  A* Star Equivalency

# A Star GCSE Maths Equivalency Revision

Here you will find the answers for the GCSE Equivalency Maths revision emails

## GCSE Equivalency Maths Revision Example Questions

$126 = 2 \times 3 \times 3 \times 7 = 2 \times 3^2 \times 7$

$234 = 2 \times 3 \times 3 \times 13 = 2 \times 3^2 \times 13$

$HCF = 2 \times 3 \times 3 = 18$

Found by multiplying the factors from the centre of the Venn diagram.

$LCM = 2\times 3 \times 3 \times 7 \times 13 = 1638$

Found by multiplying all the factors from the Venn diagram. $x^2+ 3x – 18 = (x+6)(x-3)$

$(x+6)(x-3)=0$

$x=-6$

$x=3$

$m =\dfrac{\text{change in} \, y}{\text{change in} \, x} = \dfrac{22 – 10}{11 – 5} = \dfrac{12}{6} = 2$

$y = 2x + c$

Substituting in values of $x$ and $y$ gives

$10 = 2(5) + c$

$10 = 10 + c$

$c = 0$

So, the equation of the line is:

$y = 2x$

Ratio $=$ Terry : Alisha : Ella $= 3:6:2$

$3 + 6 + 2 = 11$

$\dfrac{176}{11} = 16$

Terry: $16 \times 3 = 48$ km

Alisha: $16 \times 6 = 96$ km

Ella: $16 \times 2 = 32$ km

Volume of a cone is given by $\dfrac{1}{3}\pi r^2h$

Substituting in the given values gives the volume of the cone as $\dfrac{1}{3}\times \pi \times 3^2 \times 10 = 30\pi$ cm$^3$

Volume of a hemisphere is given by $\dfrac{1}{2}\times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3$

Substituting in the given values gives the volume of the hemisphere as $\dfrac{2}{3}\times \pi \times 3^3 = 18\pi$

Adding the two volumes together gives $30\pi + 18\pi = 48\pi$ cm$^3$

$4.9^2 – 3.7^2 = 10.32$

$\sqrt{10.32} = 3.212…$ cm

$\text{Area} = \dfrac{1}{2} \times 3.7 \times 3.212… = 5.9$ cm$^2$ ($1$ dp)

a)

Probabilities add up to $1$, so the missing value is

$1 – \dfrac{1}{3} – \dfrac{1}{6} = \dfrac{1}{2}$

b)

$180 \times \dfrac{1}{3} = 60$

$6 \times 9 \times 10^3 \times 10^4 = 54 \times 10^7 = 5.4 \times 10^8$

$(15 \times 4) + (35\times 28) + (45\times 37) + (55\times 10) + (65\times 6) = 60 + 980 + 1665 + 550 + 390 = 3645$

$n = 4 + 28 + 37 + 10 + 6 = 85$

$\dfrac{3645}{85} = 42.88… = 43$ mph to the nearest whole number

Using the equation for sum of interior angles

$(6-2)\times 180\degree = 720\degree$

This is equal to the sum of the given angles, so

$x+100+x+5+135+x+x-2=4x+238=720 \degree$

Rearranging gives $482=4x$

$x=120.5\degree$