Question

Test II. Solve the following problems on a separate sheet of paper: 1. An amount of P5,000 is deposited at 6% interest compounded annually. Find the total amount after 3 years. 2. The population of Barangay Punta Princesa at the start of year 2016 is 3, 200,000 and is steadily increasing at the rate of 2% annually. At this rate, what is the population of the city at the end of year 2023? 3. The half-life of a certain radioactive substance is 2 years. If this substance has an initial amount of 128 grams, how much of the substance will be left after a decade? 4. Suppose a culture of 100 bacteria is put in a Petri dish and the culture doubles every minute. How many bacteria will there be after half an hour? 5. The present population of Talisay City is 10 million with a growth rate of 2.5% annually. At this rate, in what year will the city reach the 100 million populations? 1​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The answers are 1] 5955, 2] 234331, 3] 4 g, 4] $\bf 1.07*10^{11}$, and 5] 93.25 years.

Given:

1. 5000 is deposited at 6% interest for 3 years

2. The population is 200000 at the start of 2016 and increasing by 2% every year

3. The half-life is 2 years and the initial amount is 128 g.

4. 100 bacteria double every minute kept in a petri dish

5. The population of Talisay City is 10 million increasing by 2.5% annually

To Find:

1. Total amount after 3 years

2. The population at end of 2023

3. How much substance will be left after a decade

4. How many bacteria will be there after 30 minutes

5. What year will the city reach 100 million population
Solution:

1. The formula of compound interest is

$P_t =P[1+r]^t$

Here P = 5000, r = 6%, and t = 3.

Therefore $P_t=5000[1+0.06]^3 = 5955$.

2.  The formula of compound interest is

$P_t =P[1+r]^t$

Here P = 200000, r = 2%, and $t=2023-2016+1=8$.

Therefore $P_t=200000[1+0.02]^8 = 234331$.

3. The formula of an amount of substance is given by

$A=\frac{A_o}{2^{\frac{t}{\lambda} }}$

Here $A_o$ is the initial amount = 128 g, $\lambda$ is the half-life = 2, and t is the time = 10.

Therefore $A=\frac{128}{2^\frac{10}{2} } = 4 g$.

4. The number of bacteria that doubles every minute after time t is given by

$A=A_o2^t$

Hence for t = 30, $A = 100*2^{30}=1.07*10^{11}$

5. The formula of compound interest is

$P_t =P[1+r]^t$

Here P = 10 million, r = 2.5%, and $P_t = 100 million$

Therefore

$100=10[1+0.025]^t \\1.025^t=10\\t\ln1.025=\ln 10\\t=93.25$

Hence the total amount after 3 years is 5955, the population of Barangay Punta is 234331, the substance left is 4 g, the number of bacteria is $\bf 1.07*10^{11}$, and the population will become 100 million after 93.25 years.

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