Higher June 2019 Paper 2 Q11

EdexcelHigherCurrent spec6 marksPercentages Standard Form

11 The table gives the average crowd attendance per match for each of five football clubs for one season.

Football club	Average crowd attendance
Monaco	\(9.5 \times 10^3\)
Chelsea	\(4.2 \times 10^4\)
Juventus	\(3.9 \times 10^4\)
Oxford United	\(8.3 \times 10^3\)
Barcelona	\(7.7 \times 10^4\)

(a) Find the difference between the average crowd attendance for Barcelona and the average crowd attendance for Monaco.
Give your answer in standard form. (2)

Antonio says,

“The average crowd attendance for Chelsea is approximately 50 times that for Oxford United.”

(b) Is Antonio correct?
You must give a reason for your answer. (2)

During last season the cost of a ticket to watch Seapron United increased by 15% and then decreased by 8%

Mark scheme (a)

Scheme	Marks
\(\pm(7.7 \times 10^4 - 9.5 \times 10^3)\) or \(\pm(7.7 \times 10^4 - 0.95 \times 10^4)\) or ±(77 000 – 9 500) or ±67 500 oe	M1
\(6.75 \times 10^4\)	A1
	(2)

Notes

M1: for clearly subtracting the correct values

A1: allow \(-6.75 \times 10^4\) allow \(\pm 6.8 \times 10^4\)

Mark scheme (b)

Scheme	Marks
\((8.3 \times 10^3) \times 50\) (= 415 000 or \(4.15 \times 10^5\)) or \((4.2 \times 10^4) \div 50\) (= 840 or \(8.4 \times 10^2\)) or \((4.2 \times 10^4) \div (8.3 \times 10^3)\) (= 5(.060…))	M1
No supported by correct comparable figures in the same form	A1
	(2)

Notes

M1: for a relevant calculation

A1: for NO and 415 000 and 42 000 or
NO and \(4.15 \times 10^5\)
NO and 840 and 8 300 or
NO and \(8.4 \times 10^2\)
NO and 5(.060…)

Mark scheme (c)

Scheme	Marks
1.15 × 0.92 (= 1.058) oe or 105.8 \(\dfrac{n \times 1.15 \times 0.92}{n}\) where \(n\) is a number or variable e.g. \(\dfrac{200 \times 1.15 \times 0.92}{200}\)	M1
5.8	A1
	(2)
	(6 marks)

Notes

M1: condone \(x \times 1.15 \times 0.92\) oe

A1: NB. −5.8 (M1A0)
decrease of 5.8% (M1A0)