2024 Cambridge IGCSE Mathematics (0580) Practice Paper with Answers

Expand the bracket:
\( 4x - 12 = 18 \)

Add 12 to both sides:
\( 4x = 30 \)

Divide by 4:
\( x = 7.5 \)

M1 for \( 4x - 12 = 18 \) or \( x - 3 = 4.5 \)
A1 for \( 7.5 \)

Convert all values to decimals:
\( \frac{3}{8} = 0.375 \)
\( 35\% = 0.35 \)
\( 0.38 = 0.38 \)
\( \frac{2}{5} = 0.4 \)

Comparing the decimals: \( 0.35 < 0.375 < 0.38 < 0.4 \).

Therefore, the order starting with the smallest is:
\( 35\% \), \( \frac{3}{8} \), \( 0.38 \), \( \frac{2}{5} \)

M1 for converting at least two to a common format (e.g. decimals: 0.375, 0.35, 0.4)
A1 for correct ordered list: \( 35\% \), \( \frac{3}{8} \), \( 0.38 \), \( \frac{2}{5} \)

The sum of the angles in a triangle is \( 180^\circ \).
\( 3y + (y + 10) + 50 = 180 \)
\( 4y + 60 = 180 \)
\( 4y = 120 \)
\( y = 30 \)

M1 for \( 3y + y + 10 + 50 = 180 \) or better
M1 for \( 4y = 120 \) or \( 4y + 60 = 180 \)
A1 for \( 30 \)

From 19:45 to 20:00 is 15 minutes.
From 20:00 to 22:00 is 2 hours.
From 22:00 to 22:18 is 18 minutes.

Total time = \( 2\text{ hours} + 15\text{ minutes} + 18\text{ minutes} = 2\text{ hours } 33\text{ minutes} \).

M1 for a correct method of counting on or subtracting times, e.g. showing 2 hours or 153 minutes
A1 for 2 hours 33 minutes (or 2 h 33 m)

The formula for the area of a trapezium is:
\( A = \frac{1}{2}(a + b)h \)

Substitute the given values:
\( 54 = \frac{1}{2}(7 + 11)h \)
\( 54 = 9h \)
\( h = 6\text{ cm} \)

M1 for substituting correctly into formula: \( 54 = \frac{1}{2}(7 + 11)h \) or \( 9h = 54 \)
A1 for 6

Let \( N \) be the total number of pens.
The probability of picking a blue pen is:
\( P(\text{Blue}) = \frac{5}{N} = \frac{1}{4} \)

This gives:
\( N = 20 \)

So the total number of pens is 20.
The number of green pens is:
\( 20 - 8 - 5 = 7 \)

M1 for \( \frac{5}{\text{total}} = \frac{1}{4} \) or showing total number of pens is 20
M1 for subtracting 8 and 5 from their total
A1 for 7

The sum of the five numbers is:
\( 5 \times 8 = 40 \)

The sum of the four given numbers is:
\( 5 + 11 + 6 + 10 = 32 \)

The fifth number is:
\( 40 - 32 = 8 \)

M1 for \( 5 \times 8 \) or 40 seen, or \( 5 + 11 + 6 + 10 + x = 40 \)
A1 for 8

Multiply the numbers:
\( 3 \times 8 = 24 \)

Multiply the powers of 10:
\( 10^5 \times 10^{-2} = 10^3 \)

Combine and convert to standard form:
\( 24 \times 10^3 = 2.4 \times 10^4 \)

M1 for \( 24 \times 10^3 \) or \( 24000 \)
A1 for \( 2.4 \times 10^4 \)

First, convert each value into a decimal:
- \(\frac{3}{8} = 0.375\)
- \(0.35 = 0.35\)
- \(36\% = 0.36\)

Comparing the decimals, we get \(0.35 < 0.36 < 0.375\).

Therefore, the correct order starting with the smallest is \(0.35\), \(36\%\), \(\frac{3}{8}\).

B1 for converting at least two numbers to a common format (e.g. decimals: 0.375, 0.35, 0.36 or percentages: 37.5%, 35%, 36%)
B1 for correct order: 0.35, 36%, 3/8

Expand the bracket first:
\(12x - 20 = 18\)

Add 20 to both sides:
\(12x = 38\)

Divide both sides by 12:
\(x = \frac{38}{12} = \frac{19}{6} = 3\frac{1}{6}\) (or \(3.17\) correct to 3 significant figures).

