Question: Answer the following situations
1. A collection is taken up to support a family in need. The initial amount in the collection is P10,000. Everyone places P5000 in the collection. When the 35th person puts their P5000 in the collection, how much is present in the collection?
2. Ana, an investor, begins a start-up to revitalize Camela Homes. She begins with P1,000,000, making her investor number 1. Each investor that joins will invest P500,000 more than the previous investor. How much does the 10th investor invest in the project? With that 10th investor, what is the total amount invested in the project?
3. Alex and Jill deposit P4,000 in an account bearing 5% interest compounded yearly. If they do not deposit any more money in that account, how much will it be worth in 30 years?
4. A car was bought for P23,000. Its value is expected to depreciate by 2.5% annually. How much would the car cost after 3 years?
5. You buy a car through a P700,000 loan at 6% annual interest for 5 years. Meanwhile, the car depreciates by 10% each year. What is the car’s value after 5 years? How much will you have paid in total for the loan by then?
1. Problem: Calculate the total amount in the collection after the 35th person contributes P5000 each, starting with P10,000.
Formula: Total amount = Initial amount + (Number of people × amount per person)
Calculation:
$$\text{Total} = 10000 + 35 \times 5000$$
$$= 10000 + 175000 = 185000$$
Answer: The collection has P185000 after the 35th person contributes.
2. Problem: Find the 10th investor's investment and total investment when each investor invests P500,000 more than the previous, starting at P1,000,000.
Formula for nth term of arithmetic sequence:
$$a_n = a_1 + (n-1)d$$
Sum of first n terms:
$$S_n = \frac{n}{2} (a_1 + a_n)$$
Calculation:
$$a_{10} = 1000000 + (10-1) \times 500000 = 1000000 + 4500000 = 5500000$$
$$S_{10} = \frac{10}{2} (1000000 + 5500000) = 5 \times 6500000 = 32500000$$
Answer: The 10th investor invests P5500000, total investment is P32500000.
3. Problem: Calculate the value of P4000 after 30 years at 5% interest compounded yearly.
Formula for compound interest:
$$A = P(1 + r)^n$$
Where $P=4000$, $r=0.05$, $n=30$
Calculation:
$$A = 4000 (1 + 0.05)^{30} = 4000 (1.05)^{30}$$
Using approximation:
$$A \approx 4000 \times 4.321942 = 17287.77$$
Answer: The account will be worth approximately P17287.77 after 30 years.
4. Problem: Find the car's value after 3 years with 2.5% annual depreciation from P23000.
Formula for depreciation:
$$V = P(1 - r)^n$$
Where $P=23000$, $r=0.025$, $n=3$
Calculation:
$$V = 23000 (1 - 0.025)^3 = 23000 (0.975)^3$$
$$V = 23000 \times 0.9259 = 21295.7$$
Answer: The car will cost approximately P21295.70 after 3 years.
5. Problem: Calculate the car's value after 5 years with 10% annual depreciation and total loan payment on P700000 at 6% annual interest for 5 years.
Car value depreciation:
$$V = 700000 (1 - 0.10)^5 = 700000 (0.90)^5$$
$$V = 700000 \times 0.59049 = 413343$$
Loan total payment (simple interest assumed):
$$\text{Total paid} = P + P \times r \times t = 700000 + 700000 \times 0.06 \times 5$$
$$= 700000 + 210000 = 910000$$
Answer: Car value after 5 years is approximately P413343. Total loan paid is P910000.