Loading....
|
Press & Hold to Drag Around |
|||
|
Click Here to Close |
|||
Question 1 Report
Solve the inequality \(-9 \le 4x + 3 \lt 15\).
A double inequality such as \(-9\le 4x+3\lt 15\) is really two statements about the same expression at once. The efficient method is to operate on all three parts simultaneously, so that \(4x+3\) is reduced to \(x\) while the outer numbers change with it.
Subtract 3 from each of the three parts:
\[-9-3\le 4x\lt 15-3\]\[-12\le 4x\lt 12\][M1]
Divide all three parts by the positive number 4, so both inequality signs keep their direction:
\[-3\le x\lt 3\]That is \(-3\le x\) [A1] and \(x\lt 3\) [A1].
The two ends behave differently and this must be preserved: \(x=-3\) is included because the original sign is \(\le\), giving \(4(-3)+3=-9\), while \(x=3\) is excluded because \(4(3)+3=15\) is not less than 15. Losing the distinction between \(\le\) and \(\lt\) is the usual way marks go here.
A double inequality such as \(-9\le 4x+3\lt 15\) is really two statements about the same expression at once. The efficient method is to operate on all three parts simultaneously, so that \(4x+3\) is reduced to \(x\) while the outer numbers change with it.
Subtract 3 from each of the three parts:
\[-9-3\le 4x\lt 15-3\]\[-12\le 4x\lt 12\][M1]
Divide all three parts by the positive number 4, so both inequality signs keep their direction:
\[-3\le x\lt 3\]That is \(-3\le x\) [A1] and \(x\lt 3\) [A1].
The two ends behave differently and this must be preserved: \(x=-3\) is included because the original sign is \(\le\), giving \(4(-3)+3=-9\), while \(x=3\) is excluded because \(4(3)+3=15\) is not less than 15. Losing the distinction between \(\le\) and \(\lt\) is the usual way marks go here.
Question 2 Report
Solve.
\(\frac{x+3}{4} - \frac{x-2}{3} = 1\)
When an equation contains fractions, clear them first by multiplying every term by the lowest common multiple of the denominators. Here the denominators are 4 and 3, so the LCM is 12.
\[ 12 \times \frac{x+3}{4} - 12 \times \frac{x-2}{3} = 12 \times 1 \] \[ 3(x+3) - 4(x-2) = 12 \] [M1]Now expand both brackets, taking care with the subtraction: the \(-4\) multiplies both terms inside the second bracket, so \(-4 \times (-2) = +8\).
\[ 3x + 9 - 4x + 8 = 12 \] \[ -x + 17 = 12 \] [M1]Solve the resulting linear equation:
\[ -x = 12 - 17 = -5 \;\Rightarrow\; x = 5. \] [A1]Check in the original equation: \(\frac{5+3}{4} - \frac{5-2}{3} = \frac{8}{4} - \frac{3}{3} = 2 - 1 = 1\), which is correct.
Two errors dominate here. The first is forgetting to multiply the right-hand side by 12, leaving \(3(x+3) - 4(x-2) = 1\) and giving \(x = 16\). The second is the sign inside the second bracket: writing \(-4x - 8\) instead of \(-4x + 8\) leads to \(x = -1\). Both are caught immediately by substituting the answer back into the original fractions.
When an equation contains fractions, clear them first by multiplying every term by the lowest common multiple of the denominators. Here the denominators are 4 and 3, so the LCM is 12.
\[ 12 \times \frac{x+3}{4} - 12 \times \frac{x-2}{3} = 12 \times 1 \] \[ 3(x+3) - 4(x-2) = 12 \] [M1]Now expand both brackets, taking care with the subtraction: the \(-4\) multiplies both terms inside the second bracket, so \(-4 \times (-2) = +8\).
\[ 3x + 9 - 4x + 8 = 12 \] \[ -x + 17 = 12 \] [M1]Solve the resulting linear equation:
\[ -x = 12 - 17 = -5 \;\Rightarrow\; x = 5. \] [A1]Check in the original equation: \(\frac{5+3}{4} - \frac{5-2}{3} = \frac{8}{4} - \frac{3}{3} = 2 - 1 = 1\), which is correct.
Two errors dominate here. The first is forgetting to multiply the right-hand side by 12, leaving \(3(x+3) - 4(x-2) = 1\) and giving \(x = 16\). The second is the sign inside the second bracket: writing \(-4x - 8\) instead of \(-4x + 8\) leads to \(x = -1\). Both are caught immediately by substituting the answer back into the original fractions.
Question 3 Report
\(\mathcal{E}=\{x:x\text{ is an integer and }1\le x\le 15\}\)
\(A=\{\text{prime numbers}\}\) and \(B=\{\text{odd numbers}\}\)
(a) List the elements of \(A'\cap B\). [2]
(b) Find \(n(A\cup B)\). [1]
The dash in \(A'\) denotes the complement: everything in the universal set that is not in \(A\). Here the universal set is the integers from \(1\) to \(15\).
(a) \(A\) is the set of prime numbers in that range. A prime has exactly two factors, itself and \(1\), so \(1\) is not prime and \(2\) is.
\(A = \{2,\ 3,\ 5,\ 7,\ 11,\ 13\}\) [M1]
\(B\) is the odd numbers, \(\{1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15\}\). The set \(A' \cap B\) means numbers that are odd but not prime, so remove the odd primes \(3\), \(5\), \(7\), \(11\) and \(13\) from \(B\):
\(A' \cap B = \{1,\ 9,\ 15\}\) [A1] cao
Each of these is genuinely composite or a unit: \(9 = 3 \times 3\), \(15 = 3 \times 5\), and \(1\) has only one factor so it is not prime.
(b) The union contains every number that is prime or odd or both, namely \(\{1,\ 2,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15\}\). Counting them gives
\(n(A \cup B) = 9\) [B1]
Equivalently \(8 + 6 - 5 = 9\), since \(A\) and \(B\) share the five odd primes. The number \(2\) is the only even element of the union, and forgetting that it is prime is a frequent error.
The dash in \(A'\) denotes the complement: everything in the universal set that is not in \(A\). Here the universal set is the integers from \(1\) to \(15\).
(a) \(A\) is the set of prime numbers in that range. A prime has exactly two factors, itself and \(1\), so \(1\) is not prime and \(2\) is.
\(A = \{2,\ 3,\ 5,\ 7,\ 11,\ 13\}\) [M1]
\(B\) is the odd numbers, \(\{1,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15\}\). The set \(A' \cap B\) means numbers that are odd but not prime, so remove the odd primes \(3\), \(5\), \(7\), \(11\) and \(13\) from \(B\):
\(A' \cap B = \{1,\ 9,\ 15\}\) [A1] cao
Each of these is genuinely composite or a unit: \(9 = 3 \times 3\), \(15 = 3 \times 5\), and \(1\) has only one factor so it is not prime.
(b) The union contains every number that is prime or odd or both, namely \(\{1,\ 2,\ 3,\ 5,\ 7,\ 9,\ 11,\ 13,\ 15\}\). Counting them gives
\(n(A \cup B) = 9\) [B1]
Equivalently \(8 + 6 - 5 = 9\), since \(A\) and \(B\) share the five odd primes. The number \(2\) is the only even element of the union, and forgetting that it is prime is a frequent error.
Question 4 Report
The depth of water in a harbour, \(d\) metres, at \(t\) hours after midnight is modelled by \(d = 6 + 2.5\sin(30t)^\circ\).
The diagram shows the graph of this model for one day.
(a) Calculate the depth of the water at 05 00. [2]
(b) Calculate the depth of the water at 20 00, correct to 3 significant figures. [2]
In this model the number of hours after midnight is converted into an angle by the multiplier 30, since \(30\times 24=720\), meaning the sine completes two full cycles in a day. That matches the two high tides and two low tides a day shown on the graph. The 6 is the mean depth and the 2.5 is the amplitude, so the depth always lies between 3.5 m and 8.5 m.
(a) At 05 00, \(t=5\), so the angle is \(30\times 5=150^\circ\):
\[d=6+2.5\sin 150^\circ\][M1]
Since \(\sin 150^\circ=0.5\), the depth is \(6+2.5\times 0.5=6+1.25=\) 7.25 m cao [A1].
(b) At 20 00, \(t=20\), so the angle is \(30\times 20=600^\circ\). Angles beyond \(360^\circ\) are perfectly acceptable in the calculator, but note \(600^\circ-360^\circ=240^\circ\), which is in the third quadrant where sine is negative:
\[d=6+2.5\sin 600^\circ\quad\text{oe}\quad 6+2.5\times(-0.866)\][M1]
This gives \(6-2.165...=3.8349...\), so the depth is 3.83 m [A1] (3.8349...).
