Math Problem Statement

  1. Jimmy invested P55,490 where after 36 months it reached to P68,211.38 compounded semiannually. Find the interest rate?

  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?

Solution

Let's solve the problems step-by-step:


1. Find the interest rate when compounded semiannually

The formula for compound interest is:
A=P(1+rn)ntA = P \left( 1 + \frac{r}{n} \right)^{n \cdot t}

Where:

  • A=68,211.38A = 68,211.38 (future value)
  • P=55,490P = 55,490 (initial principal)
  • t=36months=3yearst = 36 \, \text{months} = 3 \, \text{years}
  • n=2n = 2 (compounding semiannually per year)
  • r=interest rate (unknown)r = \text{interest rate (unknown)}

Substituting values into the formula:
68,211.38=55,490(1+r2)2368,211.38 = 55,490 \left( 1 + \frac{r}{2} \right)^{2 \cdot 3}

Simplify:
68,211.3855,490=(1+r2)6\frac{68,211.38}{55,490} = \left( 1 + \frac{r}{2} \right)^6

1.22925=(1+r2)61.22925 = \left( 1 + \frac{r}{2} \right)^6

Take the 6th root (or raise to the power of 16\frac{1}{6}):
1+r2=1.22925161 + \frac{r}{2} = 1.22925^{\frac{1}{6}}

1+r2=1.03431 + \frac{r}{2} = 1.0343

Solve for rr:
r2=1.03431=0.0343\frac{r}{2} = 1.0343 - 1 = 0.0343

r=0.0343×2=0.0686or6.86%r = 0.0343 \times 2 = 0.0686 \, \text{or} \, 6.86\%

The interest rate is approximately 6.86% per year.


2. Find the time required when compounded quarterly

The formula remains:
A=P(1+rn)ntA = P \left( 1 + \frac{r}{n} \right)^{n \cdot t}

Where:

  • A=50,692.17A = 50,692.17
  • P=45,000P = 45,000
  • r = 0.06 \, \text{(6% annual interest rate)}
  • n=4(quarterly)n = 4 \, \text{(quarterly)}
  • t=time in years (unknown)t = \text{time in years (unknown)}

Substitute the known values:
50,692.17=45,000(1+0.064)4t50,692.17 = 45,000 \left( 1 + \frac{0.06}{4} \right)^{4 \cdot t}

Simplify:
50,692.1745,000=(1+0.015)4t\frac{50,692.17}{45,000} = \left( 1 + 0.015 \right)^{4 \cdot t}

1.12649=(1.015)4t1.12649 = (1.015)^{4 \cdot t}

Take the natural logarithm on both sides:
ln(1.12649)=ln((1.015)4t)\ln(1.12649) = \ln((1.015)^{4 \cdot t})

Apply the logarithmic rule ln(ab)=bln(a)\ln(a^b) = b \ln(a):
ln(1.12649)=(4t)ln(1.015)\ln(1.12649) = (4 \cdot t) \ln(1.015)

t=ln(1.12649)4ln(1.015)t = \frac{\ln(1.12649)}{4 \cdot \ln(1.015)}

Solve:
t=0.119140.0149=0.11910.0596=2t = \frac{0.1191}{4 \cdot 0.0149} = \frac{0.1191}{0.0596} = 2

It will take 2 years for the amount to grow from $45,000 to $50,692.17.


Do you want further details or have any questions?

Here are 5 related questions to expand your understanding:

  1. What happens to the time if the interest rate increases in the second problem?
  2. How does the compounding frequency affect the interest rate in the first problem?
  3. How is semiannual compounding different from quarterly compounding in terms of formula adjustments?
  4. How would the answers change if tt was given in months instead of years?
  5. Can you derive a similar formula for simple interest instead of compound interest?

Tip: Always double-check units (years, months, etc.) to ensure the formula is applied correctly!

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth
Logarithms

Formulas

A = P (1 + r/n)^(nt)
ln(A/P) = nt * ln(1 + r/n)

Theorems

Logarithmic properties
Compound interest formula

Suitable Grade Level

Grades 10-12