D1 June 2016 Q5

EdexcelOld spec7 marksAlgorithms

Figure 4: flow chart of the algorithm — Figure 4

An algorithm is described by the flow chart shown in Figure 4.

Given that \(x = 27\) and \(y = 5\),

(a) complete the table in the answer book to show the results obtained at each step when the algorithm is applied. Give the final output. (4)

The numbers 122 and \(\dfrac{1}{2}\) are to be used as inputs for the algorithm described by the flow chart.

(b)

(i) State, giving a reason, which number should be input as \(x\).

(ii) State the output. (3)

Mark scheme (a)

Scheme

Marks

\(x\)	\(y\)	\(t\)	Is \(x\) odd?	Is \(x = 0\)?
27	5	0	Yes
26	(5)	5		No	Row 1
13	10	(5)	Yes		Row 2
12	(10)	15		No	Row 3
6	20	(15)	No	No	Row 4
3	40	(15)	Yes		Row 5
2	(40)	55		No	Row 6
1	80	(55)	Yes
0	(80)	135		Yes
		(135)

M1 (3 rows + 1^st correct)
A1 (2^nd and 3^rd rows correct)
A1 (4^th, 5^th and 6^th rows correct)

Output = 135

A1 (CSO)

(4)

Notes

Candidates may write each changed value/statement in a new row which is fine. Assume that each row begins and ends when a value in \(x\) is changed. For example, the values in row 1 in the table above consists of the \(x\) values going from the 26 to the 13

a1M1: At least three rows of cells in columns \(x\), \(y\) and \(t\) completed with a correct first row (so 26 for \(x\) and 5 for \(t\))

a1A1: CAO – second and third rows correct (for just the columns in \(x\), \(y\) and \(t\))

a2A1: CAO – fourth, fifth and sixth rows correct (for just the columns in \(x\), \(y\) and \(t\))

a3A1: CSO – including the output of 135 either on the given line in the answer book or clearly stated in the table but it must be absolutely clear that the output is the final \(t\) value (no bod). Furthermore, all ‘yes’ and ‘no’ comments must be present in the 4^th and 5^th columns with no additional/incorrect ‘yes’ or ‘no’

Mark scheme (b)

Scheme	Marks
(i) \(x\) must be a (positive) integer and therefore \(x = 122\)	B1 DB1
(ii) 61	B1
	(3)
	(7 marks)

Notes

bi1B1: \(x\) must be 122 and any attempt at a reason

bi2DB1: Dependent on previous B mark (so B0B1 is not possible) – 122 and a correct valid reason – e.g. \(x\) must be an integer/whole number or ½ is not odd or even or if you input ½ then you can never get to \(x = 0\) when halving, etc. Just saying that the algorithm ‘won’t work’ or that the algorithm ‘will get stuck in a loop’ or ‘not terminate’ is not sufficient for this second mark neither is the argument of subtracting 1 from a ½. It must be clear why the algorithm won’t output a value for \(t\) with \(x = \frac{1}{2}\) - so essentially there needs to be some indication of why \(x\) will never become 0. Furthermore, just saying that \(x\) will never reach 0 is insufficient – we need an indication of why \(x = 0\) is not possible with a starting value of \(x = \frac{1}{2}\)

bii3B1: CAO