Check both answers against the range 3.5 m to 8.5 m: 7.25 m is above the mean, as expected shortly after a high tide, and 3.83 m is near the low end, close to a low tide. Make sure the calculator is in degree mode, since the formula carries the degree symbol; in radian mode the answers would be meaningless.
In this model the number of hours after midnight is converted into an angle by the multiplier 30, since \(30\times 24=720\), meaning the sine completes two full cycles in a day. That matches the two high tides and two low tides a day shown on the graph. The 6 is the mean depth and the 2.5 is the amplitude, so the depth always lies between 3.5 m and 8.5 m.
(a) At 05 00, \(t=5\), so the angle is \(30\times 5=150^\circ\):
\[d=6+2.5\sin 150^\circ\][M1]
Since \(\sin 150^\circ=0.5\), the depth is \(6+2.5\times 0.5=6+1.25=\) 7.25 m cao [A1].
(b) At 20 00, \(t=20\), so the angle is \(30\times 20=600^\circ\). Angles beyond \(360^\circ\) are perfectly acceptable in the calculator, but note \(600^\circ-360^\circ=240^\circ\), which is in the third quadrant where sine is negative:
\[d=6+2.5\sin 600^\circ\quad\text{oe}\quad 6+2.5\times(-0.866)\][M1]
This gives \(6-2.165...=3.8349...\), so the depth is 3.83 m [A1] (3.8349...).
Check both answers against the range 3.5 m to 8.5 m: 7.25 m is above the mean, as expected shortly after a high tide, and 3.83 m is near the low end, close to a low tide. Make sure the calculator is in degree mode, since the formula carries the degree symbol; in radian mode the answers would be meaningless.
Question 5 Report
A rectangular garden is drawn on a plan using a scale of \(1:200\).
On the plan the garden measures \(6.5\) cm by \(4.2\) cm.
Calculate the actual area of the garden, giving your answer in m\(^2\).
A scale of \(1:200\) means one unit on the plan represents 200 of the same units in reality, so every plan length is multiplied by 200 to get the real length. The trap in this question is the area: the scale factor applies to lengths, so areas scale by \(200^{2}\), not by 200.
The safest route is to convert both lengths first and then find the area.
\[6.5 \times 200 = 1300 \text{ cm} = 13 \text{ m}\]
\[4.2 \times 200 = 840 \text{ cm} = 8.4 \text{ m}\]
Obtaining both real lengths scores [M1]. Recall that \(100\) cm \(= 1\) m, so dividing the centimetre values by 100 gives metres.
The actual area is then
\[13 \times 8.4 = 109.2 \text{ m}^{2}\]
[A1] cao.
As a check by the area factor: the plan area is \(6.5 \times 4.2 = 27.3\) cm\(^2\), and \(27.3 \times 200^{2} = 27.3 \times 40000 = 1\,092\,000\) cm\(^2\). Since \(1\) m\(^2 = 10\,000\) cm\(^2\), this is \(1\,092\,000 \div 10\,000 = 109.2\) m\(^2\), which agrees.
The common error is to multiply the plan area by 200 once and then convert, giving a hundredfold error. Converting the lengths first avoids the issue entirely.
A scale of \(1:200\) means one unit on the plan represents 200 of the same units in reality, so every plan length is multiplied by 200 to get the real length. The trap in this question is the area: the scale factor applies to lengths, so areas scale by \(200^{2}\), not by 200.
The safest route is to convert both lengths first and then find the area.
\[6.5 \times 200 = 1300 \text{ cm} = 13 \text{ m}\]
\[4.2 \times 200 = 840 \text{ cm} = 8.4 \text{ m}\]
Obtaining both real lengths scores [M1]. Recall that \(100\) cm \(= 1\) m, so dividing the centimetre values by 100 gives metres.
The actual area is then
\[13 \times 8.4 = 109.2 \text{ m}^{2}\]
[A1] cao.
As a check by the area factor: the plan area is \(6.5 \times 4.2 = 27.3\) cm\(^2\), and \(27.3 \times 200^{2} = 27.3 \times 40000 = 1\,092\,000\) cm\(^2\). Since \(1\) m\(^2 = 10\,000\) cm\(^2\), this is \(1\,092\,000 \div 10\,000 = 109.2\) m\(^2\), which agrees.
The common error is to multiply the plan area by 200 once and then convert, giving a hundredfold error. Converting the lengths first avoids the issue entirely.
Question 6 Report
The diagram shows a rectangular field measuring \(62\) m by \(48\) m.
Each measurement is correct to the nearest metre.
(a) Calculate the upper bound of the area of the field. [2]
(b) Calculate the lower bound of the perimeter of the field. [2]
(c) Fencing costs \(\$4.75\) for each metre. Calculate the upper bound of the cost of fencing the perimeter of the field. [2]
Each measurement is correct to the nearest metre, so the half-unit is \(0.5\) m. The length lies between \(61.5\) m and \(62.5\) m, and the width between \(47.5\) m and \(48.5\) m. Every part below picks the bounds that push the required quantity in the stated direction.
(a) Area is a product, so the greatest area uses both upper bounds:
\(62.5 \times 48.5\) [M1]
\(= 3031.25\) m\(^2\) [A1]
(b) Perimeter is a sum, so the least perimeter uses both lower bounds:
\(2 \times (61.5 + 47.5)\) [M1]
\(= 2 \times 109 = 218\) m [A1]
(c) The cost is the perimeter multiplied by a fixed price per metre, and since the price is exact the cost is greatest when the perimeter is greatest. That means both upper bounds again:
\(2 \times (62.5 + 48.5) \times 4.75\) [M1]
\(= 222 \times 4.75 = 1054.5\), so the upper bound of the cost is \(\$1054.50\). [A1]
Money answers are written to 2 decimal places, so \(\$1054.50\) rather than \(\$1054.5\). Note that part (c) does not reuse the answer to part (b): part (b) asked for the lower bound of the perimeter, whereas the greatest cost needs the upper bound.
Each measurement is correct to the nearest metre, so the half-unit is \(0.5\) m. The length lies between \(61.5\) m and \(62.5\) m, and the width between \(47.5\) m and \(48.5\) m. Every part below picks the bounds that push the required quantity in the stated direction.
(a) Area is a product, so the greatest area uses both upper bounds:
\(62.5 \times 48.5\) [M1]
\(= 3031.25\) m\(^2\) [A1]
(b) Perimeter is a sum, so the least perimeter uses both lower bounds:
\(2 \times (61.5 + 47.5)\) [M1]
\(= 2 \times 109 = 218\) m [A1]
(c) The cost is the perimeter multiplied by a fixed price per metre, and since the price is exact the cost is greatest when the perimeter is greatest. That means both upper bounds again:
\(2 \times (62.5 + 48.5) \times 4.75\) [M1]
\(= 222 \times 4.75 = 1054.5\), so the upper bound of the cost is \(\$1054.50\). [A1]
Money answers are written to 2 decimal places, so \(\$1054.50\) rather than \(\$1054.5\). Note that part (c) does not reuse the answer to part (b): part (b) asked for the lower bound of the perimeter, whereas the greatest cost needs the upper bound.
Question 7 Report
A machine starts working at 21 50 and runs without stopping for 640 minutes.
Write down the time, on the 24-hour clock, at which the machine stops.
The duration is given in minutes, so convert it to hours and minutes before adding it to a clock time. Dividing by 60 does this: \(640 \div 60 = 10\) remainder \(40\).
640 minutes \(=\) 10 hours 40 minutes [M1]
Now add to the start time in two steps. Adding 10 hours to 21 50 gives 31 50 on a 24-hour count, which is past midnight, so subtract 24 hours to get 07 50 the next day. Adding the remaining 40 minutes to 07 50 gives 08 30, because \(50 + 40 = 90\) minutes \(=\) 1 hour 30 minutes.
The machine stops at 08 30. [A1] cao
Two things are easy to get wrong. First, \(640 \div 60\) on a calculator shows \(10.6\overline{6}\), and the decimal part \(0.6\overline{6}\) is \(40\) minutes, not \(66\) minutes; multiplying \(0.6\overline{6} \times 60\) recovers the \(40\). Second, the answer must be written in 24-hour form with four digits, and here it falls on the following day.
The duration is given in minutes, so convert it to hours and minutes before adding it to a clock time. Dividing by 60 does this: \(640 \div 60 = 10\) remainder \(40\).
