2025 OCR GCSE Mathematics - J560 Practice Paper with Answers

Subtract \(3x\) from both sides to get \(2x+7=23\).
Subtract \(7\) from both sides: \(2x=16\).
Divide by \(2\): \(x=8\).

1 | M1 | Correctly collect the variable terms on one side and the constants on the other, giving \(2x=16\) or equivalent, seen or implied.
1 | A1 | \(x=8\), dependent on M1.
Accept any equivalent valid method. A correct answer without working earns both marks. Do not award the accuracy mark for an incorrect final value of \(x\). Total: 2 marks.

(a) Each ratio part represents \(28\div7=4\) tiles. There are \(4\times4=16\) square tiles.
(b) The fraction of tiles that are triangular is \(\frac{7}{4+7}=\frac{7}{11}\). Compare this with \(60\%=\frac{3}{5}\): \(\frac{7}{11}=\frac{35}{55}\) and \(\frac{3}{5}=\frac{33}{55}\). Since \(35>33\), the student is correct.

2(a) | B1 | \(16\).
2(b) | B1 | Yes, with a valid comparison establishing that the triangular proportion exceeds \(60\%\), for example \(\frac{7}{11}>\frac{3}{5}\), or \(28>0.6\times44\).
The marks are independent; no separate method marks. Accept an equivalent numerical explanation, including a correctly rounded percentage of about \(63.6\%\). Do not accept a bare decision or merely stating that there are more triangular than square tiles. No FT from an incorrect answer in (a). Total: 2 marks.

There are \(5+2=7\) volunteers who attended at least two sessions.
The total number of volunteers is \(3+10+5+2=20\).
The required fraction is \(\frac{7}{20}\).

3 | M1 | Uses \(5+2\) as the numerator and \(3+10+5+2\) as the denominator of a fraction, seen or implied; allow arithmetic errors in evaluating these sums.
3 | A1 | \(\frac{7}{20}\) or an equivalent fraction, dependent on M1.
A correct fraction without working earns both marks. A decimal or percentage alone does not earn A1; \(0.35\) or \(35\%\) with no working earns M1 by implication. Do not accept \(\frac{5}{20}\), which excludes volunteers attending three sessions. Total: 2 marks.

(a) Each of the \(3\) cards can be paired with either of the \(2\) coin results, giving \(3\times2=6\) possible outcomes.
(b) These six outcomes are equally likely. Only the outcome consisting of card \(5\) and tails satisfies both conditions, so the probability is \(\frac{1}{6}\).

4(a) | B1 | \(6\).
4(b) | B1 | \(\frac{1}{6}\) or equivalent exact probability.
The marks are independent; no separate method marks. Accept correct answers without working. Accept equivalent fractions or an exact recurring decimal. Do not accept odds notation or \(\frac{1}{3}\), which accounts only for the card. No FT from an incorrect answer in (a). Total: 2 marks.

(a) Multiply each term inside the brackets by \(3\): \(3\times 2x-3\times 5=6x-15\).
(b) The highest common factor of \(12\) and \(18\) is \(6\). Taking this outside the brackets gives \(12x+18=6(2x+3)\).

Reference | Answer or method | Marks | Guidance
1(a) | \(6x-15\) | B1 | Accept terms in either order. Do not accept an expression still containing brackets.
1(b) | Takes out a common integer factor greater than \(1\), correctly | M1 | For example, \(2(6x+9)\), \(3(4x+6)\) or \(6(2x+3)\), seen or implied.
1(b) | \(6(2x+3)\) | A1 | Dependent on M1. Accept an equivalent fully factorised form, including \(-6(-2x-3)\). An incompletely factorised answer earns M1 only. Correct final answer alone earns both marks.
Total: 3 marks.

