A* Star Equivalency

# A Star GCSE Maths Equivalency Revision

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

## Question 1:

Find the HCF and LCM of 126 and 234.

[4 marks]

SOLUTION:

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. ## Question 2:

Factorise and solve the following quadratic equation: x^2+ 3x - 18=0

[3 marks]

SOLUTION:

x^2+ 3x - 18 = (x+6)(x-3)

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

x=-6

x=3

## Question 3:

Find the equation of the line AB where, A=(5,10) , B=(11,22).

[3 marks]

SOLUTION:

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

## Question 4:

Terry, Alisha and Ella run on a weekly basis.

In total, they average 176 km a week.

If Alisha runs twice as far as Terry and Ella runs one third of Alisha’s distance, how far do each of them run in km?

[3 marks]

SOLUTION:

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

## Question 5:

The diagram below shows an ice-cream cone.

This consists of a cone of radius 3 cm and height 10 cm with an attached hemisphere of ice cream of radius 3 cm, as shown below. Assuming the cone is completely filled, calculate the total volume of ice-cream.

[5 marks]

SOLUTION:

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

## Question 6:

XYZ is a right-angled triangle with

XY = 3.7 cm

YZ = 4.9 cm Calculate the area of the triangle.

[4 marks]

SOLUTION:

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)

## Question 7:

7a)

The probabilities of a spinner landing on each of its three colours are shown in the table below.

Find the missing value from the table below. [2 marks]

SOLUTION:

Probabilities add up to 1, so the missing value is

1 - \dfrac{1}{3} - \dfrac{1}{6} = \dfrac{1}{2}

7b)

If the spinner is spun 180 times, how many times would you expect the spinner to land on blue?

[1 mark]

SOLUTION:

180  \times \dfrac{1}{3} = 60

## Question 8:

Calculate (6 \times 10^3) \times (9 \times 10^4) and write the answer in standard form.

[2 marks]

SOLUTION:

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

## Question 9:

Anna records the speed of cars going past on a busy road.

Use this table to estimate the average speed of the cars. [4 marks]

SOLUTION:

(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

## Question 10:

The diagram below shows an irregular hexagon. Find the value of x to 1 dp

[3 marks]

SOLUTION:

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