A supermarket sold all of its toilet rolls at a price of $ $x$ per pack in January. The revenue made from selling the packs of toilet rolls in January was $5940.

(a)
Write down an expression in x, for the number of packets of toilet rolls sold in January.

Answer ...
(b)
In February, the supermarket ordered an additional 600 packs to the number sold in January and sold them at 50 cents more per pack.

Write down an expression in x, for the total amount of money received in dollars, if all the packs of toilet rolls were sold in February.

Answer $ ...
(c)
The supermarket received $3870 more from the sales of toilet rolls in February as compared to January.
Write down an equation in x to represent this information, and show that it reduces to

$20x^2 - 119x + 99 = 0.$

Answer
(d)
Solve the equation $20x^2 - 119x + 99 = 0.$

Answer x = ... or x = ...
(e)
If each pack of toilet rolls was sold for more than $1, find the number of packs of toilet rolls sold by the supermarket in February.

Answer ... packets

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Worked solution

3(a)

1. Form an expression for the number of packs sold in January from revenue and price per pack.

Revenue in January = $5940, price per pack = $ $x$ .
$\text{Number of packs sold}=\frac{5940}{x}$

Answer

$\dfrac{5940}{x}$

3(b)

1. Form the February revenue expression from number of packs and price per pack.

Number of packs in February $=\dfrac{5940}{x}+600$ .

Price per pack in February $=x+0.50$ dollars.

Total amount received in February
$=\left(\dfrac{5940}{x}+600\right)(x+0.50).$

Answer

$\left(\dfrac{5940}{x}+600\right)(x+0.50)$

3(c)

1. Write an equation showing that February revenue is \$3870 more than January revenue.

From part (b), February revenue is
$\left(\dfrac{5940}{x}+600\right)(x+0.50).$
January revenue is $5940.

Since February revenue is $3870 more than January revenue,
$\left(\dfrac{5940}{x}+600\right)(x+0.50)-5940=3870.$

2. Expand and simplify the equation.

$\left(\dfrac{5940}{x}+600\right)(x+0.50)-5940=3870$
$5940+\dfrac{2970}{x}+600x+300-5940=3870$
$600x+\dfrac{2970}{x}+300=3870$

3. Clear the denominator and reduce to the required quadratic equation.

$600x+\dfrac{2970}{x}+300=3870$
$600x+\dfrac{2970}{x}=3570$
Multiplying by $x$ ,
$600x^2+2970=3570x$
$600x^2-3570x+2970=0$
Divide throughout by $30$ :
$20x^2-119x+99=0.$

Answer

$\left(\dfrac{5940}{x}+600\right)(x+0.50)-5940=3870$

Hence, $20x^2-119x+99=0$ .

3(d)

1. Factorise the quadratic expression.

$20x^2-119x+99=0$

Find two numbers with product $20\times 99=1980$ and sum $-119$ : $-44$ and $-75$ .

$20x^2-44x-75x+99=0$

$4x(5x-11)-9(5x-11)=0$

$(5x-11)(4x-9)=0$

2. Use the zero product property to find the two values of $x$.

$(5x-11)(4x-9)=0$

So,

$5x-11=0 \quad \text{or} \quad 4x-9=0$

$x=\frac{11}{5} \quad \text{or} \quad x=\frac{9}{4}$

Answer

$x=\dfrac{9}{10}$ or $x=\dfrac{11}{2}$

3(e)

1. Select the valid value of $x$ using the condition that each pack costs more than \$1.

From part (d), $x=\dfrac{9}{10}$ or $x=\dfrac{11}{2}$ .

Since each pack was sold for more than $1, $x>1$ .

Hence, $x=\dfrac{11}{2}=5.50$ .

2. Find the number of packs sold in February.

Number of packs sold in February
$=\frac{5940}{x}+600$
$=\frac{5940}{5.5}+600$
$=1080+600$
$=1680$

Answer

$1680$ packets

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A supermarket sold all of its toilet rolls at a price of $xxx per pack in January. The revenue made from selling the packs of toilet rolls in January was $5940.

Worked solution

3(a)

3(b)

3(c)

3(d)

3(e)

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A supermarket sold all of its toilet rolls at a price of $ $x$ per pack in January. The revenue made from selling the packs of toilet rolls in January was $5940.