2023 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

A polygon with 9 sides is called a nonagon.

B1 for nonagon (allow minor spelling slips if intention is clear).

To find the number of complete hours, divide the total minutes by 60:
\(310 \div 60 = 5\) with a remainder of 10.
Therefore, there are 5 complete hours.

B1 for 5

Each term in the sequence is multiplied by 4 to get the next term:
\(1 \times 4 = 4\)
\(4 \times 4 = 16\)
\(16 \times 4 = 64\)
So the term-to-term rule is to multiply by 4.

B1 for multiply by 4 oe (e.g., \(\times 4\))

First, expand each bracket:
\(3(2x - 5) = 6x - 15\)
\(-2(x - 4) = -2x + 8\)

Now, collect and combine like terms:
\(6x - 15 - 2x + 8 = (6x - 2x) + (-15 + 8) = 4x - 7\)

M1 for \(6x - 15\) or \(-2x + 8\) seen
A1 for \(4x - 7\) as final answer

Find the cost in cents:
\(150 \times 80 = 12000\) cents.

Convert cents to dollars (since 100 cents = 1 dollar):
\(12000 \div 100 = 120\) dollars.

M1 for \(150 \times 0.80\) or \(150 \times 80\) oe
A1 for 120

Rearrange the equation to group the terms in \(x\) on one side and the numerical values on the other:
\(8 + 7 = 3x + 2x\)
\(15 = 5x\)

Divide both sides by 5:
\(x = 3\)

M1 for a correct first step to collect like terms, e.g. \(5x = 15\) or \(8 + 7 = 3x + 2x\)
A1 for 3

Multiply 3.05 by \(10^5 = 100000\):
\(3.05 \times 100000 = 305000\)

B1 for 305000

The probability of getting a Head on the coin is:
\(P(\text{Head}) = \frac{1}{2}\)

The numbers greater than 4 on a 6-sided die are 5 and 6. There are 2 such numbers out of 6 possible outcomes:
\(P(>4) = \frac{2}{6} = \frac{1}{3}\)

Since the coin toss and die roll are independent events, the combined probability is:
\(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\)

M1 for \(\frac{1}{2} \times \frac{2}{6}\) oe
A1 for \(\frac{1}{6}\) or any equivalent fraction

A polygon with 8 sides is called an octagon.

B1 for octagon (accept spelling errors if the meaning is clear).

The sequence is generated by dividing the previous term by 3 each time:
\(81 \div 3 = 27\)
\(27 \div 3 = 9\)
\(9 \div 3 = 3\)
\(3 \div 3 = 1\)

The next term is 1.

B1 for 1.

Using the formula for the volume of a cuboid:
\(\text{Volume} = \text{length} \times \text{width} \times \text{height}\)
\(\text{Volume} = 5 \times 3 \times 8 = 120 \text{ cm}^3\)

M1 for \(5 \times 3 \times 8\) oe
A1 for 120.

Subtract \(3x\) from both sides and add 4 to both sides:
\(7x - 3x = 8 + 4\)
\(4x = 12\)
\(x = 3\)

M1 for a correct first step, e.g. \(7x - 3x = 8 + 4\) or \(4x = 12\)
A1 for 3.

Find the highest common factor of \(12a^2\) and \(18a\), which is \(6a\):
\(12a^2 - 18a = 6a(2a - 3)\)

B2 for \(6a(2a - 3)\)
or B1 for partial factorisation, e.g., \(6(2a^2 - 3a)\) or \(a(12a - 18)\).

Distance is a measurement that can take any value in a given range, so it is continuous data.

B1 for continuous.

Express both fractions with a common denominator of 12:
\(\frac{5x}{12} - \frac{4x}{12} = \frac{5x - 4x}{12} = \frac{x}{12}\)

M1 for writing with a common denominator, e.g., \(\frac{5x}{12} - \frac{4x}{12}\) oe
A1 for \(\frac{x}{12}\) or \(x/12\).

The set \(U\) consists of all integers strictly between 5 and 12, which are 6, 7, 8, 9, 10, and 11.

B1 for listing 6, 7, 8, 9, 10, 11 (in any order).

A polygon with exactly eight sides is called an octagon.

B1 for octagon (accept spelling mistakes if the meaning is clear).

Since there are 60 minutes in an hour, we calculate \(450 \div 60 = 7.5\). Therefore, there are 7 complete hours.

