Question

nina made two investments:

• investment a has a value of $50 at the end of the first year and increases by 8% per year.
• investment b has a value of $60 at the end of the first year and increases by $3 per year.

nina checks the value of her investments once a year, at the end of the year.
what is the first year in which nina sees that investment as value exceeded investment bs value?

Explanation:

Step1: Define value formulas

Let $n$ = number of years after Year 1.
Investment A (compound growth): $V_A = 50(1+0.08)^n = 50(1.08)^n$
Investment B (linear growth): $V_B = 60 + 3n$
We need $V_A > V_B$, so $50(1.08)^n > 60 + 3n$

Step2: Test n=1 (Year 2)

$V_A = 50(1.08)^1 = 54$
$V_B = 60 + 3(1) = 63$
$54 < 63$, not true.

Step3: Test n=2 (Year 3)

$V_A = 50(1.08)^2 = 50(1.1664) = 58.32$
$V_B = 60 + 3(2) = 66$
$58.32 < 66$, not true.

Step4: Test n=3 (Year 4)

$V_A = 50(1.08)^3 = 50(1.259712) = 62.9856$
$V_B = 60 + 3(3) = 69$
$62.9856 < 69$, not true.

Step5: Test n=4 (Year 5)

$V_A = 50(1.08)^4 = 50(1.36048896) = 68.024448$
$V_B = 60 + 3(4) = 72$
$68.0244 < 72$, not true.

Step6: Test n=5 (Year 6)

$V_A = 50(1.08)^5 = 50(1.4693280768) = 73.46640384$
$V_B = 60 + 3(5) = 75$
$73.4664 < 75$, not true.

Step7: Test n=6 (Year 7)

$V_A = 50(1.08)^6 = 50(1.586874322944) = 79.3437161472$
$V_B = 60 + 3(6) = 78$
$79.3437 > 78$, true.

Answer:

7