Ric made two investments: Investment \text{Q}Q has a value of \$500$500 at the end of the first year and increases by \$45$45 pe
r year. Investment \text{R}R has a value of \$400$400 at the end of the first year and increases by 10\%10% per year. Eric checks the value of his investments once a year, at the end of the year. What is the first year in which Eric sees that investment \text{R}R's value exceeded investment \text{Q}Q's value?
2 answers:
Answer:
10th year
Step-by-step explanation:
Ric made two investments.
Investment Q
Amount = $500 at end of the year
Increase $45 per year.
Value of investment after x year
we take time (x-1) because investment was end of first year.
Investment R
Amount = $400 at end of the year
Increase 10% per year.
Value of investment after x year
we take time (x-1) because investment was end of first year.
Eric checks the value of his investments once a year, at the end of the year.
Eric sees that investment R's value exceeded investment Q's value
Investment R > Investment Q
Using graphing calculator
Hence, In 10th year Investment R could be greater than Investment Q
Answer: 9 year (approx)
Step-by-step explanation:
Since, In the Investment R,
The initial amount = $500
Which is increasing by $45 per year,
Thus, the total amount after x years in investment R = 500 + 45 x
In the investment Q,
The initial amount = $400
Which is increasing by 10% per year,
Thus, the total amount after x years in investment Q =
Since, the intersection point of the equations y =500+45x and y = 400(1.1)^x
are (-6.178, 221.992) and (8.069, 863.12)
But we can not take a negative number as a number of year,
Thus, the year in which both investment will equal is 8.069
After that the investment Q will be exceed the investment R,
⇒ The first year at which the investment Q will be exceed the investment R is approx 9th year.
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