640 minutes \(=\) 10 hours 40 minutes [M1]
Now add to the start time in two steps. Adding 10 hours to 21 50 gives 31 50 on a 24-hour count, which is past midnight, so subtract 24 hours to get 07 50 the next day. Adding the remaining 40 minutes to 07 50 gives 08 30, because \(50 + 40 = 90\) minutes \(=\) 1 hour 30 minutes.
The machine stops at 08 30. [A1] cao
Two things are easy to get wrong. First, \(640 \div 60\) on a calculator shows \(10.6\overline{6}\), and the decimal part \(0.6\overline{6}\) is \(40\) minutes, not \(66\) minutes; multiplying \(0.6\overline{6} \times 60\) recovers the \(40\). Second, the answer must be written in 24-hour form with four digits, and here it falls on the following day.
Question 8 Report
\(A\), \(B\), \(C\) and \(D\) are points on a circle and \(AB\) is a diameter.
Angle \(BAC = 29^\circ\) and angle \(CBD = 43^\circ\).
(a) Write down angle \(ACB\) and give a reason for your answer. [2]
(b) Work out angle \(ABC\). [1]
(c) Work out angle \(ADC\), giving a reason. [2]
(d) Work out angle \(CAD\), giving a reason. [2]
(e) Write down angle \(ADB\). [1]
The diameter \(AB\) drives most of this question, because any angle subtended by a diameter at a point on the circle is a right angle. The rest uses the cyclic quadrilateral property and the same-segment property.
angle \(ACB = 90^\circ\) [B1], because the angle in a semicircle is a right angle [B1]
angle \(ABC = 180 - 90 - 29 = 61^\circ\) [B1]
angle \(ADC = 180 - 61 = 119^\circ\) [M1] [A1]
Reason: opposite angles of a cyclic quadrilateral add to \(180^\circ\).
angle \(CAD = 43^\circ\) [B1], angles in the same segment standing on arc \(CD\) [B1]
angle \(ADB = 90^\circ\) [B1]
Check the quadrilateral: angle \(BAD = 29 + 43 = 72^\circ\) at \(A\), and \(72 + 108 = 180^\circ\) requires angle \(BCD = 108^\circ\), which is consistent with angle \(BCA = 90^\circ\) plus angle \(ACD = 180 - 119 - 43 = 18^\circ\). Part (d) is where the same-segment theorem is easily confused with the centre-circumference theorem; the test is that both angles sit on the circumference and stand on the same chord, which \(CAD\) and \(CBD\) do.
The diameter \(AB\) drives most of this question, because any angle subtended by a diameter at a point on the circle is a right angle. The rest uses the cyclic quadrilateral property and the same-segment property.
angle \(ACB = 90^\circ\) [B1], because the angle in a semicircle is a right angle [B1]
angle \(ABC = 180 - 90 - 29 = 61^\circ\) [B1]
angle \(ADC = 180 - 61 = 119^\circ\) [M1] [A1]
Reason: opposite angles of a cyclic quadrilateral add to \(180^\circ\).
angle \(CAD = 43^\circ\) [B1], angles in the same segment standing on arc \(CD\) [B1]
angle \(ADB = 90^\circ\) [B1]
Check the quadrilateral: angle \(BAD = 29 + 43 = 72^\circ\) at \(A\), and \(72 + 108 = 180^\circ\) requires angle \(BCD = 108^\circ\), which is consistent with angle \(BCA = 90^\circ\) plus angle \(ACD = 180 - 119 - 43 = 18^\circ\). Part (d) is where the same-segment theorem is easily confused with the centre-circumference theorem; the test is that both angles sit on the circumference and stand on the same chord, which \(CAD\) and \(CBD\) do.
Question 9 Report
Solve \(2x^2+7x-15=0\).
A quadratic equation is solved by factorising whenever integer factors exist. Look for two numbers multiplying to \(a \times c = 2 \times (-15) = -30\) and adding to \(b = 7\); these are \(10\) and \(-3\). Splitting the middle term gives \(2x^2 + 10x - 3x - 15 = 2x(x + 5) - 3(x + 5)\), so
\((2x - 3)(x + 5) = 0\) [M1]
If a product of two factors is zero then at least one factor must be zero, which is the principle that turns one quadratic into two simple equations.
Check both: \(2(1.5)^2 + 7(1.5) - 15 = 4.5 + 10.5 - 15 = 0\), and \(2(25) - 35 - 15 = 0\). The answer \(x = 1.5\) may also be written as \(\dfrac{3}{2}\). Reading the roots straight off the brackets without changing sign, giving \(x = 3\) and \(x = 5\), is the standard error: the root is the value that makes the bracket vanish, so \((x + 5)\) gives \(x = -5\).
A quadratic equation is solved by factorising whenever integer factors exist. Look for two numbers multiplying to \(a \times c = 2 \times (-15) = -30\) and adding to \(b = 7\); these are \(10\) and \(-3\). Splitting the middle term gives \(2x^2 + 10x - 3x - 15 = 2x(x + 5) - 3(x + 5)\), so
\((2x - 3)(x + 5) = 0\) [M1]
If a product of two factors is zero then at least one factor must be zero, which is the principle that turns one quadratic into two simple equations.
Check both: \(2(1.5)^2 + 7(1.5) - 15 = 4.5 + 10.5 - 15 = 0\), and \(2(25) - 35 - 15 = 0\). The answer \(x = 1.5\) may also be written as \(\dfrac{3}{2}\). Reading the roots straight off the brackets without changing sign, giving \(x = 3\) and \(x = 5\), is the standard error: the root is the value that makes the bracket vanish, so \((x + 5)\) gives \(x = -5\).
Question 10 Report
In triangle \(ABC\), angle \(ABC = 90^\circ\), \(AB = 9\) cm and \(BC = 12\) cm.
Write down the length of \(AC\).
The right angle is at \(B\), so the side opposite it, \(AC\), is the hypotenuse and is the longest side of the triangle. "Write down" signals that the answer is expected to be recognised or found in one short step.
By Pythagoras' theorem, \[AC^{2}=AB^{2}+BC^{2}=9^{2}+12^{2}=81+144=225\] \[AC=\sqrt{225}=15\text{ cm}\] [B1] cao
This is the 3, 4, 5 right-angled triangle enlarged by a scale factor of 3, which is worth recognising on sight since it appears constantly. Check the answer is plausible: the hypotenuse must exceed both shorter sides but be less than their sum, and \(9\lt 15\lt 21\). Subtracting instead of adding would give \(\sqrt{63}=7.94\) cm, shorter than one of the other sides, which is impossible for a hypotenuse.
The right angle is at \(B\), so the side opposite it, \(AC\), is the hypotenuse and is the longest side of the triangle. "Write down" signals that the answer is expected to be recognised or found in one short step.
By Pythagoras' theorem, \[AC^{2}=AB^{2}+BC^{2}=9^{2}+12^{2}=81+144=225\] \[AC=\sqrt{225}=15\text{ cm}\] [B1] cao
This is the 3, 4, 5 right-angled triangle enlarged by a scale factor of 3, which is worth recognising on sight since it appears constantly. Check the answer is plausible: the hypotenuse must exceed both shorter sides but be less than their sum, and \(9\lt 15\lt 21\). Subtracting instead of adding would give \(\sqrt{63}=7.94\) cm, shorter than one of the other sides, which is impossible for a hypotenuse.
Question 11 Report
Solve the equation, giving your answers correct to 2 decimal places.
\(\frac{5}{x} + \frac{3}{x+2} = 2\)
Fractions with \(x\) in the denominator are cleared by multiplying every term by the product of the denominators, here \(x(x+2)\). That turns the equation into a quadratic.
Multiplying \(\dfrac{5}{x} + \dfrac{3}{x+2} = 2\) through by \(x(x+2)\):
\(5(x+2) + 3x = 2x(x+2)\) [M1]
Expand each side: \(5x + 10 + 3x = 2x^2 + 4x\), that is \(8x + 10 = 2x^2 + 4x\). Collect everything on one side so the quadratic equals zero:
\(2x^2 - 4x - 10 = 0\) oe, which divides by 2 to give \(x^2 - 2x - 5 = 0\) [M1]
This does not factorise, so use the quadratic formula with \(a = 1\), \(b = -2\), \(c = -5\):
\(x = \dfrac{2 \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2} = \dfrac{2 \pm \sqrt{24}}{2}\)
\(\sqrt{24} = 4.8989\ldots\), so \(x = \dfrac{2 + 4.8989}{2} = 3.4494\ldots\) or \(x = \dfrac{2 - 4.8989}{2} = -1.4494\ldots\)
\(x = 3.45\) and \(x = -1.45\) (2 d.p.) [A2]
[A1 is available for one correct value, or for both values seen to greater accuracy.]
