2023 Edexcel IGCSE Mathematics (Specification A) Practice Paper with Answers

First, multiply both sides of the equation by the denominator:
\[y(5 - x) = 3x + 2\]

Expand the left-hand side:
\[5y - xy = 3x + 2\]

Rearrange the equation to group all terms containing \(x\) on one side and other terms on the opposite side:
\[5y - 2 = 3x + xy\]

Factorize \(x\) out of the right-hand side:
\[5y - 2 = x(3 + y)\]

Finally, divide both sides by \(3 + y\) to make \(x\) the subject:
\[x = \frac{5y - 2}{y + 3}\]

M1: For multiplying by the denominator and expanding, e.g., \(5y - xy = 3x + 2\)
M1: For gathering terms in \(x\) on one side and factorizing, e.g., \(x(3 + y) = 5y - 2\)
A1: Correct final expression, e.g., \(x = \frac{5y - 2}{y + 3}\) (or equivalent algebraic form)

Let \(\theta\) be the angle the ladder makes with the ground.
The ladder forms a right-angled triangle with the wall and the ground.

The side adjacent to the angle \(\theta\) is \(2.5\text{ m}\) (distance along the ground).
The hypotenuse is \(6.5\text{ m}\) (length of the ladder).

Using the cosine trigonometric ratio:
\[\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6.5}\]

Solve for \(\theta\):
\[\theta = \arccos\left(\frac{2.5}{6.5}\right) \approx 67.3801...^\circ\]

Rounding to 1 decimal place gives:
\[\theta \approx 67.4^\circ\]

M1: For a correct trigonometric equation, e.g., \(\cos(\theta) = \frac{2.5}{6.5}\) or \(\cos^{-1}\left(\frac{2.5}{6.5}\right)\)
M1: For completing the inverse calculation, e.g., \(\theta = 67.38...\)
A1: For correct answer of \(67.4\) (accept \(67.4^\circ\))

The total number of counters in the bag is \(8 + n\).

The probability of drawing a red counter on the first attempt is:
\[\text{P(1st Red)} = \frac{8}{8+n}\]

Since the counters are not replaced, there are now \(7\) red counters left out of a total of \(7 + n\) counters.
The probability of drawing a second red counter is:
\[\text{P(2nd Red | 1st Red)} = \frac{7}{7+n}\]

Set up the probability equation for both being red:
\[\frac{8}{8+n} \times \frac{7}{7+n} = \frac{14}{39}\]
\[\frac{56}{(8+n)(7+n)} = \frac{14}{39}\]

Divide both sides of the equation by \(14\):
\[\frac{4}{(8+n)(7+n)} = \frac{1}{39}\]

Cross-multiply to solve:
\[(8+n)(7+n) = 156\]
\[56 + 15n + n^2 = 156\]
\[n^2 + 15n - 100 = 0\]

Factorize the quadratic equation:
\[(n + 20)(n - 5) = 0\]

Since \(n\) represents a count of counters, \(n\) must be positive, so we reject \(n = -20\).
Therefore, \(n = 5\).

M1: For a correct probability product equation, e.g., \(\frac{8}{8+n} \times \frac{7}{7+n} = \frac{14}{39}\)
M1: For algebraic expansion and forming a quadratic equation, e.g., \(n^2 + 15n - 100 = 0\) (or equivalent)
A1: For the correct value \(5\) (accept \(n = 5\))

Let \(P\) be the original price of the laptop.
Since the laptop depreciates by \(12\%\) each year, the multiplier is:
\[1 - 0.12 = 0.88\]

After \(3\) years, the value of the laptop is given by:
\[P \times (0.88)^3 = 436.48\]

Calculate \(0.88^3\):
\[0.88^3 = 0.681472\]

Solve for \(P\):
\[P = \frac{436.48}{0.681472} = 640\]

Thus, the original price of the laptop was \(£640\).