(a) Tray B contains \(20\) seedlings, of which \(8\) have not flowered. The missing number is \(20-8=12\).
(b) The proportion that have flowered in tray A is \(18/30=3/5\). For tray B it is \(12/20=3/5\). Both proportions are \(60\%\), so the grower's statement is incorrect.

Reference | Answer or method | Marks | Guidance
2(a) | \(12\) | B1 | No working required.
2(b) | Compares \(18/30\) with \(12/20\) using a common basis | M1 | Accept equivalent fractions, percentages, decimals or a valid cross-multiplication comparison. FT the candidate's answer to (a) only if it is an integer from \(0\) to \(20\), using it as the numerator over \(20\).
2(b) | Incorrect, with equal proportions demonstrated | A1 | Dependent on M1. Accept \(3/5\), \(0.6\) or \(60\%\) for both trays. FT a correct decision and comparison based on the permitted value from (a). A decision without numerical support earns no marks in (b). Comparing the counts \(18\) and \(12\) alone earns no marks in (b).
Total: 3 marks.

(a) \(10\%\) of \(80\) is \(8\), and \(5\%\) is \(4\). Therefore \(15\%\) of \(80\) is \(8+4=12\).
(b) Keeping \(12\) tiles as spares leaves \(80-12=68\) tiles. Since \(68<72\), the design cannot be made while keeping all the spares. It is short of \(72-68=4\) tiles.

Reference | Answer or method | Marks | Guidance
3(a) | \(12\) | B1 | Accept correct answer without working.
3(b) | Calculates \(80-12\), or compares the \(8\) tiles left after making the design with the \(12\) required spares | M1 | FT the candidate's spare count from (a), provided it is an integer from \(0\) to \(80\).
3(b) | No, supported by \(68<72\), a shortage of \(4\), or \(8<12\) | A1 | Dependent on M1. FT a correct decision supported by the permitted spare count from (a). Do not award for an unsupported decision. Equivalent valid calculations earn the same credit.
Total: 3 marks.

(a) There are \(5+3=8\) parts in total. Yellow counters account for \(3\) of these parts, so the fraction is \(3/8\).
(b) The difference of \(14\) counters represents \(5-3=2\) parts. One part is \(14\div 2=7\) counters. There are \(5\times 7=35\) blue counters.

Reference | Answer or method | Marks | Guidance
4(a) | \(3/8\) | B1 | Accept any equivalent fraction. Do not accept \(3/5\).
4(b) | Uses \(14\div(5-3)\) to find the value of one part | M1 | Accept \(7\) seen or implied, or an equivalent valid method such as \(5k-3k=14\).
4(b) | \(35\) | A1 | Dependent on M1. Correct answer alone earns both marks. No FT from (a); (b) uses the original ratio.
Total: 3 marks.

(a) Each of the \(3\) results on A can occur with either of the \(2\) results on B. There are \(3\times 2=6\) ordered pairs.
(b) The equally likely pairs \((1,2),(1,4),(2,2),(2,4),(3,2),(3,4)\) give products \(2,4,4,8,6,12\), respectively. Four of the six products are multiples of \(4\). The probability is \(4/6=2/3\).

Reference | Answer or method | Marks | Guidance
5(a) | \(6\) | B1 | No working required.
5(b) | Identifies the four favourable pairs \((1,4),(2,2),(2,4),(3,4)\), or the four favourable product occurrences \(4,4,8,12\) | M1 | Accept a correct sample-space table with favourable outcomes indicated, or the equivalent calculation \(1/2+(1/2\times1/3)\). The repeated product \(4\) represents two distinct equally likely outcomes.
5(b) | \(2/3\) | A1 | Dependent on M1. Accept \(4/6\) or any exact equivalent. Correct answer alone earns both marks. No FT from an incorrect answer to (a). Do not treat the five distinct product values as equally likely.
Total: 3 marks.

