IB Questionbank Test · Web viewIn part (d) the sum of the arithmetic progression was done better than the geometric series. Many candidates calculated the 15th term of the progression

1.8

1a. [1 mark]

Markscheme

wn (A1) (C1)

[1 mark]

Examiners report

[N/A]

1b. [1 mark]

Markscheme

un (A1) (C1)

[1 mark]

Examiners report

[N/A]

1c. [2 marks]

Markscheme

10 (2)11−1 (M1)

Note: Award (M1) for correct substitutions into geometric sequence formula.

= 10 240 (A1)(ft) (C2)

1

Note: Exact answer only. Accuracy rules do not apply in this question part; do not accept the 3 sf

answer of 10 200.

[2 marks]

Examiners report

[N/A]

1d. [2 marks]

Markscheme

OR (M1)

Note: Award (M1) for correct substitutions into arithmetic series formula.

= 2100 (A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

2a. [3 marks]

Markscheme

54 × (0.94)50 (M1)(A1)

Note: Award (M1) for substitution into geometric sequence formula, (A1) for correct substitution.

2.45 (cm) (2.44785… cm) (A1) (C3)

[3 marks]

2

Examiners report

[N/A]

2b. [3 marks]

Markscheme

(or equivalent) (M1)(A1)(ft)

Note: Award (M1) for substitution into geometric series formula, (A1)(ft) for correct substitution using

their common ratio from part (a).

= 862 (cm) (861.650…(cm)) (A1)(ft) (C3)

[3 marks]

Examiners report

[N/A]

3a. [3 marks]

Markscheme

60 + 10 × 10 (M1)(A1)

Note: Award (M1) for substitution into the arithmetic sequence formula, (A1) for correct substitution.

= ($) 160 (A1)(G3)

[3 marks]

Examiners report

[N/A]

3b. [3 marks]

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substituting the arithmetic series formula, (A1)(ft) for correct substitution.

Follow through from their first term and common difference in part (a).

3

= ($) 1380 (A1)(ft)(G2)

[3 marks]

Examiners report

[N/A]

3c. [3 marks]

Markscheme

60 × 1.110 (M1)(A1)

Note: Award (M1) for substituting the geometric progression nth term formula, (A1) for correct

substitution.

= ($) 156 (155.624…) (A1)(G3)

Note: Accept the answer if it rounds correctly to 3 sf, as per the accuracy instructions.

[3 marks]

Examiners report

[N/A]

3d. [3 marks]

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substituting the geometric series formula, (A1)(ft) for correct substitution.

Follow through from part (c) for their first term and common ratio.

= ($)1280 (1283.05…) (A1)(ft)(G2)

[3 marks]

Examiners report

4

[N/A]

3e. [4 marks]

Markscheme

(M1)(M1)

Note: Award (M1) for correctly substituted geometric and arithmetic series formula with n (accept

other variable for “n”), (M1) for comparing their expressions consistent with their part (b) and part (d).

OR

(M1)(M1)

Note: Award (M1) for two curves with approximately correct shape drawn in the first quadrant, (M1)

for one point of intersection with approximate correct position.

Accept alternative correct sketches, such as

Award (M1) for a curve with approximate correct shape drawn in the 1st (or 4th) quadrant and all above

(or below) the x-axis, (M1) for one point of intersection with the x-axis with approximate correct

position.

17 (A2)(ft)(G3)

5

Note: Follow through from parts (b) and (d).

An answer of 16 is incorrect. Award at most (M1)(M1)(A0)(A0) with working seen. Award (G0) if final

answer is 16 without working seen.

[4 marks]

Examiners report

[N/A]

4a. [1 mark]

Markscheme

3800 m (A1)

[1 mark]

Examiners report

[N/A]

4b. [2 marks]

Markscheme

OR (M1)(A1)

Note: Award (M1) for substitution into arithmetic sequence formula, (A1) for correct substitution.

[2 marks]

Examiners report

[N/A]

4c. [2 marks]

Markscheme

(M1)

6

Notes: Award (M1) for their correct inequality. Accept .

Accept OR . Award (M0) for .

(A1)(ft)(G2)

Note: Follow through from part (a)(ii), but only if is a positive integer.

[2 marks]

Examiners report

[N/A]

4d. [4 marks]

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct

substitution.

(A1)

Note: Award (A1) for their seen.

(A1)(ft)(G3)

7

Note: Award (A1)(ft) for correctly converting their answer in metres to km; this can be awarded

independently from previous marks.

OR

(M1)(A1)(ft)(A1)

Note: Award (M1) for substitution into sum of an arithmetic series formula, (A1)(ft) for correct

substitution, (A1) for correctly converting 3000 m and 400 m into km.

(A1)(G3)

[4 marks]

Examiners report

[N/A]

4e. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

(A1)(G3)

OR

(M1)(A1)

Note: Award (M1) for substitution into geometric series formula, (A1) for correct substitutions.