M1: For a correct method to represent the value after 3 years using a multiplier, e.g., \(P \times 0.88^3 = 436.48\)
M1: For arranging to solve for \(P\), e.g., \(P = \frac{436.48}{0.88^3}\) or \(P = \frac{436.48}{0.681...}\)
A1: Correct answer \(640\)

First, find the gradient of line \(\mathbf{L}_1\) by rearranging its equation into the form \(y = mx + c\):
\[2y = 3x - 8 \implies y = \frac{3}{2}x - 4\]

The gradient of \(\mathbf{L}_1\) is \(m_1 = \frac{3}{2}\).

Since line \(\mathbf{L}_2\) is perpendicular to \(\mathbf{L}_1\), its gradient \(m_2\) must satisfy:
\[m_1 \times m_2 = -1 \implies \frac{3}{2} \times m_2 = -1 \implies m_2 = -\frac{2}{3}\]

Now, use the point-slope form with the gradient \(-\frac{2}{3}\) and point \((6, -1)\):
\[y - y_1 = m(x - x_1)\]
\[y - (-1) = -\frac{2}{3}(x - 6)\]
\[y + 1 = -\frac{2}{3}x + 4\]
\[y = -\frac{2}{3}x + 3\]

M1: For finding the gradient of \(\mathbf{L}_1\) as \(\frac{3}{2}\) (or \(1.5\)) and calculating the perpendicular gradient as \(-\frac{2}{3}\)
M1: For substituting gradient \(-\frac{2}{3}\) and coordinates \((6, -1)\) into a straight line equation method, e.g., \(-1 = -\frac{2}{3}(6) + c\)
A1: For the correct equation \(y = -\frac{2}{3}x + 3\) (or any exact equivalent form like \(y = -0.67x + 3\) where the fraction is exact)

Let \(r\) be the radius of the circle.
The area of a sector is given by:
\[\text{Area} = \frac{\theta}{360} \pi r^2\]

Substitute the given values into the formula:
\[\frac{120}{360} \pi r^2 = 27\pi\]
\[\frac{1}{3} \pi r^2 = 27\pi\]

Divide both sides by \(\pi\) and multiply by \(3\):
\[r^2 = 81 \implies r = 9\text{ cm}\]

Next, calculate the arc length of the sector:
\[\text{Arc length} = \frac{\theta}{360} \times 2\pi r\]
\[\text{Arc length} = \frac{120}{360} \times 2\pi \times 9 = \frac{1}{3} \times 18\pi = 6\pi\text{ cm}\]

The perimeter of the sector includes the arc length plus two radii:
\[\text{Perimeter} = 2r + \text{Arc length}\]
\[\text{Perimeter} = 2(9) + 6\pi = 18 + 6\pi\text{ cm}\]

Comparing this to the form \(a + b\pi\), we have \(a = 18\) and \(b = 6\).

M1: For setting up the area equation to find the radius, e.g., \(\frac{1}{3} \pi r^2 = 27\pi\) leading to \(r = 9\)
M1: For a correct method to find the arc length, e.g., \(\frac{120}{360} \times 2 \times \pi \times 9\) (or \(6\pi\))
A1: For the correct expression \(18 + 6\pi\)

First, find the ratio of the surface areas of vase \(A\) to vase \(B\):
\[\text{Area ratio } A : B = 180 : 405\]

Simplify the area ratio (dividing both by \(45\)):
\[\text{Area ratio} = 4 : 9\]

The linear scale factor \(k\) is the square root of the area scale factor:
\[k = \sqrt{\frac{4}{9}} = \frac{2}{3}\]

The scale factor for volume is the cube of the linear scale factor:
\[k^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27}\]

Now, calculate the volume of the smaller vase \(A\):
\[\text{Volume of } A = \text{Volume of } B \times k^3\]
\[\text{Volume of } A = 1350 \times \frac{8}{27} = 50 \times 8 = 400\text{ cm}^3\]

M1: For finding the linear scale factor ratio, e.g., \(\sqrt{\frac{180}{405}}\) or \(\frac{2}{3}\) (or \(1.5\))
M1: For cubed relation used to find the volume, e.g., \(1350 \times \left(\frac{2}{3}\right)^3\) or \(1350 \div 1.5^3\)
A1: Correct answer \(400\)

To find the stationary points, we first find the derivative \(\frac{dy}{dx}\):
\[\frac{dy}{dx} = 6x^2 - 18x - 24\]

Set \(\frac{dy}{dx} = 0\) for stationary points:
\[6x^2 - 18x - 24 = 0\]

Divide the entire equation by \(6\):
\[x^2 - 3x - 4 = 0\]

Factorize the quadratic:
\[(x - 4)(x + 1) = 0\]

This gives \(x = 4\) and \(x = -1\).
The question specifies that the stationary point has a negative \(x\)-coordinate, so we select \(x = -1\).

