2025 CCEA GCSE Mathematics 2210 Practice Paper with Answers

Using \( \tan(40^{\circ}) = \dfrac{\text{opposite}}{12} \), so opposite \( = 12 \times \tan(40^{\circ}) = 12 \times 0.8391\ldots = 10.069\ldots \), which rounds to 10.1 cm (3 s.f.).

1 mark for identifying \( \tan \) as the correct ratio; 1 mark for the correct method \( 12\tan(40^{\circ}) \); 1 mark for the final answer 10.1 cm (3 s.f.). Max 3.

\( 8.4 \div 2.1 = 4 \) and \( 10^{7} \div 10^{3} = 10^{4} \), so the answer is \( 4 \times 10^{4} \).

1 mark for correctly dividing 8.4 by 2.1 to give 4; 1 mark for correctly dividing the powers of 10 (\( 10^{4} \)); 1 mark for the correct final answer \( 4\times10^{4} \). Max 3.

\( (2x+3)(x-5) = 2x^{2} - 10x + 3x - 15 = 2x^{2} - 7x - 15 \).

1 mark for correctly expanding to four terms (\( 2x^{2}-10x+3x-15 \)); 1 mark for correctly collecting the \( x \) terms; 1 mark for the fully simplified answer \( 2x^{2}-7x-15 \). Max 3.

Midpoint \( = \left(\dfrac{-2+4}{2}, \dfrac{5+(-3)}{2}\right) = (1, 1) \). Gradient \( = \dfrac{-3-5}{4-(-2)} = \dfrac{-8}{6} = -\dfrac{4}{3} \).

1 mark for the correct midpoint (1, 1); 1 mark for the correct method for gradient; 1 mark for the correct gradient \( -\frac{4}{3} \). Max 3.

We need two numbers that multiply to give \( -15 \) and add to give 2: these are 5 and \( -3 \). So \( x^{2}+2x-15 = (x+5)(x-3) = 0 \), giving \( x=-5 \) or \( x=3 \). Check: \( (-5)^{2}+2(-5)-15 = 25-10-15=0 \); \( 3^{2}+2(3)-15=9+6-15=0 \).

1 mark for the correct factorisation \( (x+5)(x-3) \); 1 mark for \( x=-5 \); 1 mark for \( x=3 \). Max 3.

The sale price represents \( 100\%-15\% = 85\% \) of the original price. So original price \( = £102 \div 0.85 = £120 \).

1 mark for recognising the sale price is 85% of the original; 1 mark for the correct method £102 \( \div \) 0.85; 1 mark for the correct answer £120. Max 3.

Class width \( = 35-20 = 15 \). Frequency density \( = \dfrac{\text{frequency}}{\text{class width}} = \dfrac{21}{15} = 1.4 \).

1 mark for correctly identifying the class width of 15; 1 mark for the correct method \( 21\div15 \); 1 mark for the correct answer 1.4. Max 3.

The angle in a semicircle is always a right angle. Since AC is a diameter, angle ABC (the angle at the circumference standing on the diameter) must be 90°.

1 mark for stating the correct reason (angle in a semicircle is 90°); 2 marks for the correct answer of 90°, clearly linked to AC being a diameter. Max 3.

Total parts \( = 2+3+5 = 10 \). Value of 1 part \( = £200 \div 10 = £20 \). Aiden \( = 2\times£20=£40 \); Bea \( = 3\times£20=£60 \); Carys \( = 5\times£20=£100 \).

1 mark for finding the value of one part (£20); 1 mark for all three correct shares (£40, £60, £100). Max 2.

Simple interest \( = P \times r \times t = 800 \times 0.035 \times 4 = £112 \).

1 mark for the correct method \( 800 \times 0.035 \times 4 \); 1 mark for the correct final answer £112. Max 2.

The sum of the interior angles of a polygon with n sides is \( (n-2)\times180^{\circ} \). For a heptagon, \( n=7 \): \( (7-2)\times180 = 5\times180 = 900^{\circ} \).

1 mark for the correct method \( (7-2)\times180 \); 1 mark for the correct answer 900°. Max 2.

