A2 June 2022 Q1

EdexcelCurrent spec5 marksAlgorithms

1. A gardener needs the following lengths of string. All lengths are in metres.

4.3

6.1

5.1

4.7

2.5

5.9

3.4

1.7

2.1

0.4

1.3

She cuts the lengths from balls of string. Each ball contains 10 m of string.

(a) Calculate a lower bound for the number of balls of string the gardener needs.
You must make your method clear. (2)

(b) Use the first-fit bin packing algorithm to determine how the lengths could be cut from the balls of string. (3)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{4.3 + 6.1 + \ldots + 1.3}{10} = \dfrac{37.5}{10} = 3.75\) so lower bound is four (balls of string)	M1 A1	1.1b 2.2a
	(2)

Notes

(a) M1: Attempt to find the lower bound \((37.5 \pm 6.1)/10\) (a value of 3.75 seen with no working can imply this mark)

A1: cso – a lower bound of 4 with either a correct calculation seen or 3.75 or ‘total is 37.5 and if each ball contains 10 this gives a lower bound of 4’. An answer of 4 with no working (or from part (b)) scores M0A0. Any incorrect working loses this mark e.g. a correct calculation followed by an incorrect value followed by 4 is A0

Mark scheme (b)

Scheme

Marks

Bin 1:	\(\boxed{4.3}\) \(\boxed{5.1}\) 0.4
Bin 2:	\(\boxed{6.1}\) 2.5 1.3
Bin 3:	\(\boxed{4.7}\) 3.4 1.7
Bin 4:	5.9 2.1

M1
A1
A1

1.1b
1.1b
1.1b

(3)

(5 marks)

Notes

(b) M1: First four items placed correctly and at least eight items placed in bins – condone cumulative totals for M1 only (the boxed values)

A1: First eight items placed correctly (the boxed and bold values), and all eleven correct values only placed in bins (so no additional/repeated values)

A1: cso (no additional/repeated values)

Condone working in cm provided consistent

No MR in this question – mark according to the scheme