(a) The square's perimeter is \(4\times11=44\) cm. The rectangle's perimeter is \(2(x+5)+2(x-1)\) cm. Therefore \(2(x+5)+2(x-1)=44\).
(b) Expanding and collecting terms gives \(4x+8=44\). Subtracting \(8\) gives \(4x=36\), and dividing by \(4\) gives \(x=9\). The rectangle then has sides \(14\) cm and \(8\) cm, giving perimeter \(44\) cm.

Reference | Answer or method | Marks | Guidance
6(a) | \(2(x+5)+2(x-1)=44\) | B1 | Accept any equivalent equation, including \(4x+8=44\) or \(2x+4=22\). An expression without an equality is insufficient.
6(b) | Correct inverse-operation step towards isolating \(x\), such as \(4x=36\) | M1 | FT the candidate's equation from (a) if it is a linear equation in \(x\) requiring at least two inverse operations to solve. Expanding brackets alone is insufficient.
6(b) | \(x=9\) | A1 | Dependent on M1; FT the correct solution of an eligible equation from (a). Working is required: an unsupported answer earns no marks in (b). If a valid equation is first written in (b), award the mark for (a) as well.
Total: 3 marks.

(a) Expanding the bracket gives \(12x-8-5x\). Collecting like terms gives \(7x-8\).
(b) The highest common factor of \(6y^2\) and \(15y\) is \(3y\). Dividing both terms by \(3y\) gives \(6y^2+15y=3y(2y+5)\).

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | \(12x-8-5x\) | M1 | Correct expansion, seen or implied. |
| (a) | \(7x-8\) | A1 | Dependent on M1; accept equivalent simplified form. |
| (b) | Takes out a common factor, for example \(3(2y^2+5y)\) or \(y(6y+15)\) | M1 | A correct factorisation with a non-trivial common factor, seen or implied. |
| (b) | \(3y(2y+5)\) | A1 | Dependent on M1. Must be fully factorised; accept equivalent order of factors. |
Total: 4 marks. Correct final answers without working earn full marks. No FT between parts.

(a) The known frequencies total \(6+9+4=19\), so \(k=32-19=13\).
(b) The greatest frequency is \(13\), which corresponds to \(2\) journeys.
(c) Half of \(32\) is \(16\). Only \(13\) people made exactly two journeys. Since \(13<16\), the student is incorrect.

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | \(13\) | B1 | Correct missing frequency. |
| (b) | \(2\) journeys | B1 | FT a positive integer value of \(k\) from (a): accept the corresponding mode, or both modes if tied. Do not accept a frequency as the mode. |
| (c) | Compares \(13\) with \(32\div2=16\), or compares \(13/32\) with \(1/2\) | M1 | Accept equivalent numerical comparison. |
| (c) | No, supported by \(13<16\) or equivalent | A1 | Dependent on M1. A bare decision earns no marks. FT a positive integer \(k\leq32\) from (a), using the stated total of \(32\), for a consistent comparison and decision. |
Total: 4 marks. In (c), accept an explanation that two journeys is the mode but its frequency is less than half the group, with numerical support.

(a) Convert the distance: \(1.8\text{ km}=1800\text{ m}\).
Time in seconds is \(1800\div1.2=1500\).
Time in minutes is \(1500\div60=25\).
(b) Adding \(25\) minutes to 14:38 gives 15:03. This is before 15:05, so the walker finishes in time.

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | \(1800\) m, or an equivalent correct conversion to compatible units | B1 | Seen or implied; for example, speed \(72\) m/min or \(0.072\) km/min. |
| (a) | Uses time = distance divided by speed with compatible units | M1 | Accept \(1800\div1.2\), \(1800\div72\) or \(1.8\div0.072\). FT an incorrect distance conversion if the correct time method is used with the intended units. |
| (a) | \(25\) minutes | A1 | Dependent on M1; answer must be in minutes. A correct unsupported answer earns all 3 marks. |
| (b) | Yes, with finish time 15:03 or comparison \(25<27\) minutes | B1 | Decision and supporting evidence required. FT a positive numerical time in minutes from (a), applied consistently. Finishing exactly at 15:05 counts as in time. |
Total: 4 marks. Do not accept \(1500\) as the final answer in minutes.

