Paper 42, worked through. Cambridge IGCSE Mathematics, May/June 2026: worked solutions

A hundred marks in two hours, with a calculator. Unless a question says otherwise, give answers to three significant figures and angles to one decimal place, and keep the full calculator value until the last step: rounding half-way through is the quiet way marks go missing on this paper. The formula list is on page 2.

1. 1Finding a base angle of an isosceles triangle from the angle between its two equal sides.[2]
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   1. The two marked sides, AB and AC, are equal. The angles opposite equal sides are equal, so the angle at B and the angle at C are the same: both are x.
   2. The 30° angle is at A, between the two equal sides. Angles in a triangle add up to 180°, so the other two share 180 − 30 = 150°.
   3. They are equal, so each is 150 ÷ 2.

   Answerx = 75

   The equal angles are the ones opposite the equal sides, which here are at B and C. Don't assume the 30° is one of the pair.

   Revise this: Angles
2. 2(a)Reading the mode from a frequency table.[1]
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   1. The mode is the value that happens most often. In a frequency table that is the value with the biggest frequency.
   2. The biggest frequency is 14, and it belongs to 1 book.

   Answer1

   The mode is the number of books, 1, not the frequency, 14.

   Revise this: Data, averages and range
3. 2(b)Finding the median from a frequency table.[1]
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   1. There are 40 students, so the median is halfway between the 20th and 21st values when they are put in order.
   2. Count up through the table. The first 5 students borrowed 0 books. The next 14 borrowed 1, which takes the count to 19. The next 10 borrowed 2, which covers the 20th to the 29th.
   3. So the 20th and the 21st are both 2.

   Answer2

   Count through the frequencies. The median is not the middle of the top row, and not the middle frequency.

   Revise this: Data, averages and range
4. 2(c)Calculating the mean from a frequency table.[3]
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   1. The mean is the total number of books divided by the number of students.
   2. Total books: multiply each number of books by its frequency, then add. 0 × 5 + 1 × 14 + 2 × 10 + 3 × 8 + 4 × 1 + 5 × 2 = 0 + 14 + 20 + 24 + 4 + 10 = 72.
   3. Number of students: 40.
   4. 72 ÷ 40 = 1.8.

   Answer1.8

   Divide by the number of students, 40, not by the number of columns, 6.

   Leave it as 1.8. A mean does not have to be a whole number, and rounding it to 2 loses the last mark.

   Revise this: Data, averages and range
5. 3Constructing a triangle with ruler and compasses when all three sides are known.[2]
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   1. C has to be 10 cm from A and 8.5 cm from B. Compasses find every point at a given distance, so use them twice.
   2. Open the compasses to 10 cm against your ruler. Put the point on A and draw an arc on the side where the triangle will go.
   3. Open them to 8.5 cm. Put the point on B and draw a second arc that crosses the first.
   4. Where the two arcs cross is C. Join it to A and to B with a ruler.

   AnswerTriangle ABC with AC = 10 cm and BC = 8.5 cm, with both construction arcs left showing.

   Leave the arcs in. They are the proof that you constructed it, and a triangle without them gets one mark of the two.

   Check which length goes from which end: the 10 cm is from A. Swapping the two gives a mirror-image triangle, which earns only one mark.

   Revise this: Constructions
6. 4Working out how many monthly payments clear what is left after a percentage deposit.[3]
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   1. Deposit: 35% of 1260 = 0.35 × 1260 = $441.
   2. Still to pay: 1260 − 441 = $819. (Or in one go: 65% of 1260 = 0.65 × 1260 = 819.)
   3. Number of payments: 819 ÷ 45.50.

   Answer18

   Divide what is left after the deposit. Not the full price, and not the deposit.

   Revise this: Percentages
7. 5Finding where a point ends up after moving along a column vector.[2]
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   1. The vector from P to Q says how to get from P to Q: the top number is the move across, the bottom number is the move up.
   2. Across: −3 + 8 = 5.
   3. Up: 5 + (−3) = 2.

