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Date May 2012 Marks available 3 Reference code 12M.2.sl.TZ2.2
Level SL only Paper 2 Time zone TZ2
Command term Calculate Question number 2 Adapted from N/A

Question

Cedric wants to buy an €8000 car. The car salesman offers him a loan repayment option of a 25 % deposit followed by 12 equal monthly payments of €600 .

Write down the amount of the deposit.

[1]
a.

Calculate the total cost of the loan under this repayment scheme.

[2]
b.

Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repay the loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.

The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The total amount that Cedric’s mother receives after 24 months is €7100.

Write down a second equation involving x and y.

[1]
c.

Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repay the loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.

The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The total amount that Cedric’s mother receives after 24 months is €7100.

Write down the value of x and the value of y.

[2]
d.

Cedric’s mother decides to help him by giving him an interest free loan of €8000 to buy the car. She arranges for him to repay the loan by paying her €x in the first month and €y in every following month until the €8000 is repaid.

The total amount that Cedric’s mother receives after 12 months is €3500. This can be written using the equation x +11y = 3500. The total amount that Cedric’s mother receives after 24 months is €7100.

Calculate the number of months it will take Cedric’s mother to receive the €8000.

[3]
e.

Cedric decides to buy a cheaper car for €6000 and invests the remaining €2000 at his bank. The bank offers two investment options over three years.

Option A: Compound interest at an annual rate of 8 %.

Option B: Compound interest at a nominal annual rate of 7.5 % , compounded monthly.

Express each answer in part (f) to the nearest euro.

Calculate the value of his investment at the end of three years if he chooses

(i) Option A;

(ii) Option B.

[5]
f.

Markscheme

2000 (euros)     (A1)

[1 mark]

a.

\(2000 + 12 \times 600\)     (M1)

Note: Award (M1) for addition of two correct terms.


9200 (euros)     (A1)(ft)(G2)

Note: Follow through from their part (a).

[2 marks]

b.

x + 23y = 7100     (A1)

[1 mark]

c.

x = 200, y = 300     (A1)(ft)(A1)(ft)(G2)

[2 marks]

d.

\(200 + n \times 300 = 8000\)     (M1)

Note: Award (M1) for setting up the equation. Follow through from their x and y found in part (d).


n = 26     (A1)(ft)

26 + 1 = 27 (months)     (A1)(ft)(G3)

Notes: Middle line n = 26 may be implied if correct answer given. The final (A1)(ft) is for adding 1 to their value of n (even if it is incorrect). Follow through from their part (d). If the final answer is not a positive integer award at most (M1)(A1)(ft)(A0). Award (G2) for final answer of 26.


OR

\(\frac{{8000 - 7100}}{{300}} + 24\)     (M1)(A1)

Note: Award (M1) for division of difference by their value of y, (A1) for 24 seen.


27 (months)     (A1)(ft)(G3)

Note: Follow through from their value of y.

[3 marks]

 

e.

(i) \(2000{\left( {1 + \frac{8}{{100}}} \right)^3}\)     (M1)

Note: Award (M1) for correct substitution in compound interest formula.


2519 (euros)     (A1)(G2)

Note: If the answer is not given to the nearest euro award at most (M1)(A0).


(ii)
\(2000{\left( {1 + \frac{7.5}{{100 \times 12}}} \right)^{3\times12}}\)     (M1)(A1)

Note: Award (M1) for substitution in compound interest formula, (A1) for correct substitutions.


2503 (euros)     (A1)(G2)

Note: If the answer is not given to the nearest euro, award at most (M1)(A1)(A0), provided this has not been penalized in part (f)(i).

 

[5 marks]

f.

Examiners report

(a) Most candidates managed to answer this correctly.

 

a.

(b) On the whole this was well answered but some candidates gave 7200 as their final answer.

 

b.

(c) Some candidates found this surprisingly difficult, others gave the answer as x + 24y = 7100.

 

c.

(d) Many managed to find the correct answers for x and y even though their answer to part (c) was not correct. Others received follow through marks.

 

d.

(e) The most common answer here was 26 months.

 

e.

(f) Part (i) was well done but there were fewer correct answers seen for part (ii). Some candidates used 6000 instead of 2000, others did not give their answer to the nearest euro and others kept the same interest rate for both parts of the question.

f.

Syllabus sections

Topic 1 - Number and algebra » 1.0 » Solving linear equations in one variable.

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