M1 for correct expansion of brackets: \(12x - 20 = 18\) or division of both sides by 4: \(3x - 5 = 4.5\)
M1 for isolating the \(x\) term: \(12x = 38\) or \(3x = 9.5\)
A1 for \(3.17\) or \(3\frac{1}{6}\) or \(\frac{19}{6}\)

A regular octagon has \(8\) sides.

Method 1: Using exterior angles.
- The sum of the exterior angles is \(360^{\circ}\).
- One exterior angle \(= 360^{\circ} \div 8 = 45^{\circ}\).
- Since the interior and exterior angles lie on a straight line, the interior angle \(= 180^{\circ} - 45^{\circ} = 135^{\circ}\).

Method 2: Using the sum of interior angles.
- Sum of interior angles \(= (8 - 2) \times 180^{\circ} = 6 \times 180^{\circ} = 1080^{\circ}\).
- One interior angle \(= 1080^{\circ} \div 8 = 135^{\circ}\).

M1 for \(360 \div 8\) [= 45] or \((8 - 2) \times 180\) [= 1080]
M1 for \(180 - \text{their } 45\) or \(\text{their } 1080 \div 8\)
A1 for 135

First, calculate the area of the triangular cross-section:
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 6 \times 4 = 12\text{ cm}^2\)

Now, multiply by the length of the prism to find the volume:
\(\text{Volume} = \text{Area} \times \text{length} = 12 \times 11 = 132\text{ cm}^3\).

M1 for \(\frac{1}{2} \times 6 \times 4 \times 11\) oe
A1 for 132

The formula for the midpoint coordinates is \(\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)\).

Substitute the coordinates:
- \(x\)-coordinate \(= \frac{-3 + 5}{2} = \frac{2}{2} = 1\)
- \(y\)-coordinate \(= \frac{8 + (-2)}{2} = \frac{6}{2} = 3\)

The midpoint coordinates are \((1, 3)\).

M1 for \(\frac{-3 + 5}{2}\) or \(\frac{8 + (-2)}{2}\) seen or implied by one correct coordinate in the final answer
A1 for \((1, 3)\)

Using Pythagoras' theorem (\(a^2 + b^2 = c^2\)):
\(c^2 = 5.4^2 + 7.2^2\)
\(c^2 = 29.16 + 51.84 = 81\)
\(c = \sqrt{81} = 9\text{ cm}\)

The length of the hypotenuse is 9 cm.

M1 for \(5.4^2 + 7.2^2\) oe
A1 for 9

Calculate the actual reduction in price:
\(\text{Reduction} = 240 - 198 = 42\)

Calculate the percentage reduction based on the original price:
\(\text{Percentage Reduction} = \frac{42}{240} \times 100 = 17.5\%\).

M1 for \(240 - 198\) [= 42]
M1 for \(\frac{\text{their } 42}{240} \times 100\)
A1 for 17.5

Using the area of a triangle formula \(\text{Area} = \frac{1}{2}ab \sin C\):

\(\text{Area} = \frac{1}{2} \times 8 \times 11 \times \sin(48^{\circ})\)
\(\text{Area} = 44 \times \sin(48^{\circ})\)
\(\text{Area} \approx 44 \times 0.7431 = 32.698...\text{ cm}^2\)

Rounding to 3 significant figures gives \(32.7\text{ cm}^2\).

M1 for \(\frac{1}{2} \times 8 \times 11 \times \sin(48)\) oe
A1 for 32.69... to 32.7
A1 for 32.7

Convert the mixed number to an improper fraction: \(1\frac{3}{4} = \frac{7}{4}\). Now divide the fractions: \(\frac{7}{8} \div \frac{7}{4} = \frac{7}{8} \times \frac{4}{7}\). Multiply and simplify: \(\frac{7 \times 4}{8 \times 7} = \frac{4}{8} = \frac{1}{2}\).

M1 for converting to improper fraction \(\frac{7}{4}\) M1 for multiplying by reciprocal \(\frac{7}{8} \times \frac{4}{7}\) A1 for \(\frac{1}{2}\) cao

Group the like terms together: \(9x - 3x - 4y + 11y\). Simplify each part: \(9x - 3x = 6x\) and \(-4y + 11y = 7y\). Combining them gives \(6x + 7y\).

