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All specs Current spec All series June 2025 November 2024 Any marks 1 to 4 marks 5 to 8 marks 9+ marks
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Higher June 2025 Paper 2 Q26
26 A and B are two similar solids.
The height of solid A is 31 cmB is 18.6 cm
Given that
volume of solid A – volume of solid B = 735 cm3
find the volume of solid A
(4)
Mark scheme
Mark scheme Scheme Marks A : B = 31 : 18.6 (= 5 : 3) oe or A 3 : B 3 = 313 : 18.63 (= 53 : 33 ) oe or
\(\dfrac{31}{18.6}\left(= \dfrac{5}{3}\right)\) oe or \(\dfrac{18.6}{31}\left(= \dfrac{3}{5}\right)\) oe
or \(\left(\dfrac{18.6}{31}\right)^3\) oe or \(\dfrac{27}{125}\) oe or \(\left(\dfrac{31}{18.6}\right)^3\) oe or \(\dfrac{125}{27}\) oe
M1 \(V_A - \left(\dfrac{3}{5}\right)^3 V_A\;(= 735)\) oe or \(V_A - \dfrac{27}{125}V_A\;(= 735)\) oe or
\(\dfrac{98}{125}V_A\;(= 735)\) oe or \(\dfrac{V_A}{V_A - 735} = \dfrac{31^3}{18.6^3}\) oe or
\(\left(\dfrac{5}{3}\right)^3 V_B - V_B\;(= 735)\) oe or \(\dfrac{125}{27}V_B - V_B\;(= 735)\) oe or
\(\dfrac{98}{27}V_B\;(= 735)\) oe or \(\dfrac{V_B + 735}{V_B} = \dfrac{31^3}{18.6^3}\) oe or
\(1 - \left(\dfrac{3}{5}\right)^3\left(= \dfrac{98}{125} = 0.784\right)\) oe or \(\left(\dfrac{5}{3}\right)^3 - 1\left(= \dfrac{98}{27} = 3.62(962\ldots)\right)\) oe or
53 – 33 (= 125 – 27 = 98)
M1 \((V_A =)735 \times \dfrac{125}{98}\) oe or \((V_A =)735 \div \dfrac{98}{125}\) oe or \((V_A =)\dfrac{31^3 \times 735}{31^3 - 18.6^3}\) oe or
\((V_B =)735 \times \dfrac{27}{98}(= 202.5)\) oe or \((V_B =)735 \div \dfrac{98}{27}(= 202.5)\) or
\((V_B =)\dfrac{18.6^3 \times 735}{31^3 - 18.6^3}(= 202.5)\) oe or 735 ÷ 98 × 53 oe or 7.5 × 125 oe or
735 ÷ 98 × 33 (= 202.5) oe or 7.5 × 27 (= 202.5) oe
M1 Working not required, so correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: for correct linear SF or volume SF either as a fraction or ratio. Allow \(\dfrac{5}{3} = 1.6(6\ldots.)\) truncated or rounded
M1: Note: 735 is given in the equation
Allow any letter for \(V_A\) or for \(V_B\)
\(V_A - \left(\dfrac{3}{5}\right)^3 V_A\;(= 735)\) can be written as \(V_A - \dfrac{V_A}{\left(\frac{5}{3}\right)^3}\;(= 735)\) or
\(\left(\dfrac{5}{3}\right)^3 V_B - V_B\;(= 735)\) can be written as \(\dfrac{V_B}{\left(\frac{3}{5}\right)^3} - V_B\;(= 735)\)
M1: for a correct method to find \(V_A\) or \(V_B\)
A1: oe allow 938 from correct working
Higher June 2025 Paper 1R Q26
26 R , S and T are three similar vases.
Diagram NOT accurately drawn
The volume of vase S is 72.8% more than the volume of vase R
The height of vase R is \(h\) cmT is \(6h\) cm
The surface area of vase S is \(A\) cm2 T is \(kA\) cm2
Work out the value of \(k\)
(4)
Mark scheme
Mark scheme Scheme Marks eg \(\sqrt[3]{1.728}(= 1.2)\) oe or
(length R : S =) 1 : 1.2 oe eg 5 : 6 or
(length R : S : T =) 1 : 1.2 : 6 oe eg 5 : 6 : 30 or
(volume S : T = ) 1.728 : 216 or 216 ÷ 1.728 (= 125)
M1 eg 6 ÷ “1.2” (= 5) or \(\sqrt[3]{\dfrac{216}{1.728}}(= 5)\)
(length S : T =) 1 : 5 or \(\sqrt[3]{1.728} : \sqrt[3]{216}\)
(area S : T =) “1.2”2 : 62 oe or
(area S : T =) 62 : 302 oe or
(area R : S : T =) 1 : “1.2”2 : 62 oe or
(area R : S : T =) 52 : 62 : 302 oe
M1 eg “5”2 or
(area S : T =) 12 : “5”2 or
(area S : T =) 1 : 25 or
\((k =)\dfrac{36}{1.44}\) oe
M1 Correct answer scores full marks (unless from obvious incorrect working) A1 (4) (4 marks)
Notes M1: for method to find the scale factor between the heights of R and S or R : S or R : S : T or for a correct ratio or scale factor for the volumes S : T
M1: for method to find the scale factor between the heights of S and T or S and T in the form 1 : \(n\) or S : T ft their ratio of lengthsor for a correct method to find the ratio for the areas R : S : T ft their ratio of lengths (maybe seen as the two separate ratios of R : S and R : T )
