2022 OCR GCSE Mathematics - J560 模擬試題連答案詳解

Group like terms together:

\(6a + 3a = 9a\)

\(-2b + 7b = +5b\)

Combining the terms gives \(9a + 5b\).

B1 for 9a or +5b seen in the final expression
B1 for 9a + 5b as final answer

Method 1 (multiplier):
\(0.15 \times 340 = 51\)

Method 2 (chunking):
\(10\%\text{ of } 340 = 34\)
\(5\%\text{ of } 340 = 17\)
\(15\%\text{ of } 340 = 34 + 17 = 51\)

Answer is £51.

M1 for 340 × 0.15 oe or for a complete correct chunking method (e.g. 10% = 34 and 5% = 17)
A1 for 51

To write 0.000305 in standard form \(A \times 10^n\), where \(1 \le A < 10\):

Move the decimal point 4 places to the right to get \(3.05\).

Since the original number is less than 1, the power is negative: \(3.05 \times 10^{-4}\).

B1 for 3.05 × 10^-4 (condone 3.05 · 10^-4)

Total number of parts = \(3 + 4 = 7\)

Value of one part = \(84 \div 7 = 12\)

\(3\text{ parts} = 3 \times 12 = 36\)
\(4\text{ parts} = 4 \times 12 = 48\)

So the shared amounts are £36 and £48.

M1 for 84 ÷ (3 + 4) or 84 ÷ 7 (soi by 12)
A1 for £36 and £48 in the correct order (accept 36 : 48 or 36, 48)

Multiples of 12: 12, 24, 36, 48, 60, 72, ...
Multiples of 15: 15, 30, 45, 60, 75, ...
The lowest common multiple is 60.

M1 for listing multiples of both 12 and 15 with at least 3 multiples of each, or correct prime factor trees/decomposition (\(12 = 2^2 \times 3\), \(15 = 3 \times 5\))
A1 for 60

Collect the \(p\) terms: \(4p - 9p = -5p\).
Collect the \(q\) terms: \(7q + 2q = 9q\).
Combine together: \(-5p + 9q\) (or \(9q - 5p\)).

B1 for \(-5p\) or \(+9q\) in a two-term linear expression
B1 for \(-5p + 9q\) or \(9q - 5p\) as final answer

10% of £140 = £14
5% of £140 = £7
15% of £140 = 14 + 7 = £21.

B1 for 21 (condone £21)

Move the decimal point 4 places to the right to obtain a number between 1 and 10: 3.08.
Since the original number is less than 1, the index is negative: \(3.08 \times 10^{-4}\).

B1 for \(3.08 \times 10^{-4}\)

1. Find the difference in ratio parts between green and red counters:
\(5 - 3 = 2\text{ parts}\)

2. Find the value of one part:
\(2\text{ parts} = 28\)
\(1\text{ part} = 28 \div 2 = 14\)

3. Calculate the total number of parts:
\(3 + 5 + 4 = 12\text{ parts}\)

4. Calculate the total number of counters:
\(12 \times 14 = 168\)

M1 for \(5 - 3 [= 2]\) or \(28 \div (5 - 3)\) oe (soi by 14)
M1 for \((3 + 5 + 4) \times \text{their } 14\) oe
A1 for 168

1. Express the sale price as a percentage of the original price:
\(100\% - 15\% = 85\%\)

2. Set up the equation:
\(85\% = £459\)

3. Divide by 0.85 to find the original (100%) price:
\(459 \div 0.85 = 540\)

The original price is £540.

