IP Practice SET – 1
SI / CI / Growth: Level 1
1.
What is the simple interest earned on $2000 invested at 5% annual interest for 3 years?
Solution (C):
Formula: \( I = P \times r \times t \).
\( I = 2000 \times 0.05 \times 3 = 100 \times 3 = 300 \).
2.
If $1000 is invested at a compound interest rate of 10% per year, what is the total amount after 2 years?
Solution (B):
Year 1: \( 1000 \times 1.1 = 1100 \).
Year 2: \( 1100 \times 1.1 = 1210 \).
3.
A population of bacteria doubles every hour. If there are 100 bacteria now, how many will there be in 4 hours?
Solution (A):
\( \text{Final} = \text{Initial} \times 2^t \).
\( 100 \times 2^4 = 100 \times 16 = 1600 \).
4.
If $500 grows to $600 in 2 years at simple interest, what is the annual interest rate?
Solution (D):
Interest = \( 600 – 500 = 100 \).
\( I = Prt \implies 100 = 500 \times r \times 2 \).
\( 100 = 1000r \implies r = 0.1 = 10\% \).
5.
A car depreciates by 10% each year. If it costs $10,000 today, what is its value after 1 year?
Solution (E):
Value = \( 10,000 \times (1 – 0.10) = 10,000 \times 0.9 = 9,000 \).
6.
Which yields more interest in 1 year: $1000 at 5% Simple Interest or $1000 at 5% Compound Interest (compounded annually)?
Solution (B):
For the **first year**, simple interest and annually compounded interest are exactly the same.
SI: \( 1000 \times 0.05 = 50 \).
CI: \( 1000 \times 1.05 – 1000 = 50 \).
7.
How many years will it take for an investment to double at a simple interest rate of 10%?
Solution (C):
To double, Interest must equal Principal (\(I=P\)).
\( P = P \times 0.10 \times t \).
\( 1 = 0.10t \implies t = 10 \) years.
8.
What is the total amount if $500 is invested at 4% simple interest for 6 months?
Solution (A):
Time = 6 months = 0.5 years.
\( I = 500 \times 0.04 \times 0.5 = 10 \).
Amount = \( 500 + 10 = 510 \).
9.
A bank offers 5% interest per year. If you deposit $100, what is the interest after 2 years? (Simple Interest)
Solution (D):
\( I = 100 \times 0.05 \times 2 = 10 \).
10.
If a population grows from 1000 to 1210 in 2 years at a constant annual percentage rate, this is an example of:
Solution (B):
Population growth generally follows a percentage increase on the *current* population, which is the definition of compound growth.
(Check: \(1000 \times 1.1 = 1100\). \(1100 \times 1.1 = 1210\). This is 10% compound growth).
Score: 0 / 0