M1: for squaring the scale factor of the heights of S and T or S : T ft their ratio of areas, may be implied by their final answer or S : T to find the value of \(k\)
A1: cao
Higher June 2025 Paper 2 Q5
5 The diagram shows two similar quadrilaterals, \(ABCD\) and \(EFGH\)
Diagram NOT accurately drawn
\(AB = 5\) cm \(BC = 4\) cm \(CD = y\) cm
(a) Work out the value of \(x\) (2)
(b) Work out the value of \(y\) (2)
Mark scheme (a) Mark scheme (b)
Mark scheme (a) Scheme Marks \(\dfrac{10}{4}\left(= \dfrac{5}{2} = 2.5\right)\) or
\(\dfrac{4}{10}\left(= \dfrac{2}{5} = 0.4\right)\) or
\(\dfrac{x}{5} = \dfrac{10}{4}\) oe or
\(\dfrac{x}{10} = \dfrac{5}{4}\) oe
M1 Working not required, so correct answer scores full marks (unless from obvious incorrect working) A1 (2)
Notes M1: for a correct SF can be expressed as a fraction, decimal or ratio (may or may not be used)
A1: oe eg \(\dfrac{50}{4}\) or \(\dfrac{25}{2}\) or \(12\dfrac{1}{2}\) or \(12\dfrac{2}{4}\)
Mark scheme (b) Scheme Marks 24 ÷ [2.5] oe or
\(\dfrac{y}{24} = \dfrac{4}{10}\) oe or
\(\dfrac{y}{24} = \dfrac{5}{[12.5]}\) oe or
\(\dfrac{y}{4} = \dfrac{24}{10}\) oe or
\(\dfrac{y}{5} = \dfrac{24}{[12.5]}\) oe
M1 Working not required, so correct answer scores full marks (unless from obvious incorrect working) A1 (2) (4 marks)
Notes M1: ft ie [2.5] is their SF from (a)
A1: oe eg \(\dfrac{48}{5}\) or \(9\dfrac{3}{5}\)
If (a) \(x = 9.6\) and (b) \(y = 12.5\) then M1A0M1A0
Higher November 2024 Paper 1 Q10
10 Here are two similar triangles.
Diagrams NOT accurately drawn
Work out the value of \(x\)
(5)
Mark scheme
Mark scheme Scheme Marks eg \(51^2 = (DE)^2 + 24^2\) oe or \(2601 = (DE)^2 + 576\) oe
or \((DE^2 =)\;51^2 - 24^2\;(= 2025)\) oe or \((DE^2 =)\;2601 - 576\;(= 2025)\) oe or
\(\cos(DFE) = \dfrac{24}{51}\) or \(\sin(DEF) = \dfrac{24}{51}\)
M1 \((DE =)\sqrt{51^2 - 24^2}\;\left(= \sqrt{2025} = 45\right)\) or \((DE =)\sqrt{2601 - 576}\;\left(= \sqrt{2025} = 45\right)\)
or \((DFE =)\cos^{-1}\left(\dfrac{24}{51}\right)(= 61.9\ldots)\) or \((DEF =)\sin^{-1}\left(\dfrac{24}{51}\right)(= 28.0\ldots)\)
M1 \(\dfrac{\textit{their } DE}{7.5}(= 6)\) oe or \(\dfrac{7.5}{\textit{their } DE}\left(= \dfrac{1}{6}\right)\) or \(\dfrac{x}{24} = \dfrac{7.5}{\textit{their } DE}\) oe
or \(\tan\text{“}\textit{their } 61.9\ldots\text{”} = \dfrac{7.5}{(x)}\) or \(\tan\text{“}\textit{their } 28.0\ldots\text{”} = \dfrac{(x)}{7.5}\) or
\(\dfrac{(x)}{\sin(\textit{their } 28.0)} = \dfrac{7.5}{\sin(\textit{their } 61.9)}\)
NB Their \(ED\) or their 61.9 or their 28.0 must be clearly identified
Their 61.9.. or their 28.0.. cannot be used as lengths of the triangle
Their 45 cannot be used as an angle of the triangle
M1 \(24 \div\) “6” oe or \(24 \times \text{“}\dfrac{1}{6}\text{”}\) or 24 × “1.67” or \((x =)\;\dfrac{7.5}{\textit{their } ED} \times 24\)
or \((x =)\;\dfrac{7.5}{\tan\text{“}\textit{their } 61.9\ldots\text{”}}\) oe or \((x =)\;7.5 \times \tan\text{“}\textit{their } 28.0\ldots\text{”}\) oe
or \((x =)\;\dfrac{7.5}{\sin(\textit{their } 61.9)} \times \sin(\textit{their } 28.0)\) oe
or \(51 \times \text{“}\dfrac{1}{6}\text{”}\;(= 8.5)\) and \((x =)\sqrt{8.5^2 - 7.5^2}\;\left(= \sqrt{16}\right)\)
M1 Working required A1 (5) (5 marks)
Notes M1: for applying Pythagoras theorem correctly
M1: for square rooting
M1: for a correct method to find the scale factor
Allow correct use of sine rule/cosine rule/Pythagoras theorem
Allow 0.17 or better for \(\dfrac{1}{6}\)
Special case Allow \((DE =)\sqrt{51^2 + 24^2}\;\left(= \sqrt{3177} = 3\sqrt{353} = 56.3\ldots\right)\) for “45” for this mark
M1: dep on previous M1 for a correct method to find \(x\) or for finding \(BC\) and using Pythagoras theorem to find \(x\)
Allow \(24 \times \text{“}\dfrac{7.5}{56(.3\ldots)}\text{”}\) or \(24 \div \text{“}\dfrac{56(.3\ldots)}{7.5}\text{”}\) for scale factor for this mark
A1: dep on M2
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