   Answer(5, 2)

   Add the vector to P. Subtracting it finds where you would have come from, not where you get to.

   The top number goes with x, the bottom number goes with y.

   Revise this: Vectors
8. 6Finding the perimeter of a semicircle from its diameter.[3]
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   1. The perimeter is all the way round the edge: the curved part and the straight part.
   2. Curved part: half the circumference of the full circle. Circumference = π × diameter, so half of it is π × 15.8 ÷ 2 = 24.819…
   3. Straight part: the diameter, 15.8.
   4. 24.819… + 15.8 = 40.619…

   Answer40.6 cm

   The straight edge counts. Stopping at 24.8, the curved part alone, is the classic slip and costs a mark.

   You are given the diameter, so use π × d. Using 2 × π × 15.8 treats it as a radius.

   Revise this: Circles, arcs and sectors
9. 7(a)Working out what an investment is worth after several years of compound interest.[2]
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   1. Each year the money is multiplied by the same number. A rise of 4.4% is a multiplier of 1 + 4.4 ÷ 100 = 1.044.
   2. Six years means multiplying six times: 2500 × 1.044⁶.
   3. On the calculator: 2500 × 1.044⁶ = 3237.00…

   Answer$3237 (3240, to three significant figures, is accepted too)

   Compound interest is not six lots of 4.4% of 2500. That is simple interest, and it gives $3160.

   Give the value of the investment, not just the interest it earned.

   Revise this: Compound interest, growth and decay
10. 7(b)Finding the yearly rate of compound interest from the start value, the end value and the number of years.[3]
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    1. Call the yearly multiplier m. Five years turn 4500 into 5420, so 4500 × m⁵ = 5420.
    2. Divide by 4500: m⁵ = 5420 ÷ 4500 = 1.20444…
    3. Take the fifth root to undo the power of 5: m = 1.03790…
    4. A multiplier of 1.0379 means a rise of 3.79%.

    Answerr = 3.79

    Do not divide the total rise by 5. That treats it as simple interest and gives 4.09.

    Keep the full calculator value until the end. Rounding 1.20444 early changes the third figure.

    Revise this: Compound interest, growth and decay
11. 8(a)Filling in the missing values in a table for a cubic curve.[3]
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    1. Put each x into y = x³ − 2x² − 2 and work it out, taking care with the negative one.
    2. x = −1: (−1)³ − 2 × (−1)² − 2 = −1 − 2 − 2 = −5.
    3. x = 1: 1 − 2 − 2 = −3.
    4. x = 2.5: 15.625 − 12.5 − 2 = 1.125, which is 1.1 to one decimal place.

    Answer−5, −3 and 1.1

    (−1)² is +1, so −2 × (−1)² is −2, not +2. Put brackets round the negative number on your calculator.

    Revise this: Graphs of curves
12. 8(b)Plotting the points and drawing the curve.[4]
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    1. Check the scales first. Across, one small square is 0.1. Up, one small square is 0.2.
    2. Plot all nine points: (−1, −5), (−0.5, −2.6), (0, −2), (0.5, −2.4), (1, −3), (1.5, −3.1), (2, −2), (2.5, 1.1) and (3, 7).
    3. Join them with one smooth curve, drawn freehand in a single line. It rises to a gentle peak at (0, −2), dips to its lowest point a little after x = 1, then climbs steeply to (3, 7).

    AnswerA smooth curve through all nine points, from (−1, −5) to (3, 7), with a small peak at (0, −2) and a dip near x = 1.3.

    No ruler: straight lines between the points lose the curve mark. And no thick or feathery line.

    The lowest point is not at (1.5, −3.1). The curve dips just below that, between x = 1 and x = 1.5, so let it round off smoothly there.

    Stop at the first and last points. The question gives the range of x.

