IBDP Maths analysis and approaches Topic: SL 1.3 : Geometric sequences and series HL Paper 2

Detailed Solution

(a) Value of the car at the end of the first year
 Original value of the car = $35,000
 

Decrease  in value in the first year =  15%
 Amount of decrease = 15% of $35,000 = 0.15 × 35,000 = $5,250
 Value at the end of the first year = Original value – Decrease = $35,000 – $5,250 = $29,750

Alternatively, we can use the multiplier method:
 After a 15% decrease, the car retains 100% – 15% = 85% of its value.
 Value = $35,000 × 0.85 = $29,750

So, the value of the car at the end of the first year is $29,750.

(b) Value of the car 10 years after purchase
After the first year, the value is $29,750.
From the second year onward, the value decreases by 11% each year.
An 11% decrease means the car retains 100% – 11% = 89% of its value each year, so the multiplier per year is 0.89.
Number of years after the first year = 10 – 1 = 9 years (since the first year is already accounted for).

The value after 10 years can be calculated as:
Value after 1st year = $29,750

Value after 9 more years with an 11% annual decrease = $29,750 × (0.89)^9

Let’s compute this:
 (0.89)^9 ≈ 0.3513 (using a calculator for precision)
 Value = $29,750 × 0.3513 ≈ $10,451.17

Rounding to the nearest dollar:
 $10,451.17 ≈ $10,451

So, the value of Darren’s car 10 years after he buys it is $10,451.

(c) Least value of n where the car’s value is less than 10% of its original value
Original value = $35,000 

10% of original value =  0.10 ×  $35,000 =  $3,500 
We need the value after n complete years to be less than $3,500.

The value of the car after n years can be expressed as:
First year:  $35,000 ×  0.85   (15% decrease) 

Each subsequent year:  Multiply by 0.89  (11% decrease)
Total years = 1 (first year) + (n – 1) (subsequent years) = n years
Formula: Value after n years = $35,000 × 0.85 × (0.89)^(n-1)

We need:
$35,000 × 0.85 × (0.89)^(n-1) < $3,500

Simplify:
$35,000 × 0.85 = $29,750
So, $29,750 × (0.89)^(n-1) < $3,500

Divide both sides by $29,750:
(0.89)^(n-1) < $3,500 / $29,750
$3,500 / $29,750 ≈ 0.1176

Thus:
(0.89)^(n-1) < 0.1176

To find the smallest integer n, take the natural logarithm:
ln[(0.89)^(n-1)] < ln(0.1176)
(n – 1) × ln(0.89) < ln(0.1176)
ln(0.89) ≈ -0.1165, ln(0.1176) ≈ -2.1400
(n – 1) × (-0.1165) < -2.1400

Since the coefficient is negative, reverse the inequality when dividing:
n – 1 > -2.1400 / -0.1165
n – 1 > 18.37
n > 19.37

Since n must be an integer, the smallest possible value is n = 20.

……………………..Markscheme……………………..

Solution: –

(a) recognition that a 15% loss leaves 85% OR finding 15% and subtracting from original
$0.85 \times 35000$ OR $35000-0.15 \times 35000$
= ($29750)

(b) EITHER
$29750 \times 0.89^9$
OR
N=9
I% = -11
PV =+29750
THEN
value (F1) = ($10423)

(c) METHOD 1

attempt to solve the inequality (or equation) $29750 \times 0.89^{x-1} < 3500$ OR table of values (M1)
19.3643… OR $(n=19) \Rightarrow 3651.80…$ OR $(n=20) \Rightarrow 3250.10…$

Note: For candidates using ($29800), n>19.3787…, (n=19) \Rightarrow 3657.93…, (n=20) \Rightarrow 3255.56…

$n=20$

METHOD 2

use of the finance app with I% = -11, PV=+29750, FV =±3500
OR $29750 \times 0.89^{N}<3500$ (condone the use of n oг x)
(N =)18.3643…

Note: For candidates using ($29800), N=18.3787…

$n=20$