(a) The change in bearing from the first leg (065°) to the second leg (155°) is \( 155-65=90^{\circ} \), so the second leg of the journey is at right angles to the first leg. (b) Since the two legs meet at a right angle, Pythagoras' theorem can be used: distance from port \( = \sqrt{40^{2}+30^{2}} = \sqrt{1600+900} = \sqrt{2500} = 50 \) km.

(a) 1 mark for correctly calculating \( 155-65=90^{\circ} \); 1 mark for correctly concluding the legs are at right angles. (b) 1 mark for identifying that Pythagoras' theorem applies here; 1 mark for correct substitution \( \sqrt{40^{2}+30^{2}} \); 1 mark for correct working \( \sqrt{2500} \); 1 mark for the final answer 50 km. Max 6.

(a) Time \( = \dfrac{\text{distance}}{\text{speed}} = \dfrac{4.0\times10^{16}}{3\times10^{8}} = 1.333\ldots \times 10^{8} \), which rounds to \( 1.33\times10^{8} \) s (3 s.f.). (b) Time in years \( = \dfrac{1.333\ldots\times10^{8}}{3.156\times10^{7}} = 4.2247\ldots \), which rounds to 4.22 years (3 s.f.).

(a) 1 mark for the correct method (distance ÷ speed); 1 mark for correct working; 1 mark for the final answer \( 1.33\times10^{8} \) s. (b) 1 mark for the correct method (dividing by \( 3.156\times10^{7} \)); 1 mark for correct working; 1 mark for the final answer 4.22 years. Max 6.

(a) We need two numbers multiplying to \( -14 \) and adding to \( -5 \): these are \( -7 \) and \( 2 \). So \( x^{2}-5x-14 = (x-7)(x+2) \). (b) Setting \( (x-7)(x+2)=0 \) gives \( x=7 \) or \( x=-2 \). (c) Using the factorisation from (a): \( \dfrac{x^{2}-5x-14}{x+2} = \dfrac{(x-7)(x+2)}{x+2} = x-7 \) (for \( x \ne -2 \)).

(a) 1 mark for the correct factorisation \( (x-7)(x+2) \). (b) 1 mark for \( x=7 \); 1 mark for \( x=-2 \). (c) 1 mark for using the factorisation from (a) as the method; 1 mark for correctly cancelling the common factor \( (x+2) \); 1 mark for the final simplified answer \( x-7 \). Max 6.

(a) Gradient \( m = \dfrac{14-2}{5-1} = \dfrac{12}{4} = 3 \). Substituting \( (1,2) \): \( 2 = 3(1)+c \Rightarrow c=-1 \). So the equation is \( y=3x-1 \). (b) Substituting \( x=3 \) into \( y=3x-1 \): \( y = 3(3)-1 = 8 \), which matches the given y-coordinate, so \( (3,8) \) does lie on the line.

(a) 1 mark for the correct gradient method; 1 mark for gradient \( m=3 \); 1 mark for correctly finding \( c=-1 \); 1 mark for the final equation \( y=3x-1 \). (b) 1 mark for correct substitution of \( x=3 \); 1 mark for the correct conclusion that (3, 8) lies on the line, with \( y=8 \) shown to match. Max 6.

(a) Area \( = \text{width} \times \text{length} = x(x+3) = 54 \), which expands to \( x^{2}+3x=54 \), i.e. \( x^{2}+3x-54=0 \), as required. (b) Using the quadratic formula with \( a=1,b=3,c=-54 \): \( x = \dfrac{-3\pm\sqrt{9+216}}{2} = \dfrac{-3\pm\sqrt{225}}{2} = \dfrac{-3\pm15}{2} \). So \( x=6 \) or \( x=-9 \). Since x is a length, it must be positive, so \( x=-9 \) is rejected and \( x=6 \). (c) Width \( = 6 \) cm, length \( = 6+3=9 \) cm. Check: \( 6\times9=54 \), correct.

