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Either paper 3 marks
A solid cone of radius \(r\), height \(h\) and density \(\rho\) has mass \(M\), where
\(M = \frac{ 1 }{ 3 }\pi \rho r^2 h\)
Make \(r\) the subject of the formula.
Show answer
\(r = \sqrt{ \frac{ 3M }{ \pi \rho h } }\)Start with the formula:
\(M = \frac{ 1 }{ 3 }\pi \rho r^2 h\)
Multiply both sides by 3 to clear the fraction:
\(3M = \pi \rho r^2 h\)
Divide both sides by \(\pi \rho h\), so that the power of \(r\) is on its own:
\(r^2 = \frac{ 3M }{ \pi \rho h }\)
Take the square root of both sides. The root must cover the bottom of the fraction as well as the top, because it is the whole of the right-hand side that has been squared (rooting only the top would be wrong). Writing the square root separately over the top and over the bottom gives the same value:
\(r = \sqrt{ \frac{ 3M }{ \pi \rho h } }\)
- M1both sides multiplied by 3 to clear the fraction: \(3M = \pi \rho r^2 h\)
- M1both sides divided by \(\pi \rho h\) to leave the power of \(r\) alone: \(r^2 = \frac{ 3M }{ \pi \rho h }\)
- A1the square root taken of the whole right-hand side, so that it covers the bottom as well as the top: \(r = \sqrt{ \frac{ 3M }{ \pi \rho h } }\). The square root written separately over the top and over the bottom is the same answer and also earns the mark
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Either paper 6 marks(a)
Make \(x\) the subject of \(8x + y = 2x + 9\)
(b)Make \(x\) the subject of \(y\left(x - 3\right) = 3x + 8\)
Show answer
(a) \(x = \frac{ 9 - y}{ 6 }\)
(b) \(x = \frac{ 8 + 3y }{y - 3 }\)Get both \(x\) terms onto the same side. Subtract \(2x\) from both sides:
\(8x - 2x + y = 9\)
\(6x + y = 9\)
Subtract \(y\) from both sides:
\(6x = 9 - y\)
Divide both sides by 6:
\(x = \frac{ 9 - y}{ 6 }\)
Multiply out the bracket first:
\(xy - 3y = 3x + 8\)
Now get every term containing \(x\) onto one side and everything else onto the other:
\(xy - 3x = 8 + 3y\)
\(x\) is a factor of both terms on the left, so take it outside a bracket. This is the step that makes the rearrangement possible at all:
\(x\left(y - 3\right) = 8 + 3y\)
Divide both sides by \(\left(y - 3\right)\):
\(x = \frac{ 8 + 3y }{y - 3 }\)
- (a) M1\(x\) terms collected: \(6x + y = 9\)
- (a) M1\(6x = 9 - y\)
- (a) A1\(x = \frac{ 9 - y}{ 6 }\)
- (b) M1bracket expanded: \(xy - 3y = 3x + 8\)
- (b) M1factorised: \(x\left(y - 3\right) = 8 + 3y\)
- (b) A1\(x = \frac{ 8 + 3y }{y - 3 }\)
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Either paper 4 marks
Make \(x\) the subject of \(y = \frac{ x - 3 }{x - 5 }\)
Show answer
\(x = \frac{ -3 + 5y }{y - 1 }\)Multiply both sides by \(\left(x - 5\right)\) to get rid of the fraction:
\(y\left(x - 5\right) = x - 3\)
Multiply out the bracket:
\(xy - 5y = x - 3\)
Collect the terms containing \(x\) on the left and everything else on the right:
\(xy - x = -3 + 5y\)
Take \(x\) out as a common factor:
\(x\left(y - 1\right) = -3 + 5y\)
Divide both sides by \(\left(y - 1\right)\):
\(x = \frac{ -3 + 5y }{y - 1 }\)
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Either paper 6 marks(a)
Rearrange \(y = \sqrt{ 5x - 14 }\) to make \(x\) the subject.
(b)Rearrange \(y = 2\sqrt{x} + 3\) to make \(x\) the subject.
Show answer
(a) \(x = \frac{y^2 + 14 }{ 5 }\)
(b) \(x = \frac{\left(y - 3\right)^2}{ 4 }\)Square both sides. Squaring undoes the square root, and everything under it comes out as it stands:
\(y^2 = 5x - 14\)
Add 14 to both sides:
\(5x = y^2 + 14\)
Divide both sides by 5:
\(x = \frac{y^2 + 14 }{ 5 }\)
Squaring now would square the whole right-hand side, brackets and all, so get the square root on its own first.
