Problem 1 :
Mark invests $1,500 at a rate of 6% interest compounded annually. How much is the investment worth after 5 years ?
Problem 2 :
The price of a new automobile is $28,000. If the value of the automobile decreases 12% per year, what will be the price of the automobile after 5 years?
Problem 3 :
David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007?
Problem 4 :
An investment worth $2500 made in a bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years?
Problem 5 :
Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours?
Problem 6 :
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture initially, how many bacteria will be present at the end of 8th hour?
Problem 7 :
A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself?
1. Answer :
Compound Interest Formula :
A = P(1 + r)^{n}
Substitute.
P = 1500
r = 6% or 0.06
n = 5
Then,
A = 1500(1 + 0.06)^{5}
= 1500(1.06)^{5}
Use a calculator.
= 2,007.34
So, the worth of the investment after 5 years is about $2,007.34.
2. Answer :
Exponential Decay Formula :
A = P(1 - r)^{n}
Substitute.
P = 28000
r = 12% or 0.12
n = 5
Then,
A = 2800(1 - 0.12)^{5}
= 1500(0.88)^{5}
Use a calculator.
= 14,776.49
So, the price of the automobile will be about $14,776.49.
3. Answer :
Number of years between 1999 and 2007 is
n = 2007 - 1999
n = 8
No. of stores in the year 2007 = P(1 + r)ⁿ
Substitute P = 200, r = 8% or 0.08 and n = 8.
No. of stores in the year 2007 = 200(1 + 0.08)^{8}
= 200(1.08)^{8}
= 200(1.8509)
= 370.18
So, the number of stores in the year 2007 is about 370.
4. Answer :
We have to use the formula given below to know the value of the investment after 3 years.
A = Pe^{rt}
Substitute.
P = 2500
r = 10% or 0.1
t = 10
e = 2.71828
Then,
A = 2500(2.71828)^{(0.1)10}
A = 6795.70
So, the value of the investment after 10 years is $6795.70.
5. Answer :
Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100.
We have to use the formula given below to find the percent of substance after 6 hours.
A = P(1 + r)^{n}
Substitute.
P = 100
r = -3.5% or -0.035
t = 6
(Here, the value of r is taken in negative sign. because the substance decays)
A = 100(1-0.035)^{6}
= 100(0.935)^{6}
= 100(0.8075)
= 80.75
Because the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75%.
6. Answer :
Note that the number of bacteria present in the culture doubles at the end of successive hours.
Since it grows at the constant ratio 2, the growth is based is on geometric progression.
We have to use the formula given below to find the no. of bacteria present at the end of 8th hour.
A = ab^{x}
Substitute.
a = 30
b = 2
x = 8
Then, we have
A = 30(2^{8})
= 30(256)
= 7680
So, the number of bacteria at the end of 8th hour is 7680.
7. Answer :
Let P be the amount invested initially.
From the given information, P becomes 2P in 3 years.
Since the investment is in compound interest, for the 4th year, the principal will be 2P.
And 2P becomes 4P (it doubles itself) in the next 3 years.
Therefore, at the end of 6 years accumulated value will be 4P.
So, the amount deposited will amount to 4 times itself in 6 years.
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