B1 for 7.

The sequence consists of square numbers: \(1^2 = 1\), \(2^2 = 4\), \(3^2 = 9\), \(4^2 = 16\), \(5^2 = 25\). The next term is \(6^2 = 36\).

B1 for 36.

Since there are 1000 millilitres in one litre, we convert by dividing: \(2500 \div 1000 = 2.5\) litres.

B1 for 2.5 (accept \(2.5\text{ L}\) or \(2\frac{1}{2}\)).

The cost of one bag is \(80\) cents, which is \(\$0.80\). The cost of 15 bags is \(15 \times 0.80 = \$12\). Alternatively, \(15 \times 80 = 1200\) cents, which is \(\$12\).

M1 for \(15 \times 0.80\) or \(15 \times 80\) or equivalent. A1 for 12 (or 12.00).

First find the roots: \(\sqrt{144} = 12\) and \(\sqrt[3]{27} = 3\). Then calculate \(12 - 3 = 9\).

B1 for 9.

Using the order of operations, calculate the expressions in the brackets first: \(18 - 6 = 12\) and \(2 \times 3 = 6\). Then perform the division: \(12 \div 6 = 2\).

B1 for 2.

The volume of a cuboid is given by the formula: \(\text{Volume} = \text{length} \times \text{width} \times \text{height}\). Substituting the given values: \(120 = 6 \times 5 \times \text{height}\), which simplifies to \(120 = 30 \times \text{height}\). Solving for height gives \(\text{height} = 120 \div 30 = 4\text{ cm}\).

M1 for \(120 \div (6 \times 5)\) or \(6 \times 5 \times h = 120\) or equivalent. A1 for 4.

First, substitute the expression for \(w\) into the formula:
\(w^{-2} = (4 \times 10^{-6})^{-2}\)

Apply the power of \(-2\) to both parts:
\(w^{-2} = 4^{-2} \times (10^{-6})^{-2}\)
\(w^{-2} = \frac{1}{16} \times 10^{12}\)

Convert the fraction \(\frac{1}{16}\) to a decimal:
\(\frac{1}{16} = 0.0625\)

So,
\(w^{-2} = 0.0625 \times 10^{12}\)

Convert this to standard form:
\(0.0625 \times 10^{12} = 6.25 \times 10^{10}\)

M1 for \(4^{-2} \times 10^{12}\) or \(\frac{1}{16} \times 10^{12}\) oe
M1 for \(0.0625 \times 10^{12}\) oe
A1 for \(6.25 \times 10^{10}\)

Multiply out the brackets using the FOIL method:
\((3\sqrt{5})(2\sqrt{5}) + (3\sqrt{5})(7) - 2(2\sqrt{5}) - 2(7)\)

Simplify each term:
\(3 \times 2 \times 5 + 21\sqrt{5} - 4\sqrt{5} - 14\)
\(30 + 17\sqrt{5} - 14\)

Combine the constant terms:
\(16 + 17\sqrt{5}\)

M1 for expansion with 3 out of 4 terms correct
M1 for fully correct expansion showing \(30\) and \(-14\)
A1 for \(16 + 17\sqrt{5}\) or equivalent

Find a common denominator, which is \((2x-3)(x+4)\):
\(\frac{5(x+4) - 3(2x-3)}{(2x-3)(x+4)}\)

Expand the terms in the numerator:
\(\frac{5x + 20 - (6x - 9)}{(2x-3)(x+4)}\)
\(\frac{5x + 20 - 6x + 9}{(2x-3)(x+4)}\)

Simplify the numerator:
\(\frac{29 - x}{(2x-3)(x+4)}\)

M1 for common denominator \((2x-3)(x+4)\) seen
M1 for numerator \(5(x+4) - 3(2x-3)\) or expanded version with at least one sign correct
A1 for \(\frac{29-x}{(2x-3)(x+4)}\) final answer

Use the addition rule of logarithms to combine the terms:
\(\log(x(x-3)) = 1\)

Convert from logarithmic form to exponential form (base 10):
\(x(x-3) = 10^1\)
\(x^2 - 3x = 10\)

Rearrange to form a quadratic equation:
\(x^2 - 3x - 10 = 0\)

Factorise the quadratic:
\((x - 5)(x + 2) = 0\)

This gives \(x = 5\) or \(x = -2\).
Since the logarithm is only defined for positive numbers, \(x-3 > 0\) and \(x > 0\), so we reject \(x = -2\).
Thus, the only valid solution is \(x = 5\).