    Revise this: Graphs of curves
13. 8(c)Solving a different cubic equation by drawing a straight line across the curve.[4]
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    1. The curve is y = x³ − 2x² − 2. The equation to solve is x³ − 2x² − x + 1 = 0. Rearrange it so that its left side is exactly the curve's formula.
    2. Add x to both sides and take away 3: x³ − 2x² − 2 = x − 3. (Check: moving everything back gives x³ − 2x² − x + 1 = 0.)
    3. So the solutions are where the curve meets the straight line y = x − 3.
    4. Draw y = x − 3 with a ruler. It passes through (0, −3), (2, −1) and (3, 0).
    5. Read off the x-coordinate of each of the three crossing points.

    AnswerThe line y = x − 3, and x ≈ −0.8, x ≈ 0.55 and x ≈ 2.25. Accepted: −0.9 to −0.7, 0.5 to 0.6, and 2.2 to 2.35.

    Two of the four marks are for the line and two are for the readings. The readings only count if the line is right, so check the rearranging before you draw.

    The answers are x values only. Do not give coordinates.

    The line has to be long enough to cross the curve in all three places.

    Revise this: Graphs of curves
14. 9Solving an equation with a fraction on each side.[3]
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    1. Get rid of the fractions first. Multiply both sides by 12, which is 4 × 3: the left becomes 3(w + 8) and the right becomes 4(2w − 3). This is the same as cross-multiplying.
    2. Expand the brackets: 3w + 24 = 8w − 12.
    3. Collect the w terms on the side that has more of them. Take 3w from both sides and add 12 to both sides: 36 = 5w.
    4. Divide by 5.

    Answerw = 7.2 (or 36/5)

    Keep the brackets when you multiply up. 3 × w + 8 is not 3(w + 8).

    Check: the left side is (7.2 + 8) ÷ 4 = 3.8, and the right side is (14.4 − 3) ÷ 3 = 3.8. They match.

    Revise this: Solving linear equations
15. 10(a)Finding one share when the percentages are taken one after the other.[2]
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    1. The 25% is not of all the coins. It is 25% of what is left after the gold ones.
    2. So among the coins that are not gold, 25% are silver and the other 75% are copper.
    3. 75% of the non-gold coins is 54, so 25% of them is 54 ÷ 3.

    Answer18

    Silver is a third of copper here, because 25% is a third of 75%. Spotting that makes this a one-line answer.

    Taking 25% of the whole collection is the trap. The question says "of the remaining coins".

    Revise this: Percentages
16. 10(b)Working back to the whole collection.[2]
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    1. The coins that are not gold: 18 silver + 54 copper = 72.
    2. Gold is 10% of all the coins, so those 72 are the other 90%.
    3. 90% of the total is 72, so the total is 72 ÷ 0.9.

    Answer80

    Check it forwards: 10% of 80 is 8 gold, leaving 72. 25% of 72 is 18 silver, leaving 54 copper. It fits.

    Adding 10% of 72 on to 72 gives 79.2, which is wrong: the 10% is of the total, not of the 72.

    Revise this: Percentages
17. 11Finding a side of a right-angled triangle from one side and an angle.[2]
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    1. Label the sides from the 29° angle at W. XY is opposite it. WX, the 12.4 cm, is next to it (adjacent). The hypotenuse, WY, is not involved.
    2. Opposite and adjacent means tan: tan 29° = XY ÷ 12.4.
    3. Multiply both sides by 12.4: XY = 12.4 × tan 29° = 6.873…

    Answer6.87 cm

    Check your calculator is in degrees. In radians this comes out as 11.0, which looks believable and is wrong. A quick test: tan 45 should give 1.

    The side you want is on top of the fraction, so you multiply. If it were underneath, you would divide.

    Revise this: Trigonometry in right-angled triangles
18. 12Finding the day on which a mass that falls by 5% a day first drops below a given value.[3]
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    1. A fall of 5% a day means each day's mass is 0.95 times the day before. After n days the mass is 150 × 0.95ⁿ.
    2. Try values of n until it goes below 96. After 8 days: 150 × 0.95⁸ = 99.5…, still above. After 9 days: 150 × 0.95⁹ = 94.5…, below.
    3. So the first reading under 96 grams comes 9 days after 10 January.

    Answer19 January

    Count 10 January as day 0. Nine days later is the 19th.

