Edexcel International GCSE Maths Solutions: 4WM1H/01 Specimen Paper (Unit 1 Higher)
Full worked solutions to the Pearson Edexcel International GCSE Mathematics A (Modular) Unit 1 Higher Tier specimen paper (4WM1H/01, Sample Assessment Materials, February 2024): 23 questions, 100 marks, 2 hours, calculator allowed. Every final answer below matches the official Pearson mark scheme. Attempt each question yourself first, then click “Show solution” to check your working step by step.
The weight of a cake is 2.75 kg, correct to 2 decimal places. Write down the lower bound of the weight of the cake.
Show solution
- Correct to 2 decimal places means the true value is within half of 0.01 (= 0.005) either side of 2.75.
Lower bound = 2.75 − 0.005 = 2.745 kgA value rounded to 2 d.p. could be up to half a unit of 0.01 below or above. The lower bound is the smallest value that would still round to 2.75.
Show that 513 − 267 = 21021
Show solution
- Step 1: convert to improper fractions. 513 = 163 and 267 = 207Subtracting mixed numbers is easiest as improper fractions. Multiply the whole number by the denominator and add the numerator: 5 × 3 + 1 = 16 and 2 × 7 + 6 = 20.
- Step 2: use a common denominator of 21. 163 = 11221 and 207 = 6021Fractions can only be subtracted when their denominators match. 21 is the lowest common multiple of 3 and 7. Multiply top and bottom by the same number: × 7 for thirds, × 3 for sevenths.
- Step 3: subtract. 11221 − 6021 = 5221 = 21021 ✓ (since 52 = 2 × 21 + 10)With equal denominators, subtract the numerators only. Then convert back: 21 goes into 52 twice (42) with 10 left over.
Here is a biased 4-sided spinner. The table gives the probabilities that, when the spinner is spun once, it will land on 1 or it will land on 3.
| Number | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Probability | 0.26 | 0.18 |
The probability that the spinner will land on 2 is equal to the probability that the spinner will land on 4. Priya is going to spin the spinner 250 times. Work out an estimate for the number of times the spinner will land on a 1 or a 2.
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- Step 1: All probabilities add to 1, so P(2) + P(4) = 1 − (0.26 + 0.18) = 0.56The spinner must land on exactly one of the four numbers, so the four probabilities add up to 1. Whatever is not used by 1 and 3 is shared between 2 and 4.
- Step 2: P(2) = P(4), so P(2) = 0.56 ÷ 2 = 0.28The question says the two probabilities are equal, so split 0.56 into two equal halves.
- Step 3: P(1 or 2) = 0.26 + 0.28 = 0.54Landing on 1 and landing on 2 can’t both happen in one spin, so P(1 or 2) = P(1) + P(2) (‘or’ means add).
- Step 4: Expected number = 0.54 × 250 = 135Expected frequency = probability × number of trials. It is an estimate of what should happen, not a guarantee.
(a) Expand and simplify (n − 6)(n + 4)
(b) Solve 2x − 3 = 3x − 54 Show clear algebraic working.
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- (a) (n − 6)(n + 4) = n² + 4n − 6n − 24 = n² − 2n − 24Multiply every term in the first bracket by every term in the second (n × n, n × 4, −6 × n, −6 × 4), then collect like terms: 4n − 6n = −2n.
- (b) Multiply both sides by 4: 4(2x − 3) = 3x − 5 → 8x − 12 = 3x − 5
Collect terms: 8x − 3x = −5 + 12 → 5x = 7
x = 7/5 (= 1.4)Multiplying both sides by 4 clears the fraction. Put 2x − 3 in brackets so both terms are multiplied by 4. Then subtract 3x and add 12 to both sides to get the x terms on one side and the numbers on the other.
A plane flies from New York City to Los Angeles. The plane flies a distance of 3980 kilometres in 5 hours 24 minutes. Work out the average speed of the plane. Give your answer in kilometres per hour correct to the nearest whole number.