Now, find the corresponding \(y\)-coordinate by substituting \(x = -1\) into the original equation:
\[y = 2(-1)^3 - 9(-1)^2 - 24(-1) + 7\]
\[y = 2(-1) - 9(1) + 24 + 7\]
\[y = -2 - 9 + 24 + 7 = 20\]

Thus, the coordinates of the stationary point are \((-1, 20)\).

M1: For differentiating correctly, getting \(6x^2 - 18x - 24\) (allow 1 error)
M1: For setting their derivative to 0, solving to find \(x = -1\), and substituting into the original equation to find \(y\)
A1: Correct coordinates \((-1, 20)\)

First, factorise the numerator by taking out the common factor of 3 and using the difference of two squares: \(3x^2 - 12 = 3(x^2 - 4) = 3(x - 2)(x + 2)\). Next, factorise the quadratic expression in the denominator: \(2x^2 + 5x + 2 = (2x + 1)(x + 2)\). Now write the simplified expression by cancelling the common factor of \((x + 2)\) from both the numerator and the denominator: \(\frac{3(x - 2)(x + 2)}{(2x + 1)(x + 2)} = \frac{3(x - 2)}{2x + 1}\) (or \(\frac{3x - 6}{2x + 1}\)).

M1: for factorising the numerator \(3(x^2 - 4)\) or \(3(x - 2)(x + 2)\) OR for factorising the denominator into two linear brackets where the product of the constant terms is 2 and the product of the \(x\) coefficients is 2, e.g. \((2x + 1)(x + 2)\). M1: for \(\frac{3(x - 2)(x + 2)}{(2x + 1)(x + 2)}\). A1: for \(\frac{3(x - 2)}{2x + 1}\) or \(\frac{3x - 6}{2x + 1}\).

Let \(P\) be the price of the house at the start of 2021. The price after an \(8\%\) increase is \(P \times 1.08\). The price after a subsequent \(5\%\) decrease is \(P \times 1.08 \times 0.95\). We are given that this final price is £246,240, so: \(P \times 1.08 \times 0.95 = 246240\). Since \(1.08 \times 0.95 = 1.026\), we have: \(1.026P = 246240\). Dividing both sides by 1.026 gives: \(P = \frac{246240}{1.026} = 240000\). The price of the house at the start of 2021 was £240,000.

M1: for using a correct multiplier, e.g., \(1.08\) or \(0.95\), or for a correct equation, e.g., \(P \times 1.08 \times 0.95 = 246240\). M1: for \(\frac{246240}{1.08 \times 0.95}\) or \(\frac{246240}{1.026}\). A1: for 240000 (accept £240,000 or 240,000).

At the end of year 1, the value of the machine is \(24000 \times 0.92 = 22080\). At the end of year 2, the value is \(22080 \times 0.92 = 20313.6\). At the end of year 3 (1st January 2024), the value is \(20313.6 \times 1.05 = 21329.28\). Correct to the nearest pound, the value is £21329.

M1 for a correct method to find the value at the end of 2 years, e.g., \(24000 \times 0.92^2\) (or equivalent step-by-step calculation showing 20313.6). M1 for multiplying their year 2 value by 1.05. A1 for 21329.28. A1 for rounding correctly to the nearest pound (21329).

First, find the length of the diagonal \(AC\) of the base using Pythagoras' Theorem: \(AC = \sqrt{8^2 + 6^2} = \sqrt{100} = 10\text{ cm}\). Next, consider the right-angled triangle \(EAC\), where \(AE = 5\text{ cm}\) is perpendicular to the base. The angle \(\theta\) between \(EC\) and the base is \(\angle ECA\). Using trigonometry: \(\tan \theta = \frac{AE}{AC} = \frac{5}{10} = 0.5\). Therefore, \(\theta = \arctan(0.5) \approx 26.565^\circ\). Correct to 1 decimal place, the angle is \(26.6^\circ\).