Subtract 3 from both sides:
\(2\sqrt{x} = y - 3\)
Divide both sides by 2:
\(\sqrt{x} = \frac{y - 3 }{ 2 }\)
Now square both sides. The top and the bottom of the fraction are both squared:
\(x = \left(\frac{y - 3 }{ 2 }\right)^2 = \frac{\left(y - 3\right)^2}{ 4 }\)
- (a) M1squares both sides: \(y^2 = 5x - 14\)
- (a) M1isolates the term in \(x\): \(5x = y^2 + 14\)
- (a) A1\(x = \frac{y^2 + 14 }{ 5 }\)
- (b) M1isolates the root term: \(2\sqrt{x} = y - 3\)
- (b) M1divides by \(2\): \(\sqrt{x} = \frac{y - 3 }{ 2 }\)
- (b) A1squares both sides: \(x = \frac{\left(y - 3\right)^2}{ 4 }\)
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Either paper 6 marks(a)
Rearrange \(y = \frac{ 2 }{ x } + 1\) to make \(x\) the subject.
(b)Rearrange \(y = \frac{ 17 }{x - 2 }\) to make \(x\) the subject.
Show answer
(a) \(x = \frac{ 2 }{y - 1 }\)
(b) \(x = \frac{ 17 }{ y } + 2\)Get the fraction on its own. Subtract 1 from both sides:
\(y - 1 = \frac{ 2 }{ x }\)
\(x\) is on the bottom, so multiply both sides by \(x\) to bring it up:
\(x\left(y - 1\right) = 2\)
Divide both sides by \(\left(y - 1\right)\):
\(x = \frac{ 2 }{y - 1 }\)
Multiply both sides by \(\left(x - 2\right)\):
\(y\left(x - 2\right) = 17\)
Divide both sides by \(y\):
\(x - 2 = \frac{ 17 }{ y }\)
Add 2 to both sides:
\(x = \frac{ 17 }{ y } + 2\)
- (a) M1fraction on its own: \(y - 1 = \frac{ 2 }{ x }\)
- (a) M1\(x\left(y - 1\right) = 2\)
- (a) A1\(x = \frac{ 2 }{y - 1 }\)
- (b) M1\(y\left(x - 2\right) = 17\)
- (b) M1\(x - 2 = \frac{ 17 }{ y }\)
- (b) A1\(x = \frac{ 17 }{ y } + 2\)
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Either paper 5 marks
A car moving in a straight line starts at a speed of \(u\) metres per second and finishes at a speed of \(v\) metres per second. Its acceleration is \(a\) metres per second squared and it travels a distance of \(s\) metres.
Those four are connected by the formula
\(v^2 = u^2 + 2as\)
(a)Make \(u\) the subject of the formula.
(b)Work out \(u\) when \(v = 11\), \(a = 8\) and \(s = 7\).
Show answer
(a) \(u = \sqrt{v^2 - 2as}\)
(b) \(3\ \text{m/s}\)Subtract \(2as\) from both sides so that \(u^2\) is on its own:
\(v^2 - 2as = u^2\)
Take the square root of both sides. A speed here is positive, so it is the positive root that is wanted:
\(u = \sqrt{v^2 - 2as}\)
The whole of \(v^2 - 2as\) goes under the root — squaring and rooting undo each other only when they act on the whole side.
Use the rearranged formula \(u = \sqrt{v^2 - 2as}\).
\(v^2 = 121\)
\(2as = 2 \times 8 \times 7 = 112\)
\(u = \sqrt{ 121 - 112 } = \sqrt{ 9 } = 3\)
The car started at 3 metres per second.
- (a) M1\(u^2 = v^2 - 2as\)
- (a) A1\(u = \sqrt{v^2 - 2as}\)
- (b) M1substituting: \(v^2 = 121\) and \(2as = 112\)
- (b) M1\(u = \sqrt{ 121 - 112 } = \sqrt{ 9 }\)
- (b) A1\(u = 3\)
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Either paper 4 marks
The time period \(T\) of a pendulum of length \(l\) is given by
\(T = 11\pi\sqrt{ \frac{ l }{ g } }\)
Make \(g\) the subject of the formula.