(a) The difference between the ratio parts is \(8-5=3\).
Each part represents \(21\div3=7\) beads.
There are \(5\times7=35\) small beads and \(8\times7=56\) large beads.
The total is \(35+56=91\).
(b) There are now \(35+14=49\) small beads and \(56\) large beads.
The ratio is \(49:56=7:8\).

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | Associates the difference \(8-5=3\) parts with \(21\) beads | M1 | Seen or implied by \(21\div3\). |
| (a) | Uses \(21\div3\) to find one part and scales to \(13\) parts, or finds both bead counts | M1 | Dependent on the first M1. Accept \((21\div3)\times(5+8)\) or equivalent; allow an arithmetic slip. |
| (a) | \(91\) | A1 | Dependent on both M marks. A correct unsupported answer earns all 3 marks. |
| (b) | \(7:8\) | B1 | Must be simplest form. FT incorrect positive whole-number bead counts explicitly found in (a), provided they are in the ratio \(5:8\): add \(14\) to the small count and simplify correctly. |
Total: 4 marks. Do not accept \(8:7\) in (b), since the order is specified.

(a) The length is \((x+9)\) cm, so the perimeter gives \(2x+2(x+9)=74\).
Simplifying gives \(4x+18=74\), then \(4x=56\), so \(x=14\).
(b) The length is \(14+9=23\) cm. The area is \(14\times23=322\text{ cm}^2\).

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | \(2x+2(x+9)=74\) | M1 | Accept an equivalent equation, such as \(x+(x+9)=37\). An equation must be shown for this mark. |
| (a) | Correct linear-equation method, for example \(4x=56\), followed by division by \(4\) | M1 | FT an incorrect linear equation from the context, provided it requires an additive and a multiplicative inverse operation. Allow arithmetic slips. |
| (a) | \(x=14\) | A1 | Dependent on both M marks. SC1 for a correct width without an equation or supporting algebra; do not add SC1 to other marks in (a). |
| (b) | \(322\) | B1 | FT a positive width \(w\) from (a) for the correctly evaluated area \(w(w+9)\). |
Total: 4 marks. Accept alternative correct algebraic methods. Do not accept \(14^2\) as the area method.

(a) The reduced price is \(85\%\) of the original price.
The original price is \(68\div0.85=80\), so it was £80.
(b) Increasing £68 by \(15\%\) gives \(68\times1.15=78.20\) pounds.
Since \(78.20<80\), the increase does not restore the original price.

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | \(68\div0.85\) or \(68\times100/85\) | M1 | Accept equivalent reverse-percentage method, seen or implied. |
| (a) | £80 | A1 | Dependent on M1. A correct unsupported answer earns both marks. |
| (b) | \(68\times1.15\) or \(68+0.15\times68\) | M1 | Accept equivalent calculation of the increased price, seen or implied by £78.20. |
| (b) | No, with £78.20 compared with £80 | A1 | Dependent on M1. Accept £78.2. FT the candidate's positive price from (a) only for the comparison and decision; the new price must be correct. |
Total: 4 marks. A decision alone earns no marks in (b). Do not accept adding \(15\%\) of £68 as the method for (a).

(a) Exactly one square can occur in two orders: square then triangle, or triangle then square.
Because the first token is replaced, the probabilities remain \(2/5\) for a square and \(3/5\) for a triangle.
Hence the required probability is \((2/5)(3/5)+(3/5)(2/5)=6/25+6/25=12/25\).
(b) The probability of two triangles is \((3/5)(3/5)=9/25\).
Since \(12/25>9/25\), the student is correct.

