to (a) also led to errors.

Calculate the exact value of the volume of the cuboid, in cm³.

Write your answer to part (a) correct to

(i) one decimal place;

(ii) three significant figures.

Write your answer to part (b)(ii) in the form \(a \times 10^k\), where \(1 \leqslant a < 10 , k \in \mathbb{Z}\).

\({\text{V}} = 8.7 \times 5.6 \times 3.4\)     (M1)

Note: Award (M1) for multiplication of the 3 given values.

\(=165.648\)     (A1)     (C2)

(i) 165.6     (A1)(ft)

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

(ii) 166     (A1)(ft)     (C2)

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

\(1.66 \times 10^2\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 1.66, (A1)(ft) for \(10^2\). Follow through from their answer to part (b)(ii) only. The follow through for the index should be dependent on the value of the mantissa in part (c) and their answer to part (b)(ii).

Find the value of the \({96^{{\text{th}}}}\) term of the sequence.

The first term of a geometric sequence is \(6\). The \({6^{{\text{th}}}}\) term of the geometric sequence is equal to the \({17^{{\text{th}}}}\) term of the arithmetic sequence given above.

Write down an equation using this information.

The first term of a geometric sequence is \(6\). The \\({6^{{\text{th}}}}\) term of the geometric sequence is equal to the \({17^{{\text{th}}}}\) term of the arithmetic sequence given above.

Calculate the common ratio of the geometric sequence.

\({u_{96}} = {u_1} + 95d\)     (M1)

\( = 0 + 95 \times 12\)

\( = 1140\)     (A1)     (C2)

[2 marks]

\(6{r^5} = 16d\)     (A1)

\(6{r^5} = 16 \times 12\) (\(192\))     (A1)     (C2)

Note: (A1) only, if both terms seen without an equation.

[2 marks]

\({r^5} = 32\)     (A1)(ft)

Note: (ft) from their (b).

\(r = 2\)     (A1)(ft)     (C2)

[2 marks]

Part (a) was done well. There was some confusion in answering part (b) with many candidates unsure what they needed to write down. Often the two terms were seen somewhere in the working without the equation being written down in the answer box, or the equation was seen in the working for part (c). Part (c) was answered well, often with follow-through marks being awarded from an incorrect part (b) provided the working was seen.

Part (a) was done well. There was some confusion in answering part (b) with many candidates unsure what they needed to write down. Often the two terms were seen somewhere in the working without the equation being written down in the answer box, or the equation was seen in the working for part (c). Part (c) was answered well, often with follow-through marks being awarded from an incorrect part (b) provided the working was seen.

Part (a) was done well. There was some confusion in answering part (b) with many candidates unsure what they needed to write down. Often the two terms were seen somewhere in the working without the equation being written down in the answer box, or the equation was seen in the working for part (c). Part (c) was answered well, often with follow-through marks being awarded from an incorrect part (b) provided the working was seen.

Find the number of competitors who play in round 6 of the tournament.

Find the total number of matches played in the tournament.

\(512{\left( {\frac{1}{2}} \right)^5}\)     (M1)(A1)

Note: Award (M1) for substituted geometric progression formula, (A1) for correct substitution.

     If a list is used, award (M1) for a list of at least six terms, beginning with \(512\) and (A1) for first six terms correct.

\(16\)     (A1)     (C3)

[3 marks]

\({S_9} = 256\left( {\frac{{1 - {{\left( {\frac{1}{2}} \right)}^9}}}{{1 - \frac{1}{2}}}} \right)\)   OR   \(\frac{{({2^9} - 1)}}{{2 - 1}}\)     (M1)(A1)

Note: Award (M1) for substituted sum of a GP formula, (A1) for correct substitution.

     If a list is used, award (A1) for at least 9 correct terms, including \(1\), and (M1) for their 9 terms, including \(1\), added together.

\(511\)     (A1)     (C3)

[3 marks]

The first part of this question was answered quite well, especially by candidates who used a list. Part (b) was poorly answered. Common errors in part (b) were to find the number of rounds rather than the total number of matches played or to take the first term as 512 rather than 256.