Show solution
- Step 1: convert the time to hours. 24 minutes = 24 ÷ 60 = 0.4 hours, so the time is 5.4 hours.The speed must be in km per hour, so the time must be in hours. There are 60 minutes in an hour, so 24 minutes is 24/60 of an hour.
- Step 2: Speed = distance ÷ time = 3980 ÷ 5.4 = 737.03…Average speed = total distance ÷ total time. Keep the full calculator value until the last step.
- To the nearest whole number: 737 km/hThe first decimal digit of 737.03… is 0, so it rounds down to 737.
The diagram shows an 8-sided shape ABCDEFGH.
HG = 28 cm, AH = FG = 12 cm, AB = EF = 5 cm. The height of the shape is 20 cm. CD is parallel to HG. The area of shape ABCDEFGH is 434 cm². Find the length of CD.
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- Step 1: split the shape into the 28 cm × 12 cm rectangle at the bottom and a trapezium BCDE on top.
Rectangle area = 28 × 12 = 336 cm²This shape has no area formula of its own, so split it into shapes that do. The bottom part is a rectangle (both sides are 12 cm with right angles) and the top part is a trapezium because CD is parallel to HG. - Step 2: trapezium dimensions. Bottom side BE = 28 − 5 − 5 = 18 cm. Height = 20 − 12 = 8 cm.
Trapezium area = 434 − 336 = 98 cm²The trapezium’s bottom side is the 28 cm width minus the two 5 cm ledges AB and EF. Its height is the total height (20) minus the rectangle’s height (12). Its area is whatever is left of 434 after the rectangle. - Step 3: use the trapezium area formula. ½(18 + CD) × 8 = 98 → 4(18 + CD) = 98 → 18 + CD = 24.5Area of a trapezium = ½(a + b)h. We know the area, one parallel side and the height, so form an equation and solve for the other parallel side. ½ × 8 = 4, then divide both sides by 4.
- CD = 6.5 cmSubtract 18 from both sides: 24.5 − 18 = 6.5.
(a) Simplify 8 × (4t)0
x6 ÷ x−5 = xp
(b) Find the value of p
(c) Simplify fully (2k2m4)3
Show solution
- (a) Anything to the power 0 equals 1, so (4t)0 = 1 and 8 × 1 = 8By the index laws, aⁿ ÷ aⁿ = a⁰, and anything divided by itself is 1. The power 0 only applies to the bracket, not to the 8 outside it.
- (b) When dividing, subtract the powers: 6 − (−5) = 11, so p = 11Dividing powers of the same base means subtracting the indices. Subtracting a negative is the same as adding: 6 − (−5) = 6 + 5.
- (c) Cube every part: 2³ = 8, (k²)³ = k6, (m4)³ = m12
Answer: 8k⁶m¹²The power outside the bracket applies to every factor inside, including the number. For a power of a power, multiply the indices: 2 × 3 = 6 and 4 × 3 = 12.
Change a speed of 81 kilometres per hour to a speed in metres per second.
Show solution
- Step 1: 81 km = 81 × 1000 = 81 000 mThe answer must be in metres, and 1 km = 1000 m.
- Step 2: 1 hour = 60 × 60 = 3600 secondsThe answer must be per second. There are 60 minutes in an hour and 60 seconds in each minute.
- Step 3: 81 000 ÷ 3600 = 22.5 m/s81 km/h means 81 000 m every 3600 seconds. Dividing gives the distance covered in one second.
(a) Simplify 3a4b5 × 4a7b2
(b) Factorise fully 14x2y4 + 21x3y2
The diagram shows a straight line drawn on a grid.
(c) Find an equation of the line.
A different straight line has equation y = 3x − 5
(d) Write down the coordinates of the point at which the line crosses the y-axis.