    5% of 150 is 7.5 g, but the mass does not fall by 7.5 g every day: each day's 5% is of a smaller amount. Taking off 7.5 g a day gives 18 January, which is wrong.

    Revise this: Compound interest, growth and decay
19. 13Using inverse proportion to find one value from another pair.[3]
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    1. Inversely proportional means y = k ÷ x for some fixed number k. Said another way: x times y always comes to the same number.
    2. Find k from the pair you are given: k = 6 × 10.5 = 63.
    3. Use it with the new x: y = 63 ÷ 9.

    Answery = 7

    Inverse means that when x goes up, y comes down. x went up from 6 to 9, so y must be less than 10.5. If your answer is bigger (15.75), you have used direct proportion.

    Revise this: Direct and inverse proportion
20. 14Finding the reflex angle that has a given cosine.[2]
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    1. The calculator gives the acute angle: cos⁻¹(0.41) = 65.8°.
    2. Cosine is positive for two angles between 0° and 360°: that acute one, and the one the same distance back from 360°.
    3. 360 − 65.8 = 294.2°. A reflex angle is between 180° and 360°, so this is the one wanted.

    Answerx = 294.2°

    180 − 65.8 = 114.2° is the rule for sine, and it is not reflex anyway. For cosine, take the calculator's angle away from 360.

    Angles are given to one decimal place unless the question says otherwise.

    Revise this: Exact values and trigonometric graphs
21. 15Adding two algebraic fractions.[3]
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    1. The bottoms are different and share nothing, so the common denominator is the two multiplied together: (x + 2)(x − 4).
    2. Multiply the top of each fraction by the bottom of the other: 8(x − 4) + 3(x + 2), all over (x + 2)(x − 4).
    3. Expand the top: 8x − 32 + 3x + 6 = 11x − 26.

    Answer(11x − 26)/((x + 2)(x − 4)), which can also be written (11x − 26)/(x² − 2x − 8)

    Each top is multiplied by the other fraction's bottom, the same as with number fractions. Adding the tops and adding the bottoms, to get 11/(2x − 2), is wrong.

    Watch 8 × −4 = −32. Then −32 + 6 = −26.

    Revise this: Algebraic fractions
22. 16(a)Counting the planes of symmetry of a square-based pyramid.[1]
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    1. A plane of symmetry cuts the solid into two halves that are mirror images of each other.
    2. Every such plane here passes through the top point A, and cuts the square base along one of the square's lines of symmetry.
    3. A square has four lines of symmetry: two through the midpoints of opposite sides and two along the diagonals.

    Answer4

    A horizontal cut does not work. The top piece would be a small pyramid and the bottom piece would not be.

    Revise this: Symmetry
23. 16(b)Calculating the volume of a pyramid.[2]
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    1. Volume of a pyramid = 1/3 × base area × height. It is in the formula list.
    2. Base area: the base is a square of side 16.5, so 16.5² = 272.25 cm².
    3. Height: the perpendicular height is AF = 28 cm.
    4. 1/3 × 272.25 × 28 = 2541.

    Answer2541 cm³

    It comes out as exactly 2541, so give all four figures. Rounding it to 2540 loses the accuracy mark.

    Forgetting the 1/3 gives the volume of a box, three times too big.

    Revise this: Volume and surface area
24. 16(c)Finding the length of a sloping edge of the pyramid, which needs Pythagoras twice.[4]
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    1. AB is the long side of the right-angled triangle AFB: AF goes straight up from F, and FB lies flat in the base. So find FB first.
    2. F is the centre of the square, so FB is half of the diagonal BD. Diagonal of the square: BD² = 16.5² + 16.5² = 544.5, so BD = 23.33…
    3. FB = 23.33… ÷ 2 = 11.667…
    4. Now triangle AFB: AB² = 28² + 11.667…² = 784 + 136.125 = 920.125.
    5. AB = √920.125 = 30.33…

    Answer30.3 cm

    FB is half the diagonal, not half the side. Using 8.25 gives 29.2, the most common wrong answer.

