2025 Cambridge IGCSE International Mathematics (0607) Practice Paper with Answers

The first differences between the terms are: 11 - 5 = 6, 19 - 11 = 8, 29 - 19 = 10. The second differences are constant and equal to 2. This indicates a quadratic sequence with general term \(an^2 + bn + c\), where \(a = \frac{2}{2} = 1\). Subtracting \(n^2\) from each term gives the sequence: 5 - 1 = 4, 11 - 4 = 7, 19 - 9 = 10, 29 - 16 = 13. The sequence 4, 7, 10, 13 is a linear sequence with a common difference of 3 and a first term of 4, represented by \(3n + 1\). Therefore, the general term of the sequence is \(u_n = n^2 + 3n + 1\). To find the position of the term with value 461, we solve the quadratic equation: \(n^2 + 3n + 1 = 461 \implies n^2 + 3n - 460 = 0\). Factoring the quadratic yields: \((n - 20)(n + 23) = 0\). Since the position must be positive, we find \(n = 20\).

M1: For finding the first differences (6, 8, 10) and second difference (2)
M1: For writing the general form of the n-th term as \(n^2 + bn + c\)
M1: For establishing the equation \(n^2 + 3n + 1 = 461\)
M1: For factorising \(n^2 + 3n - 460 = 0\) as \((n - 20)(n + 23) = 0\) or using quadratic formula
A1: For 20

Using the formula \(u_n = a \cdot 2^n + b\), we can write equations for the first two terms:
For \(n = 1\): \(2a + b = 7\) (Equation 1)
For \(n = 2\): \(4a + b = 17\) (Equation 2)
Subtracting Equation 1 from Equation 2 gives: \(2a = 10 \implies a = 5\).
Substituting \(a = 5\) back into Equation 1: \(2(5) + b = 7 \implies 10 + b = 7 \implies b = -3\).
We verify with the third term (\(n = 3\)): \(u_3 = 5 \cdot 2^3 - 3 = 40 - 3 = 37\), which is correct.
To find the 6th term (\(n = 6\)): \(u_6 = 5 \cdot 2^6 - 3 = 5(64) - 3 = 320 - 3 = 317\).

M1: For creating the system of equations \(2a + b = 7\) and \(4a + b = 17\)
M1: For a valid method to solve simultaneous equations
A1: For finding \(a = 5\) and \(b = -3\)
M1: For substituting \(n = 6\) into their general term formula
A1: For 317

The vertical asymptote occurs where the denominator is zero: \(2x - b = 0 \implies x = \frac{b}{2}\). Since the vertical asymptote is given as \(x = 3\), we have: \(\frac{b}{2} = 3 \implies b = 6\).
The horizontal asymptote occurs as \(x \to \pm\infty\), where the fraction \(f(x) \to \frac{a}{2}\). Since the horizontal asymptote is given as \(y = 4\), we have: \(\frac{a}{2} = 4 \implies a = 8\).
To find \(a + b\): \(a + b = 8 + 6 = 14\).

M1: For substituting \(x = 3\) into \(2x - b = 0\) to find \(b\)
A1: For \(b = 6\)
M1: For identifying the horizontal asymptote equation as \(y = \frac{a}{2}\)
A1: For \(a = 8\)
A1: For 14

The vertical asymptote of the function occurs where the denominator is equal to zero: \(2x + 4 = 0 \implies 2x = -4 \implies x = -2\). Hence, \(p = -2\).
To find the horizontal asymptote, we analyze the limit of \(g(x)\) as \(x \to \pm\infty\): The rational part \(\frac{6x - 1}{2x + 4}\) approaches \(\frac{6}{2} = 3\). Thus, \(g(x)\) approaches \(3 + 3 = 6\). This means \(q = 6\).
Now we calculate \(p^2 + q^2\): \(p^2 + q^2 = (-2)^2 + 6^2 = 4 + 36 = 40\).