Show answer
\(g = \frac{ 121\pi^2 l }{ T^2 }\)Divide both sides by \(11\pi\) to get the root on its own:
\(\frac{ T }{ 11\pi } = \sqrt{ \frac{ l }{ g } }\)
Square both sides. The top and the bottom of the left-hand side are both squared:
\(\frac{ T^2 }{ 121\pi^2 } = \frac{ l }{ g }\)
\(g\) is on the bottom, so multiply both sides by \(g\):
\(g \times \frac{ T^2 }{ 121\pi^2 } = l\)
Multiply both sides by \(121\pi^2\) and divide both sides by \(T^2\):
\(g = \frac{ 121\pi^2 l }{ T^2 }\)
- M1divides both sides by \(11\pi\) so the root is on its own, or squares both sides first: \(T^2 = 121\pi^2 \times \frac{ l }{ g }\)
- M1squares the constant as well as the root: \(121\pi^2\), not \(11\pi\), whichever side it ends up on
- M1multiplies by \(g\) to bring it off the bottom of the fraction
- A1\(g = \frac{ 121\pi^2 l }{ T^2 }\)
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Either paper 3 marks
The volume, \(V\), of a solid hemisphere of radius \(r\) is given by
\(V = \frac{ 2 }{ 3 }\pi r^3\)
Find a formula for \(r\).
Show answer
\(r = \sqrt[3]{ \frac{ 3V }{ 2\pi } }\)Start with the formula:
\(V = \frac{ 2 }{ 3 }\pi r^3\)
Multiply both sides by 3 to clear the fraction:
\(3V = 2\pi r^3\)
Divide both sides by \(2\pi\), so that the power of \(r\) is on its own:
\(r^3 = \frac{ 3V }{ 2\pi }\)
Take the cube root of both sides. The root must cover the bottom of the fraction as well as the top, because it is the whole of the right-hand side that has been cubed (rooting only the top would be wrong). Writing the cube root separately over the top and over the bottom gives the same value:
\(r = \sqrt[3]{ \frac{ 3V }{ 2\pi } }\)
- M1both sides multiplied by 3 to clear the fraction: \(3V = 2\pi r^3\)
- M1both sides divided by \(2\pi\) to leave the power of \(r\) alone: \(r^3 = \frac{ 3V }{ 2\pi }\)
- A1the cube root taken of the whole right-hand side, so that it covers the bottom as well as the top: \(r = \sqrt[3]{ \frac{ 3V }{ 2\pi } }\). The cube root written separately over the top and over the bottom is the same answer and also earns the mark
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Either paper 6 marks(a)
Rearrange \(6x + y = 4x - 2\) to make \(x\) the subject.
(b)Rearrange \(y\left(x + 1\right) = 4x - 5\) to make \(x\) the subject.
Show answer
(a) \(x = \frac{ -2 - y}{ 2 }\)
(b) \(x = \frac{ -5 -y }{y - 4 }\)Get both \(x\) terms onto the same side. Subtract \(4x\) from both sides:
\(6x - 4x + y = -2\)
\(2x + y = -2\)
Subtract \(y\) from both sides:
\(2x = -2 - y\)
Divide both sides by 2:
\(x = \frac{ -2 - y}{ 2 }\)
Multiply out the bracket first:
\(xy +y = 4x - 5\)
Now get every term containing \(x\) onto one side and everything else onto the other:
\(xy - 4x = -5 -y\)
\(x\) is a factor of both terms on the left, so take it outside a bracket. This is the step that makes the rearrangement possible at all:
\(x\left(y - 4\right) = -5 -y\)
Divide both sides by \(\left(y - 4\right)\):
\(x = \frac{ -5 -y }{y - 4 }\)
- (a) M1\(x\) terms collected: \(2x + y = -2\)
- (a) M1\(2x = -2 - y\)
- (a) A1\(x = \frac{ -2 - y}{ 2 }\)
- (b) M1bracket expanded: \(xy +y = 4x - 5\)
- (b) M1factorised: \(x\left(y - 4\right) = -5 -y\)
- (b) A1\(x = \frac{ -5 -y }{y - 4 }\)
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Either paper 4 marks
Make \(x\) the subject of \(y = \frac{ 3x + 5 }{x - 8 }\)
Show answer
\(x = \frac{ 5 + 8y }{y - 3 }\)Multiply both sides by \(\left(x - 8\right)\) to get rid of the fraction:
\(y\left(x - 8\right) = 3x + 5\)
Multiply out the bracket:
\(xy - 8y = 3x + 5\)
Collect the terms containing \(x\) on the left and everything else on the right:
\(xy - 3x = 5 + 8y\)
Take \(x\) out as a common factor:
\(x\left(y - 3\right) = 5 + 8y\)
Divide both sides by \(\left(y - 3\right)\):
\(x = \frac{ 5 + 8y }{y - 3 }\)