(a) 1 mark for setting up \( x(x+3)=54 \); 1 mark for correctly expanding to \( x^{2}+3x-54=0 \). (b) 1 mark for correct substitution into the quadratic formula; 1 mark for correct discriminant/working (\( \sqrt{225}=15 \)); 1 mark for \( x=6 \) with \( x=-9 \) correctly rejected. (c) 1 mark for stating the dimensions 6 cm by 9 cm. Max 6.

(a) Price after increase \( = £250 \times 1.20 = £300 \). (b) Final price \( = £300 \times 0.85 = £255 \). (c) Overall change \( = £255-£250 = £5 \) increase. As a percentage of the original: \( \dfrac{5}{250}\times100 = 2\% \). So the overall change is an increase of 2%.

(a) 1 mark for correct method \( 250\times1.20 \); 1 mark for £300. (b) 1 mark for correct method \( 300\times0.85 \); 1 mark for £255. (c) 1 mark for correct method to find % change from original; 1 mark for the correct final answer of a 2% increase. Max 6.

(a) Frequency density = frequency ÷ class width. \( 0\text{–}5 \): \( 20\div5=4.0 \). \( 5\text{–}15 \): \( 35\div10=3.5 \). \( 15\text{–}30 \): \( 30\div15=2.0 \). \( 30\text{–}60 \): \( 15\div30=0.5 \). (b) Using midpoints 2.5, 10, 22.5, 45: estimated mean \( = \dfrac{(2.5\times20)+(10\times35)+(22.5\times30)+(45\times15)}{100} = \dfrac{50+350+675+675}{100} = \dfrac{1750}{100} = 17.5 \) minutes.

(a) 1 mark for each pair of correct frequency densities, up to 4 marks total for all four correct (4.0, 3.5, 2.0, 0.5). (b) 1 mark for using correct midpoints; 1 mark for the correct final estimated mean of 17.5 minutes. Max 6.

(a) Since PT is a tangent and OT is a radius, the angle between them is 90° (tangent is perpendicular to the radius at the point of contact), so angle OTP = 90°. In triangle OPT, the angles sum to 180°, so angle POT \( = 180-90-35 = 55^{\circ} \). (b) By the alternate segment theorem, the angle between a tangent and a chord equals the angle in the alternate segment. So the angle in the alternate segment \( = 48^{\circ} \).

(a) 1 mark for stating that the tangent is perpendicular to the radius (angle OTP = 90°); 1 mark for the correct method (angle sum of triangle OPT); 1 mark for the correct answer 55°. (b) 1 mark for correctly stating the alternate segment theorem; 1 mark for the correct answer 48°. Max 6.

P(rain) = 0.25 and P(no rain) = 0.75 each day, independently. (a) P(rain on both days) \( = 0.25 \times 0.25 = 0.0625 \). (b) P(exactly one day of rain) = P(rain, no rain) + P(no rain, rain) \( = (0.25\times0.75)+(0.75\times0.25) = 0.1875+0.1875 = 0.375 \).

(a) 1 mark for the correct method \( 0.25\times0.25 \); 1 mark for the correct answer 0.0625. (b) 1 mark for identifying both orders (rain-then-no-rain and no-rain-then-rain); 1 mark for correctly summing to give 0.375. Max 5.

For an enlargement centre the origin with scale factor 3, each coordinate is multiplied by 3: \( (1,1) \to (3,3) \); \( (3,1) \to (9,3) \); \( (1,2) \to (3,6) \).

1 mark for correctly applying scale factor 3 to each coordinate; 1 mark each for the three correct vertices (3,3), (9,3) and (3,6), max 3 further marks. Max 5.

(a) IQR \( = \text{upper quartile} - \text{lower quartile} = 70 - 35 = 35 \). (b) The distance from the lower quartile to the median is \( 42-35=7 \), while the distance from the median to the upper quartile is \( 70-42=28 \). Since the upper gap (28) is much larger than the lower gap (7), the distribution is positively skewed: most students' scores are clustered towards the lower end, with a smaller number of students achieving much higher scores, stretching the upper tail of the distribution. (c) Class 1 has a median of 42 and Class 2 has a median of 45, so Class 2's typical (middle) score is slightly higher than Class 1's. However, Class 1's IQR of 35 is much larger than Class 2's IQR of 20, meaning Class 1's scores are far more spread out (less consistent), while Class 2's students achieved more similar scores to one another, clustered more closely around their median. (d) From the table, 70 out of 80 students scored 80 or less, so the number of students who scored more than 80 \( = 80-70 = 10 \).

