June 2019 Paper 2 Q7

EdexcelCurrent spec7 marksCo-ordinate Geometry Modelling

7. A small factory makes bars of soap.

On any day, the total cost to the factory, £\(y\), of making \(x\) bars of soap is modelled to be the sum of two separate elements:

a fixed cost
a cost that is proportional to the number of bars of soap that are made that day

(a) Write down a general equation linking \(y\) with \(x\), for this model. (1)

The bars of soap are sold for £2 each.

On a day when 800 bars of soap are made and sold, the factory makes a profit of £500

On a day when 300 bars of soap are made and sold, the factory makes a loss of £80

Using the above information,

(b) show that \(y = 0.84x + 428\) (3)

Assuming that each bar of soap is sold on the day it is made,

(d) find the least number of bars of soap that must be made on any given day for the factory to make a profit that day. (2)

Mark scheme (a)

Scheme	Marks	AO
£\(y\) is the total cost of making \(x\) bars of soap Bars of soap are sold for £2 each
\(y = kx + c\) {where \(k\) and \(c\) are constants}	B1	3.3
Note: Work for (a) cannot be recovered in (b) or (c)
	(1)

Notes

B1: Obtains a correct form of the equation. E.g. \(y = kx + c;\ \ k \neq 0,\ c \neq 0\). Note: Must be seen in (a)

Note: Ignore how the constants are labelled – as long as they appear to be constants. e.g. \(k, c, m\) etc.

Mark scheme (b)

Scheme	Marks	AO
Way 1 Either \(x = 800 \Rightarrow y = 2(800) - 500\ \{= 1100 \Rightarrow (x, y) = (800, 1100)\}\) \(x = 300 \Rightarrow y = 2(300) + 80\ \{= 680 \Rightarrow (x, y) = (300, 680)\}\)	M1	3.1b
Applies \((800,\ \text{their } 1100)\) and \((300,\ \text{their } 680)\) to give two equations \(1100 = 800k + c\) and \(680 = 300k + c \Rightarrow k, c = \ldots\)	dM1	1.1b
Solves correctly to find \(k = 0.84,\ c = 428\) and states \(y = 0.84x + 428\) *	A1*	2.1
Note: the answer \(y = 0.84x + 428\) must be stated in (b)
	(3)

Notes

(b) Way 2

Scheme	Marks	AO
Either \(x = 800 \Rightarrow y = 2(800) - 500\ \{= 1100 \Rightarrow (x, y) = (800, 1100)\}\) \(x = 300 \Rightarrow y = 2(300) + 80\ \{= 680 \Rightarrow (x, y) = (300, 680)\}\)	M1	3.1b
Complete method for finding both \(k = \ldots\) and \(c = \ldots\) e.g. \(k = \dfrac{1100 - 680}{800 - 300}\ \{= 0.84\}\) \((800, 1100) \Rightarrow 1100 = 800(0.84) + c \Rightarrow c = \ldots\)	dM1	1.1b
Solves to find \(k = 0.84,\ c = 428\) and states \(y = 0.84x + 428\) *	A1*	2.1
Note: the answer \(y = 0.84x + 428\) must be stated in (b)
	(3)

(b) Way 3

Scheme	Marks	AO
Either \(x = 800 \Rightarrow y = 2(800) - 500\ \{= 1100 \Rightarrow (x, y) = (800, 1100)\}\) \(x = 300 \Rightarrow y = 2(300) + 80\ \{= 680 \Rightarrow (x, y) = (300, 680)\}\)	M1	3.1b
\(\{y = 0.84x + 428 \Rightarrow\}\ \ x = 800 \Rightarrow y = (0.84)(800) + 428 = 1100\) \(x = 300 \Rightarrow y = (0.84)(300) + 428 = 680\)	dM1	1.1b
Hence \(y = 0.84x + 428\) *	A1*	2.1
	(3)

Way 1

M1: Translates the problem into the model by finding either

\(y = 2(800) - 500\) for \(x = 800\)
\(y = 2(300) + 80\) for \(x = 300\)

dM1: dependent on the previous M mark
See scheme

A1: See scheme – no errors in their working

Note Allow 1^st M1 for any of

\(1600 - y = 500\)
\(600 - y = -80\)