The first part of this question was answered quite well, especially by candidates who used a list. Part (b) was poorly answered. Common errors in part (b) were to find the number of rounds rather than the total number of matches played or to take the first term as 512 rather than 256.

Only one of the following four sequences is arithmetic and only one of them is geometric.

     \({a_n} = 1,{\text{ }}2,{\text{ }}3,{\text{ }}5,{\text{ }} \ldots \)

     \({b_n} = 1,{\text{ }}\frac{3}{2},{\text{ }}\frac{9}{4},{\text{ }}\frac{{27}}{8},{\text{ }} \ldots \)

     \({c_n} = 1,{\text{ }}\frac{1}{2},{\text{ }}\frac{1}{3},{\text{ }}\frac{1}{4},{\text{ }} \ldots \)

     \({d_n} = 1,{\text{ }}0.95,{\text{ }}0.90,{\text{ }}0.85,{\text{ }} \ldots \)

State which sequence is

(i)     arithmetic;

(ii)     geometric.

For another geometric sequence \({e_n} =  - 6,{\text{ }} - 3,{\text{ }} - \frac{3}{2},{\text{ }} - \frac{3}{4},{\text{ }} \ldots \)

write down the common ratio;

For another geometric sequence \({e_n} =  - 6,{\text{ }} - 3,{\text{ }} - \frac{3}{2},{\text{ }} - \frac{3}{4},{\text{ }} \ldots \)

find the exact value of the tenth term. Give your answer as a fraction.

(i)     \({d_n}\;\;\;\;\;\)OR\(\;\;\;1,{\text{ }}0.95,{\text{ }}0.90,{\text{ }}0.85,{\text{ }} \ldots \)     (A1)     (C1)

(ii)     \({b_n}\;\;\;\)OR\(\;\;\;1,{\text{ }}\frac{3}{2},{\text{ }}\frac{9}{4},{\text{ }}\frac{{27}}{8},{\text{ }} \ldots \)     (A1)     (C1)

\(\frac{1}{2}\;\;\;\)OR\(\;\;\;0.5\)     (A1)     (C1)

Note: Accept ‘divide by 2’ for (A1).

\( - 6{\left( {\frac{1}{2}} \right)^{10 - 1}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution in the GP \({n^{{\text{th}}}}\) term formula, (A1)(ft) for their correct substitution.

Follow through from their common ratio in part (b)(i).

OR

\(\left( { - 6,{\text{ }} - 3,{\text{ }} - \frac{3}{2},{\text{ }} - \frac{3}{4},} \right) - \frac{3}{8},{\text{ }} - \frac{3}{{16}},{\text{ }} - \frac{3}{{32}},{\text{ }} - \frac{3}{{64}},{\text{ }} - \frac{3}{{128}}\)     (M1)(A1)(ft)

Notes: Award (M1) for terms 5 and 6 correct (using their ratio).

Award (A1)(ft) for terms 7, 8 and 9 correct (using their ratio).

\( - \frac{3}{{256}}\;\;\;\left( { - \frac{6}{{512}}} \right)\)     (A1)(ft)     (C3)

Find the mean number of goals scored per match.

Find the median number of goals scored per match.

A local newspaper claims that the mean number of goals scored per match is two. Calculate the percentage error in the local newspaper’s claim.

\(\frac{{0 \times 16 + 1 \times 22 + 2 \times 19 \ldots }}{{80}}\)     (M1)

Note: Award (M1) for substituting correct values into mean formula.

1.75     (A1)     (C2)

[2 marks]

An attempt to enumerate the number of goals scored.     (M1)

\(2\)     (A1)     (C2)

[2 marks]

\(\frac{{2 - 1.75}}{{1.75}} \times 100\)     (M1)
\(14.3 \% \)     (A1)(ft)     (C2)

Notes: Award (M1) for correctly substituted \(\% \) error formula. \(\% \) sign not required. Follow through from their answer to part (a). If \(100\) is missing and answer incorrect award (M0)(A0). If \(100\) is missing and answer incorrectly rounded award (M1)(A1)(ft)(AP).