Show solution
- (a) Multiply the numbers and add the powers: 3 × 4 = 12, a4+7 = a11, b5+2 = b7
Answer: 12a¹¹b⁷Multiply the coefficients (the numbers) as normal. For letters with the same base, multiplying means adding the indices. - (b) The HCF of 14 and 21 is 7; the lowest powers are x² and y². Take out 7x²y²:
14x²y4 ÷ 7x²y² = 2y², and 21x³y² ÷ 7x²y² = 3x
Answer: 7x²y²(2y² + 3x)‘Factorise fully’ means take out the highest common factor of every part: the biggest number that divides both (7) and the lowest power of each letter that appears in both terms (x² and y²). Then divide each term by it to find what goes inside the bracket. - (c) The line crosses the y-axis at (0, 4), so c = 4. It passes through (2, 0), so the gradient = (0 − 4) ÷ (2 − 0) = −2.
Equation: y = −2x + 4Any straight line can be written as y = mx + c, where c is the y-intercept and m is the gradient. Gradient = change in y ÷ change in x. The line goes down from left to right, so the gradient is negative. - (d) In y = mx + c, the line crosses the y-axis at (0, c): (0, −5)On the y-axis x = 0, so y = 3(0) − 5 = −5.
The diagram shows a quadrilateral ABCD.
In the diagram, ABC and DAC are right-angled triangles (right angles at B and A). BC = 6 cm, AC = 7.5 cm. The area of quadrilateral ABCD is 31.5 cm².
(a) Work out the length of AB.
(b) Work out the length of AD.
Show solution
- (a) Triangle ABC is right-angled at B, and AC is the hypotenuse. By Pythagoras:
AB = √(7.5² − 6²) = √(56.25 − 36) = √20.25 = 4.5 cmPythagoras: hypotenuse² = sum of the squares of the other two sides. To find a shorter side, subtract: AB² = AC² − BC². - (b) Step 1: Area of triangle ABC = ½ × 6 × 4.5 = 13.5 cm²AD is in triangle DAC, and the only thing we can find about that triangle is its area. So first find the area of the other triangle, ABC. Its right angle is at B, so AB and BC are the base and height.
- Step 2: Area of triangle DAC = 31.5 − 13.5 = 18 cm²The quadrilateral is made of the two triangles, so the area of DAC is the total area minus the area of ABC.
- Step 3: Triangle DAC is right-angled at A, so AC and AD are the base and height: ½ × 7.5 × AD = 18 → AD = 18 ÷ 3.75 = 4.8 cmIn a right-angled triangle, the two sides at the right angle are the base and height. Substitute into ½ × base × height = area, then divide by 3.75 (which is ½ × 7.5).
Tess has a bag containing 11 beads. 4 of the beads are red. 7 of the beads are blue. Tess takes at random a bead from the bag and keeps it. Tess takes at random another bead from the bag and keeps it.
(a) Complete the probability tree diagram.
(b) Work out the probability that both of the beads are red.
(c) Work out the probability that the two beads are of different colours.
Show solution
- (a) If the first bead is blue, 10 beads remain: 4 red and 6 blue.
Missing branches: red = 4/10 blue = 6/10Tess keeps the first bead, so there are only 10 beads left for the second pick. If the first was blue, all 4 reds are still there but only 6 blues. - (b) P(red, red) = 411 × 310 = 12110 = 6/55‘Red AND then red’ means multiply along the branches. The second fraction is 3/10 because one red bead has already been taken.
- (c) “Different colours” means red then blue or blue then red:
P(R, B) = 411 × 710 = 28110 P(B, R) = 711 × 410 = 28110
Total = 56110 = 28/55There are two separate ways to get different colours. Find the probability of each by multiplying along its branches, then add them, because either way counts (‘or’ means add).
(a) Factorise 9x2 − 4y2
(b) Express 78 − x + 34x as a single fraction in its simplest form.
Show solution
- (a) This is a difference of two squares: (3x)² − (2y)²
Answer: (3x + 2y)(3x − 2y)9x² is (3x)² and 4y² is (2y)², with a minus between them. Any a² − b² factorises as (a + b)(a − b). - (b) The lowest common denominator of 8 and 4x is 8x.