M1: For solving \(2x + 4 = 0\)
A1: For vertical asymptote \(p = -2\)
M1: For evaluating the limit of the rational expression as \(3\)
A1: For horizontal asymptote \(q = 6\)
A1: For 40

Using the Sine Rule in triangle \(ABC\): \(\frac{\sin(BCA)}{AB} = \frac{\sin(BAC)}{BC}\).
Substitute the given values into the formula: \(\frac{\sin(BCA)}{12} = \frac{\sin(74^\circ)}{15}\).
Solve for \(\sin(BCA)\): \(\sin(BCA) = \frac{12 \cdot \sin(74^\circ)}{15}\).
Using a calculator: \(\sin(BCA) \approx \frac{12 \cdot 0.96126}{15} \approx 0.76901\).
Since angle \(BCA\) is acute, we calculate the inverse sine: \(BCA = \arcsin(0.76901) \approx 50.3^\circ\).

M1: For correctly using the Sine Rule formula: \(\frac{\sin(BCA)}{AB} = \frac{\sin(BAC)}{BC}\)
M1: For substituting the correct values: \(\frac{\sin(BCA)}{12} = \frac{\sin(74^\circ)}{15}\)
M1: For isolating \(\sin(BCA)\)
A1: For obtaining \(\sin(BCA) \approx 0.769\)
A1: For \(50.3\) (accept \(50.27\) to \(50.3\))

The largest angle in any triangle is always opposite the longest side. Since \(PR = 14\text{ cm}\) is the longest side, the largest angle is angle \(PQR\).
We apply the Cosine Rule to find angle \(PQR\): \(PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(PQR)\).
Substitute the given side lengths: \(14^2 = 8^2 + 11^2 - 2(8)(11)\cos(PQR)\).
Simplify the equation: \(196 = 64 + 121 - 176\cos(PQR) \implies 196 = 185 - 176\cos(PQR)\).
Rearrange to solve for \(\cos(PQR)\): \(176\cos(PQR) = 185 - 196 \implies 176\cos(PQR) = -11 \implies \cos(PQR) = -\frac{11}{176} = -0.0625\).
Using the inverse cosine function: \(PQR = \arccos(-0.0625) \approx 93.6^\circ\) (correct to 1 decimal place).

M1: For identifying that the largest angle is opposite the longest side \(PR\) (angle \(PQR\))
M1: For correctly stating the Cosine Rule: \(14^2 = 8^2 + 11^2 - 2 \cdot 8 \cdot 11 \cdot \cos(PQR)\)
M1: For simplifying to \(196 = 185 - 176\cos(PQR)\)
A1: For finding \(\cos(PQR) = -0.0625\) or equivalent fraction
A1: For \(93.6\) (accept \(93.59\) to \(93.6\))

The exponential depreciation formula is given by: \(V = V_0 \cdot \left(1 - \frac{r}{100}\right)^t\).
The time period between 2020 and 2025 is \(t = 5\) years. Substitute the given values: \(14750 = 25000 \cdot \left(1 - \frac{r}{100}\right)^5\).
Divide both sides by 25000: \(\left(1 - \frac{r}{100}\right)^5 = \frac{14750}{25000} = 0.59\).
Take the 5th root of both sides: \(1 - \frac{r}{100} = 0.59^{1/5} \approx 0.900174\).
Solve for \(r\): \(\frac{r}{100} = 1 - 0.900174 \implies \frac{r}{100} = 0.099826 \implies r \approx 9.98\%\).

M1: For using the exponential decay formula with \(t = 5\)
M1: For substituting correctly: \(14750 = 25000 (1 - r/100)^5\)
M1: For isolating the exponential term: \((1 - r/100)^5 = 0.59\)
A1: For finding \(1 - r/100 \approx 0.900\)
A1: For 9.98

First, find the gradient of line \(L_1\): \(m_1 = \frac{7 - 5}{4 - (-2)} = \frac{2}{6} = \frac{1}{3}\).
Since line \(L_2\) is perpendicular to \(L_1\), its gradient \(m_2\) is the negative reciprocal of \(m_1\): \(m_2 = -3\).
Next, find the midpoint \(M\) of the line segment joining the two given points: \(M = \left(\frac{-2 + 4}{2}, \frac{5 + 7}{2}\right) = (1, 6)\).
Now, use the point-gradient formula \(y - y_1 = m(x - x_1)\) to find the equation of \(L_2\) with point \((1, 6)\) and gradient \(-3\): \(y - 6 = -3(x - 1) \implies y - 6 = -3x + 3\).
Rearrange this equation into the requested form: \(3x + y = 9\).

