Math Problem Statement

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  1. Jimmy invested P55,490 where after 36 months it reached to P68,211.38 compounded semiannually. Find the interest rate? r=n[(A/P) ^1/nt -1]
  2. How long will it take for Rina's borrowed amount of $45,000 reach to $50,692.17 with an interest rate of 6% compounded quarterly? t= log(A/P) / (n) log(1+r/n)

Solution

Let's break down each problem using the formulas provided.


Problem 1: Finding the Interest Rate

The formula for compound interest, as given, is:

r=n[(AP)1nt1]r = n \left[ \left( \frac{A}{P} \right)^{\frac{1}{nt}} - 1 \right]

Where:

  • A=68,211.38A = 68,211.38 (final amount)
  • P=55,490P = 55,490 (initial investment)
  • n=2n = 2 (since the compounding is semiannual, so twice per year)
  • t=36t = 36 months = 3 years (time in years)

Now, substitute the values into the formula:

r=2[(68,211.3855,490)12×31]r = 2 \left[ \left( \frac{68,211.38}{55,490} \right)^{\frac{1}{2 \times 3}} - 1 \right]

First, calculate AP\frac{A}{P}:

68,211.3855,4901.2296\frac{68,211.38}{55,490} \approx 1.2296

Now, raise this value to the power of 16\frac{1}{6} (since n×t=6n \times t = 6):

1.2296161.03551.2296^{\frac{1}{6}} \approx 1.0355

Now, subtract 1:

1.03551=0.03551.0355 - 1 = 0.0355

Finally, multiply by 2 (since n=2n = 2):

r=2×0.03550.0710r = 2 \times 0.0355 \approx 0.0710

So, the interest rate is approximately 7.10% per year.


Problem 2: Finding Time for Investment to Grow

The formula for compound interest time is:

t=log(A/P)nlog(1+rn)t = \frac{\log(A / P)}{n \log(1 + \frac{r}{n})}

Where:

  • A=50,692.17A = 50,692.17 (final amount)
  • P=45,000P = 45,000 (borrowed amount)
  • r=6%=0.06r = 6\% = 0.06 (annual interest rate)
  • n=4n = 4 (since the interest is compounded quarterly)

Now, substitute the values into the formula:

t=log(50,692.17/45,000)4log(1+0.064)t = \frac{\log(50,692.17 / 45,000)}{4 \log(1 + \frac{0.06}{4})}

First, calculate AP\frac{A}{P}:

50,692.1745,0001.1276\frac{50,692.17}{45,000} \approx 1.1276

Now, take the logarithm of this value:

log(1.1276)0.0512\log(1.1276) \approx 0.0512

Next, calculate (1+0.064)=1+0.015=1.015\left(1 + \frac{0.06}{4}\right) = 1 + 0.015 = 1.015, and then take the logarithm:

log(1.015)0.00647\log(1.015) \approx 0.00647

Now substitute these into the equation:

t=0.05124×0.00647=0.05120.025881.98 yearst = \frac{0.0512}{4 \times 0.00647} = \frac{0.0512}{0.02588} \approx 1.98 \text{ years}

So, it will take approximately 1.98 years or about 2 years for the amount to grow.


Summary of Answers:

  1. Interest Rate = 7.10% per year.
  2. Time to Reach Final Amount = 2 years.

Would you like a detailed explanation of any step or additional assistance with similar problems?


Follow-up Questions:

  1. How do you calculate compound interest when interest is compounded monthly?
  2. What is the difference between compound interest and simple interest?
  3. Can you explain how to adjust the formula when compounding occurs annually?
  4. How can you calculate the future value of an investment using continuous compounding?
  5. What happens to the interest rate if the compounding frequency increases?

Tip: When solving compound interest problems, always ensure that the time and interest rate are in the correct units relative to the compounding frequency.

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth
Logarithms

Formulas

r = n[(A/P)^(1/nt) - 1]
t = log(A/P) / [n log(1 + r/n)]

Theorems

Compound Interest Formula
Logarithmic Properties

Suitable Grade Level

Grades 10-12