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| (a) | Multiplies \(2/5\) by \(3/5\) for one suitable order | M1 | Accept either order or an equivalent correctly weighted sample space. |
| (a) | Adds probabilities for both suitable orders, or doubles the probability of one order | M1 | Independent method mark; allow incorrect numerical probabilities if the two distinct orders are identified and combined correctly. |
| (a) | \(12/25\) | A1 | Dependent on both M marks. Accept \(0.48\), \(48\%\), or an equivalent fraction. SC1 for the correct answer without working; do not add SC1 to other marks in (a). |
| (b) | Yes, supported by \(9/25<12/25\) | B1 | Accept decimals or percentages. FT a probability between \(0\) and \(1\) from (a), compared correctly with \(9/25\), with a consistent decision. |
Total: 4 marks. Replacement is essential: do not award the first M1 for a product using denominator \(4\) on the second draw. In (b), a decision alone earns no marks.

Alternative for (a): M1 for a complete sample space of \(25\) equally likely ordered token pairs; M1 for identifying the \(12\) pairs containing exactly one square; A1 for \(12/25\).

(a) There are \(3+4+5=12\) ratio parts. Each part represents \(96\div12=8\) tiles, so the number of teal tiles is \(4\times8=32\).

(b) There are \(3\times8=24\) copper tiles and \(5\times8=40\) white tiles. Removing eight teal tiles leaves \(32-8=24\) teal tiles. The new ratio is \(24:24:40=3:3:5\). The student's ratio is incorrect.

| Part | Acceptable answer or stage | Marks | Guidance |
|---|---|---|---|
| (a) | \(96\div(3+4+5)\), or equivalent proportional method | M1 | Seen or implied by a correct answer. |
| (a) | \(32\) | A1 | Dependent on M1. Correct unsupported answer earns both marks. |
| (b) | Copper \(24\) and white \(40\) | M1 | Correct use of the original ratio to obtain both unchanged counts, seen or implied. |
| (b) | Teal \(32-8=24\) | M1 | Accept subtracting \(8\) from the candidate's positive numerical answer to (a), provided that answer exceeds \(8\). Independent of the other M1 in (b). |
| (b) | No, with \(24:24:40\) or \(3:3:5\) as justification | B1 | Accept an equivalent correct comparison, such as the copper:white ratio staying \(3:5\), whereas the claimed ratio has \(3:4\). No FT on this mark. A bare “No” earns no marks. |

Total: 5 marks. Award equivalent valid methods equally. A correct final ratio implies both method stages in (b); a correct decision supported only by the unchanged copper:white ratio earns B1 unless the two method stages are also shown.

(a) The number who chose modelling in the afternoon is \(25-15=10\). The number who chose painting in the morning is \(24-15=9\). The total who chose painting is \(9+6=15\).

(b) There are \(16\) afternoon attendees and a further \(9\) morning attendees who chose painting. This gives \(16+9=25\) eligible people. The probability is \(\frac{25}{40}=\frac{5}{8}\). Equivalently, subtract the \(15\) morning modelling attendees from \(40\).

| Part | Acceptable answer or stage | Marks | Guidance |
|---|---|---|---|
| (a) | Modelling, afternoon: \(10\) | B1 | Independent mark for the correct entry. |
| (a) | Painting, morning: \(9\) | B1 | Independent mark for the correct entry. |
| (a) | Painting, total: \(15\) | B1 | Independent mark for the correct entry; no FT. |
| (b) | \(16+9=25\), \(40-15=25\), or \(16+15-6=25\) | M1 | A method that counts the required people once, seen or implied. FT for \(16+p\), where \(p\) is the candidate's morning painting entry, or \(16+t-6\), where \(t\) is their painting total. |
| (b) | \(\frac{25}{40}=\frac{5}{8}\) | A1 | Dependent on M1. Accept equivalent fractions, \(0.625\) or \(62.5\%\). A1 FT for a probability obtained correctly from an allowed FT method, with denominator \(40\), provided the resulting eligible count is between \(0\) and \(40\). |

Total: 5 marks. A correct unsupported answer in (b) earns both marks. Do not accept \(31/40\) as a correct probability: it counts afternoon painting attendees twice.

