Math Problem Statement

Simple interest problems covering calculations involving rate, time, principal, and interest.

Solution

Let’s solve each question in the image step-by-step.

1. In what time will ₱400,000.00 amount to ₱512,000.00 if the simple interest is at 14%?

Using the formula for simple interest: A=P(1+rt)A = P(1 + rt) where:

  • A=512,000A = 512,000 (final amount)
  • P=400,000P = 400,000 (principal)
  • r=0.14r = 0.14 (interest rate)
  • t=?t = ?

Rearranging for tt: t=APPr=512,000400,000400,000×0.14t = \frac{A - P}{P \cdot r} = \frac{512,000 - 400,000}{400,000 \times 0.14}

Calculating: t=112,00056,000=2 yearst = \frac{112,000}{56,000} = 2 \text{ years} So, t=2t = 2.


2. At what rate of simple interest will ₱50,000 amount to ₱60,500 in 3 years, 4 months?

Convert 4 months to years: 412=0.3333\frac{4}{12} = 0.3333 years. Total time, t=3+0.3333=3.3333t = 3 + 0.3333 = 3.3333 years.

Using the formula: A=P(1+rt)A = P(1 + rt) where:

  • A=60,500A = 60,500
  • P=50,000P = 50,000
  • t=3.3333t = 3.3333 years
  • r=?r = ?

Rearranging for rr: r=APPt=60,50050,00050,000×3.3333r = \frac{A - P}{P \cdot t} = \frac{60,500 - 50,000}{50,000 \times 3.3333}

Calculating: r=10,500166,6650.063=6.3%r = \frac{10,500}{166,665} \approx 0.063 = 6.3\%


3. If Richard invests ₱200,000 in the bank at a rate of 5.7% for 6 years, how much interest will he make?

Using the formula for simple interest: I=PrtI = P \cdot r \cdot t where:

  • P=200,000P = 200,000
  • r=0.057r = 0.057
  • t=6t = 6

Calculating: I=200,000×0.057×6=68,400I = 200,000 \times 0.057 \times 6 = 68,400 So, the interest I=68,400I = 68,400.


4. Clinton earned ₱1,155 in interest from his simple interest account over the last 5 years. The account pays 1.25% interest annually. How much did Clinton originally deposit in his account?

Using the formula: I=PrtI = P \cdot r \cdot t Rearrange to solve for PP: P=IrtP = \frac{I}{r \cdot t} where:

  • I=1,155I = 1,155
  • r=0.0125r = 0.0125
  • t=5t = 5

Calculating: P=1,1550.0125×5=1,1550.0625=18,480P = \frac{1,155}{0.0125 \times 5} = \frac{1,155}{0.0625} = 18,480 So, P=18,480P = 18,480.


5. Cyrus deposits ₱15,000 in a simple interest account that pays 4.3% interest annually. After 15 years, how much total money would be in the account?

Using the formula: A=P(1+rt)A = P(1 + rt) where:

  • P=15,000P = 15,000
  • r=0.043r = 0.043
  • t=15t = 15

Calculating: A=15,000×(1+0.043×15)=15,000×(1+0.645)=15,000×1.645=24,675A = 15,000 \times (1 + 0.043 \times 15) = 15,000 \times (1 + 0.645) = 15,000 \times 1.645 = 24,675 So, the final amount A=24,675A = 24,675.


Would you like more details on any of these solutions, or do you have any questions?


Additional Questions to Expand Understanding:

  1. How is simple interest different from compound interest?
  2. What is the effect of increasing the interest rate on the total interest earned?
  3. Why is it important to convert months to years in these calculations?
  4. How would you calculate the time needed for an amount to double with simple interest?
  5. How can simple interest formulas be applied in real-life financial decisions?

Tip: Remember that simple interest is linear, meaning the interest amount grows steadily over time, unlike compound interest which grows exponentially.

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Math Problem Analysis

Mathematical Concepts

Simple Interest
Percentage
Interest Calculation

Formulas

Simple Interest Formula: I = P * r * t
Amount Formula: A = P * (1 + r * t)

Theorems

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Suitable Grade Level

Grades 9-11