M1: For finding the gradient of \(L_1\) as \(\frac{1}{3}\)
M1: For finding the perpendicular gradient as \(-3\)
M1: For finding the midpoint as \((1, 6)\)
M1: For substituting their midpoint and perpendicular gradient into the line equation formula
A1: For \(3x + y = 9\) (or any integer multiple with \(a > 0\))

1. Let the sequence be \(u_n = an^2 + bn + c\).
2. The first differences are:
- \(12 - 3 = 9\)
- \(27 - 12 = 15\)
- \(48 - 27 = 21\)
3. The second differences are:
- \(15 - 9 = 6\)
- \(21 - 15 = 6\)
4. Since the second difference is constant and equal to \(6\), we have \(2a = 6 \implies a = 3\).
5. Subtract \(3n^2\) from the terms to find the linear part:
- For \(n=1\): \(3 - 3(1)^2 = 0\)
- For \(n=2\): \(12 - 3(2)^2 = 0\)
- For \(n=3\): \(27 - 3(3)^2 = 0\)
- For \(n=4\): \(48 - 3(4)^2 = 0\)
6. The remaining sequence is constant at \(0\), so both \(b = 0\) and \(c = 0\).
7. Therefore, the \(nth\) term of the sequence is \(3n^2\).

M1 for attempting to find the first differences (9, 15, 21) or second differences (6, 6).
M1 for setting up \(2a = 6\) to find \(a = 3\).
A1 for obtaining the coefficient of \(n^2\) as 3.
A1.7 for the final correct expression \(3n^2\).

1. Calculate the 5th term, \(T_5\):
\[ T_5 = 5^3 - 2(5) + 5 = 125 - 10 + 5 = 120 \]
2. Calculate the 3rd term, \(T_3\):
\[ T_3 = 3^3 - 2(3) + 5 = 27 - 6 + 5 = 26 \]
3. Find the difference between the two terms:
\[ T_5 - T_3 = 120 - 26 = 94 \]

M1 for substituting \(n = 5\) into the formula for \(T_n\).
A1 for finding \(T_5 = 120\).
M1 for substituting \(n = 3\) into the formula for \(T_n\).
A1 for finding \(T_3 = 26\).
A0.7 for the final correct difference of \(94\).

1. To find the vertical asymptote, we set the denominator to zero:
\[ 2x + 6 = 0 \implies 2x = -6 \implies x = -3 \]
2. To find the horizontal asymptote, we determine the limit of \(f(x)\) as \(x\) approaches positive or negative infinity:
\[ y = \lim_{x \to \infty} \frac{4x - 7}{2x + 6} = \frac{4}{2} = 2 \]
3. Therefore, the equations of the asymptotes are \(x = -3\) and \(y = 2\).

M1 for setting the denominator to 0: \(2x + 6 = 0\).
A1 for obtaining the vertical asymptote \(x = -3\).
M1 for comparing the coefficients of \(x\) to find the horizontal asymptote.
A1 for obtaining the horizontal asymptote \(y = 2\).
A0.7 for expressing both answers as correct linear equations.

1. Factorise both the numerator and the denominator of the function:
\[ x^2 + 5x + 6 = (x + 2)(x + 3) \]
\[ x^2 - 9 = (x - 3)(x + 3) \]
2. Simplify \(g(x)\) for all values where \(x \neq -3\):
\[ g(x) = \frac{(x + 2)(x + 3)}{(x - 3)(x + 3)} = \frac{x + 2}{x - 3} \]
3. Set the simplified denominator to zero to find the vertical asymptote:
\[ x - 3 = 0 \implies x = 3 \]
Note: There is a hole at \(x = -3\), not a vertical asymptote.

M1 for factorising the numerator to \((x + 2)(x + 3)\).
M1 for factorising the denominator to \((x - 3)(x + 3)\).
M1 for cancelling the common factor \((x + 3)\).
A1.7 for the correct final equation \(x = 3\) (award only A1 if \(x = -3\) is also listed as an asymptote).