Both roots are valid: neither makes a denominator zero, since \(x \ne 0\) and \(x \ne -2\). Two warnings. First, the instruction "correct to 2 decimal places" signals that the quadratic will not factorise, so reach for the formula rather than hunting for factors. Second, keep the unrounded values in the calculator until the final step; rounding \(\sqrt{24}\) to 4.9 early shifts the second decimal place.
Fractions with \(x\) in the denominator are cleared by multiplying every term by the product of the denominators, here \(x(x+2)\). That turns the equation into a quadratic.
Multiplying \(\dfrac{5}{x} + \dfrac{3}{x+2} = 2\) through by \(x(x+2)\):
\(5(x+2) + 3x = 2x(x+2)\) [M1]
Expand each side: \(5x + 10 + 3x = 2x^2 + 4x\), that is \(8x + 10 = 2x^2 + 4x\). Collect everything on one side so the quadratic equals zero:
\(2x^2 - 4x - 10 = 0\) oe, which divides by 2 to give \(x^2 - 2x - 5 = 0\) [M1]
This does not factorise, so use the quadratic formula with \(a = 1\), \(b = -2\), \(c = -5\):
\(x = \dfrac{2 \pm \sqrt{(-2)^2 - 4(1)(-5)}}{2} = \dfrac{2 \pm \sqrt{24}}{2}\)
\(\sqrt{24} = 4.8989\ldots\), so \(x = \dfrac{2 + 4.8989}{2} = 3.4494\ldots\) or \(x = \dfrac{2 - 4.8989}{2} = -1.4494\ldots\)
\(x = 3.45\) and \(x = -1.45\) (2 d.p.) [A2]
[A1 is available for one correct value, or for both values seen to greater accuracy.]
Both roots are valid: neither makes a denominator zero, since \(x \ne 0\) and \(x \ne -2\). Two warnings. First, the instruction "correct to 2 decimal places" signals that the quadratic will not factorise, so reach for the formula rather than hunting for factors. Second, keep the unrounded values in the calculator until the final step; rounding \(\sqrt{24}\) to 4.9 early shifts the second decimal place.
Question 12 Report
Write down the gradient of the line \(y = 7 - 3x\).
When a straight line is written in the form \(y=mx+c\), the number multiplying \(x\) is the gradient \(m\) and the constant is the \(y\)-intercept \(c\). Here the equation is written with the terms in the other order, \(y=7-3x\), so rearranging it mentally gives \(y=-3x+7\).
The coefficient of \(x\) is \(-3\), so the gradient is \(-3\) cao [B1].
The minus sign is part of the answer and must be kept: a negative gradient means the line slopes downwards from left to right, falling 3 units for every 1 unit moved to the right. Answering 7 confuses the gradient with the intercept, and answering 3 loses the direction of the slope.
When a straight line is written in the form \(y=mx+c\), the number multiplying \(x\) is the gradient \(m\) and the constant is the \(y\)-intercept \(c\). Here the equation is written with the terms in the other order, \(y=7-3x\), so rearranging it mentally gives \(y=-3x+7\).
The coefficient of \(x\) is \(-3\), so the gradient is \(-3\) cao [B1].
The minus sign is part of the answer and must be kept: a negative gradient means the line slopes downwards from left to right, falling 3 units for every 1 unit moved to the right. Answering 7 confuses the gradient with the intercept, and answering 3 loses the direction of the slope.
Question 13 Report
\(P\) is the point \((-3, 5)\) and \(Q\) is the point \((9, -2)\).
Find the length of \(PQ\), correct to 1 decimal place.
The distance between two points comes from Pythagoras' theorem applied to the right-angled triangle whose horizontal and vertical sides are the differences in the coordinates:
\(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
From \(P(-3, 5)\) to \(Q(9, -2)\), the horizontal change is \(9 - (-3) = 12\) and the vertical change is \(-2 - 5 = -7\). Squaring removes the sign, so only the sizes matter:
\(PQ = \sqrt{12^2 + 7^2} = \sqrt{144 + 49} = \sqrt{193} = 13.892\ldots = 13.9\) (1 d.p.) [B1]
The step where marks are most often lost is the horizontal difference: subtracting a negative gives \(9 + 3 = 12\), not 6. Because both differences are squared, it makes no difference whether you work from \(P\) to \(Q\) or from \(Q\) to \(P\). A rough check confirms the size: the distance must be longer than either separate side, so it must exceed 12, and it must be less than \(12 + 7 = 19\), the path taken by going across then up. A value of 13.9 sits comfortably between them.
The distance between two points comes from Pythagoras' theorem applied to the right-angled triangle whose horizontal and vertical sides are the differences in the coordinates:
\(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
From \(P(-3, 5)\) to \(Q(9, -2)\), the horizontal change is \(9 - (-3) = 12\) and the vertical change is \(-2 - 5 = -7\). Squaring removes the sign, so only the sizes matter:
\(PQ = \sqrt{12^2 + 7^2} = \sqrt{144 + 49} = \sqrt{193} = 13.892\ldots = 13.9\) (1 d.p.) [B1]
The step where marks are most often lost is the horizontal difference: subtracting a negative gives \(9 + 3 = 12\), not 6. Because both differences are squared, it makes no difference whether you work from \(P\) to \(Q\) or from \(Q\) to \(P\). A rough check confirms the size: the distance must be longer than either separate side, so it must exceed 12, and it must be less than \(12 + 7 = 19\), the path taken by going across then up. A value of 13.9 sits comfortably between them.
Question 14 Report
The diagram shows a rectangular field measuring 240 m by 175 m.
(a) One hectare is 10 000 m\(^2\). Calculate the area of the field, in hectares. [2]
(b) A model of a building is made using a scale of 1 : 50. The model has a volume of 640 cm\(^3\). Calculate the volume of the building, in m\(^3\). [2]
Both parts are unit conversions, but they differ in an important way: part (a) converts an area and part (b) converts a volume through a scale factor. The rule is that a length scale factor \(k\) becomes \(k^2\) for areas and \(k^3\) for volumes.
(a) The area of the field in square metres is
\(240 \times 175 = 42\,000\) m\(^2\) [M1]
Since one hectare is \(10\,000\) m\(^2\), divide by \(10\,000\):
\(42\,000 \div 10\,000 = 4.2\) hectares [A1]
(b) The scale \(1 : 50\) is a length ratio, so every length on the building is 50 times the corresponding length on the model. Volume involves three lengths, so the volume scale factor is \(50^3 = 125\,000\):
\(640 \times 50^3 = 80\,000\,000\) cm\(^3\) [M1]
Converting to cubic metres uses the same cubing idea: \(1\) m \(= 100\) cm, so \(1\) m\(^3\) \(= 100^3 = 1\,000\,000\) cm\(^3\). Hence
\(80\,000\,000 \div 1\,000\,000 = 80\) m\(^3\) [A1]
Multiplying the model volume by 50 rather than \(50^3\) gives \(32\,000\) cm\(^3\), which is the single most common error on scale questions. Whenever a ratio is applied, ask whether the quantity is a length, an area or a volume, and raise the ratio to the matching power.
Both parts are unit conversions, but they differ in an important way: part (a) converts an area and part (b) converts a volume through a scale factor. The rule is that a length scale factor \(k\) becomes \(k^2\) for areas and \(k^3\) for volumes.
(a) The area of the field in square metres is
\(240 \times 175 = 42\,000\) m\(^2\) [M1]
Since one hectare is \(10\,000\) m\(^2\), divide by \(10\,000\):
\(42\,000 \div 10\,000 = 4.2\) hectares [A1]
(b) The scale \(1 : 50\) is a length ratio, so every length on the building is 50 times the corresponding length on the model. Volume involves three lengths, so the volume scale factor is \(50^3 = 125\,000\):
\(640 \times 50^3 = 80\,000\,000\) cm\(^3\) [M1]
Converting to cubic metres uses the same cubing idea: \(1\) m \(= 100\) cm, so \(1\) m\(^3\) \(= 100^3 = 1\,000\,000\) cm\(^3\). Hence
\(80\,000\,000 \div 1\,000\,000 = 80\) m\(^3\) [A1]
Multiplying the model volume by 50 rather than \(50^3\) gives \(32\,000\) cm\(^3\), which is the single most common error on scale questions. Whenever a ratio is applied, ask whether the quantity is a length, an area or a volume, and raise the ratio to the matching power.