8

(A1)(G3)

[3 marks]

Examiners report

[N/A]

4f. [3 marks]

Markscheme

(M1)(A1)

Notes: Award (M1) for substitution into sum of a geometric series formula, (A1) for correct

substitutions. Follow through from their ratio ( ) in part (d). If (distance does not increase) or

the final answer is unrealistic (eg ), do not award the final (A1).

(A1)(G2)

[3 marks]

Examiners report

[N/A]

5a. [2 marks]

Markscheme

OR (M1)

Note: Award (M1) for dividing any by .

9

(A1) (C2)

[2 marks]

Examiners report

[N/A]

5b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for their correct substitution into geometric sequence formula.

(A1)(ft) (C2)

Note: Follow through from part (a).

Award (A1)(A0) for or with or without working.

[2 marks]

Examiners report

[N/A]

5c. [2 marks]

Markscheme

(M1)

10

Note: Award (M1) for correct substitution into geometric series formula.

(A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

6a. [1 mark]

Markscheme

(A1) (C1)

[1 mark]

Examiners report

[N/A]

6b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for their correct substitution into the geometric sequence formula. Accept a list of

their five correct terms.

(A1)(ft) (C2)

Note: Follow through from their common ratio from part (a).

11

[2 marks]

Examiners report

[N/A]

6c. [3 marks]

Markscheme

(M1)(M1)

Notes: Award (M1) for their correct substitution into the geometric sequence formula with a variable

in the exponent, (M1) for comparing their expression with .

Accept an equation.

(A1)(ft) (C3)

Note: Follow through from their common ratio from part (a). “ ” must be a positive integer for the

(A1) to be awarded.

[3 marks]

Examiners report

[N/A]

7a. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into geometric sequence formula.

12

(A1) (C2)

[2 marks]

Examiners report

[N/A]

7b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into geometric sequence formula or list of eight values

using their . Follow through from part (a), only if answer is positive.

(A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

7c. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into geometric series formula. Follow through from part

(a), only if answer is positive.

13

(A1)(ft) (C2)

[2 marks]

Examiners report

[N/A]

8a. [3 marks]

Markscheme

i) (A1)

ii) (M1)

Note: Award (M1) for OR

(A1)(G3)

Examiners report

Question 5: Arithmetic and Geometric progression

Most candidates calculated the salaries in the second year correctly. The most common error was to

calculate the salaries for the third instead of the second year. In part (b) the use of instead of

was very common. For the geometric sequence often a ratio of 0.05 instead of 1.05 was used. Also many

of the expressions given did not represent a geometric sequence. Candidates who used a list for part (c)

did usually better than the ones that tried to solve an equation. In part (d) the sum of the arithmetic

progression was done better than the geometric series. Many candidates calculated the 15th term of

the progression and not the series. In general this question part was not answered well.

8b. [4 marks]

Markscheme

i) (M1)(A1)

Note: Award (M1) for substitution in arithmetic sequence formula; (A1) for correct substitutions.

14

ii) (M1)(A1)

Note: Award (M1) for substitution in arithmetic sequence formula; (A1) for correct substitutions.

Examiners report









8c. [2 marks]

Markscheme

(M1)

Note: Award (M1) for setting a correct inequality using their expressions for (b)(i) and (b)(ii). Accept

an equation.

OR

list of at least 4 correct terms of each sequence (M1)

Note: Award (M1) for correct lists corresponding to their answers for parts (b)(i) and (b)(ii).

(A1)(ft)(G2)

Note: Value must be an integer for the final (A1) to be awarded. Follow through from parts (b)(i) and

(b)(ii). Award (G1) for a final answer of seen without working.

Examiners report


15








8d. [7 marks]

Markscheme

i) (M1)(A1)(ft)

Note: Award (M1) for substitution into geometric series formula and (A1) for correct substitution of

and their from part (b)(ii). Follow through from part (b)(ii).

OR

(M1)(A1)(ft)

Note: Follow through from part (b)(ii).

(A1)(ft)(G2)

ii) (M1)(A1)(ft)

Note: Award (M1) for substitution into arithmetic series formula and (A1) for correct substitution,

using their first term and their last term from part (b)(i), or their and . Follow through from part

(b)(i).

OR

(M1)(A1)(ft)

Note: Follow through from part (b)(i).

(A1)(ft)(G2)

Antonio does not earn more than Barbara

16

(his total salary will be less than Barbara’s) (A1)(ft)

Note: Award (A1)(ft) for a final answer that is consistent with their part (d)(i) and (d)(ii). Accept

“Barbara earns more”. The final (A1) can only be awarded if two total salaries are seen.

Examiners report









9a. [2 marks]

Markscheme

(i) (A1)

(ii) (A1)(ft)

Note: Follow through from part (a)(i).