(a) 1 mark for the correct method (upper quartile − lower quartile); 1 mark for the correct answer 35. (b) 1 mark for correctly calculating both gaps (7 and 28); 1 mark for correctly identifying the larger upper gap; 1 mark for a valid, correctly reasoned conclusion of positive skew with most scores clustered lower and a tail of higher scores. (c) 1 mark for a valid comparison of the medians; 1 mark for a valid comparison of the IQRs; 1 mark for a correct overall conclusion (similar typical performance, but Class 2 more consistent/less spread than Class 1). (d) 1 mark for correct method (80 − 70); 1 mark for correct reasoning that this represents scores above the ≤80 boundary; 1 mark for the correct final answer of 10 students. Total: (a) 2, (b) 3, (c) 3, (d) 4. Max 12.

\( \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2} \).

1 mark for correctly splitting 72 into \( 36\times2 \); 1 mark for the final answer \( 6\sqrt{2} \). Max 2.

Let \( x = 0.454545\ldots \). Then \( 100x = 45.4545\ldots \). Subtracting: \( 100x - x = 45.4545\ldots - 0.4545\ldots \), so \( 99x = 45 \), giving \( x = \dfrac{45}{99} = \dfrac{5}{11} \) in simplest form.

1 mark for the correct method (multiplying by 100 and subtracting); 1 mark for the correct simplified fraction \( \frac{5}{11} \). Max 2.

This is the difference of two squares: \( x^{2}-49 = x^{2}-7^{2} = (x-7)(x+7) \).

1 mark for recognising the difference of two squares; 1 mark for the correct final factorisation \( (x-7)(x+7) \). Max 2.

The common difference is \( 5-2=3 \), so the sequence has the form \( 3n+c \). When \( n=1 \), the term is 2, so \( 3(1)+c=2 \Rightarrow c=-1 \). So the nth term is \( 3n-1 \). Check: \( n=2 \): \( 3(2)-1=5 \), correct.

1 mark for correctly identifying the common difference as 3; 1 mark for the correct final expression \( 3n-1 \). Max 2.

Starting with \( y=x^{2}-3 \), add 3 to both sides: \( y+3=x^{2} \). Taking the square root of both sides (taking the positive root since \( x>0 \)): \( x=\sqrt{y+3} \).

1 mark for correctly adding 3 to both sides (\( y+3=x^{2} \)); 1 mark for correctly taking the square root to give \( x=\sqrt{y+3} \). Max 2.

Using the index law \( a^{m}\times a^{n}=a^{m+n} \): \( 3^{-2}\times3^{4} = 3^{-2+4} = 3^{2} = 9 \).

1 mark for correctly applying the index law to give \( 3^{2} \); 1 mark for the correct final answer 9. Max 2.

Substituting the gradient \( m=-2 \) and the point \( (3,1) \) into \( y=mx+c \): \( 1 = -2(3)+c \Rightarrow 1=-6+c \Rightarrow c=7 \). So the equation is \( y=-2x+7 \). Check: at \( x=3 \), \( y=-2(3)+7=1 \), correct.

1 mark for correct substitution into \( y=mx+c \); 1 mark for correctly finding \( c=7 \); 1 mark for the final equation \( y=-2x+7 \). Max 3.

Adding the two equations together eliminates y: \( (2x+y)+(x-y) = 9+3 \Rightarrow 3x=12 \Rightarrow x=4 \). Substituting into \( x-y=3 \): \( 4-y=3 \Rightarrow y=1 \). Check: \( 2(4)+1=9 \), correct; \( 4-1=3 \), correct.

1 mark for the correct method of adding the equations to eliminate y; 1 mark for correctly finding \( x=4 \); 1 mark for correctly finding \( y=1 \). Max 3.