B1 for \(6x\) or \(7y\) in the final answer B1 for \(6x + 7y\) final answer

The sum of angles in a triangle is \(180^\circ\). The sum of the two equal angles is \(54^\circ + 54^\circ = 108^\circ\). The third angle is \(180^\circ - 108^\circ = 72^\circ\).

M1 for \(180 - 2 \times 54\) or \(180 - 108\) A1 for \(72\)

First, find the increase in wage: \(13.50 - 12.50 = 1.00\). Next, calculate the percentage increase: \(\frac{1.00}{12.50} \times 100 = 8\%\).

M1 for \(\frac{13.50 - 12.50}{12.50} \times 100\) or \(\frac{1.00}{12.50} \times 100\) A1 for 8 or 8%

The volume of a cuboid is calculated using the formula: Volume = length \(\times\) width \(\times\) height. Here, Volume = \(8 \times 5 \times 4.5 = 40 \times 4.5 = 180 \text{ cm}^3\).

M1 for \(8 \times 5 \times 4.5\) A1 for 180

Subtract the second equation from the first to eliminate \(y\): \((3x + 2y) - (x + 2y) = 19 - 9\) which simplifies to \(2x = 10\), so \(x = 5\). Substitute \(x = 5\) into the second equation: \(5 + 2y = 9\) which gives \(2y = 4\), so \(y = 2\).

M1 for a correct method to eliminate one variable (e.g. subtracting equations to get \(2x = 10\)) A1 for \(x = 5\) A1 for \(y = 2\)

First, multiply the decimal parts:
\(3.2 \times 4.5 = 14.4\).

Next, multiply the powers of 10:
\(10^5 \times 10^{-8} = 10^{5 + (-8)} = 10^{-3}\).

Combine these to get:
\(14.4 \times 10^{-3}\).

To write this in standard form (where the first number must be between 1 and 10):
\(1.44 \times 10^1 \times 10^{-3} = 1.44 \times 10^{-2}\).

B1 for \(14.4 \times 10^{-3}\) or \(0.0144\) seen
M1 for converting their non-standard value to correct standard form
A1 for \(1.44 \times 10^{-2}\) cao

Using the Cosine Rule to find the side \(PR\):
\(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\)

Substitute the given values:
\(PR^2 = 8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(54^\circ)\)
\(PR^2 = 70.56 + 42.25 - 109.2 \times 0.587785...\)
\(PR^2 = 112.81 - 64.186...\)
\(PR^2 = 48.6238...\)
\(PR = \sqrt{48.6238...} \approx 6.97\text{ cm}\) (to 3 significant figures).

M1 for correct substitution into the Cosine Rule: \(8.4^2 + 6.5^2 - 2 \times 8.4 \times 6.5 \times \cos(54)\)
M1 for \(PR = \sqrt{48.6...}\)
A1 for \(6.97\) or \(6.973...\)

The total surface area of a cone is the sum of the base area and the curved surface area:
\(\text{Total Area} = \pi r^2 + \pi r l\)

Substitute the values \(r = 3.5\) and \(l = 9.1\):
\(\text{Base Area} = \pi \times 3.5^2 = 12.25\pi \approx 38.48\text{ cm}^2\)
\(\text{Curved Area} = \pi \times 3.5 \times 9.1 = 31.85\pi \approx 100.06\text{ cm}^2\)

\(\text{Total Area} = 12.25\pi + 31.85\pi = 44.1\pi \approx 138.54\text{ cm}^2\).
To 3 significant figures, this is \(139\text{ cm}^2\).

M1 for base area \(\pi \times 3.5^2\) or curved surface area \(\pi \times 3.5 \times 9.1\) calculated
M1 for adding base area and curved area: \(\pi \times 3.5^2 + \pi \times 3.5 \times 9.1\) oe
A1 for \(139\) or \(138.5\) to \(138.6\)

Multiply the second equation by 2 to align the y-coefficients:
\(10x + 4y = 22\)

Now add this equation to the first equation:
\((3x - 4y) + (10x + 4y) = 17 + 22\)
\(13x = 39\)
\(x = 3\)

Substitute \(x = 3\) back into the second equation:
\(5(3) + 2y = 11\)
\(15 + 2y = 11\)
\(2y = -4\)
\(y = -2\)

M1 for a correct method to eliminate one variable (e.g. multiplying the second equation by 2 and adding)
A1 for \(x = 3\) or \(y = -2\)
A1 for both \(x = 3\) and \(y = -2\)

Let \(P\) be the original price of the bicycle.
An increase of \(15\%\) means the new price is \(115\%\) of the original price:
\(1.15 \times P = 414\)

Solve for \(P\):
\(P = \frac{414}{1.15} = 360\).