Examiners report

Question 2: Arithmetic and geometric sequences and series

Parts (a), (b), (c) and (e) were well done. Quite a few forgot to convert their answer to km in part (c).

The main problem with part (d) was that candidates chose to equate the term formula to 1800

rather than the sum of the first n terms formula. Some of those who managed to write the correct

equation were not always successful at solving it. Some candidates made use of the trial and error

method to reach the correct answer. Part (e) was obvious to some, others put it into a formula with

little understanding and a surprising number of candidates had place value issues (stating 10% of

17000 was 170). Many candidates used the compound interest formula in both parts (e) and (f). In part

17

(f) many candidates did not realize that they needed to use the sum of a geometric series formula. They

either used the sum of an arithmetic series or as previously mentioned, the compound interest formula.

9b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into arithmetic sequence formula. A list of their correct

terms (excluding those given in question and the from part (a)(ii)) must be seen for the (M1) to be

awarded.

(A1)(ft)(G2)

Note: Follow through from their value for .

If a list is used, award (A1) for their term.

Examiners report











9c. [3 marks]

Markscheme

OR (M1)

Note: Award (M1) for correct substitution into arithmetic series formula. Follow through from their

part (a)(i). Accept a list added together until the term.

18

(A1)(ft)

Note: Follow through from parts (a) and (b).

(A1)(ft)(G2)

Note: Award (A1)(ft) for correctly converting their metres to kilometres, irrespective of method used.

To award the last (A1)(ft) in follow through, the candidate’s answer in metres must be seen.

Examiners report











9d. [3 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution into arithmetic series formula equated to . Follow

through from their part (a)(i). Accept a list of terms that shows clearly the second and

second distances.

Correct use of kinematics equations is a valid method.

(A1)(ft)

(seconds) (A1)(ft)(G2)

Note: Award (A1)(ft) for correct unrounded value for . The second (A1)(ft) is awarded for the correct

rounding off of their value for to the nearest second if their unrounded value is seen.

19

Award (M1)(A2)(ft) for their if method is shown. Unrounded value for may not be seen. Follow

through from their and only if workings are shown.

OR

(M1)

Note: Award (M1) for adding the terms until reaching .

(A2)(ft)

Note: In this method, follow through from their from part (a) and their from part (c).

Examiners report











9e. [2 marks]

Markscheme

(or equivalent) (M1)

Note: Award (M1) for multiplying by or equivalent.

(A1)(G2)

Examiners report



The main problem with part (d) was that candidates chose to equate the term formula to 1800 20








9f. [3 marks]

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substitution into the geometric series formula, (A1)(ft) for correct substitution.

Award (A1)(ft) for a list of their correct terms, (M1) for adding their terms.

(A1)(ft)(G2)

Note: Follow through from their in part (e).

Examiners report











10a. [3 marks]

Markscheme

(i) (or equivalent) (M1)

21

Note: Award (M1) for one correct equation. Accept a list of at least 5 correct terms.

(A1)

(ii) (A1)(ft) (C3)

Note: Follow through from (a)(i), irrespective of working shown if OR

Examiners report

Part (a) was answered correctly by many candidates, but working using equations was rarely seen. A

“trial and error” method, based upon a list of terms was the most seen method.

10b. [3 marks]

Markscheme

OR (M1)(A1)(ft)

Note: Award (M1) for substituted geometric sequence formula, (A1)(ft) for their correct substitutions.

OR

(M1)(A1)(ft)

Note: Award (M1) for a list of at least 5 consecutive terms of a geometric sequence, (A1)(ft) for terms

corresponding to their answers in part (a).

(A1)(ft) (C3)

Note: Follow through from part (a).

Examiners report

In part (b) many found the correct answer, but many others did not. Some gave the seventh term of the

arithmetic sequence, some gave a term of an incorrect order and some a completely incorrect answer. 22

Finding the correct ratio was the most common problem. Often repeated multiplication was used to

find the answer, but also the formula for the nth term of a geometric sequence was used. Several did not

use the correct three terms from the question.

11a. [2 marks]

Markscheme

(i) OR (A1) (C1)

(ii) OR (A1) (C1)

Examiners report

[N/A]

11b. [1 mark]

Markscheme

OR (A1) (C1)

Note: Accept ‘divide by 2’ for (A1).

Examiners report

[N/A]

11c. [3 marks]

Markscheme

(M1)(A1)(ft)

Notes: Award (M1) for substitution in the GP term formula, (A1)(ft) for their correct substitution.

Follow through from their common ratio in part (b)(i).

OR

(M1)(A1)(ft)

23

Notes: Award (M1) for terms 5 and 6 correct (using their ratio).

Award (A1)(ft) for terms 7, 8 and 9 correct (using their ratio).