[2 marks]

In parts (a) and (b), \(2.5\) was a common incorrect error for both parts as some candidates were confused as to the concept of both the mean and the median from tabular data and simply looked at the mean and median of the Number of goals, ignoring the weighting of the number of matches. Candidates faired a little better with part (c) and many correct answers (many as follow through answers) were seen in this part of the question.

In parts (a) and (b), \(2.5\) was a common incorrect error for both parts as some candidates were confused as to the concept of both the mean and the median from tabular data and simply looked at the mean and median of the Number of goals, ignoring the weighting of the number of matches. Candidates faired a little better with part (c) and many correct answers (many as follow through answers) were seen in this part of the question.

In parts (a) and (b), \(2.5\) was a common incorrect error for both parts as some candidates were confused as to the concept of both the mean and the median from tabular data and simply looked at the mean and median of the Number of goals, ignoring the weighting of the number of matches. Candidates faired a little better with part (c) and many correct answers (many as follow through answers) were seen in this part of the question.

Calculate \(\frac{{77.2 \times {3^3}}}{{3.60 \times {2^2}}}\).

Express your answer to part (a) in the form \(a \times 10^k\), where \(1 \leqslant a < 10\) and \(k \in {\mathbb{Z}}\).

Juan estimates the length of a carpet to be 12 metres and the width to be 8 metres. He then estimates the area of the carpet.

(i) Write down his estimated area of the carpet.

When the carpet is accurately measured it is found to have an area of 90 square metres.

(ii) Calculate the percentage error made by Juan.

\(144.75\left( { = \frac{{579}}{4}} \right)\)     (A1)

accept 145     (C1)

[1 mark]

\(1.4475 \times 10^2\)     (A1)(ft)(A1)(ft)

accept \(1.45 \times 10^2\)     (C2)

[2 marks]

Unit penalty (UP) is applicable in question part (c)(i) only.

(UP) (i) Area = 96 m²     (A1)

(ii) \(\% {\text{ error}} = \frac{{(96 - 90)}}{{90}} \times 100\)     (M1)

\( = \frac{{6 \times 100}}{{90}}\)

\(\frac{{20}}{3}\% \) or 6.67 %     (A1)(ft)     (C3)

[3 marks]

(a) This was answered correctly by the majority of the candidates however some candidates entered the numbers without using brackets and arrived at the wrong answer.

(b) Most made a successful attempt to change their answer to part (a) into scientific notation.

(c) (i) Many candidates managed to find the answer but then lost the mark by not adding the units.

(ii) Several candidates are still having a problem finding the percentage error. The formula is given in their information booklet and they should have had practice using all the formulae that they are given. There are some schools that are still using the incorrect formula sheet for percentage error.

Write down the area of the rectangle in the form a × 10^k, where 1 ≤ a < 10, k ∈ \(\mathbb{Z}\).

Helen’s estimate of the area of the rectangle is \(1\,600\,000{\text{ }}{{\text{m}}^2}\).

Find the percentage error in Helen’s estimate.

UP applies in part (a).

\(9.5 \times 10^2 \times 1.6 \times 10^3\)     (M1)
(UP)     \( = 1.52 \times {10^6}{\text{ }}{{\text{m}}^2}\)     (A1)(A1)     (C3)

Notes: Award (M1) for multiplication of the two numbers.

Award (A1) for \(1.52\), (A1) for \(10^6\).

[3 marks]

\(\frac{{1600000 - 1520000}}{{1520000}} \times 100\)     (M1)(A1)(ft)

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

= 5.26 % (percent sign not required).     (A1)(ft)     (C3)

Note: Accept positive or negative answer.

[3 marks]

This was generally well done except for missing or incorrect units. Most candidates could give their answer in standard form and find the percentage error.

This was generally well done except for missing or incorrect units. Most candidates could give their answer in standard form and find the percentage error.

Find the value of z when x = 12.5, a = 0.572 and b = 0.447. Write down your full calculator display.