So the original price was \(\$360\).

M2 for \(\frac{414}{1.15}\) oe
(or M1 for \(1.15 \times P = 414\) or equivalent)
A1 for \(360\) cao

The interior angle and exterior angle of any polygon sum to \(180^\circ\).
Therefore, the exterior angle is:
\(180^\circ - 162^\circ = 18^\circ\).

The sum of exterior angles in any regular polygon is always \(360^\circ\).
Therefore, the number of sides, \(n\), is:
\(n = \frac{360^\circ}{18^\circ} = 20\).

M1 for finding the exterior angle: \(180 - 162 = 18\)
M1 for \(\frac{360}{\text{their } 18}\) oe
A1 for \(20\) cao

Because \(y\) is inversely proportional to the square of \(x\), we can write:
\(y = \frac{k}{x^2}\)

Substitute \(x = 4\) and \(y = 9\) to find the constant \(k\):
\(9 = \frac{k}{4^2}\)
\(9 = \frac{k}{16}\)
\(k = 9 \times 16 = 144\)

So, the formula is:
\(y = \frac{144}{x^2}\)

Now, substitute \(x = 6\) to find \(y\):
\(y = \frac{144}{6^2} = \frac{144}{36} = 4\).

M1 for set up of proportionality equation: \(y = \frac{k}{x^2}\) oe
M1 for finding constant of proportionality \(k = 144\) or using \(y_1 x_1^2 = y_2 x_2^2\)
A1 for \(4\) cao

The probability of winning a match is \(P(W) = 0.7\).
The probability of losing a match is \(P(L) = 1 - 0.7 = 0.3\).

To win exactly one of the two matches, Maya can either:
1. Win the first match and lose the second match (WL):
\(P(WL) = 0.7 \times 0.3 = 0.21\)
2. Lose the first match and win the second match (LW):
\(P(LW) = 0.3 \times 0.7 = 0.21\)

Adding these two mutually exclusive probabilities together:
\(\text{Total Probability} = 0.21 + 0.21 = 0.42\).

M1 for finding probability of losing \(P(L) = 0.3\) soi
M1 for \(0.7 \times 0.3 + 0.3 \times 0.7\) oe
A1 for \(0.42\) or \(\frac{21}{50}\)

Using the cosine rule: \(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\). Substituting the given values: \(PR^2 = 8.4^2 + 11.2^2 - 2 \times 8.4 \times 11.2 \times \cos(54^\circ) = 70.56 + 125.44 - 188.16 \times 0.5878 = 85.40\). Thus, \(PR = \sqrt{85.40} \approx 9.24\text{ cm}\).

M1 for correct substitution into the cosine rule: \(8.4^2 + 11.2^2 - 2(8.4)(11.2)\cos(54)\). A1 for \(85.4\dots\) or \(PR^2 = 85.4\dots\). A1 for 9.24 or 9.241...

The first differences are 5, 7, 9, 11. The second differences are constant at 2. Since the second difference is 2, the coefficient of \(n^2\) is 1. Subtracting \(n^2\) from the sequence terms yields: 3 - 1 = 2, 8 - 4 = 4, 15 - 9 = 6, 24 - 16 = 8, 35 - 25 = 10. The remaining linear sequence is 2, 4, 6, 8, 10, which has the general term \(2n\). Therefore, the \(nth\) term of the sequence is \(n^2 + 2n\).

M1 for identifying that the second difference is 2 or that the term involves \(n^2\). M1 for subtracting \(n^2\) to obtain the linear sequence 2, 4, 6, 8, ... or setting up simultaneous equations. A1 for \(n^2 + 2n\) or equivalent.