(A1)(ft) (C3)

Examiners report

[N/A]

12a. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitutions. If a list is

used, award (M1) for at least 6 correct terms seen, award (A1) for at least 20 correct terms seen.

(A1)(G3)

[3 marks]

Examiners report

[N/A]

12b. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substituted arithmetic series formula, (A1) for their correct substitutions. Follow

through from part (a). For consistent use of geometric series formula in part (b) with the geometric

24

sequence formula in part (a) award a maximum of (M1)(A1)(A0) since their final answer cannot be an

integer.

OR

(M1)

(M1)

Note: Award (M1) for their correctly substituted arithmetic sequence formula, (M1) for their correctly

substituted arithmetic series formula. Follow through from part (a) and within part (b).

Note: If a list is used, award (M1) for at least 10 correct terms seen, award (A1) for these terms being

added.

(accept ) (A1)(ft)(G2)

[3 marks]

Examiners report

[N/A]

12c. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions. If a list is

used, award (M1) for at least 6 correct terms seen, award (A1) for at least 8 correct terms seen.

(A1)(G3)

25

Note: Exact answer only. If both exact and rounded answer seen, award the final (A1).

[3 marks]

Examiners report

[N/A]

12d. [3 marks]

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substituted geometric series formula, (A1) for their correct substitutions. Follow

through from part (c). If a list is used, award (M1) for at least 8 correct terms seen, award (A1) for

these 8 correct terms being added. For consistent use of arithmetic series formula in part (d) with the

arithmetic sequence formula in part (c) award a maximum of (M1)(A1)(A1).

(A1)(ft)(G2)

[3 marks]

Examiners report

[N/A]

12e. [3 marks]

Markscheme

(M1)

Note: Award (M1) for their correct inequality; allow equation.

26

Follow through from parts (a) and (c). Accept sketches of the two functions as a valid method.

(may be implied) (A1)(ft)

Note: Award (A1) for seen. The GDC gives answers of and to the inequality;

award (M1)(A1) if these are seen with working shown.

OR

(M1)

(M1)

Note: Award (M1) for and both seen, (M1) for and both seen.

(A1)(ft)(G2)

Note: Award (G1) for and seen as final answer without working. Accept use of .

[3 marks]

Examiners report

[N/A]

13a. [2 marks]

Markscheme

The first time an answer is not given to the nearest dollar in parts (a) to (e), the final (A1) in that

part is not awarded.

27

(M1)

Note: Award (M1) for correct product.

(A1)(G2)

[2 marks]

Examiners report

[N/A]

13b. [5 marks]

Markscheme



(i) (M1)(A1)

Note: Award (M1) for substituted arithmetic sequence formula, (A1) for correct substitution.

(A1)(G2)

(ii) (M1)

OR

(M1)

Note: Award (M1) for correct substitution in arithmetic series formula.

(A1)(ft)(G1)28

Note: Follow through from part (b)(i).

[5 marks]

Examiners report

[N/A]

13c. [5 marks]

Markscheme



(i) (M1)(A1)

Note: Award (M1) for substituted geometric sequence formula, (A1) for correct substitutions.

(A1)(G2)

Note: Award (M1)(A1)(A0) for .

Award (G1) for if workings are not shown.

(ii) (M1)

Note: Award (M1) for correct substitution in geometric series formula.

(A1)(ft)(G1)

29

Note: Follow through from part (c)(i).

[5 marks]

Examiners report

[N/A]

13d. [3 marks]

Markscheme



(M1)(A1)

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

OR

(A1)(M1)

Note: Award (A1) for seen, (M1) for other correct entries.

30

OR

(A1)(M1)


(A1)(G2)

[3 marks]

Examiners report

[N/A]

13e. [1 mark]

Markscheme



Option D (A1)(ft)

Note: Follow through from their parts (a), (b), (c) and (d). Award (A1)(ft) only if values for the four

options are seen and only if their answer is consistent with their parts (a), (b), (c) and (d).

31

[1 mark]

Examiners report

[N/A]

13f. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substituted compound interest formula equated to , (A1) for correct

substitutions into formula.

OR

(A1)(M1)


(A1)(G2)

[3 marks]

Examiners report

[N/A]

32

14a. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substituted geometric progression formula, (A1) for correct substitution.

If a list is used, award (M1) for a list of at least six terms, beginning with and (A1) for first six

terms correct.

(A1) (C3)

[3 marks]

Examiners report

The first part of this question was answered quite well, especially by candidates who used a list. Part

(b) was poorly answered. Common errors in part (b) were to find the number of rounds rather than the

total number of matches played or to take the first term as 512 rather than 256.

14b. [3 marks]

Markscheme

OR (M1)(A1)

Note: Award (M1) for substituted sum of a GP formula, (A1) for correct substitution.

If a list is used, award (A1) for at least 9 correct terms, including , and (M1) for their 9 terms,

including , added together.

