Calculate, in GBP, the price that Arthur paid for the car.

Write down, in \({\text{GBP}}\), the amount of money Arthur saved by buying the car in Germany.

Between \({1^{{\text{st}}}}\) August 2008 and \({1^{{\text{st}}}}\) August 2012 Arthur’s car depreciated at an annual rate of \(9\%\) of its current value.

Calculate the value, in \({\text{GBP}}\), of Arthur’s car on \({1^{{\text{st}}}}\) August 2009.

Between \({1^{{\text{st}}}}\) August 2008 and \({1^{{\text{st}}}}\) August 2012 Arthur’s car depreciated at an annual rate of \(9\%\) of its current value.

Show that the value of Arthur’s car on \({1^{{\text{st}}}}\) August 2012 was \({\text{18}}\,{\text{600}}\,{\text{GBP}}\), correct to the nearest \({\text{100}}\,{\text{GBP}}\).

The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter. 

\(37\,500 \times 0.7234\)     (M1)

\( = 27\,127.50\)     (A1)(G2)

[2 marks]

The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter. 

\(6947.50\)     (A1)(ft)(G1)

 

Note: Follow through from part (a) irrespective of whether working is seen.

 

[1 mark]

The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.

\(27\,127.50 \times 0.91\)     (A1)(M1)

 

Note: Award (A1) for \(0.91\) seen or equivalent, (M1) for their \({\text{27}}\,{\text{127.50}}\) multiplied by \(0.91\)

 

OR

\(27\,127.50 - 0.09 \times 27\,127.50\)     (A1)(M1)

 

Note: Award (A1) for \(0.09 \times 27\,127.50\) seen, and (M1) for \(27\,127.50 - 0.09 \times 27\,127.50\).

 

\( = 24\,686.03\)     (A1)(ft)(G2)

 

Note: Follow through from part (a).

 

[3 marks]

The first answer not given to two decimal places is not awarded the final (A1). Incorrect rounding is not penalized thereafter.

\(27\,127.50 \times {\left( {1 - \frac{9}{{100}}} \right)^4}\)     (M1)(A1)(ft)

 

Notes: Award (M1) for substituted compound interest formula, (A1)(ft) for correct substitution.

     Follow through from part (a).

 

OR

\(27\,127.50 \times {(0.91)^4}\)     (M1)(A1)(ft)

 

Notes: Award (M1) for substituted geometric sequence formula, (A1)(ft) for correct substitution.

     Follow through from part (a).

 

OR (lists (i))

\({\text{24}}\,{\text{686.03, 22}}\,{\text{464.28..., 20}}\,{\text{442.50..., 18}}\,{\text{602.67...}}\)     (M1)(A1)(ft)

 

Notes: Award (M1) for at least the \({{\text{2}}^{{\text{nd}}}}\) term correct (calculated from their \(({\text{a}}) \times 0.91\)). Award (A1)(ft) for four correct terms (rounded or unrounded).

     Follow through from part (a).

     Accept list containing the last three terms only (\({\text{24}}\,{\text{686.03}}\) may be implied).

 

OR (lists(ii))

\(27\,127.50 - (2441.47... + 2221.74... + 2021.79... + 1839.82…)\)     (M1)(A1)(ft)

 

Notes: Award (M1) for subtraction of four terms from \({\text{27}}\,{\text{127.50}}\).

     Award (A1) for four correct terms (rounded or unrounded).

     Follow through from part (a).

 

\( = 18\,602.67\)     (A1) 

\( = 18\,600\)     (AG)

 

Note: The final (A1) is not awarded unless both the unrounded and rounded answers are seen.

 

[3 marks]

Despite the fact that “Give all answers in this question correct to two decimal places” was written in bold at the top of the question, many candidates lost one (and only one) mark for giving at least one answer to only a single decimal place. There was a lot of reading in this question and some candidates seemed to lose their way as their solution developed and, as a consequence, lost marks in the latter part of the question. A significant number of candidates obtained nearly full marks for parts (a) through to (d). The marks which tended to not be awarded were not giving the required answer to two decimal places and not adding the amount invested onto the interest earned in part (c). Indeed, many candidates were able to correctly determine the depreciated value of the car on \({1^{{\text{st}}}}\) August 2009 by simply finding 91% of the original price. However, part (e) proved to be elusive for many candidates as some simply treated the problem as a ‘reverse simple interest problem’ and subtracted 9% for each of a further 3 years. As a consequence, erroneous answers of the form 17,361.60, from \(\left( {27127.50 \times (1 - 0.09 \times 4)} \right)\), were often conveniently ignored and rounded to the required answer of 18,600 GBP. Such a method earned no marks at all. There was a lot of information given in the stem to the last part of the question and, as a consequence, many candidates were unable to achieve full marks here. There was certainly a great deal of confusion as to what to divide by 0.8694 (seeing \(\frac{{18\,600 + 8198.05 - 30\,500}}{{0.86944}} =  - 4258.05\) was not uncommon) and even introducing the original exchange rate of 0.7234 caused confusion. As a further example, an incorrect value carried forward from part (c) (1,250.55) led to a negative result. Provided the method was correct (despite an incorrect value carried forward), the three method marks were awarded. However, the negative result of –7,667.53 should have flagged to the candidate that something was wrong somewhere and this could only be in the current part of the question or part (c).