First, find the perpendicular height \(h\) of the cone using Pythagoras' theorem: \(h = \sqrt{13^2 - 5^2} = 12\text{ cm}\). Next, calculate the volume of the cone: \(V = \frac{1}{3} \pi \times 5^2 \times 12 = 100\pi\text{ cm}^3\). Equating the volume of the sphere to the volume of the cone: \( \frac{4}{3}\pi r^3 = 100\pi \implies r^3 = 75 \implies r = \sqrt[3]{75} \approx 4.22\text{ cm}\).

M1 for finding height of the cone \(h = 12\). M1 for setting up equation \(\frac{4}{3}\pi r^3 = \frac{1}{3}\pi \times 5^2 \times 12\) or \(\frac{4}{3}\pi r^3 = 100\pi\). A1 for 4.22 or 4.217...

Using the compound interest formula: \(4500 \left(1 + \frac{r}{100}\right)^6 = 5390\). Dividing by 4500: \(\left(1 + \frac{r}{100}\right)^6 = \frac{5390}{4500} \approx 1.1978\). Taking the 6th root of both sides: \(1 + \frac{r}{100} = 1.03049\). Thus, \(\frac{r}{100} = 0.03049 \implies r \approx 3.05\).

M1 for \(4500(1 + \frac{r}{100})^6 = 5390\). M1 for \(1 + \frac{r}{100} = \sqrt[6]{\frac{5390}{4500}}\) or equivalent. A1 for 3.05

The magnitude of vector \(\mathbf{p}\) is calculated as \(|\mathbf{p}| = \sqrt{(2k)^2 + (-3)^2}\). Given that this is equal to \(\sqrt{73}\), we can write: \((2k)^2 + (-3)^2 = 73 \implies 4k^2 + 9 = 73 \implies 4k^2 = 64 \implies k^2 = 16\). Since \(k > 0\), we take the positive root, which gives \(k = 4\).

M1 for writing \((2k)^2 + (-3)^2 = 73\) or \(\sqrt{(2k)^2 + (-3)^2} = \sqrt{73}\). M1 for simplifying to \(4k^2 = 64\) or \(k^2 = 16\). A1 for \(k = 4\).

To subtract the fractions, find a common denominator: \((x+2)(2x-1)\). Express each fraction over the common denominator: \(\frac{5(2x-1) - 3(x+2)}{(x+2)(2x-1)}\). Expand the numerator: \(10x - 5 - 3x - 6 = 7x - 11\). Thus, the simplified single fraction is \(\frac{7x-11}{(x+2)(2x-1)}\).

M1 for a common denominator of \((x+2)(2x-1)\) or \(2x^2+3x-2\). M1 for expanding numerator to \(5(2x-1) - 3(x+2)\) or \(10x-5 - 3x-6\). A1 for \(\frac{7x-11}{(x+2)(2x-1)}\) or equivalent.

The relationship can be written as \(y = \frac{k}{(x-1)^2}\). Substitute the given values to find \(k\): \(2 = \frac{k}{(4-1)^2} \implies 2 = \frac{k}{9} \implies k = 18\). So the formula is \(y = \frac{18}{(x-1)^2}\). Substitute \(x = 7\) to find \(y\): \(y = \frac{18}{(7-1)^2} = \frac{18}{36} = 0.5\).

M1 for \(y = \frac{k}{(x-1)^2}\). M1 for finding \(k = 18\). A1 for 0.5 or \(\frac{1}{2}\).

The probability of getting two counters of different colours is the sum of the probabilities of getting (Red, Blue) and (Blue, Red). \(P(\text{different}) = P(R, B) + P(B, R) = \left(\frac{6}{10} \times \frac{4}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right) = \frac{24}{90} + \frac{24}{90} = \frac{48}{90} = \frac{8}{15}\).

M1 for \(\frac{6}{10} \times \frac{4}{9}\) or \(\frac{4}{10} \times \frac{6}{9}\) seen. M1 for adding the two different permutations: \(\left(\frac{6}{10} \times \frac{4}{9}\right) + \left(\frac{4}{10} \times \frac{6}{9}\right)\). A1 for \(\frac{8}{15}\) or equivalent decimal \(0.533\) (or 0.533...)