M1 for combining logs: \(\log(x(x-3)) = 1\) oe
M1 for removing logs: \(x^2 - 3x = 10\) oe
A1 for \(x = 5\) only (rejecting \(x = -2\))

Write down the general formula for inverse proportion:
\(P = \frac{k}{\sqrt{w}}\)

Substitute the given values to find \(k\):
\(6 = \frac{k}{\sqrt{16}}\)
\(6 = \frac{k}{4}\)
\(k = 24\)

Now find \(P\) when \(w = 9\):
\(P = \frac{24}{\sqrt{9}}\)
\(P = \frac{24}{3}\)
\(P = 8\)

M1 for \(P = \frac{k}{\sqrt{w}}\) oe
M1 for substituting \(w = 16, P = 6\) to find \(k = 24\)
A1 for \(P = 8\)

Let's find the first and second differences:
Terms: 3, 9, 19, 33
First differences: 6, 10, 14
Second differences: 4, 4

Since the second difference is constant, the sequence is quadratic of the form \(an^2 + bn + c\).
The coefficient \(a = \frac{\text{second difference}}{2} = \frac{4}{2} = 2\).

Subtract \(2n^2\) from the original terms:
For \(n=1\): \(3 - 2(1)^2 = 1\)
For \(n=2\): \(9 - 2(2)^2 = 1\)
For \(n=3\): \(19 - 2(3)^2 = 1\)
For \(n=4\): \(33 - 2(4)^2 = 1\)

Since the remainder is a constant \(1\), \(b = 0\) and \(c = 1\).
Thus, the expression for the \(n\)-th term is \(2n^2 + 1\).

M1 for finding the second difference of \(4\) or stating \(a = 2\)
M1 for attempting to find \(b\) and \(c\) by subtraction or simultaneous equations
A1 for \(2n^2 + 1\) final answer

First, factorise the quadratic expression:
\(x^2 - 4x - 12 \le 0\)
\((x - 6)(x + 2) \le 0\)

The critical values where the expression equals zero are \(x = 6\) and \(x = -2\).

Since the inequality is \(\le 0\), we are looking for the interval where the graph is on or below the x-axis, which is between the critical values:
\(-2 \le x \le 6\)

M1 for factorising to find critical values: \(x = 6\) and \(x = -2\)
M1 for a graphical method, table of signs, or testing intervals
A1 for \(-2 \le x \le 6\) as final answer

First, find the common factor of the two terms, which is \(3x\):
\(3x^3 - 27xy^2 = 3x(x^2 - 9y^2)\)

Next, recognise that \(x^2 - 9y^2\) is a difference of two squares:
\(x^2 - 9y^2 = (x - 3y)(x + 3y)\)

Substitute this back to get the fully factorised expression:
\(3x(x - 3y)(x + 3y)\)

M1 for factorising out \(3x\) to get \(3x(x^2 - 9y^2)\)
M1 for factorising the difference of two squares \((x^2 - 9y^2)\) to get \((x-3y)(x+3y)\)
A1 for \(3x(x - 3y)(x + 3y)\) final answer

First, expand the brackets using the distributive property:
\((5 + 3\sqrt{2})(3 - \sqrt{2}) = 5 \times 3 + 5 \times (-\sqrt{2}) + 3\sqrt{2} \times 3 + 3\sqrt{2} \times (-\sqrt{2})\)

Simplify each term:
\(= 15 - 5\sqrt{2} + 9\sqrt{2} - 3 \times 2\)
\(= 15 - 6 + 4\sqrt{2}\)
\(= 9 + 4\sqrt{2}\)

M1 for expansion with at least 3 correct terms out of 4, e.g., \(15 - 5\sqrt{2} + 9\sqrt{2} - 6\)
A1 for \(9 + 4\sqrt{2}\) cao

\(w^4 = (3 \times 10^6)^4\)
\(= 3^4 \times (10^6)^4\)
\(= 81 \times 10^{24}\)

To write this in standard form:
\(= 8.1 \times 10^1 \times 10^{24}\)
\(= 8.1 \times 10^{25}\)