M1 for \(100 - 15 = 85\) or 0.85 seen
M1 for \(459 \div 0.85\) oe
A1 for 540 (condone £540)

1. Set up the formula for inverse proportion:
\(y = \frac{k}{x}\)

2. Substitute the given values to find constant \(k\):
\(15 = \frac{k}{8} \implies k = 15 \times 8 = 120\)

3. Find \(y\) when \(x = 24\):
\(y = \frac{120}{24} = 5\)

M1 for \(y = \frac{k}{x}\) or \(k = 15 \times 8\) (soi by 120)
M1 for \(\text{their } 120 \div 24\) oe
A1 for 5

1. Multiply the numerical parts:
\(2.4 \times 7.5 = 18\)

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

3. Combine and convert to standard form:
\(18 \times 10^3 = (1.8 \times 10^1) \times 10^3 = 1.8 \times 10^4\)

M1 for \(2.4 \times 7.5 = 18\) or \(10^5 \times 10^{-2} = 10^3\) or \(240000 \times 0.075\)
M1 for \(18 \times 10^3\) or \(18000\) seen
A1 for \(1.8 \times 10^4\)

Since \(y\) is inversely proportional to \(x\):
\[y = \frac{k}{x}\]
Substitute \(x = 5\) and \(y = 12\) to find the constant of proportionality \(k\):
\[12 = \frac{k}{5} \implies k = 12 \times 5 = 60\]
Substitute \(x = 15\) into the formula:
\[y = \frac{60}{15} = 4\]

M1 for \(y = \frac{k}{x}\) or \(k = 12 \times 5\) soi by 60
M1 for \(y = \frac{\text{their } 60}{15}\)
A1 for 4

The sale price corresponds to \(100\% - 15\% = 85\%\) of the normal price.

\[85\% = £442\]
\[1\% = £442 \div 85 = £5.20\]
\[100\% = £5.20 \times 100 = £520\]

Alternatively:
\[\text{Normal price} = \frac{442}{0.85} = £520\]

M2 for \(442 \div 0.85\) oe
or M1 for \(100 - 15 = 85\) or \(0.85\) seen or \(442\) associated with \(85\%\)
A1 for 520

1. Find the value of 1 ratio part:
Difference in parts = \(5 - 3 = 2\) parts.
\(2\text{ parts} = £420\)
\(1\text{ part} = 420 \div 2 = £210\).

2. Calculate the total money received by Charity A:
\(\text{Charity A} = 3 \times 210 = £630\).

3. Calculate the percentage and amount spent on educational books:
Percentage on books = \(100\% - 35\% = 65\%\).
\(\text{Amount on books} = 0.65 \times 630 = £409.50\).

M1 for \(5 - 3 = 2\) parts seen or implied by \(420 \div 2\)
A1 for finding 1 part = 210
M1 for \(3 \times 210\) or 630
M1 for \(0.65 \times \text{their } 630\) or \(\text{their } 630 - (0.35 \times \text{their } 630)\) oe
A1 for £409.50 (condone 409.5)

1. Find the volume of the cylinder:
\(V_{\text{cylinder}} = \pi r^2 h = \pi \times 4^2 \times 15 = 240\pi \approx 753.982\text{ cm}^3\).

2. Find the volume of the cone:
\(V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times \pi \times 4^2 \times 9 = 48\pi \approx 150.796\text{ cm}^3\).

3. Find the volume of the remaining solid:
\(V = 240\pi - 48\pi = 192\pi \approx 603.186\text{ cm}^3\).

4. Calculate the mass in grams:
\(\text{Mass} = \text{Density} \times \text{Volume} = 7.8 \times 192\pi \approx 4704.85\text{ g}\).

5. Convert to kilograms and round to 2 decimal places:
\(\text{Mass in kg} = 4704.85 \div 1000 = 4.70485\text{ kg} \approx 4.70\text{ kg}\).

M1 for volume of cylinder \(\pi \times 4^2 \times 15\) soi \(240\pi\) or 753.98...
M1 for volume of cone \(\frac{1}{3}\pi \times 4^2 \times 9\) soi \(48\pi\) or 150.79...
M1 for subtracting cone volume from cylinder volume soi \(192\pi\) or 603.18...
M1 for their volume \(\times 7.8 \div 1000\) oe
A1 for 4.70 (accept answers in range 4.70 to 4.71)

(a) Since \(I \propto \frac{1}{d^2}\), we have \(I = \frac{k}{d^2}\).
Substitute \(d = 3\) and \(I = 80\):
\(80 = \frac{k}{3^2} \implies 80 = \frac{k}{9} \implies k = 720\).