(a) Multiplying each term by \(3\) gives \(3(2x-5)=6x-15\).

(b) The equation becomes \(6x-15=4x+9\). Subtracting \(4x\) and adding \(15\) on both sides gives \(2x=24\). Dividing by \(2\) gives \(x=12\).

(c) At \(x=12\), the left side is \(3(24-5)=57\), and the right side is \(48+9=57\). Equal values do not satisfy \(>\), so the student is incorrect.

| Part | Acceptable answer or stage | Marks | Guidance |
|---|---|---|---|
| (a) | \(6x-15\) | B1 | Accept \(-15+6x\). Do not accept \(6x-5\). |
| (b) | Correct operation on both sides to collect variable terms or constants | M1 | For example, \(2x-15=9\) or \(6x=4x+24\). Allow a consistent linear expansion error carried from (a). Merely rewriting the expanded equation earns no method mark. |
| (b) | Isolate the variable by dividing a collected equation by its non-zero coefficient of \(x\) | M1 | For example, \(x=24\div2\). May be implied by a correct solution following a collected equation. Allow a consistent earlier arithmetic or expansion error. |
| (b) | \(x=12\) | A1 | Dependent on both method marks. No FT on this accuracy mark. Because working is requested, a correct unsupported answer earns SC1 only, instead of M1M1A1. A correct alternative method with sufficient working earns all three marks. |
| (c) | No, because the two sides are equal at \(x=12\) | B1 | Accept both sides evaluated as \(57\), with the correct decision. Accept a clear reference to equality in (b). Do not require numerical evaluation if equality is explained. Do not accept a bare “No”. No FT from an incorrect answer to (b). |

Total: 5 marks. In (b), a valid trial method earns the first M1 for comparing both sides at a trial value and the second M1 for verifying equality at a candidate solution. The accurate verified value \(12\) then earns A1.

(a) There are \(5+3=8\) ratio parts. Each part represents \(64\div8=8\) beads, so there are \(5\times8=40\) dark beads.

(b) Originally there are \(64-40=24\) light beads. The \(40\) dark beads still represent \(5\) parts. The new number of light beads is \(40\times\frac65=48\). Therefore \(48-24=24\) light beads are added.

(c) The final ratio has \(5+6=11\) parts, of which \(5\) are dark. The fraction is \(\frac5{11}\).

| Part | Acceptable answer or method | Marks | Guidance |
|---|---|---|---|
| (a) | \(64\div(5+3)\times5\) | M1 | Accept an equivalent ratio method, seen or implied. |
| (a) | \(40\) | A1 | Dependent on M1. A correct answer alone earns both marks. |
| (b) | Original light count \(24\), or \(64-D\), where \(D\) is the candidate's dark count in (a) | B1 | Accept \(3\times8\), seen or implied. |
| (b) | New light count calculated as \(40\times\frac65=48\), or \(D\times\frac65\) | M1 | May use the original ratio-part size. Accept a direct method: an increase of \(6-3=3\) ratio parts. |
| (b) | \(48-24=24\) light beads added | A1 FT | Dependent on M1 and subtraction of the original light count. FT \(\frac65D-(64-D)\) only when \(D\) is a positive whole number less than \(64\), and both the resulting new light count and added count are non-negative whole numbers. The direct method \((6-3)\times8=24\) earns all three marks. SC1 for an unsupported final answer of \(24\), instead of the three listed marks. |
| (c) | \(\frac5{11}\) | B1 | Accept an equivalent fraction. Do not accept \(\frac56\) or a ratio. |
Total: 6 marks. M denotes method; A denotes dependent accuracy; B denotes independent credit; FT denotes the stated follow-through only.