M1 for a correct method to find the diagonal AC, e.g., \(\sqrt{8^2 + 6^2}\). A1 for \(AC = 10\). M1 for using a correct trigonometric ratio to find the angle, e.g., \(\tan \theta = \frac{5}{10}\) or \(\sin \theta = \frac{5}{\sqrt{125}}\). A1 for \(26.6\) (accept 26.5-26.6).

The ratio of the surface areas of cone A to cone B is \(4 : 9\). The linear scale factor \(k\) is therefore \(\sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5\). The volume scale factor (and hence mass scale factor, as they are made of the same material) is \(k^3 = 1.5^3 = 3.375\). Mass of cone B = \(360 \times 3.375 = 1215\text{ g}\).

M1 for finding the linear scale factor, e.g., \(\sqrt{\frac{9}{4}}\) or 1.5 (or 2:3). M1 for finding the volume/mass scale factor, e.g., \(1.5^3\) or 3.375 (or 8:27). M1 for a complete method to find the mass of B, e.g., \(360 \times 3.375\) or \(360 \times \frac{27}{8}\). A1 for 1215.

The total number of counters is 12. The number of green counters is 7, so the number of blue counters is \(12 - 7 = 5\). The probability of choosing at least one blue counter is equivalent to \(1 - P(\text{both are green})\). \(P(\text{both are green}) = \frac{7}{12} \times \frac{6}{11} = \frac{42}{132} = \frac{7}{22}\). Thus, \(P(\text{at least one blue}) = 1 - \frac{7}{22} = \frac{15}{22}\). Alternatively, this can be calculated as \(P(G, B) + P(B, G) + P(B, B) = \frac{7}{12}\times\frac{5}{11} + \frac{5}{12}\times\frac{7}{11} + \frac{5}{12}\times\frac{4}{11} = \frac{35+35+20}{132} = \frac{90}{132} = \frac{15}{22}\).

M1 for a correct first probability, e.g., \(\frac{7}{12}\) or \(\frac{5}{12}\). M1 for a product of two probabilities representing one of the correct paths, e.g., \(\frac{7}{12} \times \frac{6}{11}\) or \(\frac{7}{12} \times \frac{5}{11}\). M1 for a complete method to find the final probability, e.g., \(1 - \frac{7}{12} \times \frac{6}{11}\). A1 for \(\frac{15}{22}\) (accept equivalent fractions, or decimal 0.68 or 0.682).

First, find the velocity \(v\) by differentiating \(s\) with respect to \(t\): \(v = \frac{ds}{dt} = 6t^2 - 18t + 12\). Set the velocity to 0 to find the times when the particle is stationary: \(6t^2 - 18t + 12 = 0 \Rightarrow t^2 - 3t + 2 = 0 \Rightarrow (t-1)(t-2) = 0\). Thus, \(t = 1\) or \(t = 2\). The velocity is first \(0\) at \(t = 1\) second. Now, find the acceleration \(a\) by differentiating \(v\) with respect to \(t\): \(a = \frac{dv}{dt} = 12t - 18\). Substitute \(t = 1\) into the expression for acceleration: \(a = 12(1) - 18 = -6\text{ m/s}^2\).

M1 for differentiating the displacement formula to get velocity: \(6t^2 - 18t + 12\). M1 for setting their velocity expression to 0 and solving to find \(t = 1\) (and \(t = 2\)). M1 for differentiating their velocity expression to get acceleration: \(12t - 18\). A1 for -6.

Multiply each term by the common denominator \(x(x+2)\) to clear the fractions: \(2(x+2) + 3x = x(x+2)\). Expand both sides of the equation: \(2x + 4 + 3x = x^2 + 2x \Rightarrow 5x + 4 = x^2 + 2x\). Rearrange into standard quadratic form: \(x^2 - 3x - 4 = 0\). Factorise the quadratic equation: \((x - 4)(x + 1) = 0\). This gives the solutions: \(x = 4\) or \(x = -1\).