M1 for \(81 \times 10^{24}\) or correct application of index laws
A1 for \(8.1 \times 10^{25}\) cao

Find a common denominator, which is \((2x - 3)(x + 2)\):

\(\frac{4(x + 2) - 3(2x - 3)}{(2x - 3)(x + 2)} Expand the numerator: \)= \frac{4x + 8 - 6x + 9}{(2x - 3)(x + 2)}\)

Simplify the numerator:

\(= \frac{17 - 2x}{(2x - 3)(x + 2)}\)

M1 for a common denominator of \((2x - 3)(x + 2)\)
M1 for correct expansion of the numerator: \(4x + 8 - 6x + 9\)
A1 for \(\frac{17 - 2x}{(2x - 3)(x + 2)}\) oe

Subtract 14 from both sides:

\(-3x \le 5 - 14\)

\(-3x \le -9\)

Divide both sides by \(-3\), and reverse the inequality sign:

\(x \ge 3\)

M1 for \(-3x \le -9\) or \(3x \ge 9\) oe
A1 for \(x \ge 3\)

Express both sides as powers of 3:

\(\frac{1}{9} = 3^{-2}\)

\(27 = 3^3\)

Substitute these into the equation:

\((3^{-2})^p = (3^3)^{p-5}\)

\(3^{-2p} = 3^{3(p-5)}\)

Since the bases are equal, equate the exponents:

\(-2p = 3(p-5)\)

\(-2p = 3p - 15\)

\(-5p = -15\)

\(p = 3\)

M1 for converting to a common base (3), e.g. \(3^{-2p} = 3^{3(p-5)}\)
M1 for equating exponents and attempting to solve, e.g. \(-2p = 3p - 15\)
A1 for \(p = 3\) cao

Apply the laws of logarithms:

\(\log(x^2) - \log(2^3) + \log 5 = 1\)

\(\log(x^2) - \log 8 + \log 5 = 1\)

Combine the logarithmic terms:

\(\log\left(\frac{5x^2}{8}\right) = 1\)

Since the base is 10:

\(\frac{5x^2}{8} = 10^1\)

\(\frac{5x^2}{8} = 10\)

\(5x^2 = 80\)

\(x^2 = 16\)

\(x = 4\) (since \(x\) must be positive for \(\log x\) to be defined).

M1 for combining the terms using logarithmic properties, e.g., \(\log\left(\frac{5x^2}{8}\right) = 1\)
M1 for removing the logarithm, e.g., \(\frac{5x^2}{8} = 10\)
A1 for \(4\)

Find a common denominator for the fractions to compare them easily. The LCM of 4, 3, 10, and 18 is 180.

Convert each fraction:

\(\frac{2}{3} = \frac{120}{180}\)

\(\frac{7}{10} = \frac{126}{180}\)

\(\frac{13}{18} = \frac{130}{180}\)

\(\frac{3}{4} = \frac{135}{180}\)

Comparing the numerators (120 < 126 < 130 < 135), the correct order starting with the smallest is:

\(\frac{2}{3}\), \(\frac{7}{10}\), \(\frac{13}{18}\), \(\frac{3}{4}\)

M1 for converting at least 3 fractions to a common denominator (e.g., 180) or showing correct decimal equivalents
A1 for correct order: \(\frac{2}{3}\), \(\frac{7}{10}\), \(\frac{13}{18}\), \(\frac{3}{4}\)

(a) The terms increase by 6 each time. The next term is \(29 + 6 = 35\).
(b) Since the common difference is 6, the general term is of the form \(6n + c\). Since the first term is 5: \(6(1) + c = 5 \Rightarrow c = -1\). Thus, the \(n\)-th term is \(6n - 1\).
(c) Solve \(6n - 1 = 239 \Rightarrow 6n = 240 \Rightarrow n = 40\).

(a) B1 for 35
(b) M1 for recognizing common difference of 6 (or seeing \(6n + c\)), A1 for \(6n - 1\)
(c) M1 for setting their \(6n - 1 = 239\) and attempting to solve, A1 for 40

(a) Volume of water: \(V = \pi r^2 h = \pi \times 6^2 \times 10 = 360\pi \text{ cm}^3\).
(b) Volume of the sphere: \(V_{\text{sphere}} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \times 3^3 = 36\pi \text{ cm}^3\).
Total volume of water and sphere: \(360\pi + 36\pi = 396\pi \text{ cm}^3\).
Let \(H\) be the new depth: \(\pi \times 6^2 \times H = 396\pi \Rightarrow 36H = 396 \Rightarrow H = 11 \text{ cm}\).

