The orders from three shops in July 2021 were as follows:
Tasty shop ordered 100 of type A, 50 of type B and 25 of type C.
Fragrant shop ordered 120 of type A, 60 of type B and 20 of type C.
Unusual shop ordered 80 of type A, 70 of type B and 30 of type C.

  1. (a)

    When Tap A and Tap B are turned on simultaneously, it takes (2.4) hours to fill tank completely. Given that Tap A alone takes 6 hours to fill the tank completely, calculate how long it takes in hours for Tap B to fill the same tank completely on its own.

  2. (b)

    A distributor supplied rice in three different packages, type A (2 kg), type B (5 kg), and type C (10 kg) at $6.20, $13.50 and $24.00 respectively. A 3×13 \times 1 matrix P is used to represent the above information about the weight of each type and a 1×31 \times 3 matrix Q is used to represent the respective prices.

    1. (i)

      State the matrix P.

    2. (ii)

      State the matrix Q.

  3. (c)

    Write down a 3×3 matrix R to represent this information.

  4. (d)

    Calculate the matrix product RP which represents the total weight of rice ordered from each shop respectively.

  5. (e)

    Given that SRP is a 1×1 matrix and its only element represents the total weight of rice ordered from the three shops, write down the matrix S.

  6. (f)

    Using matrix multiplication, find the total revenue that the distributor earned from the sales to the three shops.

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Worked solution

9(a)

1. Find the combined filling rate and Tap A's filling rate.

Combined rate of Tap A and Tap B:
12.4=512\frac{1}{2.4}=\frac{5}{12}
tank per hour.

Rate of Tap A:
16\frac{1}{6}
tank per hour.

2. Subtract rates to find Tap B's rate, then convert it to time taken.

Rate of Tap B:
512−16=512−212=312=14\frac{5}{12}-\frac{1}{6}=\frac{5}{12}-\frac{2}{12}=\frac{3}{12}=\frac{1}{4}
tank per hour.

So time taken by Tap B alone:
114=4\frac{1}{\frac{1}{4}}=4
hours.

Answer

Tap B takes 44 hours.

9(b)(i)

1. Write the weights of type A, type B and type C as a $3\times1$ column matrix.

Type A, B and C have weights 22 kg, 55 kg and 1010 kg respectively.

Since PP is a 3×13\times1 matrix,
P=(2510)P=\begin{pmatrix}2\\5\\10\end{pmatrix}

Answer

P=(2510)P=\begin{pmatrix}2\\5\\10\end{pmatrix}

9(b)(ii)

1. Write the prices of type A, type B and type C as a $1\times3$ row matrix.

The respective prices are type A: $6.20, type B: $13.50, type C: $24.00.

Since QQ is a 1×31\times3 matrix,
Q=(6.2013.5024.00)Q=\begin{pmatrix}6.20&13.50&24.00\end{pmatrix}

Answer

Q=(6.2013.5024.00)Q=\begin{pmatrix}6.20&13.50&24.00\end{pmatrix}

9(c)

1. Write the quantities ordered as a $3\times3$ matrix with rows representing shops and columns representing types A, B and C.

Using the order Tasty, Fragrant, Unusual for the rows, and type A, type B, type C for the columns,
R=(10050251206020807030)R=\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix}

Answer

R=(10050251206020807030)R=\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix}

9(d)

1. Set up the matrix product $RP$ using the order matrix and the weight matrix.

From earlier parts,
R=(10050251206020807030),P=(2510)R=\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix},\qquad P=\begin{pmatrix}2\\5\\10\end{pmatrix}

So
RP=(10050251206020807030)(2510)RP=\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix}\begin{pmatrix}2\\5\\10\end{pmatrix}

2. Multiply each row of $R$ by the column matrix $P$ to find each shop's total weight.

RP=(100(2)+50(5)+25(10)120(2)+60(5)+20(10)80(2)+70(5)+30(10))RP=\begin{pmatrix}100(2)+50(5)+25(10)\\120(2)+60(5)+20(10)\\80(2)+70(5)+30(10)\end{pmatrix}

=(200+250+250240+300+200160+350+300)=\begin{pmatrix}200+250+250\\240+300+200\\160+350+300\end{pmatrix}

=(700740810)=\begin{pmatrix}700\\740\\810\end{pmatrix}

Answer

RP=(700740810)RP=\begin{pmatrix}700\\740\\810\end{pmatrix}

The total weights ordered by Tasty, Fragrant and Unusual shops are 700700 kg, 740740 kg and 810810 kg respectively.

9(e)

1. Choose a row matrix that adds the three shop totals in $RP$.

Since RPRP is a 3×13\times1 matrix, SS must be a 1×31\times3 matrix so that SRPSRP is 1×11\times1.

To add the three entries of RPRP,
S=(111).S=\begin{pmatrix}1&1&1\end{pmatrix}.

Answer

S=(111)S=\begin{pmatrix}1&1&1\end{pmatrix}

9(f)

1. Set up the matrix product for total revenue.

Since
S=(111),R=(10050251206020807030),Q=(6.2013.5024.00),S=\begin{pmatrix}1&1&1\end{pmatrix},\quad R=\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix},\quad Q=\begin{pmatrix}6.20&13.50&24.00\end{pmatrix},
the total revenue is given by
SRQT=(111)(10050251206020807030)(6.2013.5024.00).SRQ^T=\begin{pmatrix}1&1&1\end{pmatrix}\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix}\begin{pmatrix}6.20\\13.50\\24.00\end{pmatrix}.

2. Carry out the multiplication to find the total revenue.

First find the total number of packages of each type:
(111)(10050251206020807030)=(30018075).\begin{pmatrix}1&1&1\end{pmatrix}\begin{pmatrix}100&50&25\\120&60&20\\80&70&30\end{pmatrix}=\begin{pmatrix}300&180&75\end{pmatrix}.
Then multiply by the price column matrix:
(30018075)(6.2013.5024.00)\begin{pmatrix}300&180&75\end{pmatrix}\begin{pmatrix}6.20\\13.50\\24.00\end{pmatrix}
=300(6.20)+180(13.50)+75(24.00)=300(6.20)+180(13.50)+75(24.00)
=1860+2430+1800=6090.=1860+2430+1800=6090.

Answer

The total revenue earned was $6090\char36{}6090.

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