M1 for a correct method to clear the fractions, e.g., \(2(x+2) + 3x = x(x+2)\). M1 for expanding and rearranging to form a three-term quadratic equation, e.g., \(x^2 - 3x - 4 = 0\). M1 for factorising their quadratic equation, e.g., \((x-4)(x+1) = 0\) (or correct use of quadratic formula). A1 for both correct solutions: \(x = 4\) and \(x = -1\).

First, find the vector \(\overrightarrow{PQ}\): \(\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ} = -\mathbf{a} + \mathbf{b}\). Since \(PR : RQ = 2 : 3\), \(\overrightarrow{PR} = \frac{2}{5}\overrightarrow{PQ} = \frac{2}{5}(-\mathbf{a} + \mathbf{b})\). Now find \(\overrightarrow{OR}\): \(\overrightarrow{OR} = \overrightarrow{OP} + \overrightarrow{PR} = \mathbf{a} + \frac{2}{5}(-\mathbf{a} + \mathbf{b}) = \mathbf{a} - \frac{2}{5}\mathbf{a} + \frac{2}{5}\mathbf{b} = \frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\).

M1 for expressing \(\overrightarrow{PQ}\) in terms of \(\mathbf{a}\) and \(\mathbf{b}\): \(-\mathbf{a} + \mathbf{b}\). M1 for expressing \(\overrightarrow{PR}\) (or \(\overrightarrow{RQ}\)) in terms of \(\mathbf{a}\) and \(\mathbf{b}\): \(\frac{2}{5}(-\mathbf{a} + \mathbf{b})\). M1 for a complete vector journey for \(\overrightarrow{OR}\), e.g., \(\overrightarrow{OP} + \frac{2}{5}\overrightarrow{PQ}\) or \(\overrightarrow{OQ} - \frac{3}{5}\overrightarrow{PQ}\). A1 for \(\frac{3}{5}\mathbf{a} + \frac{2}{5}\mathbf{b}\) (or equivalent simplified form).

The formula for the area of a sector is \(\text{Area} = \frac{\theta}{360} \times \pi r^2\). Given \(\theta = 120^\circ\) and \(\text{Area} = 27\pi\): \(27\pi = \frac{120}{360} \times \pi r^2 \Rightarrow 27\pi = \frac{1}{3}\pi r^2\). Dividing both sides by \(\pi\) and multiplying by 3: \(r^2 = 81 \Rightarrow r = 9\text{ cm}\) (since radius must be positive). The arc length \(AB\) of the sector is: \(\text{Arc length} = \frac{120}{360} \times 2\pi r = \frac{1}{3} \times 2\pi(9) = 6\pi\text{ cm}\). The perimeter of the sector is the sum of the arc length and the two straight edges (radii): \(\text{Perimeter} = 2r + \text{Arc length} = 2(9) + 6\pi = 18 + 6\pi\text{ cm}\).

M1 for setting up a correct equation for the sector area: \(27\pi = \frac{120}{360} \times \pi r^2\). A1 for finding the radius \(r = 9\). M1 for a method to calculate the arc length, e.g., \(\frac{120}{360} \times 2 \times \pi \times 9\) (or \(6\pi\)). A1 for the final perimeter: \(18 + 6\pi\) (or equivalent exact form, e.g., \(6(3+\pi)\)).

To find the velocity, we differentiate the displacement with respect to \(t\):
\(v = \frac{ds}{dt} = 6t^2 - 30t + 24\)

We find the times when the velocity is \(0\text{ m/s}\) by setting \(v = 0\):
\(6t^2 - 30t + 24 = 0\)
\(t^2 - 5t + 4 = 0\)
\((t - 1)(t - 4) = 0\)

So, \(t = 1\) and \(t = 4\).