(a) Add the fixed fee to the hourly charge: the cost is \(7+3h\) pounds.

(b) Equate the costs: \(7+3h=4+4h\). Subtract \(3h\) and then subtract \(4\) to obtain \(h=3\). Both options then cost £\(16\).

(c) For \(2\) hours, option A costs \(7+3\times2=13\) pounds and option B costs \(4+4\times2=12\) pounds. The customer is incorrect because option B is cheaper for this booking.

| Part | Acceptable answer or method | Marks | Guidance |
|---|---|---|---|
| (a) | \(7+3h\) | B1 | Accept an equivalent expression. Do not accept \(10h\). |
| (b) | \(7+3h=4+4h\) | M1 | An equation equating both correct costs is required. No FT from an incorrect expression in (a). |
| (b) | A correct rearrangement towards isolating \(h\), such as \(7-4=4h-3h\) | M1 | Dependent on the first M1. This step may be implied by \(h=3\) following the correct equation. |
| (b) | \(h=3\) | A1 | Dependent on both M marks. An answer of \(3\) alone earns no marks because an equation and its solution are required. |
| (c) | Option A costs £\(13\) and option B costs £\(12\) | B1 | Both correct costs required, seen or implied. Alternatively accept the correct comparison calculation \((7+3\times2)-(4+4\times2)=1\). |
| (c) | No, supported by option B being cheaper for this booking | B1 | The comparison must be supported by the correct costs or by a correct statement that B is £\(1\) cheaper. A bare decision earns no marks. |
Total: 6 marks. Accept equivalent algebraic rearrangements.

(a) The class midpoints are \(5\), \(15\) and \(25\) minutes.
The estimated total preparation time is \(6\times5+10\times15+8\times25=30+150+200=380\) minutes.
The estimated mean is \(380\div24=15.833\ldots\), which rounds to \(15.8\) minutes.

(b) With \(24\) times, the median is the mean of the \(12\)th and \(13\)th times in order. The cumulative frequencies are \(6\), \(16\) and \(24\), so both times are in \(10

| Part | Acceptable answer or method | Marks | Guidance |
|---|---|---|---|
| (a) | Midpoints \(5,15,25\) | B1 | All three correct, seen or implied by correct products. |
| (a) | \(6\times5+10\times15+8\times25\) | M1 | Accept a sum of frequency multiplied by a representative value from each corresponding interval. Correct midpoint products are \(30,150,200\). |
| (a) | Their weighted total divided by \(24\) | M1 | Dependent on the preceding M1. Do not award for division by \(3\). |
| (a) | \(15.8\) minutes | A1 | Dependent on both M marks and use of the correct midpoints. Do not accept an unrounded final answer. SC1 for the correct answer without working, instead of the four listed marks. Full credit requires working showing midpoint weighting and division by the total frequency. |
| (b) | \(10

(a) The probability of a star on one spin is \(\frac35\). By independence, the probability of two stars is \(\frac35\times\frac35=\frac9{25}\).

(b) Exactly one star can occur as star then circle, or circle then star. These are distinct outcomes, so their probabilities are added:
\(\frac35\times\frac25+\frac25\times\frac35=\frac6{25}+\frac6{25}=\frac{12}{25}\).

(c) Different symbols occur precisely when there is exactly one star. Matching symbols therefore have probability \(1-\frac{12}{25}=\frac{13}{25}\). Since \(\frac{13}{25}>\frac{12}{25}\), matching symbols are more likely. The player is incorrect.