(A1) (C3)

[3 marks]

33

Examiners report

The first part of this question was answered quite well, especially by candidates who used a list. Part

(b) was poorly answered. Common errors in part (b) were to find the number of rounds rather than the

total number of matches played or to take the first term as 512 rather than 256.

15a. [2 marks]

Markscheme

(M1)

(A1) (C2)

Examiners report

The weakest candidates erroneously used an arithmetic sequence rather than a geometric sequence as

specified in the question.

15b. [2 marks]

Markscheme

(M1)

OR

(M1)

OR

(by list)

(M1)

Notes: Award (M1) for their correct substitution in geometric sequence formula, or stating explicitly

and .

34

(A1)(ft) (C2)

Note: Follow through from their answer to part (a).

Examiners report



15c. [2 marks]

Markscheme

(M1)

Notes: Award (M1) for their correct substitution in geometric series formula.

Accept a list of all their ten geometric terms.

= 1210 (1207.918...) (A1)(ft) (C2)

Note: Follow through from their parts (a) and (b).

Examiners report



16a. [3 marks]

Markscheme

(M1)(A1)

35

Notes: Award (M1) for substituted AP formula, (A1) for correct substitutions. Accept a list of 4 correct

terms.

= 160 (A1)(G3)

Examiners report

[N/A]

16b. [3 marks]

Markscheme

(M1)(M1)

Notes: Award (M1) for correctly substituted AP formula, (M1) for equating to 260. Accept a list of

correct terms showing at least the 14th and 15th terms.

= 15 (A1)(G2)

Examiners report

[N/A]

16c. [4 marks]

Markscheme

or (M1)(A1)(ft)

Notes: Award (M1) for substituted AP sum formula, (A1)(ft) for correct substitutions. Accept a sum of

a list of 15 correct terms. Follow through from their answer to part (b).

2850 seconds (A1)(ft)(G2)

Note: Award (G2) for 2850 seen with no working shown.

36

47.5 minutes (A1)(ft)(G3)

Notes: A final (A1)(ft) can be awarded for correct conversion from seconds into minutes of their

incorrect answer. Follow through from their answer to part (b).

Examiners report

[N/A]

16d. [3 marks]

Markscheme

(M1)(A1)

Notes: Award (M1) for substituted GP formula, (A1) for correct substitutions. Accept a list of 3 correct

terms.

= 135 (134.832) (A1)(G2)

Examiners report

[N/A]

16e. [3 marks]

Markscheme

(M1)(A1)

Notes: Award (M1) for substituted GP sum formula, (A1) for correct substitutions. Accept a sum of a

list of 4 correct terms.

= 525 (524.953...) (A1)(G2)

Examiners report37

[N/A]

16f. [3 marks]

Markscheme

(M1)(M1)

Notes: Award (M1) for correct left hand side, (M1) for correct right hand side. Accept an equation.

Follow through from their expressions given in parts (a) and (d).

OR

List of at least 2 terms for both sequences (120, 130, … and 120, 127.2, …) (M1)

List of correct 12th and 13th terms for both sequences (..., 230, 240 and …, 227.8, 241.5) (M1)

OR

A sketch with a line and an exponential curve, (M1)

An indication of the correct intersection point (M1)

13th lap (A1)(ft)(G2)

Note: Do not award the final (A1)(ft) if final answer is not a positive integer.

Examiners report

[N/A]

17a. [2 marks]

Markscheme

2r2 = 2.205 (M1)

Note: Award (M1) for correct substitution in geometric sequence formula.

r = 1.05 (A1) (C2)

38

[2 marks]

Examiners report

In part (a), 1.1025 proved to be a popular, but erroneous, answer. Similarly to question 4, such

candidates failed to find a square root. Whilst this accuracy mark was lost for such candidates, much

good work was seen in this question reflecting how well drilled the majority of candidates were in both

arithmetic and geometric sequence techniques.

17b. [2 marks]

Markscheme

2(1.05)10 (M1)

Note: Award (M1) for the correct substitution, using their answer to part (a), in geometric sequence

formula.

= 3.26 (3.25778…) (A1)(ft) (C2)

Note: Follow through from their part (a).

[2 marks]

Examiners report





17c. [2 marks]

Markscheme

(M1)

Note: Award (M1) for their correct substitution in geometric sum formula.

39

= 82.9 (82.8609…) (A1)(ft) (C2)

Notes: Accept an answer of 3.97221...if r = −1.05 is found in part (a) and used again in part (c). Follow

through from their part (a).

[2 marks]

Examiners report





18a. [1 mark]

Markscheme

1.65 (km) or 1650 (m) (A1) (C1)

[1 mark]

Examiners report

Most candidates could answer the first part of this question, although a number found it difficult to find

the total distance run after 7 days. Many gave the correct answer of 1.65 km or 1650 m for part (a).

