IB Maths AI SL 1.1 Number and Algebra Sl Content Questions

attempt to substitute into percentage error formula

3.58 (%) (3.58292…(%))

Note: Award (M1)A0 for a final answer of -3.58 (%) or 0.0358 .

278000000

$42(\mathrm{~km})$

$\quad(\varepsilon=)\left|\frac{42-42.19}{42.19}\right| \times 100 \%$
0.450 (\%) (0.450343...(\%))

Note: Award (M1) for correct substitution of their part (a)(i) into percentage error formula.

METHOD 1 attempt to substitute into arithmetic sequence formula
$42=6+(n-1) \times 2$ OR 42=2 n+4
( n= ) 19

METHOD 2
list of any consecutive terms ending on/above 42
at least two correct terms (not including 6, 8 and 10)
( n= ) 19

METHOD 1
$\left(S_{10}=\right) \frac{10}{2}(2 \times 6+(10-1) 2)$$150(\mathrm{~km})$

METHOD 2
summing (some) correct terms up to 24 OR list of correct cumulative values $6+8+10+\ldots+24$ OR $6,14,24,36 \ldots$$150(\mathrm{~km})$

(d) METHOD 1
$3(1.2)^{n-1}$

Note: Award (A1) for correct substitution into geometric sequence formula.
comparing their geometric expression to 42
e.g. $3(1.2)^{n-1}>42$ OR $3(1.2)^{n-1}=42$$(n=) 15.4747 \ldots$ seen OR $38.5175 \ldots$ and $46.2210 \ldots$ seen (from table)

16 (weeks)

METHOD 2

List of values see. At least 3, 3.6, 4.32 etc
$38.5175 \ldots$ and $46.2210 \ldots$ seen

16 (weeks)

METHOD 1 attempt to substitute into geometric series formula

Note: FT their r from (d).

14 (weeks)

METHOD 2
evidence using tables in GDC or adding the first 14 terms both crossover values seen, e.g $145.589 \ldots$ and $177.587 \ldots$

EITHER
N=72
(M1)(A1)
I %=3.4
$P V= \pm 10000$
P / Y=12
C / Y=12
Note: Award M1 for an attempt to use a financial app in their technology (i.e., at least three entries seen, but not necessarily correct), A1 for all entries correct.

OR
N=72
I %=3.4 / 12
$P V= \pm 10000$
P / Y=1
C / Y=1
Note: Award M1 for an attempt to use a financial app in their technology (i.e., at least three entries seen, but not necessarily correct), A1 for all entries correct.

OR
attempt to substitute into the compound interest formula
$F V=10000 \times\left(1+\frac{3.4}{100 \times 12}\right)^{6 \times 12}$

THEN
(FV =) 12259 (CAD)

212741 (CAD)

$\quad N=180$
F V=0
I %=6.4
$P V= \pm 212741$
P / Y=12
C / Y=2

Note: Award M1 for attempt to use a financial app with at least three entries seen, but not necessarily correct, A1 for N and I, A1 for P / Y and C / Y, A1 for their P V from (b) and F V=0. Approaches that use the compound interest formula receive no marks.
( P M T= ) 1832 (CAD)

Note: Accept -1832 (CAD).

$\quad N=36$$P V= \pm 194572$
PMT = 2200
Note: Award M1 for adjusting at least one of N, P V, or P M T;
A1 for correct entries ( P V and P M T must have opposite signs). Allow FT from incorrect entries for P / Y and/or C / Y from part (c).
( F V= )148099 (CAD)
(194572-148099)
46473 (CAD)

$2200 \times 12 \times 3(=79200)$
(79200-46473=)
32727 (CAD)

attempt to substitute 5000 for G
$0.301 p=\log _{10} 5000$
( p= ) 12.3 (bits) ( $12.2889 \ldots$ )

$(G=) 10^{0.301 p} \quad \mathbf{OR} \quad 2^{p}$

EITHER
attempt to solve the simultaneous equations
e.g., $\quad(x=) 53$ or (y=) 15 or (z=) 20
OR
elimination of a variable using algebraic method

THEN
15 (boxes)

Note: The answer 15 must be identified as the final answer to award A1. Award M1A0 if all 3 values are shown as the final answer.