78 = 7x8x and x + 34x = 2(x + 3)8x
Subtract: 7x − 2x − 68x = 5x − 68xTo subtract fractions they need the same denominator. 8x is the smallest expression that both 8 and 4x divide into. Multiply the first fraction top and bottom by x, and the second by 2.
(a) Expand and simplify (3x − 1)(x + 2)(3x + 1)
(b) Simplify fully (2x58xy2)−2
Show solution
- (a) Step 1: Multiply the two brackets that make a difference of two squares: (3x − 1)(3x + 1) = 9x² − 1You can multiply the three brackets in any order. Choosing (3x − 1)(3x + 1) first is smart because it is a difference of two squares and the middle terms cancel.
- Step 2: Multiply by the third bracket: (9x² − 1)(x + 2) = 9x³ + 18x² − x − 2
Answer: 9x³ + 18x² − x − 2Multiply each term of 9x² − 1 by each term of x + 2: 9x² × x = 9x³, 9x² × 2 = 18x², −1 × x = −x, −1 × 2 = −2. - (b) Step 1: simplify inside the bracket. 2x58xy2 = x44y2Simplifying first keeps the numbers small. 2 ÷ 8 = 1/4, x⁵ ÷ x = x⁴, and y² stays on the bottom.
- Step 2: the negative power flips the fraction. (x44y2)−2 = (4y2x4)2A negative power means the reciprocal: a⁻ⁿ = 1/aⁿ. For a fraction, (a/b)⁻ⁿ = (b/a)ⁿ.
- Step 3: square. Answer: 16y⁴x⁸Square every part: 4² = 16, (y²)² = y⁴, and (x⁴)² = x⁸.
100 farmers are asked if they have goats (G), sheep (S) or chickens (C) on their farms. Of these farmers: 31 have sheep; 53 have chickens; 6 have goats, sheep and chickens; 11 have sheep and goats; 17 have sheep and chickens; 18 have goats and chickens; 20 do not have any goats, sheep or chickens.
(a) Using this information, complete the Venn diagram to show the number of farmers in each appropriate subset.
(b) Find (i) n(G) (ii) n([G ∪ S]′) (iii) n(G′ ∩ C)
One of the farmers who has chickens is chosen at random.
(c) Find the probability that this farmer also has goats.
Show solution
- (a) Work from the centre outwards.The facts about two animals (such as 11 with sheep and goats) include the 6 farmers who have all three. Fill in the centre first, then subtract it from each overlap, then work out the ‘only’ regions.
- All three: 6
- Sheep and goats only: 11 − 6 = 5; sheep and chickens only: 17 − 6 = 11; goats and chickens only: 18 − 6 = 12
- Sheep only: 31 − (5 + 6 + 11) = 9
- Chickens only: 53 − (12 + 6 + 11) = 24
- None: 20
- Goats only: 100 − (5 + 9 + 6 + 12 + 11 + 24 + 20) = 100 − 87 = 13
- (b)(i) n(G) = 13 + 5 + 6 + 12 = 36n(G) means the total number in circle G, so add all four regions inside it, not just ‘goats only’.
- (b)(ii) [G ∪ S]′ is everything outside both G and S: 24 + 20 = 44G ∪ S is everything inside either circle. The dash (′) means the complement, so we want the regions outside both: chickens only and the 20 outside all circles.
- (b)(iii) G′ ∩ C is chickens but not goats: 11 + 24 = 35G′ means not in G, and ∩ C means also in C. So count the parts of C that are outside circle G.
- (c) The farmer is chosen only from the 53 with chickens. Of those, the ones with goats as well: 12 + 6 = 18.
Probability = 18/53The farmer is picked only from those with chickens, so the denominator is n(C) = 53, not 100. The numerator is the farmers in both C and G.
Martin and Lucia went on two different car journeys.