18b. [2 marks]

Markscheme

(M1)

Notes: Award (M1) for correct substitution of candidate’s 10 % into the correct formula. Accept a list.

14.2 (km) (A1)(ft) (C2)

[2 marks]

Examiners report

40

In part (b), stronger candidates answered correctly, however many used a list or the incorrect

arithmetic formula.

18c. [3 marks]

Markscheme

(M1)

Note: Award (M1) for setting up their inequality/equation. Accept a list.

n = 21.371... (A1)(ft)

n = 22 (A1)(ft) (C3)

Notes: Follow through from their values of 1.1 and 1.5 in part (b). The final (A1)(ft) is for rounding up

their answer for n to a whole number of days.

[3 marks]

Examiners report

In part (c), the most common mistake was to use the arithmetic formula. Many candidates rounded

their answer down rather than up.

19a. [2 marks]

Markscheme

(i) u1 + 5d = 100 (A1)

(ii) u1 + 9d = 124 (A1)

[2 marks]

Examiners report

Part A: Arithmetic

41

The contextual nature of this question posed problems for many, though there were many fine

attempts. Failure to discriminate between the sequence and series formulas was the cause of the most

errors. The final part saw many able to substitute into the formula for the series, but then unable to

continue. The use of the GDC in such situations is encouraged; either by graphing each side of the

equation and drawing the resultant sketch or by the solver function.

19b. [2 marks]

Markscheme

(i) 6 (G1)(ft)

(ii) 70 (G1)(ft)

Notes: Follow through from their equations in parts (a) and (b) even if working not seen. Their

answers must be integers. Award (M1)(A0) for an attempt to solve two equations analytically.

[2 marks]

Examiners report

Part A: Arithmetic






19c. [3 marks]

Markscheme

(M1)(A1)(ft)

Note: Award (M1) for substituted sum of AP formula, (A1)(ft) for their correct substituted values.

42

= 2540 (A1)(ft)(G2)

Note: Follow through from their part (b).

[3 marks]

Examiners report

Part A: Arithmetic






19d. [4 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substituted sum of AP formula, (A1) for their correct substituted values.

4n2 +136n – 3200 = 0 (M1)

Note: Award (M1) for this equation (or other equivalent expanded quadratic) seen, may be implied if

correct final answer seen.

n = 16 (A1)(G3)

Note: Do not award the final (A1) for n = 16, – 50 given as final answer, award (G2) if n = 16, – 50 given

as final answer without working.

[4 marks]

Examiners report

43

Part A: Arithmetic






19e. [1 mark]

Markscheme

9, 27 (A1)

[1 mark]

Examiners report

Part B: Geometric

The early straightforward parts were accessible to the majority. The context caused the problems with

many choosing the incorrect value of n when using the formulas. Weaker candidates were more

successful via counting. The context again proved challenging in the final part, with the incorrect time

being determined from the correct value of n. Here, as in Part A, the use of the GDC by graphing each

side of the equation is encouraged; however, if teachers feel that such questions require the use (and

teaching) of logarithms, such an approach is, of course, given full credit.

19f. [1 mark]

Markscheme

3 (A1)

[1 mark]

Examiners report

Part B: Geometric



44





19g. [2 marks]

Markscheme

1 × 36 (M1)

= 729 (A1)(ft)(G2)

Note: Award (M1) for correctly substituted GP formula. Follow through from their answer to part (b).

[2 marks]

Examiners report

Part B: Geometric







19h. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correctly substituted GP formula. Accept sum 1+ 3 + 9 + 27 + ... + 729. If lists are

used, award (M1) for correct list that includes 1093. (1, 4, 13, 40, 121, 364, 1093, 3280…)

= 1093 (A1)(ft)(G2)

45

Note: Follow through from their answer to part (b). For consistent use of n = 6 from part (c) (243) to

part (d) leading

to an answer of 364, treat as double penalty and award (M1)(A1)(ft) if working is shown.

[2 marks]

Examiners report

Part B: Geometric







19i. [3 marks]

Markscheme

(M1)

Note: Award (M1) for correctly substituted GP formula. If lists are used, award (M1) for correct list that

includes 29524. (1, 4, 13, 40, 121, 364, 1093, 3280, 9841, 29524, 88573...). Accept alternative methods,

for example continuation of sum in part (d).

n = 10 (A1)(ft)

Note: Follow through from their answer to part (b).

Exact time = 12:45 (A1)(ft)(G2)

[3 marks]

Examiners report

Part B: Geometric

46







20a. [3 marks]

Markscheme

(M1)(A1)

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

OR

If a list is used, award (M1) for a list of 9 terms, (A1) for all 9 terms correct. (M1)(A1)

( ) (A1) (C3)

Note: Award (A1) for exact answer only.

