IP Practice SET – 3
SI / CI / Growth: Level 2
1.
Invest $1000 at 8% annual interest compounded **semi-annually**. What is the amount after 1 year?
Solution (C):
Semi-annual means rate is \(8\%/2 = 4\%\) per period, and there are 2 periods in 1 year.
\( A = 1000 (1 + 0.04)^2 \).
\( A = 1000 (1.0816) = 1081.60 \).
2.
What principal amount is needed to earn $60 in interest in 3 years at 5% Simple Interest?
Solution (D):
\( I = Prt \implies 60 = P \times 0.05 \times 3 \).
\( 60 = 0.15P \).
\( P = 60 / 0.15 = 400 \).
3.
The population of a city is 100,000. It increases by 10% in the first year and by 20% in the second year. What is the population after 2 years?
Solution (B):
Year 1: \( 100,000 \times 1.1 = 110,000 \).
Year 2: \( 110,000 \times 1.2 = 132,000 \).
4.
At what simple interest rate will a sum of money triple itself in 20 years?
Solution (A):
To triple, the Interest must be \( 2P \) (so Total = \( P + 2P = 3P \)).
\( 2P = P \times r \times 20 \).
\( 2 = 20r \implies r = 2/20 = 0.1 = 10\% \).
5.
John borrows $2000 at 10% Simple Interest for 3 years. Mary borrows $2000 at 10% Compound Interest (annual) for 3 years. How much more interest does Mary pay?
Solution (E):
Simple Int = \( 2000 \times 0.10 \times 3 = 600 \).
Compound Amt = \( 2000 \times (1.1)^3 = 2000 \times 1.331 = 2662 \).
Compound Int = \( 2662 – 2000 = 662 \).
Difference = \( 662 – 600 = 62 \).
6.
A value depreciates by \(20\%\) in the first year and then increases by \(20\%\) in the second year. The final value is what percent of the original value?
Solution (C):
Start with 100.
After Year 1: \( 100 \times 0.8 = 80 \).
After Year 2: \( 80 \times 1.2 = 96 \).
96 is 96% of 100.
7.
If a population grows by a factor of 1.5 every 10 years, by what factor does it grow in 20 years?
Solution (B):
Growth factor for 10 years = 1.5.
For 20 years (two 10-year periods), factor is \( 1.5 \times 1.5 = 2.25 \).
8.
In how many years will $500 grow to $600 at 5% simple interest?
Solution (A):
Interest = 100.
\( 100 = 500 \times 0.05 \times t \).
\( 100 = 25t \implies t = 4 \).
9.
Calculate the compound interest on $2000 for 1 year at 12% per annum, compounded quarterly.
Solution (D):
Rate per quarter = \(12\%/4 = 3\% = 0.03\).
Periods = \(1 \times 4 = 4\).
Amount = \( 2000(1.03)^4 \).
\( 1.03^4 \approx 1.1255 \).
\( 2000 \times 1.1255 = 2251 \).
Interest = \( 2251 – 2000 = 251 \). (Closest choice is D).
10.
A sum of money doubles itself in 10 years at simple interest. In how many years will it triple itself?
Solution (C):
To double, Interest = P. Time = 10 yrs. So rate is \(P/10\) per year.
To triple, Interest must be 2P.
Time = \( 2P / (P/10) = 20 \) years.
Score: 0 / 0