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[Maximum number: 4]

The diameter of a spherical planet is $6 \times 10^{4} \mathrm{~km}$ .

(a)

Write down the radius of the planet.

The volume of the planet can be expressed in the form $\pi\left(a \times 10^{k}\right) \mathrm{km}^{3}$ where $1 \leq a<10$ and $k \in \mathbb{Z}$ .

[ 1 ]

$3\times10^4$
or $30000\,\mathrm{km}$ . Accept $3\bullet10^4$ .

(b)

Find the value of a and the value of k.

[ 3 ]

Use the volume of a sphere with radius $3\times10^4$ :
$\frac43\pi(3\times10^4)^3=36\pi\times10^{12}=\pi(3.6\times10^{13}).$ $a=3.6,\qquad k=13$

[Maximum number: 22]

Give your answers to parts (a)(ii), (c)(i) and (d) correct to two decimal places.
Daniela and Sorin have each recently received some money. Daniela won a cash prize and Sorin received an inheritance.
Daniela had two options to choose from to receive her winnings. In both options she receives a payment on the first day of each month for three years.
Option A Each payment is $ 5500.
Option B The first payment is $ 2000. In each month which follows, the payment is 6 % more than the previous month.

(a)

Find the total amount Daniela would receive if she chooses

[ 6 ]

(i)

Option A;

[ 1 ]

$5500 \times 36$
=($) 198000

(ii)

Option B.

Sorin received an inheritance of $ 120000. Sorin invested his inheritance in an account that pays a nominal annual interest rate of 4 % per annum, compounded monthly. The interest is added on the last day of each month.

[ 5 ]

recognizing sum of a geometric sequence is required
$\frac{2000\left(1-1.06^{36}\right)}{1-1.06}$ $=238241.7333 \ldots$
=($) 238241.73

(b)

Write down an expression for the value of Sorin's investment after n years.

Daniela chose Option B and received her first payment on $1^{\text {st }}$ January 2023. Sorin invested his inheritance on the same day.

[ 1 ]

Sorin's future value after n years $=120000\left(1+\frac{4}{100 \times 12}\right)^{12 n}$

(c)

Find the total value of Daniela's winnings and Sorin's investment on the last day of the sixth month.

[ 6 ]

Sorin's total $=120000\left(1+\frac{4}{100 \times 12}\right)^{6}(=122420.09)$

Daniela's total $=\frac{2000\left(1-1.06^{6}\right)}{1-1.06}(=13950.64)$
total value =($) 136370.73

(d)

Find the minimum number of complete months before the total value of Daniela's winnings and Sorin's investment is at least $\$ 250000$ .

At the end of the three years, Daniela invested $40000 for a further six years in a second account that pays a nominal interest rate of r % per annum compounded quarterly.

[ 6 ]

EITHER (finding number of months, m )
$120000\left(1+\frac{4}{100 \times 12}\right)^{m}+\frac{2000\left(1-1.06^{m}\right)}{1-1.06}(\geq 250000)$ $m \geq 26.0905 \ldots$ OR $(m=26 \Rightarrow) 249157 \ldots$ AND $(m=27 \Rightarrow) 258692 \ldots$

Note: Condone use of an equation or strict inequality.
OR (finding number of years, n )
$120000\left(1+\frac{4}{100 \times 12}\right)^{12 \times n}+\frac{2000\left(1-1.06^{12 \times n}\right)}{1-1.06}(\geq 250000)$ $n \geq 2.17421 \ldots$ (years)

Note: Condone use of an equation or strict inequality.

THEN
m=27 (months)

(e)

Find the value of r if this investment grows to $ 53000 after six years.

[ 3 ]

EITHER
N=24
OR
N=6
$\mathrm{PV}=\mp 40000$ $\mathrm{PV}=\mp 40000$
PMT=0
PMT=0
$\mathrm{FV}= \pm 53000$ $\mathrm{FV}= \pm 53000$
P / Y=4
P / Y=1
C / Y=4
C / Y=4
Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, and award (A1) for all entries correct. PV and FV must have opposite signs.

OR
$40000\left(1+\frac{r}{100 \times 4}\right)^{6 \times 4}=53000$
Note: Award (M1) for attempting to substitute into compound interest formula, award (A1) for correct equation.

THEN
$4.71781 \ldots$
( r= ) 4.72 (\%)

IB Maths AA SL 1.1 Number and Algebra Sl Content Topic Practice

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IB Maths AA SL 1.1 Number and Algebra Sl Content Topic Practice

Learn this topic

Question 1

Question (a)

Question (b)

Question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)

Question (e)