Percentages: Eduqas GCSE Foundation June 2025 Paper 2 Q14

Foundation June 2025 Paper 2 Q14

EduqasFoundationCurrent spec5 marksPercentages

14. Igor invests £2500 in a savings account.
The account pays 5·4% simple interest per year.
The interest is added to the account at the end of each year.

After how many years will the amount of money in the savings account be greater than £3500?

You must show all your working. [5]

Mark scheme

Scheme	Marks	Comments
0.054 × 2500 oe	M1	No marks are awarded for an unsupported answer of 8. Allow M1 for \(2500 \times 1.054^n\) oe where \(n \geqslant 1\)
= (£) 135	A1	CAO. May be implied by e.g. 1 year = (£)2635
(3500 – 2500) ÷ 135 oe OR (using multiples of 135/repeated addition) 2500 + 135 × n (working towards 3500) OR 135 × n (working towards 1000)	M1	FT ‘their 135’ provided M1 previously awarded Note: If using trials of 2500 + 135 × n or 135 × n then there must be at least two improving trials seen where n > 1. Note: If using repeated additions of 135, there must be enough additions that would take them to the value that is just below or just above 3500 or 1000
= 8 (years)	A2	Note: Award final M1 A2 for an answer of 8, provided 135 is seen and it is not from incorrect work. Award A1 for 7(.4…..) If using multiples of 135/repeated addition: Award final M1A2 for 8 years = (£) 3580 or 8 years = (£) 1080 with no further trials (unless an answer other than 8 (years) is indicated) Award final M1A1 for 7 years = (£) 3445 or 7 years = (£) 945 FT ‘their 135’ for A2 or A1, provided M1 M1 previously awarded. Note: When repeated addition is used and there are errors in accuracy: If there is 1 error which leads to a final answer of 8 award A1 If there is 1 error which leads to a different answer e.g. between 8 and 9, 9 must be chosen in order to award A1. If there are 2 errors, award A0 If no marks, or if only M1 has been awarded, award M0 SC2 for an answer of 7 years (from use of compound interest) AND either of the following: \(2500 \times 1.054^6\) = 3427.54(9..) or 3427.55 \(2500 \times 1.054^7\) = 3612.63(6..) or 3612.64
	(5)

Alternative

Alternative method 1

Scheme	Marks	Comments
(3500 – 2500) ÷ 2500 × 100	M1	No marks are awarded for an unsupported answer of 8.
= 40(%)	A1	CAO
40 ÷ 5.4 oe OR (using multiples of 5.4/repeated addition) 5.4 × n (working towards 40)	M1	FT ‘their 40’ provided M1 previously awarded Note: If using trials of 5.4 × n then there must be at least two improving trials seen where n > 1. Note: If using repeated additions of 5.4, there must be enough additions that would take them to the value that is just below or just above 40
= 8 (years)	A2	Note: Award final M1 A2 for an answer of 8, provided 40(%) is seen and it is not from incorrect work. Award A1 for 7(.4…..) If using multiples of 5.4/repeated addition: Award final M1A2 for 8 years = 43.2(%) with no further trials (unless an answer other than 8 (years) is indicated) Award final M1A1 for 7 years = 37.8(%) FT ‘their 40’ for A2 or A1, provided M1 M1 previously awarded. Note: When repeated addition is used and there are errors in accuracy: If there is 1 error which leads to a final answer of 8 award A1 If there is 1 error which leads to a different answer e.g. between 8 and 9, 9 must be chosen in order to award A1. If there are 2 errors, award A0

Alternative method 2

Scheme	Marks	Comments
3500 ÷ 2500 × 100	M1	No marks are awarded for an unsupported answer of 8.
= 140(%)	A1	CAO
(140 – 100) ÷ 5.4 oe OR (using multiples of 5.4/repeated addition) 100 + 5.4 × n (working towards 140)	M1	FT ‘their 140’ provided M1 previously awarded Note: If using trials of 100 + 5.4 × n then there must be at least two improving trials seen where n > 1. Note: If using repeated additions of 5.4, there must be enough additions that would take them to the value that is just below or just above 40
= 8 (years)	A2	Note: Award final M1 A2 for an answer of 8, provided 140(%) is seen and it is not from incorrect work. Award A1 for 7(.4…..) If using multiples of 5.4/repeated addition: Award final M1A2 for 8 years = 143.2(%) with no further trials (unless an answer other than 8 (years) is indicated) Award final M1A1 for 7 years = 137.8(%) FT ‘their 140’ for A2 or A1, provided M1 M1 previously awarded. Note: When repeated addition is used and there are errors in accuracy: If there is 1 error which leads to a final answer of 8 award A1 If there is 1 error which leads to a different answer e.g. between 8 and 9, 9 must be chosen in order to award A1. If there are 2 errors, award A0

Alternative method 3 (use of SI = PTR/100)

Scheme	Marks	Comments
(\(T\) = ) \(\dfrac{(3500-2500) \times 100}{2500 \times 5.4}\)	M3	M2 for use of (3500 – 2500) = \(\dfrac{2500 \times T \times 5.4}{100}\) with at least two improving trials where \(T \gt 1\) M1 for (3500 – 2500) = \(\dfrac{2500 \times T \times 5.4}{100}\)
8 (years)	A2	A1 for 7(.4…..)