By Pythagoras' theorem, hypotenuse\(^2\) \( = 4^{2}+8^{2} = 16+64 = 80 \). So hypotenuse \( = \sqrt{80} = \sqrt{16\times5} = \sqrt{16}\times\sqrt{5} = 4\sqrt{5} \) cm.

1 mark for the correct method \( 4^{2}+8^{2} \); 1 mark for correct working to 80; 1 mark for identifying \( \sqrt{80}=\sqrt{16\times5} \); 1 mark for correctly simplifying \( \sqrt{16}=4 \); 1 mark for the exact final answer \( 4\sqrt{5} \) cm; 1 mark for correct units and full exactness (no decimal approximation used). Max 6.

Using a common denominator of \( x(x+1) \): \( \dfrac{3}{x} + \dfrac{2}{x+1} = \dfrac{3(x+1)}{x(x+1)} + \dfrac{2x}{x(x+1)} = \dfrac{3(x+1)+2x}{x(x+1)} = \dfrac{3x+3+2x}{x(x+1)} = \dfrac{5x+3}{x(x+1)} \).

1 mark for correctly identifying the common denominator \( x(x+1) \); 1 mark for correctly rewriting \( \frac{3}{x} \) as \( \frac{3(x+1)}{x(x+1)} \); 1 mark for correctly rewriting \( \frac{2}{x+1} \) as \( \frac{2x}{x(x+1)} \); 1 mark for correctly combining the numerators; 1 mark for correctly simplifying the numerator to \( 5x+3 \); 1 mark for the final answer as a single fraction \( \frac{5x+3}{x(x+1)} \). Max 6.

(a) The angle at the centre of a circle is twice the angle at the circumference standing on the same arc, so angle ACB \( = 130 \div 2 = 65^{\circ} \). (b) Since OA and OB are both radii, triangle OAB is isosceles with \( OA=OB \), so the base angles are equal: angle OAB = angle OBA. The angles of a triangle sum to 180°, so \( 2\times\text{angle OAB} = 180-70=110 \), giving angle OAB \( = 55^{\circ} \).

(a) 1 mark for stating the correct circle theorem (angle at centre is twice angle at circumference); 1 mark for the correct answer 65°. (b) 1 mark for correctly identifying triangle OAB as isosceles (equal radii); 1 mark for the correct method (angles of a triangle sum to 180°, base angles equal); 1 mark for the correct answer 55°. Max 5.

Midpoint of AB \( = \left(\dfrac{2+8}{2}, \dfrac{6+2}{2}\right) = (5, 4) \). Gradient of AB \( = \dfrac{2-6}{8-2} = \dfrac{-4}{6} = -\dfrac{2}{3} \). The perpendicular gradient is \( \dfrac{3}{2} \) (since gradients of perpendicular lines multiply to \( -1 \)). Substituting the midpoint \( (5,4) \) and gradient \( \frac{3}{2} \) into \( y=mx+c \): \( 4 = 1.5(5)+c \Rightarrow 4=7.5+c \Rightarrow c=-3.5 \). So the perpendicular bisector is \( y=1.5x-3.5 \).

1 mark for the correct midpoint (5, 4); 1 mark for the correct gradient of AB (\( -\frac{2}{3} \)); 1 mark for the correct perpendicular gradient (\( \frac{3}{2} \)); 1 mark for correct substitution of the midpoint; 1 mark for the final equation \( y=1.5x-3.5 \). Max 5.

\( x^{2}+6x+2 = (x+3)^{2}-9+2 = (x+3)^{2}-7 \). Setting this equal to zero: \( (x+3)^{2}-7=0 \Rightarrow (x+3)^{2}=7 \Rightarrow x+3=\pm\sqrt{7} \Rightarrow x=-3\pm\sqrt{7} \).

1 mark for correctly completing the square to \( (x+3)^{2}-7 \); 1 mark for correctly setting \( (x+3)^{2}=7 \); 1 mark for taking the square root of both sides (\( x+3=\pm\sqrt{7} \)); 1 mark for the correct final answers \( x=-3+\sqrt{7} \); 1 mark for \( x=-3-\sqrt{7} \). Max 5.

