Category Archives: Year 8 Mathematics

Area Problem

Two rectangular garden beds have a combined area of 40m^2. The larger bed has twice the perimeter of the smaller and the larger side of the smaller bed is equal to the smaller side of the larger bed. If the two beds are not similar, and if all edges are a whole number of metres, what is the length, in metres, of the longer side of the larger bed?
AMC 2007 S.14

Let’s draw a diagram

From the information in the question, we know

(1)   \begin{equation*}xy+xz=40\end{equation*}

and

    \begin{equation*}2x+2y=4x+4z\end{equation}

    \begin{equation*}x+y=2x+2z\end{equation}

    \begin{equation*}x+y=2x+2z\end{equation}

(2)   \begin{equation*}y=x+2z\end{equation*}

Equation 1 becomes

    \begin{equation*}x(x+3z)=40\end{equation}

As the sides are whole numbers, consider the factors of 40.

1, 2, 4, 5, 8, 10, 20, 40

Remember z<x<y

xx+3zzyPerimeter LargePerimeter SmallComment
2206x must be greater than z
410282(4+8)=242(2+4)=12This one works
58172(5+7)=242(5+1)=12This one also works
810\frac{2}{3}z not a whole number
104z<0Not possible
202z<0
Not possible
401z<0Not possible

There are two possibilities

The large garden bed could be 4 by 8 and the smaller 4 by 2 (Area 40 Perimeters 24 and 12)

or

The large garden bed could be 5 by 7 and the smaller 5 by 1 (Area 40 Perimeters 24 and 12)

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Filed under Area, Interesting Mathematics, Measurement, Puzzles, Solving Equations, Year 8 Mathematics

Age Question (Year 8 equation solving)

Eight years ago my father was three times as old as I shall be in five years time. When I was born he was 41 years old. How old am I now?

I always find these age questions a bit weird – a bit of a riddle, and contrived (just so we can solve some equations)

Let x be my age now, and y be my fathers age now.

(1)   \begin{equation*}y=x+41\end{equation*}

Because my father was 41 when I was born.

(2)   \begin{equation*}y-8=3(x+5)\end{equation*}

y-8 for 8 years ago, and 3(x+5) for three times my age in 5 years.

Solve the equations simultaneously. Substitute y=x+41 into equation 2

    \begin{equation*}x+41-8=3(x+5)\end{equation}

    \begin{equation*}x-33=3x+15\end{equation}

    \begin{equation*}18=2x\end{equation}

    \begin{equation*}x=9\end{equation}

Hence my current age is 9.

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Filed under Algebra, Simultaneous Equations, Solving Equations, Year 8 Mathematics