Write down your answer to part (a)

(i) correct to the nearest 1000 ;

(ii) correct to three significant figures.

Write your answer to part (b)(ii) in the form a × 10^k, where 1 ≤ a < 10, \(k \in \mathbb{Z}\).

\(z = \frac{{17{{(12.5)}^2}}}{{(0.572 - 0.447)}}\)     (M1)

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

= 21250     (A1)     (C2)

[2 marks]

(i) 21000     (A1)(ft)

(ii) 21300     (A1)(ft)     (C2)

Note: Follow through from part (a).

[2 marks]

\(2.13 \times 10^4\)     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 2.13, (A1)(ft) for \(\times 10^4\). Follow through from part (b)(ii).

[ 2 marks]

Many candidates calculated \(\frac{{17{x^2}}}{a} - b\) instead of \(\frac{{17{x^2}}}{{a - b}}\) on their calculators; however they were able to get follow through points. It is important that candidates learn how to correctly input expressions into their calculators.

Many candidates calculated \(\frac{{17{x^2}}}{a} - b\) instead of \(\frac{{17{x^2}}}{{a - b}}\) on their calculators; however they were able to get follow through points. It is important that candidates learn how to correctly input expressions into their calculators.

Although the question explicitly stated in bold to use the answer to part(b)(ii) many candidates used their answer to part (a) for part (c). The general notes about rounding in the mark scheme are over-ruled if the question has explicit directions such as in this question.

Write down the boiling point of water at sea level.

Use the model to calculate the boiling point of water at a height of 1.37 km above sea level.

Water boils at the top of Mt. Everest at 70 °C.

Use the model to calculate the height above sea level of Mt. Everest.

100 °C     (A1)     (C1)

[1 mark]

\(T = -0.0034 \times 1370 + 100\)     (A1)(M1)

Note: Award (A1) for 1370 seen, (M1) for substitution of their h into the equation.

95.3 °C (95.342)     (A1)     (C3)

Notes: If their h is incorrect award at most (A0)(M1)(A0). If their h = 1.37 award at most (A0)(M1)(A1)(ft).

[3 marks]

\(70 = -0.0034h + 100\)     (M1)

Note: Award (M1) for correctly substituted equation.

h = 8820 m (8823.52...)     (A1)     (C2)

Note: The answer is 8820 m (or 8.82 km.) units are required.

[2 marks]

The majority of the candidates showed they were able to substitute values into the model. The most common mistake was to neglect converting 1.37 km into metres. Some candidates did not appreciate the practical considerations of this question; Mount Everest will never be less than one metre high. It is important to remind students to check their answers in terms of the context of the information given.

The majority of the candidates showed they were able to substitute values into the model. The most common mistake was to neglect converting 1.37 km into metres. Some candidates did not appreciate the practical considerations of this question; Mount Everest will never be less than one metre high. It is important to remind students to check their answers in terms of the context of the information given.

The majority of the candidates showed they were able to substitute values into the model. The most common mistake was to neglect converting 1.37 km into metres. Some candidates did not appreciate the practical considerations of this question; Mount Everest will never be less than one metre high. It is important to remind students to check their answers in terms of the context of the information given.

Calculate the exact value of the mean length.

Write your answer to part (a) in the form a × 10^k where 1 ≤ a < 10 and \(k \in \mathbb{Z}\).

Marian calculates the mean length and finds it to be 105 cm.

Calculate the percentage error made by Marian.

\(\left( {\frac{{104.5 + 105.1 + ...}}{6}} \right)\)     (M1)

Note: Award (M1) for use of mean formula.

= 104.9 (cm)     (A1)     (C2)

[2 marks]

1.049 × 10²     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for 1.049, (A1)(ft) for 10². Follow through from their part (a).

[2 marks]

\(\frac{{105 - 104.9}}{{104.9}} \times 100\)  (%)     (M1)

Notes: Award (M1) for their correctly substituted % error formula.