So the formula is \(I = \frac{720}{d^2}\).
When \(I = 45\):
\(45 = \frac{720}{d^2} \implies d^2 = \frac{720}{45} = 16\).
Since distance must be positive, \(d = \sqrt{16} = 4\text{ m}\).

(b) If distance \(d\) is replaced by \(\frac{d}{2}\), the new intensity is:
\(I_{\text{new}} = \frac{k}{\left(\frac{d}{2}\right)^2} = \frac{k}{\frac{d^2}{4}} = 4 \times \frac{k}{d^2} = 4I\).
Therefore, the intensity is quadrupled (multiplied by 4).

(a) M1 for \(I = \frac{k}{d^2}\) oe or \(80 \times 3^2\)
A1 for \(k = 720\)
A1 for \(d = 4\)
(b) B1 for stating it increases by a factor of 4 / quadruples / is 4 times greater oe

1. Identify the probability of taking the bus:
\(P(\text{Bus}) = 1 - 0.7 = 0.3\).

2. Find the on-time probability for each mode of transport:
- If cycling: \(P(\text{On time} \mid \text{Cycle}) = 1 - 0.08 = 0.92\).
- If taking bus: \(P(\text{On time} \mid \text{Bus}) = 1 - 0.25 = 0.75\).

3. Calculate the combined probability of arriving on time:
\(P(\text{Cycle and On time}) = 0.7 \times 0.92 = 0.644\)
\(P(\text{Bus and On time}) = 0.3 \times 0.75 = 0.225\)

4. Total probability on time:
\(P(\text{On time}) = 0.644 + 0.225 = 0.869\).

B1 for \(P(\text{Bus}) = 0.3\) or \(P(\text{On time} \mid \text{Cycle}) = 0.92\) or \(P(\text{On time} \mid \text{Bus}) = 0.75\) soi
M1 for \(0.7 \times 0.92\) soi by 0.644 OR \(0.3 \times 0.75\) soi by 0.225
M1 for \((0.7 \times 0.92) + (0.3 \times 0.75)\) oe
OR for \(1 - [(0.7 \times 0.08) + (0.3 \times 0.25)] = 1 - [0.056 + 0.075] = 1 - 0.131\)
A1 for 0.869 oe (e.g. 86.9% or \(\frac{869}{1000}\))

1. Volume of the outer cylinder:
\[ V_1 = \pi \times 3^2 \times 8 = 72\pi \approx 226.19\text{ cm}^3 \]

2. Volume of the cylindrical hole:
\[ V_2 = \pi \times 1^2 \times 8 = 8\pi \approx 25.13\text{ cm}^3 \]

3. Volume of remaining metal:
\[ V_{\text{metal}} = 72\pi - 8\pi = 64\pi \approx 201.06\text{ cm}^3 \]

4. Volume of one sphere:
\[ V_{\text{sphere}} = \frac{4}{3}\pi \times 1.5^3 = \frac{4}{3}\pi \times 3.375 = 4.5\pi \approx 14.137\text{ cm}^3 \]

5. Maximum number of complete spheres:
\[ \frac{64\pi}{4.5\pi} = \frac{64}{4.5} = \frac{128}{9} \approx 14.22 \]
Since only complete spheres can be made, the maximum number is 14.

M1 for \(\pi \times 3^2 \times 8\) [soi by \(72\pi\) or \(226.19\ldots\)]
M1 for \(\pi \times 1^2 \times 8\) [soi by \(8\pi\) or \(25.13\ldots\)]
M1 for \(\frac{4}{3} \times \pi \times 1.5^3\) [soi by \(4.5\pi\) or \(14.13\ldots\) to \(14.14\)]
M1 for their remaining volume \(\div\) their sphere volume [soi by \(14.22\ldots\)]
A1 for 14 cao