Question 15 Report
A rectangular enclosure is made against a long straight wall. The wall forms one side of the enclosure and 60 m of fencing is used for the other three sides. The two sides at right angles to the wall are each \(x\) m and the side parallel to the wall is \(y\) m.
(a) Show that the area, \(A\) m\(^2\), of the enclosure is given by \(A = 60x - 2x^2\). [2]
(b) Write \(60x - 2x^2\) in the form \(a - 2(x - b)^2\) and hence write down the greatest possible area of the enclosure. [4]
(c) Solve \(60x - 2x^2 = 430\), giving your answers correct to 2 decimal places. [3]
The wall replaces one side of the rectangle, so only three sides are fenced. That constraint links \(x\) and \(y\), which is what allows the area to be written in terms of \(x\) alone. Parts (b) and (c) then use two different techniques on the same quadratic: completing the square to find the maximum, and the quadratic formula to solve for a given value.
(a) The fencing covers the two sides of length \(x\) and the one side of length \(y\), so
\(2x + y = 60\), giving \(y = 60 - 2x\) [M1]
The area is the product of the two dimensions:
\(A = x(60 - 2x) = 60x - 2x^2\) [A1]
(b) To complete the square, first take out the factor \(-2\) from both terms containing \(x\):
\(-2(x^2 - 30x)\) [M1]
Inside the bracket, \(x^2 - 30x = (x - 15)^2 - 225\), since halving \(-30\) gives \(-15\) and \((-15)^2 = 225\) must be subtracted back:
\(-2\left[(x - 15)^2 - 225\right]\) [M1]
Multiplying out the outer \(-2\):
\(A = 450 - 2(x - 15)^2\) [A1]
A square is never negative, so \(2(x - 15)^2 \ge 0\) and the largest \(A\) occurs when the squared bracket is zero, at \(x = 15\). The greatest possible area is \(450\) m\(^2\). [A1]
(c) Set \(60x - 2x^2 = 430\) and rearrange to the standard form:
\(2x^2 - 60x + 430 = 0\), or dividing by \(2\), \(x^2 - 30x + 215 = 0\) [M1]
Using the formula with \(a = 1\), \(b = -30\), \(c = 215\), the discriminant is \(900 - 860 = 40\):
\(x = \dfrac{30 \pm \sqrt{40}}{2}\) [M1]
With \(\sqrt{40} = 6.3245\ldots\), the roots are \(\dfrac{36.3245\ldots}{2} = 18.1622\ldots\) and \(\dfrac{23.6754\ldots}{2} = 11.8377\ldots\):
\(x = 18.16\) and \(x = 11.84\) [A1]
Both roots are valid here, because both give a positive \(y = 60 - 2x\), so there are genuinely two rectangles of area \(430\) m\(^2\). They lie either side of \(x = 15\), the value that gives the maximum, which is a useful check on the completed square in part (b).
The wall replaces one side of the rectangle, so only three sides are fenced. That constraint links \(x\) and \(y\), which is what allows the area to be written in terms of \(x\) alone. Parts (b) and (c) then use two different techniques on the same quadratic: completing the square to find the maximum, and the quadratic formula to solve for a given value.
(a) The fencing covers the two sides of length \(x\) and the one side of length \(y\), so
\(2x + y = 60\), giving \(y = 60 - 2x\) [M1]
The area is the product of the two dimensions:
\(A = x(60 - 2x) = 60x - 2x^2\) [A1]
(b) To complete the square, first take out the factor \(-2\) from both terms containing \(x\):
\(-2(x^2 - 30x)\) [M1]
Inside the bracket, \(x^2 - 30x = (x - 15)^2 - 225\), since halving \(-30\) gives \(-15\) and \((-15)^2 = 225\) must be subtracted back:
\(-2\left[(x - 15)^2 - 225\right]\) [M1]
Multiplying out the outer \(-2\):
\(A = 450 - 2(x - 15)^2\) [A1]
A square is never negative, so \(2(x - 15)^2 \ge 0\) and the largest \(A\) occurs when the squared bracket is zero, at \(x = 15\). The greatest possible area is \(450\) m\(^2\). [A1]
(c) Set \(60x - 2x^2 = 430\) and rearrange to the standard form:
\(2x^2 - 60x + 430 = 0\), or dividing by \(2\), \(x^2 - 30x + 215 = 0\) [M1]
Using the formula with \(a = 1\), \(b = -30\), \(c = 215\), the discriminant is \(900 - 860 = 40\):
\(x = \dfrac{30 \pm \sqrt{40}}{2}\) [M1]
With \(\sqrt{40} = 6.3245\ldots\), the roots are \(\dfrac{36.3245\ldots}{2} = 18.1622\ldots\) and \(\dfrac{23.6754\ldots}{2} = 11.8377\ldots\):
\(x = 18.16\) and \(x = 11.84\) [A1]
Both roots are valid here, because both give a positive \(y = 60 - 2x\), so there are genuinely two rectangles of area \(430\) m\(^2\). They lie either side of \(x = 15\), the value that gives the maximum, which is a useful check on the completed square in part (b).
Question 16 Report
Simplify \(\left(8x^6\right)^{\frac{2}{3}}\), giving your answer in the form \(ax^n\).
An index outside a bracket applies to every factor inside, and a fractional index means root then power. Deal with the number and the letter separately.
For the number: \(8^{\frac{2}{3}}\) means the cube root of 8, which is 2, then squared.
\(8^{\frac{2}{3}} = 4\) [B1]
For the letter, a power of a power multiplies the indices: \(\left(x^6\right)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\). Combining the two parts:
\(4x^4\) [B1]
The answer is in the required form \(ax^n\) with \(a = 4\) and \(n = 4\).
The error that costs the first mark is leaving the coefficient alone and writing \(8x^4\); the outer index applies to the 8 just as much as to \(x^6\). The second trap is multiplying \(6 \times \frac{3}{2} = 9\) by flipping the fraction; the index is multiplied as it stands, and dividing by 3 then multiplying by 2 gives 4.
An index outside a bracket applies to every factor inside, and a fractional index means root then power. Deal with the number and the letter separately.
For the number: \(8^{\frac{2}{3}}\) means the cube root of 8, which is 2, then squared.
\(8^{\frac{2}{3}} = 4\) [B1]
For the letter, a power of a power multiplies the indices: \(\left(x^6\right)^{\frac{2}{3}} = x^{6 \times \frac{2}{3}} = x^4\). Combining the two parts:
\(4x^4\) [B1]
The answer is in the required form \(ax^n\) with \(a = 4\) and \(n = 4\).
The error that costs the first mark is leaving the coefficient alone and writing \(8x^4\); the outer index applies to the 8 just as much as to \(x^6\). The second trap is multiplying \(6 \times \frac{3}{2} = 9\) by flipping the fraction; the index is multiplied as it stands, and dividing by 3 then multiplying by 2 gives 4.
Question 17 Report
A shop sign is a right-angled triangle \(LMN\) with angle \(LMN = 90^\circ\), \(LM = 24\) cm and angle \(MLN = 53^\circ\).
(a) Calculate \(MN\). [2]
(b) Calculate \(LN\). [2]
(c) Calculate the area of the sign. [1]
Give each answer correct to 3 significant figures.
The sign is a right-angled triangle with the right angle at \(M\), so \(LM\) and \(MN\) are the two shorter sides (the legs) and \(LN\) is the hypotenuse. Working from the \(53^\circ\) angle at \(L\): \(LM\) is the side adjacent to it, \(MN\) is the side opposite it, and \(LN\) is the hypotenuse. Choosing the right trigonometric ratio is simply a matter of naming those three sides correctly first.
(a) \(MN\) is opposite and \(LM = 24\) cm is adjacent, so use the tangent ratio:
\[\tan 53^\circ = \frac{MN}{24} \quad\Rightarrow\quad MN = 24\tan 53^\circ\]
That substitution earns [M1]. Evaluating, \(24 \times 1.32704\ldots = 31.849\ldots\), so \(MN = 31.8\) cm to 3 significant figures [A1].
(b) \(LN\) is the hypotenuse and \(LM = 24\) cm is adjacent, so use cosine:
\[\cos 53^\circ = \frac{24}{LN} \quad\Rightarrow\quad LN = \frac{24}{\cos 53^\circ}\]
Rearranging to put \(LN\) on top scores [M1]. Then \(LN = 24 \div 0.601815\ldots = 39.879\ldots\), so \(LN = 39.9\) cm [A1]. A useful check: the hypotenuse must be the longest side, and \(39.9 \gt 31.8 \gt 24\), which it is.
