June 2018 Paper 1 Q13

EdexcelCurrent spec8 marksLogs & Exponentials Modelling

13.

Figure 3: graph of log₁₀V against t: a straight line l with positive gradient and positive intercept on the vertical axis — Figure 3

The value of a rare painting, £\(V\), is modelled by the equation \(V = pq^t\), where \(p\) and \(q\) are constants and \(t\) is the number of years since the value of the painting was first recorded on 1st January 1980.

The line \(l\) shown in Figure 3 illustrates the linear relationship between \(t\) and \(\log_{10} V\) since 1st January 1980.

The equation of line \(l\) is \(\log_{10} V = 0.05t + 4.8\)

(a) Find, to 4 significant figures, the value of \(p\) and the value of \(q\). (4)

(b) With reference to the model interpret

(i) the value of the constant \(p\),

(ii) the value of the constant \(q\). (2)

(c) Find the value of the painting, as predicted by the model, on 1st January 2010, giving your answer to the nearest hundred thousand pounds. (2)

Mark scheme (a)

Scheme	Marks	AO
For a correct equation in \(p\) or \(q\) \(p = 10^{4.8}\) or \(q = 10^{0.05}\)	M1	1.1b
For \(p =\) awrt 63100 or \(q =\) awrt 1.122	A1	1.1b
For correct equations in \(p\) and \(q\) \(p = 10^{4.8}\) and \(q = 10^{0.05}\)	dM1	3.1a
For \(p =\) awrt 63100 and \(q =\) awrt 1.122	A1	1.1b
	(4)

Notes

(a) This is now being marked M1 A1 M1 A1 and in this order on e pen

M1: For a correct equation in \(p\) or \(q\) This is usually \(p = 10^{4.8}\) or \(q = 10^{0.05}\) but may be \(\log q = 0.05\) or \(\log p = 4.8\)

A1: For \(p =\) awrt 63100 or \(q =\) awrt 1.122

M1: For linking the two equations and forming correct equations in \(p\) and \(q\). This is usually \(p = 10^{4.8}\) and \(q = 10^{0.05}\) but may be \(\log q = 0.05\) and \(\log p = 4.8\)

A1: For \(p =\) awrt 63100 and \(q =\) awrt 1.122 Both these values implies M1 M1

ALT I(a)

M1: Substitutes \(t = 0\) and states that \(\log p = 4.8\)

A1: \(p =\) awrt 63100

M1: Uses their found value of \(p\) and another value of \(t\) to find form an equation in \(q\)

A1: \(p =\) awrt 63100 and \(q =\) awrt 1.122

Mark scheme (b)

Scheme	Marks	AO
(i) The value of the painting on 1st January 1980	B1	3.4
(ii) The proportional increase in value each year	B1	3.4
	(2)

Notes

(b)(i)

B1: The value of the painting on 1st January 1980 (is £63 100)
Accept the original value/cost of the painting or the initial value/cost of the painting

(b)(ii)

B1: The proportional increase in value each year. Eg Accept an explanation that explains that the value of the painting will rise 12.2% a year. (Follow through on their value of \(q\).)
Accept "the rate" by which the value is rising/price is changing. "1.122 is the decimal multiplier representing the year on year increase in value"
Do not accept "the amount" by which it is rising or "how much" it is rising by

If they are not labelled (b)(i) and (b)(ii) mark in the order given but accept any way around as long as clearly labelled " \(p\) is………… " and "\(q\) is ……………"

Mark scheme (c)

Scheme	Marks	AO
Uses \(V = 63100 \times 1.122^{30}\) or \(\log V = 0.05 \times 30 + 4.8\) leading to \(V =\)	M1	3.4
\(=\) awrt (£)2000000	A1	1.1b
	(2)
	(8 marks)

Notes

M1: For substituting \(t = 30\) into \(V = pq^t\) using their values for \(p\) and \(q\) or substituting \(t = 30\) into \(\log_{10} V = 0.05t + 4.8\) and proceeds to \(V\)

A1: For awrt either £1.99 million or £2.00 million. Condone the omission of the £ sign.
Remember to isw after a correct answer