1. Find the area of the lawn:
\[ \text{Lawn area} = 0.40 \times 360 = 144\text{ m}^2 \]

2. Find the remaining area for vegetables and flowers:
\[ 360 - 144 = 216\text{ m}^2 \]

3. Divide the remaining area in the ratio \(3 : 5\):
\[ \text{Total parts} = 3 + 5 = 8 \]
\[ 1\text{ part} = 216 \div 8 = 27\text{ m}^2 \]
\[ \text{Flower area} = 5 \times 27 = 135\text{ m}^2 \]

4. Find the number of packets of seeds needed:
\[ 135 \div 12 = 11.25 \]
Since whole packets must be purchased, 12 packets are needed.

5. Calculate the total cost:
\[ 12 \times £4.75 = £57 \]

M1 for \(360 - (0.40 \times 360)\) oe [soi by 216]
M1 for their \(216 \div (3 + 5) \times 5\) oe [soi by 135]
M1 for their \(135 \div 12\) [soi by \(11.25\)] rounded up to 12 packets
A1 for 57 or 57.00 cao

(a) Acceleration is given by the gradient of the velocity-time graph:
\[ \text{Acceleration} = \frac{\text{Change in velocity}}{\text{Time taken}} = \frac{12 - 0}{10} = 1.2\text{ m/s}^2 \]

(b) The distance travelled is the total area under the velocity-time graph.
The shape is a trapezium with parallel sides of length \(b_1 = 15\text{ s}\) (from \(t=10\) to \(t=25\)) and \(b_2 = 40\text{ s}\), and height \(h = 12\text{ m/s}\):
\[ \text{Area} = \frac{1}{2}(a + b)h = \frac{1}{2}(15 + 40) \times 12 = \frac{1}{2} \times 55 \times 12 = 330\text{ m} \]
Alternatively, split into three sections:
- Triangle 1 (0 to 10 s): \(\frac{1}{2} \times 10 \times 12 = 60\text{ m}\)
- Rectangle (10 to 25 s): \(15 \times 12 = 180\text{ m}\)
- Triangle 2 (25 to 40 s): \(\frac{1}{2} \times 15 \times 12 = 90\text{ m}\)
Total distance \(= 60 + 180 + 90 = 330\text{ m}\).

(c) Average speed is given by:
\[ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} = \frac{330}{40} = 8.25\text{ m/s} \]

(a) [1 mark]
B1 for 1.2 or \(\frac{6}{5}\) oe

(b) [2 marks]
M1 for a complete method to find the area under the graph, e.g. \(\frac{1}{2}(15 + 40) \times 12\) or \(\frac{1}{2}(10)(12) + (15)(12) + \frac{1}{2}(15)(12)\) soi by two correct sub-areas (60, 180, or 90)
A1 for 330

(c) [2 marks]
M1 for \(\text{their } 330 \div 40\)
A1 for 8.25 or \(\frac{33}{4}\) oe (FT their distance)

(a) Set \(y = 0\):
\[ x^2 - 4x - 5 = 0 \]
Factorise the quadratic:
\[ (x - 5)(x + 1) = 0 \]
So \(x = 5\) or \(x = -1\).
The coordinates are \((-1, 0)\) and \((5, 0)\).

(b) By symmetry, the turning point occurs halfway between the \(x\)-intercepts:
\[ x = \frac{-1 + 5}{2} = 2 \]
Substitute \(x = 2\) into the curve's equation:
\[ y = (2)^2 - 4(2) - 5 = 4 - 8 - 5 = -9 \]
So the coordinates of the turning point are \((2, -9)\).