Using the Cosine Rule:
\(PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(PQR)\)

\(PR^2 = 8.4^2 + 11.2^2 - 2 \times 8.4 \times 11.2 \times \cos(125^\circ)\)

\(PR^2 = 70.56 + 125.44 - 188.16 \times \cos(125^\circ)\)

\(PR^2 \approx 196 - (-107.92) = 303.92\)

\(PR = \sqrt{303.92} \approx 17.4\text{ cm}\) (to 3 significant figures)

M1 for \(8.4^2 + 11.2^2 - 2 \times 8.4 \times 11.2 \times \cos(125)\)
A1 for \(303.92...\) or better
A1 for \(17.4\) or \(17.43\) to \(17.44\)

The total surface area consists of the circular base and the curved surface area:

\(\text{Total Surface Area} = \pi r^2 + \pi r l\)

\(\text{Total Surface Area} = \pi \times 3.5^2 + \pi \times 3.5 \times 9.2\)

\(\text{Total Surface Area} = 12.25\pi + 32.2\pi = 44.45\pi\)

\(\text{Total Surface Area} \approx 139.64\text{ cm}^2\), which rounds to \(140\text{ cm}^2\) (to 3 significant figures).

M1 for finding the base area \(\pi \times 3.5^2\) or the curved surface area \(\pi \times 3.5 \times 9.2\)
M1 for adding the base area and curved surface area: \(\pi \times 3.5^2 + \pi \times 3.5 \times 9.2\)
A1 for \(140\) or \(139.6\) to \(139.7\)

Let the original price be \(x\).
Since the price is reduced by \(15\%\), \(85\%\) of the original price is equal to \(\$459\):

\(0.85x = 459\)

\(x = \frac{459}{0.85} = 540\)

The original price of the television is \(\$540\).

M2 for \(\frac{459}{0.85}\)
or M1 for associating \(85\%\) with \(459\) (e.g., \(0.85x = 459\) or \(459 \div 85\))
A1 for \(540\)

Multiply the first equation by 5 and the second equation by 2:

\(15x - 10y = 95\)
\(4x + 10y = 0\)

Add the two equations to eliminate \(y\):

\(19x = 95\)
\(x = 5\)

Substitute \(x = 5\) into the second equation:

\(2(5) + 5y = 0\)
\(10 + 5y = 0\)
\(5y = -10\)
\(y = -2\)

M1 for a correct method to equate the coefficients or express one variable in terms of the other
A1 for \(x = 5\)
A1 for \(y = -2\)

1. Find angle \(AXB\):
\(AXB = 180^\circ - 105^\circ = 75^\circ\) (angles on a straight line).

2. In triangle \(ABX\):
\(ABX = 180^\circ - 38^\circ - 75^\circ = 67^\circ\).

3. Since angle \(ACD\) and angle \(ABD\) are subtended by the same arc \(AD\), they are equal:
\(ACD = ABD = 67^\circ\).

M1 for angle \(AXB = 75^\circ\) or angle \(BXC = 105^\circ\)
M1 for angle \(ABX = 180 - 38 - \text{their } 75 = 67^\circ\)
A1 for \(67\)

First, find the derivative \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = 3x^2 - 6x - 9\)

At a turning point, \(\frac{dy}{dx} = 0\):

\(3x^2 - 6x - 9 = 0\)

Divide the entire equation by 3:

\(x^2 - 2x - 3 = 0\)

Factorise the quadratic equation:

\((x - 3)(x + 1) = 0\)

Thus, the \(x\)-coordinates are \(x = 3\) and \(x = -1\).

M1 for correct differentiation of at least two terms to get \(3x^2 - 6x - 9\)
M1 for setting their derivative to 0 and attempting to solve the resulting quadratic equation
A1 for \(3\) and \(-1\)

There are two ways the counters can be of the same colour: both are red, or both are blue.

1. Probability of both being red:
\(P(\text{Red, Red}) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90}\)

2. Probability of both being blue:
\(P(\text{Blue, Blue}) = \frac{4}{10} \times \frac{3}{9} = \frac{12}{90}\)

3. Total probability of same colour:
\(P(\text{Same Colour}) = \frac{30}{90} + \frac{12}{90} = \frac{42}{90} = \frac{7}{15}\) (or \(0.467\))

M1 for \(\frac{6}{10} \times \frac{5}{9}\) or \(\frac{4}{10} \times \frac{3}{9}\)
M1 for adding their two correct calculated probabilities
A1 for \(\frac{7}{15}\) or any equivalent fraction/decimal (such as \(0.467\) or \(\frac{42}{90}\))