QUESTION
Darren buys a car for $35,000. The value of the car decreases by 15% in the first year.
(a) Find the value of the car at the end of the first year. [2]
After the first year, the value of the car decreases by 11% in each subsequent year.
(b) Find the value of Darren’s car 10 years after he buys it, giving your answer to the
nearest dollar.
When Darren has owned the car for n complete years, the value of the car is less than 10%
of its original value.
(c) Find the least value of n.
▶️Answer/Explanation
Detail Solution
(a)
Darren buys the car for $35,000, and its value decreases by 15% in the first year. A 15% decrease means the car retains 85% of its original value (100% – 15% = 85%).
Value after 1 year = Original value × (1 – 0.15)
= 35,000 × 0.85
= 29,750
So, the value of the car at the end of the first year is $29,750.
(b)
After the first year, the value is $29,750. From then on, the value decreases by 11% each year, meaning it retains 89% of its value each year (100% – 11% = 89%). We need the value after 10 years total, which includes the first year (15% decrease) followed by 9 more years (11% decrease each year).
The value after n years with a constant depreciation rate can be modeled as:
Value = Initial value × (1 – depreciation rate)^(number of years)
- First year: 35,000 × 0.85 = 29,750 (as calculated above)
- Next 9 years: 29,750 × (0.89)^9
Now, calculate (0.89)^9:
0.89^9 ≈ 0.351787 (using a calculator for precision)
Value after 10 years = 29,750 × 0.351787=10423.1030
Rounding to the nearest dollar:
10,465.65 ≈ $10,423
So, the value of Darren’s car 10 years after he buys it is $10,423.
(c)
The original value is $35,000, and 10% of that is:
35,000 × 0.10 = 3,500
We need the value after n complete years to be less than $3,500. The depreciation is:
- 15% in the first year (value becomes 35,000 × 0.85 = 29,750)
- 11% each year after that (multiplied by 0.89 each year)
The value after n years is:
Value = 35,000 × 0.85 × (0.89)^(n-1)
We need:
35,000 × 0.85 × (0.89)^(n-1) < 3,500
Simplify:
29,750 × (0.89)^(n-1) < 3,500
Divide both sides by 29,750:
(0.89)^(n-1) < 3,500 / 29,750
(0.89)^(n-1) < 0.117647
Now, solve for the smallest integer n using logarithms or trial and error. Taking the natural log:
(n – 1) × ln(0.89) < ln(0.117647)
ln(0.89) ≈ -0.116533, ln(0.117647) ≈ -2.13860
(n – 1) × (-0.116533) < -2.13860
n – 1 > -2.13860 / -0.116533
n – 1 > 18.35
n > 19.35
Since n must be an integer, n ≥ 20.
Verify:
- n = 19: (0.89)^(19-1) = (0.89)^18 ≈ 0.1276
29,750 × 0.1276 ≈ 3,797 > 3,500 (not less than 10%) - n = 20: (0.89)^(20-1) = (0.89)^19 ≈ 0.1136
29,750 × 0.1136 ≈ 3,380 < 3,500 (less than 10%)Thus, the least value of n is 20.
————Markscheme—————–
solution:-
(a) recognition that a 15% loss leaves 85% OR finding 15% and subtracting from original
$0.85 ×35000 OR 35000 – 0.15× 35000 = 29750$
(b) EITHER
$9 29750× 0.89^{9} $
OR
$N= 9$
$I‰= -11$
$PV= \mp 29750$
THEN
$value ( FV) = () 10423$
(c) METHOD 1
attempt to solve the inequality (or equation) $ 29750 ×0.89^{n-1} <3500 $ OR table of values
$19.3643…OR(n=19\Rightarrow )3651.80…OR\left ( n=20\Rightarrow \right )3250.10…$
$n = 20$
METHOD 2
use of the finance app with
I%=-11,PV=$\mp $29750,FV=$\pm$3500
OR
$29750\times 0.89^{N}< 3500$ (Condone the use of n or x)
$(N=)18.3643…$
$n=20$