To enlarge triangle ABC by scale factor 2 from centre O using only a ruler and compasses: first draw a straight ray from O through vertex A, extending well beyond A. Open the compasses to the distance OA (placing the point on O and the pencil on A), then, keeping the compasses open to this same distance, step out from A along the same ray to mark a new point A' such that AA' = OA — this makes OA' = OA + AA' = 2 × OA, so A' lies on ray OA at exactly twice the distance from O as A. This process is then repeated for vertices B and C, drawing rays OB and OC and using the compasses to mark B' and C' at twice the distance from O along each ray. Joining A', B' and C' gives the enlarged triangle A'B'C'. All construction rays and arcs should be left visible as evidence of the method used.

1 mark for correctly describing drawing a ray from O through each vertex; 1 mark for correctly describing using compasses to measure the distance OA (or OB/OC); 1 mark for correctly describing stepping out an equal distance beyond A to double the length from O (giving OA'=2×OA); 1 mark for correctly repeating the process for all three vertices; 1 mark for correctly describing joining the image points to form triangle A'B'C', with construction arcs left visible. Max 5.

There are 4 kings in a standard pack of 52 cards. \( P(\text{king}) = \dfrac{4}{52} = \dfrac{1}{13} \).

1 mark for identifying 4 favourable outcomes out of 52; 1 mark for the correct simplified probability \( \frac{1}{13} \). Max 2.

Estimated probability \( = \dfrac{\text{number of sixes}}{\text{total rolls}} = \dfrac{28}{150} = 0.1866\ldots \), which rounds to 0.187 (3 s.f.).

1 mark for the correct method \( \frac{28}{150} \); 1 mark for the correct final answer 0.187 (3 s.f.). Max 2.

Mean \( = \dfrac{6+9+9+12+14}{5} = \dfrac{50}{5} = 10 \). Range \( = 14-6 = 8 \).

1 mark for the correct mean of 10; 1 mark for the correct range of 8. Max 2.

A mass given as 250 g to the nearest 10 g could have been rounded from any value from 245 g up to (but not including) 255 g. The lower bound is 245 g.

1 mark for correct method (half of 10 subtracted from 250); 1 mark for the correct lower bound of 245 g. Max 2.

Scale factor from 12 to 20 cupcakes \( = \dfrac{20}{12} = \dfrac{5}{3} \). Flour \( = 300 \times \frac{5}{3} = 500 \) g. Sugar \( = 200 \times \frac{5}{3} = 333.3\ldots \approx 333 \) g. Butter \( = 100 \times \frac{5}{3} = 166.6\ldots \approx 167 \) g.

1 mark for the correct scale factor \( \frac{5}{3} \); 1 mark for correctly scaling at least two ingredients; 1 mark for all three correct final amounts (500 g, 333 g, 167 g) to the nearest gram. Max 3.

Population after the 6% decrease \( = 4200 \times 0.94 = 3948 \). Population after the 6% increase \( = 3948 \times 1.06 = 4184.88 \), which rounds to 4185 (nearest whole number). Since \( 4185 < 4200 \), the final population is less than the original — a decrease followed by an equal percentage increase does not return to the original value, because the second percentage change is applied to a smaller starting number.

1 mark for correctly calculating the population after the decrease (3948); 1 mark for correctly calculating the population after the increase (4184.88, rounding to 4185); 1 mark for correctly stating this is less than the original 4200, with a valid reason. Max 3.

Area of the triangular cross-section \( = \dfrac{1}{2} \times 6 \times 4 = 12 \) cm\(^2\). Volume of the prism \( = \text{cross-sectional area} \times \text{length} = 12 \times 10 = 120 \) cm\(^3\).

1 mark for correctly calculating the cross-sectional area (12 cm²); 1 mark for the correct method (area × length); 1 mark for the correct final answer 120 cm³. Max 3.

For similar shapes, the ratio of areas is the square of the ratio of corresponding lengths. Length ratio \( 2:3 \) gives area ratio \( 2^{2}:3^{2}=4:9 \). Area of larger shape \( = 16 \times \dfrac{9}{4} = 36 \) cm\(^2\).

