Question: Question 46:
The graph below shows the growth in value of a $1000 investment over a period of four years.
value of
investment ($)
1250
1200
1150
1100
1050
1000
950
year
A different amount of money is invested under the same investment conditions for eight years.
In total, the amount of interest earned on this investment is $600.
The amount of money invested is
A. $500
B. $600
C. $1500
D. $2000
E. $2400
Graph shape: a scatter plot of investment value versus year, with points at approximately (0,1000), (1,1050), (2,1100), (3,1160), and (4,1210), rising steadily upward; position_hint = center.
1. **State the problem:**
We have an investment of $1000 growing over 4 years as shown by the graph.
We want to find the initial amount invested for a different investment that earns $600 interest over 8 years under the same conditions.
2. **Analyze the given data:**
From the graph points: (0,1000), (1,1050), (2,1100), (3,1160), (4,1210), the investment grows steadily.
3. **Determine the type of growth:**
The growth appears linear or close to linear since the increments are roughly constant.
Calculate yearly interest for the $1000 investment:
Year 0 to 1: 1050 - 1000 = 50
Year 1 to 2: 1100 - 1050 = 50
Year 2 to 3: 1160 - 1100 = 60
Year 3 to 4: 1210 - 1160 = 50
Average yearly interest $\approx \frac{50 + 50 + 60 + 50}{4} = 52.5$
4. **Calculate total interest earned on $1000 over 4 years:**
$$\text{Total interest} = 1210 - 1000 = 210$$
5. **Calculate interest rate per year:**
$$\text{Interest rate} = \frac{210}{1000 \times 4} = \frac{210}{4000} = 0.0525 = 5.25\% \text{ per year}$$
6. **Use the same interest rate for the new investment over 8 years:**
Let the initial investment be $P$.
Total interest earned is $600$.
7. **Set up the equation for total interest:**
$$600 = P \times 0.0525 \times 8$$
8. **Solve for $P$:**
$$P = \frac{600}{0.0525 \times 8} = \frac{600}{0.42} = 1428.57$$
9. **Compare with given options:**
Closest option is C. 1500
**Final answer:** C. 1500