To find the acceleration, we differentiate the velocity with respect to \(t\):
\(a = \frac{dv}{dt} = 12t - 30\)

At \(t = 1\):
\(a = 12(1) - 30 = -18\text{ m/s}^2\)

At \(t = 4\):
\(a = 12(4) - 30 = 18\text{ m/s}^2\)

M1: For differentiating \(s\) to find an expression for velocity, \(v = 6t^2 - 30t + 24\) (at least 2 terms correct).
M1: For setting their \(v = 0\) and solving for \(t\) to get \(t = 1\) and \(t = 4\).
M1: For differentiating their \(v\) to find an expression for acceleration, \(a = 12t - 30\), and substituting at least one of their values of \(t\).
A1: For both correct accelerations: \(-18\text{ m/s}^2\) and \(18\text{ m/s}^2\).

(a) The ratio of the volume of vase \(B\) to vase \(A\) is:
\(\frac{V_B}{V_A} = \frac{4050}{1200} = \frac{27}{8} = 3.375\)

Since the vases are mathematically similar, the linear scale factor is:
\(\text{SF} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = 1.5\)

Therefore, the height of vase \(B\) is:
\(\text{Height} = 16 \times 1.5 = 24\text{ cm}\)

(b) The area scale factor from vase \(B\) to vase \(A\) is:
\(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\)

Therefore, the total surface area of vase \(A\) is:
\(\text{Area} = 900 \times \frac{4}{9} = 400\text{ cm}^2\)

For part (a):
M1: For finding the linear scale factor, e.g. \(\sqrt[3]{\frac{4050}{1200}}\) (which is \(\frac{3}{2}\) or \(1.5\)) or \(\sqrt[3]{\frac{1200}{4050}}\) (which is \(\frac{2}{3}\)).
A1: For \(24\text{ cm}\).

For part (b):
M1: For finding the area scale factor, e.g. \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\) or \(\left(\frac{3}{2}\right)^2 = \frac{9}{4}\), and using it to find the area of \(A\).
A1: For \(400\text{ cm}^2\).

Let be the midpoint of . Since is a square of side , we have:

In the right-angled triangle :



Now, we find the distance from the center to the vertex :


Let be the angle between the edge and the base , which is .
In the right-angled triangle :

M1: for finding the distance and setting up the trigonometric ratio
M1: for finding the vertical height of the pyramid (accept 10.7)
M1: for calculating the diagonal distance using Pythagoras' theorem (accept 7.07)
M1: for setting up the trigonometric ratio
A1: for the correct angle in the range (typically 56.6)

The total surface area of the closed box with a square base is given by:


We rearrange this formula to express in terms of :


The volume of the box is:


To find the maximum volume, we differentiate with respect to :


We set for the stationary point:

(since must be positive)

To justify that this is a maximum, we find the second derivative:

For , , which is less than 0, confirming a maximum.

We now calculate the maximum volume:

M1: for expressing in terms of from the surface area formula
M1: for substituting this expression into the volume formula to obtain
M1: for differentiating to find and setting the derivative to 0
M1: for solving to find and verifying it is a maximum (using second derivative or sign change)
A1: for the correct maximum volume of 8000

The arc length of the outer sector is given by:

This simplifies to:
(Equation 1)

The area of the shaded region is the difference between the areas of the two sectors:

This simplifies to:


(Equation 2)

We divide Equation 1 by Equation 2:



Substitute into Equation 2:



Thus, and .

M1: for setting up the outer arc equation or equivalent
M1: for setting up the shaded area equation or equivalent
M1: for simplifying both equations to a system in terms of and
M1: for eliminating to find a linear equation in and solving for
A1: for both correct values: and

The total number of counters in Bag A is .
The probability of selecting a red counter from Bag A is .
The probability of selecting a blue counter from Bag A is .

If a red counter is transferred from Bag A to Bag B, Bag B will contain 6 red counters and 3 blue counters (total 9 counters). The probability of choosing a red counter from Bag B is .

If a blue counter is transferred from Bag A to Bag B, Bag B will contain 5 red counters and 4 blue counters (total 9 counters). The probability of choosing a red counter from Bag B is .

Using the law of total probability, the probability of selecting a red counter from Bag B is:


We are given that this probability is :

Multiply both sides by :

M1: for finding the probabilities of choosing red and blue from Bag A: and
M1: for identifying the updated contents of Bag B under both branches and their respective probabilities of choosing red ( and )
M1: for constructing the total probability expression:
M1: for setting the expression equal to and forming a linear equation in
A1: for the correct value of