For Martin’s journey: distance = 80 km correct to the nearest 5 km; time = 2.7 hours correct to 1 decimal place.
For Lucia’s journey: distance = 33 km correct to 2 significant figures; time = 1 hour correct to the nearest 0.1 hour.
Martin says, “My average speed could have been greater than Lucia’s average speed.” By considering bounds, show that Martin is correct. Show your working clearly.
Show solution
- Step 1: bounds for Martin. Distance: 77.5 ≤ d < 82.5. Time: 2.65 ≤ t < 2.75‘Nearest 5 km’ means ± 2.5 km, and ‘1 decimal place’ means ± 0.05 hours.
- Step 2: bounds for Lucia. Distance: 32.5 ≤ d < 33.5. Time: 0.95 ≤ t < 1.0533 to 2 significant figures is accurate to the nearest 1 km, so ± 0.5 km. The nearest 0.1 hour means ± 0.05 hours.
- Step 3: Martin’s fastest possible speed uses upper bound distance ÷ lower bound time:
82.5 ÷ 2.65 = 31.13… km/hSpeed = distance ÷ time. A bigger distance and a smaller time both give a bigger answer, so the upper bound of the speed uses the upper bound distance and the lower bound time. - Step 4: Lucia’s slowest possible speed uses lower bound distance ÷ upper bound time:
32.5 ÷ 1.05 = 30.95… km/hFor the smallest possible speed, do the opposite: the smallest distance divided by the largest time. - Conclusion: 31.13… > 30.95…, so Martin’s maximum speed is greater than Lucia’s minimum speed. Martin is correct ✓Martin only says his speed ‘could have been’ greater, so it’s enough to show that his fastest possible speed beats Lucia’s slowest possible speed.
The histogram shows information about the total time, m minutes, taken by each child in a school to walk to school every day for one week.
There are no children for whom m > 100. There are 10 children for whom m ≤ 20. Work out an estimate for the number of children for whom 50 < m ≤ 80.
Show solution
- Step 1: find the frequency density scale. The 0–20 bar holds 10 children, so its frequency density = 10 ÷ 20 = 0.5. Reading the rest of the graph against this scale gives: 40–60 bar FD = 2.9, 60–75 bar FD = 3.2, 75–100 bar FD = 2.The vertical axis has no numbers, so use the one known frequency to calibrate it. Frequency density = frequency ÷ class width. Once the 0–20 bar gives FD = 0.5, every other bar’s height can be read on the same scale.
- Step 2: frequency = frequency density × class widthIn a histogram, the area of a bar represents frequency. 50 and 80 fall in the middle of bars, so use only the part of each bar inside 50 to 80, assuming the children are spread evenly across each bar. for each part of 50 < m ≤ 80:
- 50–60 (part of the 40–60 bar): 10 × 2.9 = 29
- 60–75: 15 × 3.2 = 48
- 75–80 (part of the 75–100 bar): 5 × 2 = 10
- Step 3: Total = 29 + 48 + 10 = 87Add the three parts together. The question says ‘estimate’ because we assumed an even spread inside the bars that were split.
Express 3 + √8(√2 − 1)2 in the form p + √q where p and q are integers. Show clearly each stage of your working.
Show solution
- Step 1: expand the denominator. (√2 − 1)² = 2 − 2√2 + 1 = 3 − 2√2(a − b)² = a² − 2ab + b². Here (√2)² = 2 and 2 × √2 × 1 = 2√2.
- Step 2: simplify √8. √8 = 2√2, so the numerator is 3 + 2√2. The expression becomes 3 + 2√23 − 2√2√8 = √(4 × 2) = √4 × √2 = 2√2. Writing it like this shows that the top and bottom use the same surd.