1 mark for correctly squaring the length ratio to get the area ratio 4:9; 1 mark for the correct method \( 16\times\frac{9}{4} \); 1 mark for the final answer 36 cm². Max 3.

Using the cosine rule: \( BC^{2} = AB^{2}+AC^{2}-2(AB)(AC)\cos(BAC) = 9^{2}+11^{2}-2(9)(11)\cos(48^{\circ}) = 81+121-198\cos(48^{\circ}) \). \( \cos(48^{\circ})=0.6691\ldots \), so \( BC^{2} = 202-198(0.6691\ldots) = 202-132.49\ldots = 69.51\ldots \). \( BC = \sqrt{69.51\ldots} = 8.337\ldots \), which rounds to 8.34 cm (3 s.f.).

1 mark for correctly identifying the cosine rule as the appropriate method (two sides and the included angle known); 1 mark for correct substitution; 1 mark for correct working to \( BC^{2}\approx69.5 \); 1 mark for correctly taking the square root; 1 mark for the final answer 8.34 cm (3 s.f.). Max 5.

Total mass \( = (1.67\times10^{-24}) \times (5\times10^{23}) = (1.67\times5) \times 10^{-24+23} = 8.35 \times 10^{-1} \) g.

1 mark for correctly multiplying 1.67 by 5 to give 8.35; 1 mark for correctly combining the powers of 10 (\( 10^{-24}\times10^{23}=10^{-1} \)); 1 mark for correctly identifying the method (multiplying the two standard form numbers); 1 mark for correct intermediate working shown; 1 mark for the correct final answer \( 8.35\times10^{-1} \) g. Max 5.

There are 10 counters in total. \( P(\text{both green}) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3} \), since after the first green counter is removed, 5 green counters remain out of 9 total.

1 mark for correct probability of the first draw (\( \frac{6}{10} \)); 1 mark for correctly adjusting the second draw for no replacement (\( \frac{5}{9} \)); 1 mark for the correct method (multiplying); 1 mark for correct working \( \frac{30}{90} \); 1 mark for the correct simplified final answer \( \frac{1}{3} \). Max 5.

(a) Volume of a cone \( = \dfrac{1}{3}\pi r^{2}h = \dfrac{1}{3}\pi(5)^{2}(12) = \dfrac{1}{3}\pi(300) = 100\pi = 314.15\ldots \), which rounds to 314 cm\(^3\) (3 s.f.). (b) The slant height \( l = \sqrt{r^{2}+h^{2}} = \sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169} = 13 \) cm. Curved surface area \( = \pi r l = \pi(5)(13) = 65\pi = 204.20\ldots \), which rounds to 204 cm\(^2\) (3 s.f.).

(a) 1 mark for the correct formula \( \frac{1}{3}\pi r^{2}h \); 1 mark for correct substitution; 1 mark for the final answer 314 cm³. (b) 1 mark for correctly finding the slant height using Pythagoras (13 cm); 1 mark for the final curved surface area 204 cm² using \( \pi rl \). Max 5.

Using the compound interest formula: \( A = P(1+r)^{t} = 2000(1.04)^{3} = 2000 \times 1.124864 = 2249.728 \), which rounds to £2249.73 (nearest penny).

1 mark for the correct compound interest formula \( P(1+r)^{t} \); 1 mark for correct substitution \( P=2000, r=0.04, t=3 \); 1 mark for correctly calculating \( (1.04)^{3}=1.124864 \); 1 mark for correct working to £2249.728; 1 mark for the final answer £2249.73 (nearest penny). Max 5.

Actual distance \( = 8.4 \times 25\,000 = 210\,000 \) cm. Converting to kilometres: \( 210\,000 \text{ cm} = 2100 \text{ m} = 2.1 \) km (since \( 1 \text{ km} = 100\,000 \text{ cm} \)).

1 mark for the correct method \( 8.4\times25\,000 \); 1 mark for correct working to 210 000 cm; 1 mark for correctly converting cm to m; 1 mark for correctly converting m to km; 1 mark for the final answer 2.1 km. Max 5.