% error = 0.0953  (%)     (0.0953288...)     (A1)(ft)     (C2)

Notes: A 2sf answer of 0.095 following \(\frac{{105 - 104.9}}{{105}} \times 100\) working is awarded no marks. Follow through from their part (a), provided it is not 105. Do not accept a negative answer. % sign not required.

[2 marks]

Another well answered question with candidates showing a good understanding of standard form and many correct answers were seen in parts (a) and (b). Whilst the formula is given for percentage error, there were still a minority of candidates who divided by 105 rather than the required value of 104.9.

Another well answered question with candidates showing a good understanding of standard form and many correct answers were seen in parts (a) and (b). Whilst the formula is given for percentage error, there were still a minority of candidates who divided by 105 rather than the required value of 104.9.

Another well answered question with candidates showing a good understanding of standard form and many correct answers were seen in parts (a) and (b). Whilst the formula is given for percentage error, there were still a minority of candidates who divided by 105 rather than the required value of 104.9.

Calculate the percentage of the population of Ottawa that speak English but not French.

Calculate the number of people in Ottawa that speak both English and French.

Write down your answer to part (b) in the form \(a \times {10^k}\) where \(1 \leqslant a < 10\) and k \( \in \mathbb{Z}\).

\(97 - 36\)     (M1)

Note:     Award (M1) for subtracting 36 from 97.

OR

(M1)

Note:     Award (M1) for 61 and 36 seen in the correct places in the Venn diagram.

\( = 61{\text{ }}(\% )\)     (A1)     (C2)

Note:     Accept 61.0 (%).

[2 marks]

\(\frac{{36}}{{100}} \times 985\,000\)     (M1)

Note:     Award (M1) for multiplying 0.36 (or equivalent) by \(985\,000\).

\( = 355\,000{\text{ }}(354\,600)\)     (A1)     (C2)

[2 marks]

\(3.55 \times {10^5}{\text{ }}(3.546 \times {10^5})\)     (A1)(ft)(A1)(ft)     (C2)

Note:     Award (A1)(ft) for 3.55 (3.546) must match part (b), and (A1)(ft) \( \times {10^5}\).

Award (A0)(A0) for answers of the type: \(35.5 \times {10^4}\). Follow through from part (b).

[2 marks]

Use this exchange rate to calculate the amount of ARS that is equal to 350 AUD.

Calculate the total amount of ARS Jose paid to get 350 AUD.

Find the value of \(x\).

Note:     In this question, the first time an answer is not to 2 dp the final (A1) is not awarded.

\(\frac{{350}}{{0.1559}}\)     (M1)

Note:     Award (M1) for dividing 350 by 0.1559.

\( = 2245.03{\text{ (ARS)}}\)     (A1)     (C2)

[2 marks]

\(2245.03 \times 1.02\)     (M1)

Note:     Award (M1) for multiplying their answer to part (a) by 1.02.

\( = 2289.93{\text{ (ARS)}}\)     (A1)(ft)     (C2)

OR

\(2245.03 \times 0.02\)     (M1)

Note:     Award (M1) for multiplying their answer to part (a) by 0.02.

\( = 44.9006\)

\(2245.03 + 44.90\)

\( = 2289.93{\text{ (ARS)}}\)     (A1)(ft)     (C2)

Note:     Follow through from part (a).

[2 marks]

\(\frac{{4228.38}}{{585}}\)     (M1)

Note:     Award (M1) for dividing 4228.38 by 585.

\( = 7.23\)     (A1)     (C2)

[2 marks]

Write down the number of elements in \(A \cap C\) .

List the elements of \(B\) .

Complete the following Venn diagram with all the elements of \(U\) .

\(1\) (one)     (A1)     (C1)

Note: \(6\), \(\{6\} \) or \(\{1\} \) earns no marks.

[1 mark]

\(1\), \(3\), \(5\), \(7\), \(9\), \(11\)     (A1)     (C1)

Note: Do not penalise if braces, parentheses or brackets are seen.

[1 mark]

     (A1)(A1)(ft)(A1)(ft)(A1)(ft)     (C4)

Notes: Award (A1) for the empty set \(A \cap B \cap C\) .