(c) The line of symmetry is the vertical line passing through the turning point:
\[ x = 2 \]

(a) [2 marks]
M1 for factorising into \((x - 5)(x + 1)\) or use of the quadratic formula \(\frac{4 \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2}\)
A1 for \((-1, 0)\) and \((5, 0)\) or \(x = -1, x = 5\)

(b) [2 marks]
M1 for finding \(x = 2\) (midpoint of roots or \(-\frac{b}{2a}\)) and substituting into the equation
A1 for \((2, -9)\)

(c) [1 mark]
B1 for \(x = 2\) oe

(a) The gradient \(m\) is calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - (-11)}{4 - (-2)} = \frac{18}{6} = 3 \]

(b) Parallel lines have the same gradient, so line \(L_2\) has gradient \(m = 3\).
The equation of line \(L_2\) can be written as \(y = 3x + c\).
Substitute the coordinates of the point \((3, 2)\):
\[ 2 = 3(3) + c \]
\[ 2 = 9 + c \]
\[ c = 2 - 9 = -7 \]
Thus, the equation of the line is:
\[ y = 3x - 7 \]

(a) [2 marks]
M1 for \(\frac{7 - (-11)}{4 - (-2)}\) oe
A1 for 3

(b) [3 marks]
M1 for using gradient \(m = 3\) or their (a) in \(y = mx + c\) or \(y - y_1 = m(x - x_1)\)
M1 for substituting \((3, 2)\) into their linear equation
A1 for \(y = 3x - 7\) (FT their gradient for M marks only)

First, find the critical values by solving the corresponding quadratic equation:
\[2x^2 - 7x - 15 = 0\]
Factorise the quadratic expression:
\[(2x + 3)(x - 5) = 0\]
This gives critical values of:
\[x = -\frac{3}{2} = -1.5 \quad \text{and} \quad x = 5\]
Since we require the quadratic to be less than or equal to 0 (\leqslant 0), the solution lies between and includes the two critical values:
\[-1.5 \leqslant x \leqslant 5\]
Expressing this in set notation gives:
\[\{x : -1.5 \leqslant x \leqslant 5\}\]

M2 for \((2x + 3)(x - 5)\) or correct use of the quadratic formula \(\frac{7 \pm \sqrt{(-7)^2 - 4(2)(-15)}}{2(2)}\)
(M1 for brackets that expand to give two correct terms or quadratic formula with at most one error)
B1 for critical values \(-1.5\) (or \(-\frac{3}{2}\)) and \(5\)
B1 for \(-1.5 \leqslant x \leqslant 5\) (or \(-\frac{3}{2} \leqslant x \leqslant 5\)) with correct working
B1 for fully correct set notation \(\{x : -1.5 \leqslant x \leqslant 5\}\)
(Award SC2 for \(\{x : -1.5 \leqslant x \leqslant 5\}\) with no or insufficient working)

Factorise the numerator:
\[2x^2 + 5x - 12 = (2x - 3)(x + 4)\]
Factorise the denominator:
\[6x^2 - 7x - 3 = (2x - 3)(3x + 1)\]
Substitute the factorised expressions into the fraction:
\[\frac{(2x - 3)(x + 4)}{(2x - 3)(3x + 1)}\]
Cancel the common factor \((2x - 3)\):
\[\frac{x + 4}{3x + 1}\]

M2 for factorising numerator as \((2x - 3)(x + 4)\)
(M1 for brackets that expand to give two correct terms of \(2x^2 + 5x - 12\))
AND
M2 for factorising denominator as \((2x - 3)(3x + 1)\)
(M1 for brackets that expand to give two correct terms of \(6x^2 - 7x - 3\))
AND
A1 for \(\frac{x + 4}{3x + 1}\) final answer

Rearrange the inequality to make the right-hand side 0:
\[6x^2 + 7x - 5 > 0\]
Factorise the quadratic expression:
\[(2x - 1)(3x + 5) > 0\]
Find the critical values:
\[2x - 1 = 0 \implies x = \frac{1}{2} = 0.5\]
\[3x + 5 = 0 \implies x = -\frac{5}{3}\]
Since we require the quadratic expression to be greater than 0 (\(> 0\)), the solutions lie outside the critical values:
\[x < -\frac{5}{3} \quad \text{or} \quad x > \frac{1}{2}\]
In set notation, this is written as:
\[\{x : x < -\frac{5}{3}\} \cup \{x : x > \frac{1}{2}\}\]