- Step 3: rationalise by multiplying top and bottom by (3 + 2√2):
Denominator: (3 − 2√2)(3 + 2√2) = 9 − 8 = 1
Numerator: (3 + 2√2)² = 9 + 12√2 + 8 = 17 + 12√2To remove a surd from the denominator, multiply the top and bottom by the conjugate (the same terms with the opposite sign). (3 − 2√2)(3 + 2√2) = 3² − (2√2)² = 9 − 8 = 1, so the denominator becomes 1. - Step 4: write 12√2 as a single surd. 12√2 = √(144 × 2) = √288
Answer: 17 + √288 (p = 17, q = 288)The answer must be p + √q with q an integer, so move the 12 inside the root by squaring it: 12 = √144.
The diagram shows two identical circles drawn inside a rectangle.
The length of the rectangle is 4x cm and the width of the rectangle is 2x cm. Each circle has a radius of x cm. The circles touch each other and each circle touches three sides of the rectangle. The region inside the rectangle that is outside the circles, shown shaded in the diagram, has a total area of 20 cm². Work out the perimeter of the rectangle. Give your answer correct to 3 significant figures.
Show solution
- Step 1: shaded area in terms of x. Rectangle − 2 circles = 4x × 2x − 2πx² = 8x² − 2πx²The shaded region is the rectangle with the two circles taken out. Each circle has area πx², so two circles have area 2πx².
- Step 2: set equal to 20. x²(8 − 2π) = 20 → x² = 208 − 2π = 11.649… → x = 3.4131…Both terms contain x², so factorise it out. Then divide by (8 − 2π) and take the square root to find x.
- Step 3: perimeter. 2(4x + 2x) = 12x = 12 × 3.4131… = 40.957…Perimeter = 2 × (length + width) = 2(4x + 2x) = 12x. Substitute the full calculator value of x.
- To 3 significant figures: 41.0 cm40.957… to 3 significant figures is 41.0. Keep the zero: it shows the answer is accurate to 3 significant figures.
The diagram shows a triangular prism, ABCDEF, with a rectangular base ABCD.
AB = 6 cm, DE = 2.2 cm, angle DAE = 18°, angle ADE = 90°. Work out the angle that BE makes with the plane ABCD. Give your answer correct to one decimal place.
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- Step 1: find AD in triangle ADE (right angle at D): tan 18° = 2.2AD → AD = 2.2tan 18° = 6.7709… cmTriangle ADE is right-angled at D. DE is opposite the 18° angle and AD is adjacent, so use tan (TOA). AD is also the width of the rectangular base.
- Step 2: find DB, the diagonal of the rectangular base: DB = √(6.7709…² + 6²) = 9.0468… cmThe base ABCD is a rectangle, so triangle DAB is right-angled at A. Use Pythagoras with AD and AB = 6.
- Step 3: find the angle. E is directly above D, so in triangle EDB (right angle at D): tan θ = 2.29.0468… = 0.2432…E is directly above D, so ED is vertical and DB lies flat in the base. That makes EDB a right-angled triangle, and the angle at B is the angle between BE and the base. ED is opposite and DB is adjacent, so use tan.
- θ = tan⁻¹(0.2432…) = 13.67…° = 13.7°Inverse tan turns the ratio back into an angle. The question asks for 1 decimal place.
Find the values of a, b and c so that 7 + 12x − 2x2 is written as a − b(x − c)2.
Show solution
- Step 1: factor −2 out of the x terms. 7 + 12x − 2x² = 7 − 2(x² − 6x)Completing the square is easiest when the x² coefficient is 1. The target form has −b in front of the bracket, so take out −2. Check: −2 × x² = −2x² and −2 × (−6x) = +12x.
- Step 2: complete the square inside the bracket. x² − 6x = (x − 3)² − 9Halve the coefficient of x (−6 ÷ 2 = −3) to make the bracket. (x − 3)² = x² − 6x + 9 has an extra +9, so subtract it.
- Step 3: substitute back and simplify. 7 − 2[(x − 3)² − 9] = 7 + 18 − 2(x − 3)² = 25 − 2(x − 3)²Multiply −2 into the square bracket: −2 × (−9) = +18, and 7 + 18 = 25.