[3 marks]

Examiners report

The upper quartile of candidates scored well on this question with the vast majority scoring more than

4 marks. However, the lower quartile did very badly with the majority scoring less than 2 marks. A

fundamental error in part (a) resulted in many less able candidates using a common ratio of instead

of . Where lists were used in either part of the question, they were either invariably incomplete or

contained numerical errors. Indeed, using lists seem to be as problematic in this question as they were

in Q6 on arithmetic sequences. Correctly quoted and substituted formula in a correct inequality (= sign

was allowed) did earn many candidates two marks here. The required answer of however did not

always follow.

20b. [3 marks]

47

Markscheme

(M1)(A1)(ft)

Notes: Award (M1) for setting substituted geometric sum formula (A1)(ft) for correct

substitution into geometric sum formula. Follow through from their common ratio.

OR

If list is used, award (M1) for S(6) and S(7) seen, values don’t have to be correct.

(A1) for correct S(6) and S(7). (S(6) and S(7) ). (M1)(A1)

(A1)(ft) (C3)

Notes: Follow through from their common ratio. Do not award the final (A1)(ft) if is less than or if

is not an integer.

[3 marks]

Examiners report

The upper quartile of candidates scored well on this question with the vast majority scoring more than

4 marks. However, the lower quartile did very badly with the majority scoring less than 2 marks. A

fundamental error in part (a) resulted in many less able candidates using a common ratio of instead

of . Where lists were used in either part of the question, they were either invariably incomplete or

contained numerical errors. Indeed, using lists seem to be as problematic in this question as they were

in Q6 on arithmetic sequences. Correctly quoted and substituted formula in a correct inequality (= sign

was allowed) did earn many candidates two marks here. The required answer of however did not

always follow.

21a. [1 mark]

Markscheme

(A1) (C1)

48

Note: Accept .

[1 mark]

Examiners report

In part a, many candidates gave the common ratio as 3.

21b. [2 marks]

Markscheme

(M1)

Note: Award (M1) for correct substitution in formula for nth term of a GP. Accept equivalent forms.

(A1)(ft) (C2)

Notes: Accept . Follow through from their common ratio found in part (a). If used from

part (a) award (M1)(A1)(ft) for an answer of or irrespective of whether working is

shown.

[2 marks]

Examiners report

In part a, many candidates gave the common ratio as 3. While they could set up the equation for part c,

relatively few succeeded in solving it. Those who arrived at an answer did not always realize that the

answer must be an integer.

21c. [3 marks]

Markscheme

(M1)(M1)

Notes: Award (M1) for correct substitution in the sum of a GP formula, (M1) for equating their sum to

. Follow through from parts (a) and (b).

OR49

Sketch of the function (M1)

Indication of point where (M1)

OR

(M1)(M1)

Note: Award (M1) for a list of at least 8 correct terms, (M1) for the sum of the terms equated to

.

(A1)(ft) (C3)

Notes: Follow through from parts (a) and (b). If k is not an integer, do not award final (A1). Accept

alternative methods. If and used award (M1)(M1)(A1)(ft) for . If and

used award (M1)(M1)(A0).

[3 marks]

Examiners report

In part a, many candidates gave the common ratio as 3. While they could set up the equation for part c,

relatively few succeeded in solving it. Those who arrived at an answer did not always realize that the

answer must be an integer.

22a. [2 marks]

Markscheme

(M1)

or (M1)

(AG)

Notes: Award at most (M1)(M0) if last line not seen. Award (M1)(M0) if is found by repeated

multiplication (division) of by .

50

[2 marks]

Examiners report

Part A: Geometric sequences/series

The majority of the candidates were not able to offer a satisfactory justification in a) and only scored 1

mark.

22b. [2 marks]

Markscheme

(M1)

Notes: Award (M1) for correct substitution into correct formula. Accept an equivalent method.

1 (A1)(G2)

[2 marks]

Examiners report


Parts b) and c) were mostly well answered.

22c. [3 marks]

Markscheme

(M1)(A1)

Note: Award (M1) for substitution into the correct formula, (A1) for correct substitution.

OR

(A1) for complete and correct list of eight terms (A1)

51

(M1) for their eight terms added (M1)

(A1)(G2)

[3 marks]

Examiners report


Parts b) and c) were mostly well answered.

22d. [3 marks]

Markscheme

(M1)(M1)(ft)

Notes: Award (M1) for correct substitution into the correct formula for the sum, (M1) for comparing to

. Accept equation. Follow through from their expression for the sum used in part (c).

OR

If a list is used: (M1)

(M1)

(A1)(ft)(G2)

Note: Follow through from their expression for the sum used in part (c).

[3 marks]

Examiners report


The responses to part d) were often weak. Those candidates who set up the equation scored two marks

but very few of them were able to reach the correct final answer.