    Keep the unrounded values in the calculator. FB² is exactly 136.125, so you can use that and skip the square root in the middle.

    Revise this: Pythagoras and trigonometry in 3D
25. 16(d)Finding the angle between a sloping edge and the base of the pyramid.[2]
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    1. The angle between a line and a flat surface is found in the right-angled triangle made by the line, its "shadow" on the surface, and the height. Here that is triangle AFB again: AB is the line, FB is its shadow on the base, and the angle wanted is at B.
    2. From B, the height AF = 28 is opposite and FB = 11.667… is adjacent. So tan B = 28 ÷ 11.667…
    3. B = tan⁻¹(2.3998…) = 67.38…

    Answer67.4°

    The angle is at B, between AB and FB. It is not the angle at the top of the pyramid, and not the angle between AB and BC.

    Use the exact sides, 28 and 11.667. Using your rounded 30.3 from part (c) still lands inside the accepted range here, but it will not always.

    Revise this: Pythagoras and trigonometry in 3D
26. 17(a)Putting a number into a function.[1]
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    1. f(4) means replace x with 4 in f(x) = 20 − 3x.
    2. 20 − 3 × 4 = 20 − 12.

    Answer8

    Multiply before you subtract. (20 − 3) × 4 is not what it says.

    Revise this: Functions
27. 17(b)Finding the inverse of a linear function.[2]
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    1. Write the function as y = 7 + 2x. The inverse undoes it, so rearrange to get x alone.
    2. Take away 7: y − 7 = 2x.
    3. Divide by 2: x = (y − 7)/2.
    4. Write the answer with x as the letter.

    Answerh⁻¹(x) = (x − 7)/2

    h⁻¹ is the inverse, not "one over h". 1/(7 + 2x) is a different thing.

    Undo the steps in reverse order: h doubles and then adds 7, so the inverse takes away 7 and then halves.

    Check with a number: h(1) = 9, and (9 − 7)/2 = 1. It gets you back.

    Revise this: Functions
28. 17(c)Applying the same function twice.[2]
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    1. gg(5) means work out g(5), then put that answer into g again.
    2. g(5) = 64 ÷ 2⁵ = 64 ÷ 32 = 2.
    3. g(2) = 64 ÷ 2² = 64 ÷ 4.

    Answer16

    gg(5) is not g(5) × g(5). That gives 4.

    Work from the inside out.

    Revise this: Functions
29. 17(d)Solving an equation that has an inverse function in it, without finding the inverse.[2]
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    1. g⁻¹(x) = 8 says: the number that g turns into x is 8. In other words, x = g(8).
    2. g(8) = 64 ÷ 2⁸ = 64 ÷ 256.

    Answerx = 1/4 (or 0.25)

    You never need the formula for g⁻¹ here. Applying g to both sides removes it: g of g⁻¹(x) is just x.

    Solving g(x) = 8 gives x = 3, which answers a different question.

    Revise this: Functions
30. 18(a)Showing a formula for the area of a triangle from two sides and the angle between them.[1]
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    1. The area of a triangle is 1/2 × a × b × sin C, where C is the angle between the sides a and b. It is in the formula list.
    2. Here the two sides are 2x and x, and the angle between them is 30°. Area = 1/2 × 2x × x × sin 30°.
    3. sin 30° = 1/2, so the area is 1/2 × 2x² × 1/2 = 1/2 x².

    AnswerArea = 1/2 × 2x × x × sin 30° = 1/2 × 2x² × 1/2 = 1/2 x²

    In a "show that" question the marks are for the steps, because the answer is already printed. Write the sin 30° in: jumping straight to 1/2 x² earns nothing.

    Revise this: The sine and cosine rules
31. 18(b)Turning "these two areas are equal" into a quadratic equation.[3]
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    1. Triangle DEF has a right angle at E, so its area is 1/2 × base × height = 1/2 × (2x − 6) × (x − 4).
    2. Set the two areas equal: 1/2 x² = 1/2 (2x − 6)(x − 4).
    3. Double both sides to clear the halves: x² = (2x − 6)(x − 4).
    4. Expand the brackets: (2x − 6)(x − 4) = 2x² − 8x − 6x + 24 = 2x² − 14x + 24.
    5. So x² = 2x² − 14x + 24. Take x² from both sides: 0 = x² − 14x + 24.