(c) Because the angle at \(M\) is \(90^\circ\), the two legs are already perpendicular, so they act as base and height directly and no extra trigonometry is needed:
\[\text{Area} = \tfrac{1}{2} \times 24 \times 31.849\ldots = 382.1\ldots\]
giving \(382\) cm\(^2\) [B1], and this mark is awarded ft (follow through) on the candidate's own value of \(MN\).
Note the unrounded \(31.849\) is used in part (c), not the rounded \(31.8\). Rounding at every stage and then rounding again is the commonest source of a lost accuracy mark: keep full calculator accuracy in the working and round only the final answer to 3 significant figures.
The sign is a right-angled triangle with the right angle at \(M\), so \(LM\) and \(MN\) are the two shorter sides (the legs) and \(LN\) is the hypotenuse. Working from the \(53^\circ\) angle at \(L\): \(LM\) is the side adjacent to it, \(MN\) is the side opposite it, and \(LN\) is the hypotenuse. Choosing the right trigonometric ratio is simply a matter of naming those three sides correctly first.
(a) \(MN\) is opposite and \(LM = 24\) cm is adjacent, so use the tangent ratio:
\[\tan 53^\circ = \frac{MN}{24} \quad\Rightarrow\quad MN = 24\tan 53^\circ\]
That substitution earns [M1]. Evaluating, \(24 \times 1.32704\ldots = 31.849\ldots\), so \(MN = 31.8\) cm to 3 significant figures [A1].
(b) \(LN\) is the hypotenuse and \(LM = 24\) cm is adjacent, so use cosine:
\[\cos 53^\circ = \frac{24}{LN} \quad\Rightarrow\quad LN = \frac{24}{\cos 53^\circ}\]
Rearranging to put \(LN\) on top scores [M1]. Then \(LN = 24 \div 0.601815\ldots = 39.879\ldots\), so \(LN = 39.9\) cm [A1]. A useful check: the hypotenuse must be the longest side, and \(39.9 \gt 31.8 \gt 24\), which it is.
(c) Because the angle at \(M\) is \(90^\circ\), the two legs are already perpendicular, so they act as base and height directly and no extra trigonometry is needed:
\[\text{Area} = \tfrac{1}{2} \times 24 \times 31.849\ldots = 382.1\ldots\]
giving \(382\) cm\(^2\) [B1], and this mark is awarded ft (follow through) on the candidate's own value of \(MN\).
Note the unrounded \(31.849\) is used in part (c), not the rounded \(31.8\). Rounding at every stage and then rounding again is the commonest source of a lost accuracy mark: keep full calculator accuracy in the working and round only the final answer to 3 significant figures.
Question 18 Report
A lighthouse is \(65\) m tall. Two boats, \(P\) and \(Q\), are on the sea in a straight line with the foot of the lighthouse, with \(P\) further away than \(Q\). From the top of the lighthouse the angle of depression of \(P\) is \(24^\circ\) and the angle of depression of \(Q\) is \(41^\circ\).
Calculate the distance \(PQ\), correct to 3 significant figures.
An angle of depression is measured downwards from the horizontal at the observer's eye. Because the horizontal at the top of the lighthouse is parallel to the sea, each angle of depression equals the angle of elevation of the lighthouse top from the boat, by alternate angles. So each boat sits in a right-angled triangle whose vertical side is the lighthouse, 65 m, and whose horizontal side is that boat's distance from the foot.
For each boat the tangent ratio links the height to the horizontal distance, \(\tan(\text{angle})=\dfrac{65}{\text{distance}}\), so the distance is \(\dfrac{65}{\tan(\text{angle})}\).
Both boats lie on the same side in a straight line with the foot of the lighthouse, so \(PQ\) is the difference of the two distances: \[PQ=146.0\ldots-74.77\ldots=71.2\text{ m}\] [A1]
The steeper angle of depression belongs to the nearer boat, which is why \(Q\) at \(41^\circ\) is much closer than \(P\) at \(24^\circ\). Multiplying by the tangent instead of dividing is the usual error and would place both boats absurdly close to the lighthouse. Keep the unrounded distances until the subtraction, since rounding each to 3 significant figures first can shift the final answer.
An angle of depression is measured downwards from the horizontal at the observer's eye. Because the horizontal at the top of the lighthouse is parallel to the sea, each angle of depression equals the angle of elevation of the lighthouse top from the boat, by alternate angles. So each boat sits in a right-angled triangle whose vertical side is the lighthouse, 65 m, and whose horizontal side is that boat's distance from the foot.
For each boat the tangent ratio links the height to the horizontal distance, \(\tan(\text{angle})=\dfrac{65}{\text{distance}}\), so the distance is \(\dfrac{65}{\tan(\text{angle})}\).
Both boats lie on the same side in a straight line with the foot of the lighthouse, so \(PQ\) is the difference of the two distances: \[PQ=146.0\ldots-74.77\ldots=71.2\text{ m}\] [A1]
The steeper angle of depression belongs to the nearer boat, which is why \(Q\) at \(41^\circ\) is much closer than \(P\) at \(24^\circ\). Multiplying by the tangent instead of dividing is the usual error and would place both boats absurdly close to the lighthouse. Keep the unrounded distances until the subtraction, since rounding each to 3 significant figures first can shift the final answer.
Question 19 Report
The diagram shows the points \(A\) and \(B\). The point \(C\) is such that \(B\) is the midpoint of \(AC\).
Find the coordinates of \(C\).
If \(B\) is the midpoint of \(AC\), then \(B\) is the average of \(A\) and \(C\). Rearranging the midpoint relationship gives the rule for finding the far end from one end and the midpoint:
\(\dfrac{x_A + x_C}{2} = x_B\), so \(x_C = 2x_B - x_A\), and likewise \(y_C = 2y_B - y_A\)
Reading from the diagram, \(A\) is \((-3, -2)\) and \(B\) is \((2, 3)\):
\(x\)-coordinate of \(C = 2 \times 2 - (-3) = 4 + 3 = 7\) [M1]
\(y\)-coordinate of \(C = 2 \times 3 - (-2) = 6 + 2 = 8\) [M1]
\(C = (7, 8)\) [A1]
An equivalent and often clearer route is to treat the move from \(A\) to \(B\) as a step and repeat it: from \((-3, -2)\) to \((2, 3)\) is 5 across and 5 up, so from \(B\) the same step again reaches \((7, 8)\).
Check by taking the midpoint of \(A\) and \(C\): \(\left(\frac{-3+7}{2}, \frac{-2+8}{2}\right) = (2, 3)\), which is \(B\). The usual error is averaging \(A\) and \(B\), which finds the midpoint of \(AB\) rather than extending the line beyond \(B\); the wording "\(B\) is the midpoint of \(AC\)" places \(B\) in the middle, so \(C\) must lie on the far side of \(B\) from \(A\).
If \(B\) is the midpoint of \(AC\), then \(B\) is the average of \(A\) and \(C\). Rearranging the midpoint relationship gives the rule for finding the far end from one end and the midpoint:
\(\dfrac{x_A + x_C}{2} = x_B\), so \(x_C = 2x_B - x_A\), and likewise \(y_C = 2y_B - y_A\)
Reading from the diagram, \(A\) is \((-3, -2)\) and \(B\) is \((2, 3)\):
\(x\)-coordinate of \(C = 2 \times 2 - (-3) = 4 + 3 = 7\) [M1]
\(y\)-coordinate of \(C = 2 \times 3 - (-2) = 6 + 2 = 8\) [M1]
\(C = (7, 8)\) [A1]
An equivalent and often clearer route is to treat the move from \(A\) to \(B\) as a step and repeat it: from \((-3, -2)\) to \((2, 3)\) is 5 across and 5 up, so from \(B\) the same step again reaches \((7, 8)\).
Check by taking the midpoint of \(A\) and \(C\): \(\left(\frac{-3+7}{2}, \frac{-2+8}{2}\right) = (2, 3)\), which is \(B\). The usual error is averaging \(A\) and \(B\), which finds the midpoint of \(AB\) rather than extending the line beyond \(B\); the wording "\(B\) is the midpoint of \(AC\)" places \(B\) in the middle, so \(C\) must lie on the far side of \(B\) from \(A\).
Question 20 Report
A regular polygon has 24 sides, as shown.
Work out the sum of the interior angles of this polygon.
