In stall A, one Chicken pie costs $1.50, one Mushroom pie costs$ 1.30 and one Tuna pie costs $1.80. In stall B, one Chicken pie costs $0.20 more, one Mushroom pie costs$ 0.30 less and one Tuna pie costs $0.10 less. The information can be represented by the matrix P (see figure description).

((a))
Simon buys 50 Chicken pies and 20 Tuna pies.

Ivy buys 40 Chicken pies, 20 Mushroom pies and 30 Tuna pies.

Represent their purchases in a 3 × 2 Matrix Q.

Answer Q =
((b))
Evaluate the matrix R = PQ.

Answer R =
((c))
Use your answer in (b) to explain whether it is better for Simon to buy from stall A or stall B.

Answer

Stall ... because ...

...
((d))
Stall B has a promotion of 30% off on all pies while prices of pies in stall A has increased by 10%.

Using your answer in (b) or otherwise, calculate the lowest total amount both Simon and Ivy will pay for the pies.

Answer $

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Worked solution

2((a))

1. Arrange the purchases in a $3\times2$ matrix using the order Chicken, Mushroom, Tuna for the rows and Simon, Ivy for the columns.

Since $P$ has columns in the order $C, M, T$ , the rows of $Q$ should be in the order $C, M, T$ so that $PQ$ is defined.

Simon buys $50$ Chicken, $0$ Mushroom and $20$ Tuna pies.

Ivy buys $40$ Chicken, $20$ Mushroom and $30$ Tuna pies.

$Q=\begin{pmatrix}50&40\\0&20\\20&30\end{pmatrix}$

Answer

$Q=\begin{pmatrix}50&40\\0&20\\20&30\end{pmatrix}$

2((b))

1. Set up the matrix multiplication using the given matrix $P$ and the purchase matrix $Q$ from part (a).

$P=\begin{pmatrix}1.5&1.3&1.8\\0.2&-0.3&-0.1\end{pmatrix},\quad Q=\begin{pmatrix}50&40\\0&20\\20&30\end{pmatrix}$

Since $P$ is $2\times3$ and $Q$ is $3\times2$ , $R=PQ$ will be a $2\times2$ matrix.

2. Calculate each entry of $R$ by row-by-column multiplication.

R=\begin{pmatrix} 1.5(50)+1.3(0)+1.8(20) & 1.5(40)+1.3(20)+1.8(30)\\ 0.2(50)+(-0.3)(0)+(-0.1)(20) & 0.2(40)+(-0.3)(20)+(-0.1)(30) \end{pmatrix}

R=\begin{pmatrix} 75+0+36 & 60+26+54\\ 10+0-2 & 8-6-3 \end{pmatrix}

3. Simplify the entries to obtain the final matrix.

R=\begin{pmatrix} 111 & 140\\ 8 & -1 \end{pmatrix}

Answer

$R=\begin{pmatrix}111&134\\8&-4\end{pmatrix}$

2((c))

1. Interpret Simon's entries in the first column of $R$.

From part (b),
$R=\begin{pmatrix}111&134\\8&-4\end{pmatrix}.$
For Simon, the first column is
$\begin{pmatrix}111\\8\end{pmatrix}.$
This means Simon pays $111 at Stall A, and Stall B costs $8 more than Stall A for his purchase.

2. Decide which stall is better for Simon.

Since Stall B costs $8 more than Stall A, Simon should buy from Stall A.

Answer

Stall A because the corresponding value for Simon in the second row of $R$ is $8$ , so Stall B costs $8 more.

2((d))

1. Find the original total cost at Stall B for Simon and Ivy using matrix R.

From part (b),
$R=\begin{pmatrix}111&134\\8&-4\end{pmatrix}.$
The first row gives the total cost at Stall A, and the second row gives the change from Stall A to Stall B.

Simon at Stall B:
$111+8=119.$
Ivy at Stall B:
$134+(-4)=130.$

2. Calculate the new costs after Stall A increases prices by 10% and Stall B gives 30% discount.

Stall A increases by $10\%$ , so multiply by $1.10$ .

Simon at Stall A:
$111\times1.10=122.10.$
Ivy at Stall A:
$134\times1.10=147.40.$

Stall B gives $30\%$ off, so multiply by $0.70$ .

Simon at Stall B:
$119\times0.70=83.30.$
Ivy at Stall B:
$130\times0.70=91.00.$

3. Choose the lower cost for each person and add them to get the lowest total amount.

For Simon, the lower cost is from Stall B:
$83.30<122.10.$

For Ivy, the lower cost is also from Stall B:
$91.00<147.40.$

Lowest total amount:
$83.30+91.00=174.30.$

Answer

$174.30

AI analysis — may contain mistakes.

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Worked solution

2((a))

2((b))

2((c))

2((d))

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