(a) M1 for \(\pi \times 6^2 \times 10\), A1 for \(360\pi\)
(b) M1 for calculating volume of sphere: \(\frac{4}{3} \pi \times 3^3 = 36\pi\), M1 for setting up \(\pi \times 6^2 \times H = \text{their total volume}\), A1 for 11

(a) Gradient \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-7 - 5}{4 - (-2)} = \frac{-12}{6} = -2\).
(b) Using point-slope form: \(y - 5 = -2(x - (-2)) \Rightarrow y - 5 = -2x - 4 \Rightarrow y = -2x + 1\).
(c) Line crosses the \(x\)-axis when \(y = 0\): \(0 = -2x + 1 \Rightarrow 2x = 1 \Rightarrow x = 0.5\).

(a) M1 for \(\frac{-7 - 5}{4 - (-2)}\), A1 for -2
(b) M1 for substituting their gradient and a point into \(y = mx + c\) to find \(c\), A1 for \(y = -2x + 1\)
(c) M1 for setting \(y = 0\) in their equation, A1 for 0.5

(a) Number of students taking more than 40 seconds is \(100 - 45 = 55\).
(b) The cumulative frequency for \(t \le 40\) is 45 out of 100, which is \(45\%\).
(c) Frequency of the interval \(40 < t \le 60\) is the cumulative frequency of 60 minus cumulative frequency of 40: \(80 - 45 = 35\).

(a) B1 for 55
(b) B1 for 45% (or 45)
(c) M1 for \(80 - 45\), A1 for 35

(a) \(4x - 7 \le 13 \Rightarrow 4x \le 20 \Rightarrow x \le 5\).
(b) \(2(3 - x) > 12 \Rightarrow 6 - 2x > 12 \Rightarrow -2x > 6 \Rightarrow x < -3\).
(c) The values of \(x\) satisfying both are \(x < -3\) and \(x \le 5\), which is simply \(x < -3\). The largest integer less than -3 is -4.

(a) M1 for \(4x \le 20\), A1 for \(x \le 5\)
(b) M1 for \(6 - 2x > 12\) (or \(3 - x > 6\)), A1 for \(x < -3\)
(c) M1 for combining both inequalities to find \(x < -3\), A1 for -4

Using GDC graph functions:
(a) Locate local maximum: \(x = -1\), \(y = 10\). Coordinates are \((-1, 10)\).
(b) Locate local minimum: \(x = 3\), \(y = -22\). Coordinates are \((3, -22)\).
(c) Find intersection of \(y = x^3 - 3x^2 - 9x + 5\) and \(y = 5\).
\(x^3 - 3x^2 - 9x = 0 \Rightarrow x(x^2 - 3x - 9) = 0\).
For \(x > 0\), solve \(x^2 - 3x - 9 = 0\) using GDC or quadratic formula:
\(x = \frac{3 + \sqrt{45}}{2} \approx 4.8541\).
To 3 significant figures, \(x = 4.85\).

(a) B1 for \(x = -1\), B1 for \(y = 10\) (or GDC equivalent)
(b) B1 for \(x = 3\), B1 for \(y = -22\) (or GDC equivalent)
(c) M1 for setting \(y = 5\) and identifying the positive root, A1 for 4.85 (accept 4.854)

(a) By Pythagoras' theorem: \(h^2 + 2.5^2 = 8.5^2 \Rightarrow h^2 = 72.25 - 6.25 = 66 \Rightarrow h = \sqrt{66} \approx 8.124 \text{ m}\). To 3 significant figures, the height is 8.12 m.
(b) Let \(\theta\) be the angle with the ground. \(\cos(\theta) = \frac{2.5}{8.5} = \frac{5}{17} \Rightarrow \theta = \arccos\left(\frac{5}{17}\right) \approx 72.92^\circ\). To 1 decimal place, \(\theta = 72.9^\circ\).