Award (A1)(ft) for the correct placement of \(6\), \(5\), \(1\) and \(3\).

Award (A1)(ft) for the correct placement of \(2\), \(4\), \(12\), \(7\), \(9\), \(11\), \(8\).

Award (A1)(ft) for the correct placement of \(10\).

Follow through from part (b).

[4 marks]

There was much confusion amongst candidates as to the understanding of the words number of elements. Many candidates simply wrote down \(6\) or \(\{ 6\} \) and consequently lost the first mark.

There was much confusion amongst candidates as to the understanding of the words number of elements. Many candidates simply wrote down \(6\) or \(\{ 6\} \) and consequently lost the first mark. Part (b) was done well and many successful attempts were made at completing the Venn diagram in part (c). The most common error in the last part of the question was the omission of the element \(10\).

Part (b) was done well and many successful attempts were made at completing the Venn diagram in part (c). The most common error in the last part of the question was the omission of the element \(10\).

Write down the total area of the flag.

Write down the value of y.

The areas of regions A, B, and C are equal.

Write down the area of region A.

Using your answers to parts (b) and (c), find the value of x.

Units are required in this question for full marks to be awarded.

13800 cm²     (A1)     (C1)

75     (A1)     (C1)

Units are required in this question for full marks to be awarded.

4600 cm²     (A1)(ft)     (C1)

Notes: Units are required unless already penalized in part (a). Follow through from their part (a).

\(0.5(x + 92) \times 75 = 4600\)     (M1)(A1)(ft)

OR

\(0.5 \times 150 \times (92 - x) = 4600\)     (M1)(A1)(ft)

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

(= 30.7 (cm)(30.6666...(cm))     (A1)(ft)     (C3)

Note: Follow through from their parts (b) and (c).

A snack container has a cylindrical shape. The diameter of the base is \(7.84\,{\text{cm}}\). The height of the container is \(23.4\,{\text{cm}}\). This is shown in the following diagram.

Write down the radius, in \({\text{cm}}\), of the base of the container.

Calculate the area of the base of the container.

Dan is going to paint the curved surface and the base of the snack container.

Calculate the area to be painted.

\(3.92\,({\text{cm}})\)        (A1)    (C1)

\(\pi  \times {3.92^2}\)             (M1)

\( = 48.3\,{\text{c}}{{\text{m}}^2}\,\,\,(15.3664\,\pi \,{\text{c}}{{\text{m}}^2},\,\,48.2749...\,{\text{c}}{{\text{m}}^2})\)              (A1)(ft)         (C2)

Note: Award (M1) for correct substitution in area of circle formula. Follow through from their part (a). The answer is \(48.3\,{\text{c}}{{\text{m}}^2}\), units are required.

\(2 \times \pi  \times 3.92 \times 23.4 + 48.3\)        (M1)(M1)

\(625\,{\text{c}}{{\text{m}}^2}\,\,\,(624.618...\,{\text{c}}{{\text{m}}^2})\)                  (A1)(ft)    (C3)

Note: Award (M1) for correct substitution in curved surface area formula, (M1) for adding their answer to part (b). Follow through from their parts (a) and (b). The answer is \(625\,{\text{c}}{{\text{m}}^2}\), units are required.

Question 11: Cylinder base area and curved surface area.

In responses to this question, units were sometimes missing or the wrong units were given. The question explicitly asked for the base and curved surface area but many gave both the top and bottom as well as the curved surface area, or omitted the ends.

Question 11: Cylinder base area and curved surface area.

In responses to this question, units were sometimes missing or the wrong units were given. The question explicitly asked for the base and curved surface area but many gave both the top and bottom as well as the curved surface area, or omitted the ends.

Question 11: Cylinder base area and curved surface area.

In responses to this question, units were sometimes missing or the wrong units were given. The question explicitly asked for the base and curved surface area but many gave both the top and bottom as well as the curved surface area, or omitted the ends.

Find the common difference.

Find the twelfth term.

Find the sum of the first 100 terms.