The interior angles of a polygon with \(n\) sides add up to
\((n - 2) \times 180^\circ\)
The reason for the \(n - 2\) is that any \(n\)-sided polygon can be cut by diagonals from one vertex into \(n - 2\) triangles, and each triangle contributes \(180^\circ\). For \(n = 24\):
\((24 - 2) \times 180\) [M1]
\(= 22 \times 180 = 3960^\circ\) [A1]
The fact that the polygon is regular is not needed for this part, since the sum of the interior angles depends only on the number of sides; it would matter only if a single angle were asked for, which would be \(3960 \div 24 = 165^\circ\).
Two errors are common. The first is multiplying by 24 rather than 22, giving \(4320^\circ\), which forgets the two triangles lost in the dissection. The second is confusing this with the exterior angles, which always total \(360^\circ\) regardless of the number of sides. A rough check on the size: 24 interior angles each a little under \(180^\circ\) should total a little under \(24 \times 180 = 4320^\circ\), and \(3960^\circ\) fits.
The interior angles of a polygon with \(n\) sides add up to
\((n - 2) \times 180^\circ\)
The reason for the \(n - 2\) is that any \(n\)-sided polygon can be cut by diagonals from one vertex into \(n - 2\) triangles, and each triangle contributes \(180^\circ\). For \(n = 24\):
\((24 - 2) \times 180\) [M1]
\(= 22 \times 180 = 3960^\circ\) [A1]
The fact that the polygon is regular is not needed for this part, since the sum of the interior angles depends only on the number of sides; it would matter only if a single angle were asked for, which would be \(3960 \div 24 = 165^\circ\).
Two errors are common. The first is multiplying by 24 rather than 22, giving \(4320^\circ\), which forgets the two triangles lost in the dissection. The second is confusing this with the exterior angles, which always total \(360^\circ\) regardless of the number of sides. A rough check on the size: 24 interior angles each a little under \(180^\circ\) should total a little under \(24 \times 180 = 4320^\circ\), and \(3960^\circ\) fits.
Question 21 Report
(a) \(\$6000\) is invested at \(r\%\) per year compound interest. After 5 years the investment is worth \(\$7500\). Calculate the value of \(r\), correct to 2 decimal places. [3]
(b) A different account pays 4.5% per year compound interest. Calculate the value of \(\$6000\) in that account after 5 years. [2]
Part (a) reverses the usual compound interest calculation: the start value, the end value and the number of years are known, and the rate is the unknown. That requires a root, not a division.
A tempting but wrong method in part (a) is \(\dfrac{7500-6000}{6000}\times 100\div 5=5\%\). That is the simple interest rate; it is too high because it ignores the interest that itself earns interest. Part (b) confirms the reasoning: 4.5% compound over five years gives \(\$7477.09\), slightly less than \(\$7500\), so the rate needed must be slightly above 4.5%, exactly as part (a) found.
Part (a) reverses the usual compound interest calculation: the start value, the end value and the number of years are known, and the rate is the unknown. That requires a root, not a division.
A tempting but wrong method in part (a) is \(\dfrac{7500-6000}{6000}\times 100\div 5=5\%\). That is the simple interest rate; it is too high because it ignores the interest that itself earns interest. Part (b) confirms the reasoning: 4.5% compound over five years gives \(\$7477.09\), slightly less than \(\$7500\), so the rate needed must be slightly above 4.5%, exactly as part (a) found.
Question 22 Report
The diagram shows two points \(A\) and \(B\) with \(AB = 10\) cm. The locus of points 6 cm from \(A\) meets the locus of points 8 cm from \(B\) at the point \(P\).
(a) Show that angle \(APB\) is 90 degrees. [2]
(b) Write down the mathematical name of the locus of all points \(X\) for which angle \(AXB\) is 90 degrees. [1]
The locus of points a fixed distance from a point is a circle centred on that point, so \(P\) lies 6 cm from \(A\) and 8 cm from \(B\), while \(AB = 10\) cm. That gives a triangle with all three sides known.
(a) The converse of Pythagoras states that if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled, with the right angle opposite the longest side. Test the two shorter sides:
\(6^2 + 8^2 = 36 + 64 = 100\) [M1]
\(100 = 10^2 = AB^2\), so by the converse of Pythagoras angle \(APB\) is a right angle [A1]
The right angle is at \(P\) because \(AB\) is the longest side and the angle at \(P\) is the one facing it.
(b) Every point that sees the segment \(AB\) at a right angle lies on the same circle:
A circle with \(AB\) as its diameter [B1] oe
This is the circle theorem that the angle in a semicircle is \(90^\circ\), read in reverse. Since \(AB\) is 10 cm, the circle has radius 5 cm and is centred on the midpoint of \(AB\). Note that this must be quoted as a circle on \(AB\) as diameter; a circle of radius 10 cm, or one centred at \(A\), does not have the required property.
The locus of points a fixed distance from a point is a circle centred on that point, so \(P\) lies 6 cm from \(A\) and 8 cm from \(B\), while \(AB = 10\) cm. That gives a triangle with all three sides known.
(a) The converse of Pythagoras states that if the square of the longest side equals the sum of the squares of the other two, the triangle is right-angled, with the right angle opposite the longest side. Test the two shorter sides:
\(6^2 + 8^2 = 36 + 64 = 100\) [M1]
\(100 = 10^2 = AB^2\), so by the converse of Pythagoras angle \(APB\) is a right angle [A1]
The right angle is at \(P\) because \(AB\) is the longest side and the angle at \(P\) is the one facing it.
(b) Every point that sees the segment \(AB\) at a right angle lies on the same circle:
A circle with \(AB\) as its diameter [B1] oe
This is the circle theorem that the angle in a semicircle is \(90^\circ\), read in reverse. Since \(AB\) is 10 cm, the circle has radius 5 cm and is centred on the midpoint of \(AB\). Note that this must be quoted as a circle on \(AB\) as diameter; a circle of radius 10 cm, or one centred at \(A\), does not have the required property.
Question 23 Report
A rectangular garden bed measures \((2x+1)\) m by \((x-3)\) m and has an area of 60 m\(^2\).
(a) Show that \(2x^2-5x-63=0\). [2]
(b) Solve this equation to find the length of the garden bed. [1]
An area given as a product of two linear expressions leads to a quadratic. Part (a) asks for the quadratic to be established, and because the result is printed, every algebraic step must be shown. Part (b) then solves it and converts the answer back into a length.
(a) Area is length \(\times\) width, so
\((2x + 1)(x - 3) = 60\) [M1]
Expanding the left side: \(2x^2 - 6x + x - 3 = 2x^2 - 5x - 3\), so the equation is \(2x^2 - 5x - 3 = 60\). Subtracting \(60\) from both sides gives
\(2x^2 - 5x - 63 = 0\) [A1]
(b) Two numbers multiplying to \(2 \times (-63) = -126\) and adding to \(-5\) are \(-14\) and \(9\), giving \(2x^2 - 14x + 9x - 63 = 2x(x - 7) + 9(x - 7)\), so
\((2x + 9)(x - 7) = 0\), so \(x = 7\), and the length is \(2(7) + 1 = 15\) m [B1]
The other root, \(x = -4.5\), is rejected because it would make the width \(x - 3\) negative. Check the answer: with \(x = 7\) the bed is \(15\) m by \(4\) m, and \(15 \times 4 = 60\) m\(^2\) as stated. The question asks for the length, not for \(x\), so substituting back into \((2x + 1)\) is the step that earns the mark.
An area given as a product of two linear expressions leads to a quadratic. Part (a) asks for the quadratic to be established, and because the result is printed, every algebraic step must be shown. Part (b) then solves it and converts the answer back into a length.
(a) Area is length \(\times\) width, so
\((2x + 1)(x - 3) = 60\) [M1]
Expanding the left side: \(2x^2 - 6x + x - 3 = 2x^2 - 5x - 3\), so the equation is \(2x^2 - 5x - 3 = 60\). Subtracting \(60\) from both sides gives
\(2x^2 - 5x - 63 = 0\) [A1]
(b) Two numbers multiplying to \(2 \times (-63) = -126\) and adding to \(-5\) are \(-14\) and \(9\), giving \(2x^2 - 14x + 9x - 63 = 2x(x - 7) + 9(x - 7)\), so
\((2x + 9)(x - 7) = 0\), so \(x = 7\), and the length is \(2(7) + 1 = 15\) m [B1]
The other root, \(x = -4.5\), is rejected because it would make the width \(x - 3\) negative. Check the answer: with \(x = 7\) the bed is \(15\) m by \(4\) m, and \(15 \times 4 = 60\) m\(^2\) as stated. The question asks for the length, not for \(x\), so substituting back into \((2x + 1)\) is the step that earns the mark.
