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

[latex]126 = 2 \times 3 \times 3 \times 7 = 2 \times 3^2 \times 7[/latex]

[latex]234 = 2 \times 3 \times 3 \times 13 = 2 \times 3^2 \times 13[/latex]

[latex]HCF = 2 \times 3 \times 3 = 18[/latex]

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

[latex]LCM = 2\times 3 \times 3 \times 7 \times 13 = 1638[/latex]

Found by multiplying all the factors from the Venn diagram.

[latex]x^2+ 3x – 18 = (x+6)(x-3)[/latex]

[latex](x+6)(x-3)=0[/latex]

[latex]x=-6[/latex]

[latex]x=3[/latex]

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

[latex]y = 2x + c[/latex]

Substituting in values of [latex]x[/latex] and [latex]y[/latex] gives

[latex]10 = 2(5) + c[/latex]

[latex]10 = 10 + c[/latex]

[latex]c = 0[/latex]

So, the equation of the line is:

[latex]y = 2x[/latex]

Ratio [latex]=[/latex] Terry : Alisha : Ella [latex]= 3:6:2[/latex]

[latex]3 + 6 + 2 = 11[/latex]

[latex]\dfrac{176}{11} = 16[/latex]

Terry: [latex]16 \times 3 = 48[/latex] km

Alisha: [latex]16 \times 6 = 96[/latex] km

Ella: [latex]16 \times 2 = 32[/latex] km

Volume of a cone is given by [latex]\dfrac{1}{3}\pi r^2h[/latex]

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

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

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

Adding the two volumes together gives [latex]30\pi + 18\pi = 48\pi[/latex] cm[latex]^3[/latex]

[latex]4.9^2 – 3.7^2 = 10.32[/latex]

[latex]\sqrt{10.32} = 3.212…[/latex] cm

[latex]\text{Area} = \dfrac{1}{2} \times 3.7 \times 3.212… = 5.9[/latex] cm[latex]^2[/latex] ([latex]1[/latex] dp)

a)

Probabilities add up to [latex]1[/latex], so the missing value is

[latex]1 – \dfrac{1}{3} – \dfrac{1}{6} = \dfrac{1}{2}[/latex]

 

b)

[latex]180  \times \dfrac{1}{3} = 60[/latex]

[latex]6 \times 9 \times 10^3 \times 10^4 = 54 \times 10^7 = 5.4 \times 10^8[/latex]

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

[latex]n = 4 + 28 + 37 + 10 + 6 = 85[/latex]

[latex]\dfrac{3645}{85} = 42.88… = 43[/latex] mph to the nearest whole number

Using the equation for sum of interior angles

[latex](6-2)\times 180\degree = 720\degree[/latex]

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

[latex]x+100+x+5+135+x+x-2=4x+238=720 \degree[/latex]

Rearranging gives [latex]482=4x[/latex]

[latex]x=120.5\degree[/latex]