1. Apply the Sine Rule:
\[ \frac{\sin(ACB)}{AB} = \frac{\sin(BAC)}{BC} \]
2. Substitute the given values into the formula:
\[ \frac{\sin(ACB)}{15} = \frac{\sin(35^\circ)}{10} \]
3. Rearrange to find \(\sin(ACB)\):
\[ \sin(ACB) = \frac{15 \cdot \sin(35^\circ)}{10} \approx 1.5 \cdot 0.573576 = 0.860364 \]
4. Calculate the acute reference angle:
\[ \sin^{-1}(0.860364) \approx 59.36^\circ \]
5. Since the angle is specified as obtuse, find the supplementary angle:
\[ ACB = 180^\circ - 59.36^\circ = 120.64^\circ \]
6. Round to 1 decimal place to obtain \(120.6^\circ\).

M1 for correctly substituting into the Sine Rule formula: \(\frac{\sin(ACB)}{15} = \frac{\sin(35^\circ)}{10}\).
M1 for rearranging to find \(\sin(ACB) = 1.5 \sin(35^\circ)\).
A1 for \(\sin(ACB) \approx 0.860\) or \(0.8604\).
M1 for subtracting the acute reference angle from \(180^\circ\).
A0.7 for the final answer \(120.6^\circ\) (accept \(120.6\) to \(120.7\)).

1. Find the distance travelled by Ship \(P\) in 2 hours:
\[ OP = 18\text{ km/h} \times 2\text{ h} = 36\text{ km} \]
2. Find the distance travelled by Ship \(Q\) in 2 hours:
\[ OQ = 24\text{ km/h} \times 2\text{ h} = 48\text{ km} \]
3. Calculate the angle between the paths of the two ships:
\[ \angle POQ = 130^\circ - 40^\circ = 90^\circ \]
4. Since the angle is \(90^\circ\), use Pythagoras' theorem to find the distance \(PQ\):
\[ PQ^2 = OP^2 + OQ^2 = 36^2 + 48^2 \]
\[ PQ^2 = 1296 + 2304 = 3600 \]
\[ PQ = \sqrt{3600} = 60\text{ km} \]

M1 for finding the distances travelled: \(OP = 36\) and \(OQ = 48\).
M1 for calculating the angle between paths: \(\angle POQ = 90^\circ\).
M1 for correctly setting up Pythagoras' theorem (or Cosine Rule): \(PQ^2 = 36^2 + 48^2\).
A1 for \(PQ^2 = 3600\).
A0.7 for the final answer of \(60\text{ km}\) (or simply \(60\)).

1. Let \(n\) be the number of years. The formula for compound interest is:
\[ 4000(1 + 0.035)^n > 6000 \]
2. Simplify the inequality:
\[ 1.035^n > 1.5 \]
3. Solve for \(n\) using logarithms:
\[ n > \frac{\log(1.5)}{\log(1.035)} \approx \frac{0.17609}{0.01494} \approx 11.79 \]
4. Since \(n\) must be an integer representing complete years, round up to the next integer:
\[ n = 12 \]

M1 for setting up the inequality \(4000(1.035)^n > 6000\).
M1 for simplifying to \(1.035^n > 1.5\).
M1 for using logarithms or systematic trial and error to find \(n \approx 11.8\).
A1 for showing values around 11 and 12 years (e.g., \(1.035^{11} \approx 1.46\), \(1.035^{12} \approx 1.51\)).
A0.7 for the final answer of \(12\).

1. To find the average consumption per person, divide the total energy consumption by the population:
\[ \text{Average consumption} = \frac{1.8 \times 10^{11}}{4.5 \times 10^6} \]
2. Divide the coefficients:
\[ \frac{1.8}{4.5} = 0.4 \]
3. Subtract the exponents of 10:
\[ 10^{11 - 6} = 10^5 \]
4. Combine the components:
\[ 0.4 \times 10^5 \]
5. Express the result in proper standard form:
\[ 0.4 \times 10^5 = 4 \times 10^4\text{ kWh} \]

M2 for setting up the correct division expression: \(\frac{1.8 \times 10^{11}}{4.5 \times 10^6}\) (M1 for writing either value correctly or attempting division).
M1 for dividing coefficients to get \(0.4\) or calculating the value \(40000\).
A1 for obtaining \(40000\) or \(0.4 \times 10^5\).
A0.7 for expressing the final answer in correct standard form: \(4 \times 10^4\).