Question 24 Report
A line has equation \(3x - 4y = 12\).
(a) Find the gradient of this line. [2]
(b) Find the equation of the line parallel to it that passes through the point \((8,\ -1)\). Give your answer in the form \(y = mx + c\). [1]
An equation given in the form \(ax+by=c\) does not show its gradient directly. Rearranging it into \(y=mx+c\) makes the gradient visible as the coefficient of \(x\). Parallel lines then share that gradient and differ only in their intercept.
(a) Starting from \(3x-4y=12\), add \(4y\) and subtract 12 from both sides:
\[4y=3x-12\quad\text{oe}\quad y=\frac{3x-12}{4}\][M1]
Dividing every term by 4 gives \(y=\frac{3}{4}x-3\), so the gradient is
\[\frac{3}{4}\quad\text{oe}\quad 0.75\][A1]
Note the sign: dividing \(-4y\) across correctly leaves a positive \(\frac{3}{4}\), and answering \(-\frac{3}{4}\) is the usual slip.
(b) A parallel line has the same gradient \(\frac{3}{4}\), so its equation is \(y=\frac{3}{4}x+c\). Substituting the point \((8,\ -1)\):
\[-1=\frac{3}{4}\times 8+c=6+c\quad\Longrightarrow\quad c=-7\]The line is
\[y=\frac{3}{4}x-7\]oe [B1]
A quick verification: at \(x=8\), \(\frac{3}{4}\times 8-7=6-7=-1\), which is the given \(y\)-coordinate, so the point does lie on the line.
An equation given in the form \(ax+by=c\) does not show its gradient directly. Rearranging it into \(y=mx+c\) makes the gradient visible as the coefficient of \(x\). Parallel lines then share that gradient and differ only in their intercept.
(a) Starting from \(3x-4y=12\), add \(4y\) and subtract 12 from both sides:
\[4y=3x-12\quad\text{oe}\quad y=\frac{3x-12}{4}\][M1]
Dividing every term by 4 gives \(y=\frac{3}{4}x-3\), so the gradient is
\[\frac{3}{4}\quad\text{oe}\quad 0.75\][A1]
Note the sign: dividing \(-4y\) across correctly leaves a positive \(\frac{3}{4}\), and answering \(-\frac{3}{4}\) is the usual slip.
(b) A parallel line has the same gradient \(\frac{3}{4}\), so its equation is \(y=\frac{3}{4}x+c\). Substituting the point \((8,\ -1)\):
\[-1=\frac{3}{4}\times 8+c=6+c\quad\Longrightarrow\quad c=-7\]The line is
\[y=\frac{3}{4}x-7\]oe [B1]
A quick verification: at \(x=8\), \(\frac{3}{4}\times 8-7=6-7=-1\), which is the given \(y\)-coordinate, so the point does lie on the line.
Question 25 Report
Use your calculator to work out \(\frac{\sqrt{20.4}-1.9}{0.32\times 4.7}\).
Give your answer correct to 3 significant figures.
The fraction bar acts as a bracket around both the numerator and the denominator, so each must be fully evaluated before the division. Entering the expression as written from left to right, without brackets, would divide only \(1.9\) by \(0.32\) and then multiply by \(4.7\), giving a completely different value.
The answer correct to 3 significant figures is \(1.74\). [A1] cao
On the calculator, enter it in one go using the fraction template, or type it as \((\sqrt{20.4}-1.9)\div(0.32\times 4.7)\) with both sets of brackets. Rounding \(\sqrt{20.4}\) to \(4.52\) before subtracting shifts the answer to \(1.741\), which still rounds correctly here but will not always do so. Carry full accuracy to the last step.
The fraction bar acts as a bracket around both the numerator and the denominator, so each must be fully evaluated before the division. Entering the expression as written from left to right, without brackets, would divide only \(1.9\) by \(0.32\) and then multiply by \(4.7\), giving a completely different value.
The answer correct to 3 significant figures is \(1.74\). [A1] cao
On the calculator, enter it in one go using the fraction template, or type it as \((\sqrt{20.4}-1.9)\div(0.32\times 4.7)\) with both sets of brackets. Rounding \(\sqrt{20.4}\) to \(4.52\) before subtracting shifts the answer to \(1.741\), which still rounds correctly here but will not always do so. Carry full accuracy to the last step.
Question 26 Report
Three towns \(P\), \(Q\) and \(R\) are joined by straight roads. \(PQ = 63\) km, \(QR = 47\) km and \(PR = 88\) km.
(a) Calculate angle \(PQR\). [3]
(b) Calculate the area of triangle \(PQR\). [2]
(c) Calculate the shortest distance from \(Q\) to the road \(PR\). [2]
(d) Calculate angle \(QPR\). [3]
Give each answer correct to 3 significant figures.
All three sides of triangle \(PQR\) are given and no angles, so every angle must come from the cosine rule in its rearranged form. Parts (b) and (c) then reuse the results rather than starting again.
Part (c) rewards recognising that "shortest distance to a line" always means the perpendicular distance, and that the area already found is the quickest route to it. As a check on parts (a) and (d), the third angle would be \(180-105.3-31.0=43.7^\circ\), and the sides 88, 63, 47 km rank in the same order as the angles opposite them, \(105.3^\circ\), \(43.7^\circ\) and \(31.0^\circ\).
All three sides of triangle \(PQR\) are given and no angles, so every angle must come from the cosine rule in its rearranged form. Parts (b) and (c) then reuse the results rather than starting again.
Part (c) rewards recognising that "shortest distance to a line" always means the perpendicular distance, and that the area already found is the quickest route to it. As a check on parts (a) and (d), the third angle would be \(180-105.3-31.0=43.7^\circ\), and the sides 88, 63, 47 km rank in the same order as the angles opposite them, \(105.3^\circ\), \(43.7^\circ\) and \(31.0^\circ\).
Question 27 Report
\(A\) and \(B\) are independent events.
\(\text{P}(A) = x\), \(\text{P}(B) = x + 0.2\) and \(\text{P}(A \text{ and } B) = 0.15\).
Find the value of \(x\).
For independent events the probability that both occur is the product of the separate probabilities:
\[\text{P}(A\text{ and }B)=\text{P}(A)\times\text{P}(B)\]Substituting the given expressions turns a probability statement into a quadratic equation:
\[x(x+0.2)=0.15\][M1]
Expand and collect everything on one side:
\[x^{2}+0.2x-0.15=0\][M1]
Factorising, two numbers with product \(-0.15\) and sum \(0.2\) are \(-0.3\) and \(0.5\):
\[(x-0.3)(x+0.5)=0\][M1] oe (formula)
The two roots are \(x=0.3\) and \(x=-0.5\), but a probability cannot be negative, so
\[x=0.3\][A1] (\(x=-0.5\) rejected)
Rejecting the negative root with a brief reason is part of the answer, not an optional extra. Check the solution in context: \(\text{P}(A)=0.3\) and \(\text{P}(B)=0.5\), both valid probabilities, and \(0.3\times 0.5=0.15\) as required. The quadratic formula gives the same roots if the factorisation is not spotted, and multiplying the equation by 100 first, to give \(100x^{2}+20x-15=0\), avoids decimals if you prefer whole numbers.
For independent events the probability that both occur is the product of the separate probabilities:
\[\text{P}(A\text{ and }B)=\text{P}(A)\times\text{P}(B)\]Substituting the given expressions turns a probability statement into a quadratic equation:
\[x(x+0.2)=0.15\][M1]
Expand and collect everything on one side:
\[x^{2}+0.2x-0.15=0\][M1]
Factorising, two numbers with product \(-0.15\) and sum \(0.2\) are \(-0.3\) and \(0.5\):
\[(x-0.3)(x+0.5)=0\][M1] oe (formula)
The two roots are \(x=0.3\) and \(x=-0.5\), but a probability cannot be negative, so
\[x=0.3\][A1] (\(x=-0.5\) rejected)
Rejecting the negative root with a brief reason is part of the answer, not an optional extra. Check the solution in context: \(\text{P}(A)=0.3\) and \(\text{P}(B)=0.5\), both valid probabilities, and \(0.3\times 0.5=0.15\) as required. The quadratic formula gives the same roots if the factorisation is not spotted, and multiplying the equation by 100 first, to give \(100x^{2}+20x-15=0\), avoids decimals if you prefer whole numbers.
Would you like to proceed with this action?