Way 2

M1: Translates the problem into the model by finding either
\(y = 2(800) - 500\) for \(x = 800\)
\(y = 2(300) + 80\) for \(x = 300\)

dM1: dependent on the previous M mark
See scheme

A1: See scheme – no error in their working

Way 3

M1: Translates the problem into the model by finding either
\(y = 2(800) - 500\) for \(x = 800\)
\(y = 2(300) + 80\) for \(x = 300\)

dM1: dependent on the previous M mark
Uses the model to test both points \((800,\ \text{their } 1100)\) and \((300,\ \text{their } 680)\)

A1: Confirms \(y = 0.84x + 428\) is true for both \((800, 1100)\) and \((300, 680)\) and gives a conclusion

Note: Conclusion could be “\(y = 0.84x + 428\)” or “QED” or “proved”

All ways

Note: Give 1^st M0 for \(500 = 800k + c,\ 80 = 300k + c \Rightarrow k = \dfrac{500 - 80}{800 - 300} = 0.84\)

Mark scheme (c)

Scheme	Marks	AO
Allow any of {0.84, in £s} represents the *cost* of {making} each extra bar {of soap} the direct *cost* of {making} a bar {of soap} the marginal *cost* of {making} a bar {of soap} the *cost* of {making} a bar {of soap} (Condone this answer) Note: Do not allow {0.84, in £s} is the profit per bar {of soap} {0.84, in £s} is the (selling) price per bar {of soap}	B1	3.4
	(1)

Scheme

Marks

Allow any of {0.84, in £s} represents

the cost of {making} each extra bar {of soap}
the direct cost of {making} a bar {of soap}
the marginal cost of {making} a bar {of soap}
the cost of {making} a bar {of soap} (Condone this answer)

Note: Do not allow

{0.84, in £s} is the profit per bar {of soap}
{0.84, in £s} is the (selling) price per bar {of soap}

3.4

(1)

Notes

B1: see scheme

Note: Also condone B1 for “rate of change of cost”, “cost of {making} a bar”, “constant of proportionality for cost per bar of soap” or “rate of increase in cost”,

Note: Do not allow reasons such as “price increase or decrease”, “rate of change of the bar of soap” or “decrease in cost”

Note: Give B0 for incorrect use of units.
E.g. Give B0 for “the cost of making each extra bar of soap is £84”
Condone the use of £0.84p

Mark scheme (d)

Scheme	Marks	AO
Way 1 {Let \(n\) be the least number of bars required to make a profit}
\(2n = 0.84n + 428 \Rightarrow n = \ldots\) (Condone \(2x = 0.84x + 428 \Rightarrow x = \ldots\))	M1	3.4
Answer of 369 {bars}	A1	3.2a
	(2)
	(7 marks)

Notes

(d) Way 2

Scheme	Marks	AO
Trial 1: \(n = 368 \Rightarrow y = (0.84)(368) + 428 \Rightarrow y = 737.12\) {revenue \(= 2(368) = 736\) or loss \(= 1.12\)}	M1	3.4
Trial 2: \(n = 369 \Rightarrow y = (0.84)(369) + 428 \Rightarrow y = 737.96\) {revenue \(= 2(369) = 738\) or profit \(= 0.04\)} leading to an answer of 369 {bars}	A1	3.2a
	(2)

Way 1

M1: Using the model and constructing an argument leading to a critical value for the number of bars of soap sold. See scheme.

A1: 369 only. Do not accept decimal answers.

Way 2

M1: Uses either 368 or 369 to find the cost \(y = \ldots\)

A1: Attempts both trial 1 and trial 2 to find both the cost \(y = \ldots\) and arrives at an answer of 369 only. Do not accept decimal answers.

All ways

Note: You can ignore inequality symbols for the method mark in part (d)

Note: Give M1 A1 for no working leading to 369 {bars}

Note: Give final A0 for \(x \gt 369\) or \(x \gt 368\) or \(x \geqslant 369\) without \(x = 369\) or 369 stated as their final answer

Note: Condone final A1 for in words “at least 369 bars must be made/sold”

Note: Special Case:
Assuming a profit of £1 is required and achieving \(x = 370\) scores special case M1A0