IB Maths AI HL 1 Number & Algebra

[Maximum number: 5]

A bakery sells egg tarts in boxes. Each box contains 6 tarts. On a particular morning, the bakery has 400 freshly baked tarts. The number of tarts remaining, after n boxes have been sold, forms an arithmetic sequence, $u_{n}=u_{1}+(n-1) d$ .

(a)

Write down the value of d.

[ 1 ]

(b)

Find the value of $u_{1}$ .

[ 1 ]

(c)

Find the greatest number of boxes the bakery can sell that morning.

[ 3 ]

attempt to solve $394-6(n-1) \geq 0$ OR $400 \div 6$

Note: Accept equality or any inequality with zero seen, graphical approach to find zero, use of table/list to find zero.
( n< ) $66.6666 \ldots$
( n= ) 66

Note: If 400 is given as the answer in (b) then n=67 in part (c).

[Maximum number: 8]

Nanthana is completing an exploration on multiplication using complex numbers. She considers a geometric sequence where $u_{1}=9$ and $r=\frac{2}{3}+\frac{2}{3} \mathrm{i}$ .

(a)

Write down the value of

[ 2 ]

(i)

$\quad u_{2}$ .

[ 1 ]

$\left(u_{2}=\right) 6+6 \mathrm{i} \quad \boldsymbol{A} \mathbf{1}$

(ii)

$u_{3}$ .

Nanthana claims that the sequence $\left|u_{1}\right|,\left|u_{2}\right|,\left|u_{3}\right|, \ldots$ also forms a geometric sequence.

[ 1 ]

$\quad\left(u_{3}=\right) 8 \mathrm{i} \quad \boldsymbol{A} \mathbf{1}$

(b)

Show that Nanthana's claim is correct, stating the exact value of the common ratio for this sequence.

[ 4 ]

attempt to find |r|

\begin{aligned} & |r|=\sqrt{\left(\frac{2}{3}\right)^{2}+\left(\frac{2}{3}\right)^{2}} \\ & |r|=\frac{2 \sqrt{2}}{3} \end{aligned}

statement that multiplying a number by a complex number changes the modulus of the number by a factor of |r| (may be seen at any stage)

Note: Award at most (M1)A1A0R1 if common ratio is given as $0.942809 \ldots \ldots$ .
therefore it is a geometric sequence

(c)

Hence, find the sum of the infinite sequence $\left|u_{1}\right|,\left|u_{2}\right|,\left|u_{3}\right|, \ldots$ .

[ 2 ]

substitution of their r and 9 into the sum formula for infinite sequence
$S_{\infty}=\frac{9}{1-\frac{2 \sqrt{2}}{3}}$ $\left(S_{\infty}=\right) 157(157.367 \ldots, 81+54 \sqrt{2})$

Note: No FT if their $|r| \geq 1$ .

[Maximum number: 19]

Give your answers in parts (a), (d)(i), (e) and (f) to the nearest dollar.
Daisy invested 37000 Australian dollars (AUD) in a fixed deposit account with an annual interest rate of 6.4 % compounded quarterly.

(a)

Calculate the value of Daisy's investment after 2 years.

[ 3 ]

EITHER
N=2
P V=-37000
I %=6.4
P / Y=1
C / Y=4
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.

OR
N=8
P V=-37000
I %=6.4
P / Y=4
C / Y=4

Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.

OR
$F V=37000 \times\left(1+\frac{6.4}{100 \times 4}\right)^{4 \times 2}$
Note: Award M1 for substitution into compound interest formula, (A1) for correct substitution.
= 42010 AUD

Note: Award (M1)(A1)A0 for unsupported 42009.87.

(b)

After m months, the amount of money in the fixed deposit account has appreciated to more than 50000 AUD.

Find the minimum value of m, where $m \in \mathbb{N}$ .

[ 4 ]

EITHER
P V=-37000
F V=50000
I %=6.4
P / Y=1
C / Y=4
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct. The final mark can still be awarded for the correct number of months (multiple of 3 ).
OR
P V=-37000
F V=50000
I %=6.4
P / Y=4
C / Y=4
Note: Award M1 for an attempt to use a financial app in their technology, award A1 for all entries correct.
OR

50000<37000 \times\left(1+\frac{6.4}{100 \times 4}\right)^{4 \times n} \text { OR } 50000<37000 \times\left(1+\frac{6.4}{100 \times 4}\right)^{n} \quad \text { (M1)(A1) }

Note: Award M1 for the correct inequality, 50000 and substituted compound interest formula. Allow an equation. Award A1 for correct substitution.

THEN

\begin{aligned} & N=4.74(\text { years })(4.74230 \ldots) \text { OR } N=18.9692 \ldots \text { (quarters) } \\ & m=57 \text { months } \end{aligned}

Note: Award A1 for rounding their m to the correct number of months. The final answer must be a multiple of 3 . Follow through within this part.

(c)

Daisy is saving to purchase a new apartment. The price of the apartment is 200000 AUD.
Daisy makes an initial payment of 25 % and takes out a loan to pay the rest.

Write down the amount of the loan.

[ 1 ]

(d)

The loan is for 10 years, compounded monthly, with equal monthly payments of 1700 AUD made by Daisy at the end of each month.

For this loan, find

[ 5 ]

(i)

the amount of interest paid by Daisy.

[ 2 ]

$120 \times 1700-150000$
=54000 AUD

(ii)

the annual interest rate of the loan.

[ 3 ]

$\quad N=120$
P V=-150000
P M T=1700
F V=0
P / Y=12
C / Y=12
(M1)(A1)

Note: Award M1 for an attempt to use a financial app in their technology or an attempt to use an annuity formula or F V=0 seen. If a compound interest formula is equated to zero, award M1, otherwise award M0 for a substituted compound interest formula.

Award A1 for all entries correct in financial app or correct substitution in annuity formula, but award A0 for a substituted compound interest formula. Follow through marks in part (d)(ii) are contingent on working seen.

$r=6.46\%\quad(6.45779\ldots)$

(e)

After 5 years of paying off this loan, Daisy decides to pay the remainder in one final payment.

Find the amount of Daisy's final payment.

[ 3 ]

$\quad N=60$
I=6.46 (6.45779...)
P V=-150000
P M T=1700
P / Y=12
C / Y=12
(M1)(A1)
Note: Award M1 for an attempt to use a financial app in their technology or an attempt to use an annuity formula. Award (M0) for a substituted compound interest formula. Award A1 for all entries correct. Follow through marks in part (e) are contingent on working seen.

$FV=86973\ \mathrm{AUD}$

(f)

Find how much money Daisy saved by making one final payment after 5 years.

[ 3 ]

$204000-(60 \times 1700+86973)$ OR 204000-188973
(M1)(M1)

Note: Award M1 for $60 \times 1700$ . Award M1 for subtracting their $(60 \times 1700+86973)$ from their (204000). Award at most M1M0 for their 204000- ( $60 \times 1700$ ) or M0M0 for their 204000-(86973). Follow through from parts (d)(i) and (e). Follow through marks in part (f) are contingent on working seen.

15027 AUD

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IB Maths AI HL 1 Number and Algebra Questions

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question 3

Question (a)

Question (b)

Question (c)

Question (d)

Question (i)

Question (ii)

Question (e)

Question (f)