M1 for rearranging to \(6x^2 + 7x - 5 > 0\) or \(6x^2 + 7x - 5 = 0\)
M1 for factorising as \((2x - 1)(3x + 5)\) or using formula with at most one error
B1 for critical values \(-\frac{5}{3}\) and \(\frac{1}{2}\) (or \(0.5\))
B1 for \(x < -\frac{5}{3}\) or \(x > \frac{1}{2}\) (accept with 'and')
B1 for fully correct set notation \(\{x : x < -\frac{5}{3}\} \cup \{x : x > \frac{1}{2}\}\) (also accept \(\{x : x < -\frac{5}{3} \text{ or } x > 0.5\}\))
(Award SC2 for correct final set notation answer with no or insufficient working)

1. Find the gradient of the radius \(OP\):
\[\text{Gradient of } OP = \frac{6 - 0}{3 - 0} = 2\]

2. Find the gradient of the tangent line \(L\):
Since the tangent is perpendicular to the radius,
\[\text{Gradient of } L = -\frac{1}{2}\]

3. Find the equation of line \(L\):
Using the point \(P(3, 6)\):
\[y - 6 = -\frac{1}{2}(x - 3)\]
\[y = -\frac{1}{2}x + 7.5\]

4. Determine the intercepts \(A\) and \(B\):
- At point \(B\) on the \(y\)-axis, \(x = 0\):
\[y = 7.5 \implies B(0, 7.5) \implies OB = 7.5\]
- At point \(A\) on the \(x\)-axis, \(y = 0\):
\[0 = -\frac{1}{2}x + 7.5 \implies \frac{1}{2}x = 7.5 \implies x = 15 \implies A(15, 0) \implies OA = 15\]

5. Calculate the area of triangle \(OAB\):
\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times 7.5 = 56.25\]

M1 for finding gradient of radius \(OP = \frac{6}{3} = 2\)
M1 for gradient of tangent \(= -\frac{1}{\text{their } 2} = -\frac{1}{2}\)
M1 for finding the equation of the tangent or finding both coordinates \(A(15, 0)\) and \(B(0, 7.5)\)
M1 for \(\frac{1}{2} \times \text{their } 15 \times \text{their } 7.5\) oe
A1 for \(56.25\) or \(\frac{225}{4}\) or \(56\frac{1}{4}\) with correct working shown

1. Find angle \(\angle LPB\):
\[\angle LPB = 80^\circ - 35^\circ = 45^\circ\]

2. Find angle \(\angle PLB\):
The back-bearing from \(L\) to \(P\) is:
\[35^\circ + 180^\circ = 215^\circ\]
The bearing of \(B\) from \(L\) is \(115^\circ\).
Therefore, the interior angle at \(L\) is:
\[\angle PLB = 215^\circ - 115^\circ = 100^\circ\]

3. Find angle \(\angle PBL\):
\[\angle PBL = 180^\circ - (100^\circ + 45^\circ) = 35^\circ\]

4. Use the sine rule in triangle \(PLB\) to find \(PB\):
\[\frac{PB}{\sin(100^\circ)} = \frac{14}{\sin(35^\circ)}\]
\[PB = \frac{14 \times \sin(100^\circ)}{\sin(35^\circ)}\]
\[PB = \frac{14 \times 0.9848077...}{0.5735764...} \approx 24.0378...\text{ km}\]

To 3 significant figures, \(PB = 24.0\text{ km}\).

B1 for angle \(\angle LPB = 45^\circ\)
M1 for a correct method to find angle \(\angle PLB\), e.g. \(180 + 35 - 115\) or \(180 - (180 - 115 + 35)\)
A1 for \(\angle PLB = 100^\circ\)
B1 for \(\angle PBL = 35^\circ\) (or \(180 - 100 - 45\))
M1 for \(\frac{PB}{\sin(\text{their } 100^\circ)} = \frac{14}{\sin(\text{their } 35^\circ)}\) oe
A1 for answer in the range \(24.0\) to \(24.04\) with correct working shown