    Answer1/2 x² = 1/2 (2x − 6)(x − 4), so x² = 2x² − 14x + 24, which gives x² − 14x + 24 = 0.

    Show the expansion with all four terms before you collect them. A line missed out of a "show that" costs the final mark.

    Don't solve the equation here. This part only asks you to reach it.

    Revise this: Quadratic equations
32. 18(c)Solving the quadratic by factorising, throwing out the value that cannot be a length, and finding the area.[4]
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    1. Find two numbers that multiply to +24 and add to −14. Both must be negative: −2 and −12.
    2. x² − 14x + 24 = (x − 2)(x − 12) = 0, so x = 2 or x = 12.
    3. Check each against the diagram. With x = 2, the side (x − 4) would be −2, and a length cannot be negative. So x = 12.
    4. Area of triangle ABC = 1/2 x² = 1/2 × 12² = 1/2 × 144.

    Answer72 cm²

    The question asks for the area, not for x. Stopping at x = 12 loses the last marks.

    A quadratic gives two answers, but the picture decides which one is real. Always test them in the lengths.

    Check with the other triangle: 1/2 × (24 − 6) × (12 − 4) = 1/2 × 18 × 8 = 72. The same.

    Revise this: Quadratic equations
33. 19(a)Estimating the mean from a grouped frequency table.[4]
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    1. The exact times are not known, only which group each one falls in. So let the middle of each group stand for everyone in it: 2.5, 7.5, 15 and 30.
    2. Multiply each midpoint by its frequency: 2.5 × 3 = 7.5, 7.5 × 18 = 135, 15 × 13 = 195, 30 × 6 = 180.
    3. Add them: 7.5 + 135 + 195 + 180 = 517.5.
    4. Divide by the number of customers, 40: 517.5 ÷ 40 = 12.9375.

    Answer12.9 minutes (12.9375 exactly)

    The midpoint of a group is halfway between its two ends: (20 + 40) ÷ 2 = 30. Not 20, and not the width of the group.

    The groups are different widths, so the midpoints are not evenly spaced. Work out each one.

    Divide by the total frequency, 40, not by the number of groups.

    Revise this: Data, averages and range
34. 19(b)Completing a histogram for groups of unequal width.[3]
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    1. In a histogram the height of a bar is the frequency density, which is frequency ÷ class width. The bar already drawn shows the idea: 3 ÷ 5 = 0.6.
    2. 5 to 10 minutes: 18 ÷ 5 = 3.6.
    3. 10 to 20 minutes: 13 ÷ 10 = 1.3.
    4. 20 to 40 minutes: 6 ÷ 20 = 0.3.
    5. Draw each bar over its own group, with no gaps between the bars. One small square up the side is 0.1.

    AnswerThree more bars: from 5 to 10 at a height of 3.6, from 10 to 20 at 1.3, and from 20 to 40 at 0.3.

    Plotting the frequencies themselves (18, 13 and 6) is the big mistake. They would not even fit on this grid, which is a hint.

    Each bar is as wide as its group. The last bar is four times as wide as the first.

    Revise this: Histograms
35. 20(a)Reading a bearing from an angle marked at the point.[1]
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    1. A bearing is measured at the place you are standing, clockwise from North, and written with three figures. "Of A from B" means standing at B.
    2. The 103° on the diagram is the angle from North round to BA going the other way, anticlockwise.
    3. Clockwise from North to BA is the rest of the full turn: 360 − 103.

    Answer257°

    "From B" tells you where to measure. Start at the North line at B and turn clockwise until you face A.

    A is to the west of B, so the bearing has to be more than 180°. 103° would point east of south.