- a = 25 b = 2 c = 3Compare 25 − 2(x − 3)² with a − b(x − c)² term by term.
Express (20x2 − 36 − 2x − 6) × 14 − x as a single fraction in its simplest form. Show clear algebraic working.
Show solution
- Step 1: factorise. x² − 36 = (x − 6)(x + 6), so the common denominator is (x − 6)(x + 6).x² − 36 is a difference of two squares (x² − 6²). Factorising shows that the second denominator, x − 6, is already part of the first, so the common denominator is just (x − 6)(x + 6).
- Step 2: combine the bracket. 20(x − 6)(x + 6) − 2(x + 6)(x − 6)(x + 6) = 20 − 2x − 12(x − 6)(x + 6) = 8 − 2x(x − 6)(x + 6)Multiply the top and bottom of the second fraction by (x + 6) so the denominators match. Then subtract the numerators, taking care that the minus applies to both 2x and 12.
- Step 3: factorise the numerator. 8 − 2x = 2(4 − x)Taking out the common factor 2 reveals the factor (4 − x), which matches the last fraction and will cancel.
- Step 4: multiply by 14 − x and cancel (4 − x): 2(4 − x)(x − 6)(x + 6)(4 − x) = 2x² − 36Multiplying fractions: multiply tops and bottoms. (4 − x) is on both the top and the bottom, so it cancels. Multiplying (x − 6)(x + 6) back out gives x² − 36.
18 × (√27)4n + 66 × 92n + 8 = 3x
Express x in terms of n. Show your working clearly and simplify your expression.
Show solution
- Step 1: write everything as a power of 3.Powers can only be combined when they have the same base. 27 = 3³ and 9 = 3², so every number here is a power of 3. A square root is a power of ½, and (aᵐ)ⁿ = aᵐⁿ.
- 186 = 3 = 31
- √27 = 27½ = 33/2, so (√27)4n+6 = 3(3/2)(4n+6) = 36n+9
- 9 = 3², so 92n+8 = 32(2n+8) = 34n+16
- Step 2: combine using the index laws. 31 × 36n+934n+16 = 31 + 6n + 9 − 4n − 16 = 32n − 6When multiplying, add the indices. When dividing, subtract them. Collect the n terms (6n − 4n) and the numbers (1 + 9 − 16) separately.
- So x = 2n − 6 (or 2(n − 3))Both sides are now powers of 3, so the indices must be equal.
ABCD is a kite. AB = AD and CB = CD. The point B has coordinates (k, 1) where k is a negative constant. The point D has coordinates (8, 7). The straight line L passes through the points B and D and has a gradient of 35.
Find an equation of AC. Give your answer in the form px + qy = r where p, q and r are integers. Show your working clearly.
Show solution
- Key fact: in a kite, the diagonal AC is the perpendicular bisector of the other diagonal BD. So AC passes through the midpoint of BD and is perpendicular to it.This means we never need to find A or C. One point on the line (the midpoint of BD) and its gradient are enough to write its equation.
- Step 1: find k. Gradient of BD = 7 − 18 − k = 35 → 30 = 24 − 3k → k = −2. So B = (−2, 1).Gradient = change in y ÷ change in x. Set this equal to the given 3/5 and cross-multiply: 6 × 5 = 3(8 − k).
- Step 2: midpoint of BD. (−2 + 82, 1 + 72) = (3, 4)The midpoint is the average of the x-coordinates and the average of the y-coordinates.
- Step 3: gradient of AC. Perpendicular gradients multiply to −1, so m = −53If two gradients multiply to −1, the second is the negative reciprocal of the first: flip 3/5 to 5/3, then change the sign.
- Step 4: equation through (3, 4). y − 4 = −53(x − 3) → 3y − 12 = −5x + 15Use y − y₁ = m(x − x₁), then multiply both sides by 3 to clear the fraction.
- Rearrange: 5x + 3y = 27Add 5x to both sides and add 12 to both sides to get the form px + qy = r with integers.
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