A loan is to be amortized every quarter by payments of P2,500.00 for 6 years. Money is worth 0.5% converted quarterly.14. How much is the original loan? Group of answer choicesP 59,072.57P 18,815.95P 68,746.96P 51,560.59
Question
A loan is to be amortized every quarter by payments of P2,500.00 for 6 years. Money is worth 0.5% converted quarterly.14. How much is the original loan? Group of answer choicesP 59,072.57P 18,815.95P 68,746.96P 51,560.59
Solution 1
To find the original loan amount, we can use the formula for the present value of an ordinary annuity. The formula is:
P = PMT * [(1 - (1 + r)^-n) / r]
where: P = original loan amount PMT = periodic payment amount r = interest rate per period n = total number of payments
Given in the problem: PMT = P2,500.00 r = 0.5% per quarter = 0.005 n = 6 years * 4 quarters/year = 24 quarters
Substituting these values into the formula:
P = 2500 * [(1 - (1 + 0.005)^-24) / 0.005]
Calculating the above expression will give us the original loan amount.
Let's calculate it:
P = 2500 * [(1 - (1.005)^-24) / 0.005] P = 2500 * [(1 - 0.8899964482) / 0.005] P = 2500 * [0.1100035518 / 0.005] P = 2500 * 22.00071036 P = P55,001.7759
So, the original loan amount is approximately P55,001.7759. However, this value is not in the given options. There might be a mistake in the problem or in the given options.
Solution 2
To find the original loan amount, we can use the formula for the present value of an ordinary annuity. The formula is:
P = PMT * [(1 - (1 + r)^-n) / r]
where: PMT = periodic payment = P2,500 r = interest rate per period = 0.5% or 0.005 n = number of periods = 6 years * 4 quarters/year = 24 quarters
Substituting the given values into the formula, we get:
P = 2500 * [(1 - (1 + 0.005)^-24) / 0.005]
Solving this equation will give us the original loan amount.
Let's calculate it:
P = 2500 * [(1 - (1 + 0.005)^-24) / 0.005] P = 2500 * [(1 - 0.8899964482) / 0.005] P = 2500 * [0.1100035518 / 0.005] P = 2500 * 22.00071036 P = 55001.7759
So, the original loan amount is approximately P 55,001.78. However, this value is not in the provided options. There might be a mistake in the problem or the provided options.
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