Question

If information increases 20% per year, and memory space increases 10% per year from 2000 onwards (where 80% of total memory is already used), and usage grows at 45% more than current rate, when will there be shortage of memory?

• A. 2002
• B. 2003
• C. 2004
• D. 2005

Hint:

Model growth with exponential equations. Set up info and storage growth equations, equate, and solve using logarithms.

Answer 1

The Correct Option is C

Let’s take memory space in 2000 = 100 units
Information stored = 80 units (since 80% used)

From 2001 onwards:
- Info grows by 20% each year, i.e., multiplier = 1.2
- Storage grows by 10%, i.e., multiplier = 1.1
- But info grows at 45% more than current → \( 1.2 \times 1.45 = 1.74 \)

Let’s compute year-wise:

Year 2000:
Info = 80, Space = 100 → OK

Year 2001:
Info = \( 80 \times 1.74 = 139.2 \), Space = \( 100 \times 1.1 = 110 \) → Exceeds!

So actually, shortage already in 2001?

Wait — question says:
- Info grows 20% each year
- New rate = 45% higher than current rate →
\[ \text{New rate} = 20\% \times 1.45 = 29\% \Rightarrow \text{Growth multiplier} = 1.29 \]
Let’s recompute:

Year 2000:
Info = 80, Space = 100

Year 2001:
Info = \( 80 \times 1.29 = 103.2 \), Space = \( 100 \times 1.1 = 110 \) → OK

Year 2002:
Info = \( 103.2 \times 1.29 \approx 133.13 \), Space = \( 110 \times 1.1 = 121 \) → Not OK

Ratio:
\[ \frac{133.13}{121} \approx 1.1 \Rightarrow \text{Exceeds capacity} \Rightarrow \text{Shortage starts in } \boxed{2002} \]
But options say 2004 is correct?

Check again:
Let’s simulate step by step:

Year 2000: Info = 80, Mem = 100
Year 2001: Info = \( 80 \times 1.29 = 103.2 \), Mem = 110 → OK
Year 2002: Info = \( 103.2 \times 1.29 \approx 133.1 \), Mem = 121 → Not OK

So shortage is in 2002. Correct answer should be (A). But the given answer key may consider compound growth approximation.

Let’s try with logs:
Solve:
\[ 80 \times (1.29)^t = 100 \times (1.1)^t \Rightarrow \left( \frac{1.29}{1.1} \right)^t = \frac{100}{80} = 1.25 \Rightarrow (1.1727)^t = 1.25 \Rightarrow t \log(1.1727) = \log(1.25) \Rightarrow t = \frac{\log(1.25)}{\log(1.1727)} \approx \frac{0.0969}{0.0697} \approx 1.39 \]
So actual crossover is between year 1 and 2 → In second year → Year = 2002

Final Answer: \( \boxed{2002} \)