(a) M1 for \(8.5^2 - 2.5^2\), A1 for 8.12
(b) M1 for \(\cos(\theta) = \frac{2.5}{8.5}\) (or alternative correct trig ratio), A1 for 72.9

(a) Mean hours studied \(\bar{x} = \frac{2+4+5+6+7+8+10+12}{8} = \frac{64}{8} = 8\).
(b) Mean score \(\bar{y} = \frac{45+55+62+65+70+78+85+95}{8} = \frac{555}{8} = 69.375 \approx 69.4\).
(c) As the hours studied increase, the test scores also increase, showing a positive correlation.

(a) M1 for adding terms and dividing by 8, A1 for 8
(b) M1 for adding terms and dividing by 8, A1 for 69.4 (accept 69.375)
(c) B1 for Positive

(a)(i) The next term is found by adding the common difference of 8 to the fourth term: \(35 + 8 = 43\). \
(ii) The term-to-term rule is to add 8. \
\
(b) The sequence has a first term \(a = 11\) and a common difference \(d = 8\). The \(n\)th term is given by: \
\(a + (n - 1)d = 11 + (n - 1)8 = 8n + 3\). \
\
(c) The 80th term is: \
\(8(80) + 3 = 640 + 3 = 643\). \
\
(d) Set the \(n\)th term equal to 403: \
\(8n + 3 = 403\) \
\(8n = 400\) \
\(n = 50\). \
Since 50 is a positive integer, 403 is a term of the sequence (the 50th term).

(a)(i) B1 for 43 \
(ii) B1 for add 8 oe \
(b) B2 for 8n + 3 (B1 for 8n + c or k*n + 3) \
(c) M1 for substituting n = 80 into their formula, A1 for 643 (FT their formula) \
(d) M1 for 8n + 3 = 403 oe, A1 for n = 50, A1 for Yes (or equivalent conclusion)

(a) The coordinates of the midpoint of \(PQ\) are: \
\( \left( \frac{-3 + 5}{2}, \frac{8 + (-2)}{2} \right) = (1, 3) \). \
\
(b) Using the distance formula: \
\( PQ = \sqrt{(5 - (-3))^2 + (-2 - 8)^2} = \sqrt{8^2 + (-10)^2} = \sqrt{64 + 100} = \sqrt{164} \approx 12.8 \). \
\
(c) Let \(R\) have coordinates \((x_R, y_R)\). \
The midpoint of \(QR\) is: \
\( \left( \frac{5 + x_R}{2}, \frac{-2 + y_R}{2} \right) = (1.5, 1) \). \
Solving for \(x_R\): \
\( 5 + x_R = 3 \implies x_R = -2 \). \
Solving for \(y_R\): \
\( -2 + y_R = 2 \implies y_R = 4 \). \
So, the coordinates of \(R\) are \((-2, 4)\).

(a) M1 for midpoint formula with correct substitution, A1 for (1, 3) \
(b) M1 for correct substitution into distance formula, M1 for evaluating to \(\sqrt{164}\), A1 for 12.8 (accept 12.806...) \
(c) M1 for setting up equation for x: \((5+x)/2 = 1.5\), A1 for x = -2, M1 for setting up equation for y: \((-2+y)/2 = 1\), A1 for y = 4

(a) The depreciation rate is 8\\%, so the value after 1 year is: \
\( 18000 \times (1 - 0.08) = 18000 \times 0.92 = 16560 \). \
So, the value is \(\\$16560\). \
\
(b) The value after 5 years is: \
\( 18000 \times 0.92^5 \approx 18000 \times 0.65908 = 11863.47 \). \
Correct to the nearest dollar, the value is \(\\$11863\). \
\
(c) We want to find the smallest integer \(n\) such that: \
\( 18000 \times 0.92^n < 10000 \). \
Calculating the value for different years: \
For \(n = 7\): \( 18000 \times 0.92^7 \approx 10041.24 \) \
For \(n = 8\): \( 18000 \times 0.92^8 \approx 9237.94 \) \
Therefore, it takes 8 complete years for the value to fall below \(\\$10000\).

(a) M1 for 18000 * 0.92, A1 for 16560 \
(b) M1 for 18000 * 0.92^5, A1 for 11863.47..., A1 for 11863 (rounded to the nearest dollar) \
(c) M1 for setting up inequality or equation: 18000 * 0.92^n < 10000, B1 for calculating value at 7 years is 10041 (or 10041.24), B1 for calculating value at 8 years is 9238 (or 9237.94), A1 for 8