    Revise this: Scale drawings and bearings
36. 20(b)Finding the third side of a triangle from two sides and the angle between them.[3]
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    1. CD is a side of triangle ACD. In that triangle AD = 38.2, AC = 86.8, and the angle between them, at A, is 79°.
    2. Two sides and the angle between them: that is the cosine rule. CD² = 38.2² + 86.8² − 2 × 38.2 × 86.8 × cos 79°.
    3. CD² = 1459.24 + 7534.24 − 1265.35… = 7728.1…
    4. CD = √7728.1… = 87.90…

    Answer87.9 m

    Type the whole right-hand side in one go, or work out 2 × 38.2 × 86.8 × cos 79° before taking it away. Subtracting first and then multiplying by cos 79° is the famous slip with this rule.

    Remember the square root at the end. 7728 is CD², not CD.

    Revise this: The sine and cosine rules
37. 20(c)Finding the shortest distance from a corner of a triangle to the opposite side, which means finding an angle first.[6]
    Show how to do itHide the working

    1. The shortest distance from B to AC is the perpendicular from B to AC. Call the point where it meets AC, P. Then triangle ABP has a right angle at P, with AB = 50.3 as its hypotenuse. To find BP, the angle BAC is needed.
    2. Work in triangle ABC: AB = 50.3, AC = 86.8, and angle ABC = 112° is opposite AC. A side and the angle opposite it: use the sine rule. sin C ÷ 50.3 = sin 112° ÷ 86.8.
    3. sin C = 50.3 × sin 112° ÷ 86.8 = 0.5373…, so angle ACB = 32.49…°
    4. Angles in a triangle: angle BAC = 180 − 112 − 32.49… = 35.50…°
    5. In the right-angled triangle ABP, BP is opposite angle A and AB is the hypotenuse: BP = 50.3 × sin 35.50…°

    Answer29.2 m

    "Shortest distance" always means the perpendicular. Draw it on the diagram and mark the right angle: there is a mark just for showing you know that.

    Angle C must be acute, because the triangle already has an obtuse angle of 112°. So the calculator's answer for sin⁻¹ is the right one.

    Another route to the same answer: the area of triangle ABC is 1/2 × 50.3 × 86.8 × sin 35.5°, and it is also 1/2 × 86.8 × BP. Set the two equal.

    Keep the unrounded angle in your calculator. Rounding it to 32° early moves the answer.

    Revise this: The sine and cosine rules
38. 21Finding the lower bound of a density from rounded measurements.[3]
    Show how to do itHide the working

    1. Density = mass ÷ volume. A fraction is smallest when its top is as small as possible and its bottom is as large as possible.
    2. The mass is 140 g to the nearest 10 g, so it lies between 135 and 145. Smallest mass: 135.
    3. The volume is 22 cm³ to the nearest 1 cm³, so it lies between 21.5 and 22.5. Largest volume: 22.5.
    4. Lower bound of the density = 135 ÷ 22.5.

    Answer6 g/cm³

    For the lower bound of a division, use the lower bound on top and the upper bound underneath. Using both lower bounds (135 ÷ 21.5) gives 6.28, which is not the smallest it could be.

    Half the rounding unit either side: to the nearest 10 means ±5, to the nearest 1 means ±0.5.

    Revise this: Rounding, estimating and bounds
39. 22Simplifying an algebraic fraction by factorising the top and the bottom.[3]
    Show how to do itHide the working

    1. Nothing can be cancelled until both parts are factorised.
    2. Top: 8x² + 24x. Both terms have 8x in them: 8x(x + 3).
    3. Bottom: 2x² − 18. Take out the 2 first: 2(x² − 9). Then x² − 9 is a difference of two squares: 2(x − 3)(x + 3).
    4. So the fraction is 8x(x + 3) over 2(x − 3)(x + 3). Cancel the (x + 3) that is on both, and 8 ÷ 2 = 4.

    Answer4x/(x − 3)

    You can only cancel whole factors, things that are multiplied. Crossing out the x² on top with the x² underneath is not allowed.

    Look for a difference of two squares whenever you see something squared minus a square number.

    Revise this: Algebraic fractions

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