| Part | Acceptable answer or method | Marks | Guidance |
|---|---|---|---|
| (a) | \(\frac35\times\frac35\) | M1 | Accept an equivalent enumeration of equally likely outcomes, seen or implied. |
| (a) | \(\frac9{25}\) | A1 | Dependent on M1. A correct answer alone earns both marks. |
| (b) | One ordered outcome: \(\frac35\times\frac25\) or \(\frac25\times\frac35\) | M1 | Accept \(\frac6{25}\), seen or implied in working. |
| (b) | Adds the probabilities for star-circle and circle-star, or doubles the single ordered probability | M1 | Dependent on the first M1. Accept the complement method \(1-(\frac9{25}+\frac4{25})\) for both M marks. |
| (b) | \(\frac{12}{25}\) | A1 | Dependent on both M marks. SC1 for the correct answer without working, instead of the three listed marks. Full marks require visible working. Do not accept \(\frac6{25}\) as the final answer. |
| (c) | No, with \(\frac{12}{25}<\frac{13}{25}\) and probabilities correctly identified | B1 | Accept “No, the probability of different symbols is \(\frac{12}{25}<\frac12\).” A bare decision earns no marks. No FT from an incorrect answer in (b). |
Total: 6 marks. Accept equivalent fractions, exact decimals or percentages for probabilities. Equivalent valid methods receive the same credit.

(a) Area equals width multiplied by length, so \(x(x+5)=84\). Expanding and rearranging gives \(x^2+5x-84=0\).

(b) Factorise: \((x+12)(x-7)=0\). Hence \(x=-12\) or \(x=7\). A width must be positive, so reject \(x=-12\). The panel is \(7\) cm wide and \(7+5=12\) cm long.

(c) The second panel is \(7+2=9\) cm wide and \(12-2=10\) cm long. Its area is \(9\times10=90\text{ cm}^2\). The designer is incorrect about the area: it increases by \(90-84=6\text{ cm}^2\). Both perimeters are \(38\) cm, but equal perimeters do not imply equal areas.

Point-based marking scheme. Total: 7 marks. M = method; A = accuracy dependent on the associated method. Award equivalent valid methods equivalent credit.

| Part | Acceptable answer or method | Marks | Guidance |
| --- | --- | --- | --- |
| 1(a) | \(x(x+5)=84\), or equivalent area equation | M1 | Seen or implied by a correct final equation. Do not accept a perimeter equation as an area method. |
| 1(a) | \(x^2+5x-84=0\) | A1 | Dependent on M1. Accept a non-zero scalar multiple in the requested form. A correct final equation without intermediate working earns both marks. |
| 1(b) | \((x+12)(x-7)=0\), or valid use of the quadratic formula | M1 | Accept any valid method for solving the equation from (a). For the formula method, require correct substitution into the formula. |
| 1(b) | \(x=7\) and \(x=-12\) | A1 | Dependent on M1. Both roots required for this mark. FT both correctly calculated real roots of the candidate's quadratic from (a). |
| 1(b) | Width \(7\) cm, length \(12\) cm | A1 | Dependent on M1, but not on the preceding A1. Rejection of the negative root may be implied by selecting the positive dimensions. FT a unique positive root of the candidate's quadratic from (a), with length equal to width plus \(5\) cm. Correct dimensions with no solving method earn 0 marks in (b), since working is requested; correct factorisation and dimensions with the negative root omitted earn 2 marks. |
| 1(c) | \((7+2)(12-2)\), giving the new area, or equivalent comparison | M1 | Accept \(9\times10\) or \(90\text{ cm}^2\) seen. Alternatively accept expansion of \((x+2)(x+3)\) for comparison with \(x(x+5)\). FT positive dimensions from (b), provided the new length remains positive. |
| 1(c) | Incorrect: new area \(90\text{ cm}^2\), compared with \(84\text{ cm}^2\); or area increases by \(6\text{ cm}^2\) | A1 | Dependent on M1 and a supported decision. Accept a correct algebraic demonstration that the area increases by \(6\text{ cm}^2\). FT only when both original and new areas are calculated consistently from the candidate's dimensions in (b), followed by a correct comparison and decision. A bare decision or a perimeter calculation alone earns 0 marks in (c). |

No special-case marks. Do not award marks beyond each part's allocation.