Despite the fact that “Give all answers in this question correct to two decimal places” was written in bold at the top of the question, many candidates lost one (and only one) mark for giving at least one answer to only a single decimal place. There was a lot of reading in this question and some candidates seemed to lose their way as their solution developed and, as a consequence, lost marks in the latter part of the question. A significant number of candidates obtained nearly full marks for parts (a) through to (d). The marks which tended to not be awarded were not giving the required answer to two decimal places and not adding the amount invested onto the interest earned in part (c). Indeed, many candidates were able to correctly determine the depreciated value of the car on \({1^{{\text{st}}}}\) August 2009 by simply finding 91% of the original price. However, part (e) proved to be elusive for many candidates as some simply treated the problem as a ‘reverse simple interest problem’ and subtracted 9% for each of a further 3 years. As a consequence, erroneous answers of the form 17,361.60, from \(\left( {27127.50 \times (1 - 0.09 \times 4)} \right)\), were often conveniently ignored and rounded to the required answer of 18,600 GBP. Such a method earned no marks at all. There was a lot of information given in the stem to the last part of the question and, as a consequence, many candidates were unable to achieve full marks here. There was certainly a great deal of confusion as to what to divide by 0.8694 (seeing \(\frac{{18\,600 + 8198.05 - 30\,500}}{{0.86944}} =  - 4258.05\) was not uncommon) and even introducing the original exchange rate of 0.7234 caused confusion. As a further example, an incorrect value carried forward from part (c) (1,250.55) led to a negative result. Provided the method was correct (despite an incorrect value carried forward), the three method marks were awarded. However, the negative result of –7,667.53 should have flagged to the candidate that something was wrong somewhere and this could only be in the current part of the question or part (c).

Despite the fact that “Give all answers in this question correct to two decimal places” was written in bold at the top of the question, many candidates lost one (and only one) mark for giving at least one answer to only a single decimal place. There was a lot of reading in this question and some candidates seemed to lose their way as their solution developed and, as a consequence, lost marks in the latter part of the question. A significant number of candidates obtained nearly full marks for parts (a) through to (d). The marks which tended to not be awarded were not giving the required answer to two decimal places and not adding the amount invested onto the interest earned in part (c). Indeed, many candidates were able to correctly determine the depreciated value of the car on \({1^{{\text{st}}}}\) August 2009 by simply finding 91% of the original price. However, part (e) proved to be elusive for many candidates as some simply treated the problem as a ‘reverse simple interest problem’ and subtracted 9% for each of a further 3 years. As a consequence, erroneous answers of the form 17,361.60, from \(\left( {27127.50 \times (1 - 0.09 \times 4)} \right)\), were often conveniently ignored and rounded to the required answer of 18,600 GBP. Such a method earned no marks at all. There was a lot of information given in the stem to the last part of the question and, as a consequence, many candidates were unable to achieve full marks here. There was certainly a great deal of confusion as to what to divide by 0.8694 (seeing \(\frac{{18\,600 + 8198.05 - 30\,500}}{{0.86944}} =  - 4258.05\) was not uncommon) and even introducing the original exchange rate of 0.7234 caused confusion. As a further example, an incorrect value carried forward from part (c) (1,250.55) led to a negative result. Provided the method was correct (despite an incorrect value carried forward), the three method marks were awarded. However, the negative result of –7,667.53 should have flagged to the candidate that something was wrong somewhere and this could only be in the current part of the question or part (c).

Despite the fact that “Give all answers in this question correct to two decimal places” was written in bold at the top of the question, many candidates lost one (and only one) mark for giving at least one answer to only a single decimal place. There was a lot of reading in this question and some candidates seemed to lose their way as their solution developed and, as a consequence, lost marks in the latter part of the question. A significant number of candidates obtained nearly full marks for parts (a) through to (d). The marks which tended to not be awarded were not giving the required answer to two decimal places and not adding the amount invested onto the interest earned in part (c). Indeed, many candidates were able to correctly determine the depreciated value of the car on \({1^{{\text{st}}}}\) August 2009 by simply finding 91% of the original price. However, part (e) proved to be elusive for many candidates as some simply treated the problem as a ‘reverse simple interest problem’ and subtracted 9% for each of a further 3 years. As a consequence, erroneous answers of the form 17,361.60, from \(\left( {27127.50 \times (1 - 0.09 \times 4)} \right)\), were often conveniently ignored and rounded to the required answer of 18,600 GBP. Such a method earned no marks at all. There was a lot of information given in the stem to the last part of the question and, as a consequence, many candidates were unable to achieve full marks here. There was certainly a great deal of confusion as to what to divide by 0.8694 (seeing \(\frac{{18\,600 + 8198.05 - 30\,500}}{{0.86944}} =  - 4258.05\) was not uncommon) and even introducing the original exchange rate of 0.7234 caused confusion. As a further example, an incorrect value carried forward from part (c) (1,250.55) led to a negative result. Provided the method was correct (despite an incorrect value carried forward), the three method marks were awarded. However, the negative result of –7,667.53 should have flagged to the candidate that something was wrong somewhere and this could only be in the current part of the question or part (c).