52

22e. [2 marks]

Markscheme

(may be implied) (A1)

(A1)(G2)

[2 marks]

Examiners report

Part B: Arithmetic sequences/series

Parts a), and b)(i) were mostly answered correctly.

22f. [6 marks]

Markscheme

(i) OR (M1)

(A1)(G2)

(ii) (a) OR (M1)

(M1)

(AG)

Notes: Award second (M1) for correct removal of denominator or brackets and no further incorrect

working seen. Award at most

(M1)(M0) if last line not seen.

(b) (G2)

53

Note: If both solutions to the quadratic equation are seen and the correct value is not identified as the

required answer, award (G1)(G0).

[6 marks]

Examiners report

Part B: Arithmetic sequences/series

Parts a), and b)(i) were mostly answered correctly. Parts b)(ii)a) and b)(ii)b) were poorly answered.

Many candidates did not know how to approach the “show that” question. A few were able to solve the

quadratic equation using the GDC. Those who attempted to solve it without the GDC generally failed to

find the correct answer.

23a. [8 marks]

Markscheme

Option 1: Amount (M1)(A1)

= (A1)(G2)

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

full credit for use of lists.

Option 2: Amount (M1)

= (A1)(G2)

Note: Award (M1) for correct substitution in the compound interest formula. Give full credit for use of

lists.

[8 marks]

54

Examiners report

For many, this question came as a welcome relief following the previous two questions. For those with

a sound grasp of the topic, there were many very successful attempts.

A common error was to make all the comparisons using interest alone; though much credit was given

for doing this, candidates should be aware of what is being asked for in the question.

Many did not understand the notion of monthly compounding periods.

23b. [1 mark]

Markscheme

Option 1 is the best investment option. (A1)(ft)

[1 mark]

Examiners report



A common error was to make all the comparisons using interest alone; though much credit was given

for doing this, candidates should be aware of what is being asked for in the question.

Many did not understand the notion of monthly compounding periods.

23c. [2 marks]

Markscheme

u1 = 135 + 7(1) (M1)

= 142 (A1)(G2)

[2 marks]

Examiners report



55

A common weakness was seen in the “show that” parts of the question where, despite a lenient

approach to method, many were unable to communicate their thoughts on paper.

For many, finding an expression for Sn in (c) was problematical.

The final part was challenging to the great majority, with a large number not attempting it at all; only

the highly competent reached the correct answer.

23d. [2 marks]

Markscheme

u2 = 135 + 7(2) = 149 (M1)

d = 149 – 142 OR alternatives (M1)(ft)

d = 7 (AG)

[2 marks]

Examiners report








23e. [3 marks]

Markscheme

(M1)(ft)

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

56

OR equivalent (A1)

(= 3.5n2 + 138.5n) (A1)(G3)

[3 marks]

Examiners report








23f. [2 marks]

Markscheme

20r3 = 67.5 (M1)

r3 = 3.375 OR (A1)

r = 1.5 (AG)

[2 marks]

Examiners report





57




23g. [2 marks]

Markscheme

(M1)

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

= 643 (accept 643.4375) (A1)(G2)

[2 marks]

Examiners report








23h. [2 marks]

Markscheme

(M1)

Note: Award (M1) for an attempt using lists or for relevant graph.

58

n = 10 (A1)(ft)(G2)

Note: Follow through from their (c).

[2 marks]

Examiners report








24a. [2 marks]

Markscheme

(or equivalent) (M1)

Note: Award (M1) for dividing correct terms.

r = 1.04 (A1) (C2)

Notes: In (b) and (c) (ft) from candidate’s r.

Allow lists, graphs etc. as working in (b) and (c).

59

[2 marks]

Examiners report

Many marks were lost through incorrect rounding or premature rounding (if a year by year approach

was used).

This part was well attempted, errors being the use of 4% as the common ratio.

24b. [2 marks]

Markscheme

Financial penalty (FP) applies in this part

Fees = 8000 (1.04)6 (M1)

Note: Award (M1) for correct substitution into correct formula.

(FP) Fees = 10122.55 USD (USD not required) (A1)(ft) (C2)

Note: Special exception to the note above.

Award maximum of (M1)(A0) if 5 is used as the power.

[2 marks]

Examiners report


was used).

The common error here was the use of the incorrect index in the formula.

24c. [2 marks]

Markscheme

Financial penalty (FP) applies in this part

60

(M1)

Notes: Award (M1) for correct substitution into correct formula.

Give full credit for solution by lists.

(FP) Total = 73713.81 USD (USD not required) (A1)(ft) (C2)

[2 marks]

Examiners report


was used).

Attempts at calculation without use of the formula were largely unsuccessful.

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Documents

IB Questionbank Test · Web viewIn part (d) the sum of the arithmetic progression was done better than the geometric series. Many candidates calculated the 15th term of the progression