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• 1. Basic Compound Interest a.compound-interest.1

  Find the maturity value of ₱15,000 invested at 5% per year compounded annually for 3 years.

  Your answer (₱)

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  Answer: 17364.38 ₱

  1. $A = P(1+r)^t = 15{,}000(1.05)^3$.
  2. $(1.05)^3 = 1.157625$.
  3. $A = 15{,}000 \times 1.157625 = 17{,}364.38$ pesos.
• 2. Standard Compound Interest a.compound-interest.2

  ₱20,000 is invested at 8% per year compounded quarterly for 2 years. Find the maturity value to the nearest centavo.

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  Answer: 23433.19 ₱

  1. Quarterly rate $i = 0.08/4 = 0.02$; number of periods $n = 2 \times 4 = 8$.
  2. $A = 20{,}000(1.02)^8 \approx 20{,}000 \times 1.171659$.
  3. $A \approx 23{,}433.19$ pesos.
• 3. Basic Compound Interest a.compound-interest.3

  Two accounts both advertise 6% per year. Account X compounds annually; Account Y compounds monthly. After one year, which is true?

  Choices

  a. X earns more b. Y earns more c. They earn exactly the same d. It depends on the deposit size

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  Answer: b. Y earns more

  1. Monthly compounding adds interest on interest within the year: $(1+0.06/12)^{12} \approx 1.0617 > 1.06$.
  2. The effective rate of Y is about 6.17%, so Y earns more for any deposit.
• 4. Standard Compound Interest a.compound-interest.4

  Parents want ₱50,000 for enrolment in 5 years. How much must they deposit now at 4% compounded annually? Round to the nearest peso.

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  Answer: 41096 ₱

  1. Present value $P = \dfrac{A}{(1+r)^t} = \dfrac{50{,}000}{(1.04)^5}$.
  2. $(1.04)^5 \approx 1.216653$.
  3. $P \approx 41{,}096$ pesos.
• 5. Challenge Non-routine Compound Interest a.compound-interest.5

  An investment grew from ₱10,000 to ₱12,000 in 3 years with annual compounding. Find the annual rate, in percent, to two decimal places.

  Your answer (%)

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  Answer: 6.27 %

  1. $12{,}000 = 10{,}000(1+r)^3 \Rightarrow (1+r)^3 = 1.2$.
  2. $1 + r = 1.2^{1/3} \approx 1.0627$.
  3. $r \approx 0.0627 = 6.27\%$.
• 6. Basic Units and Conversion a.farm-data.1

  A farm lot measures 2.5 hectares. Express the area in square metres.

  Your answer (m²)

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  Answer: 25000 m²

  1. $1\text{ ha} = 10{,}000\text{ m}^2$.
  2. $2.5 \times 10{,}000 = 25{,}000\text{ m}^2$.
• 7. Standard Units and Conversion a.farm-data.2

  A farmer harvests 4.2 tonnes of palay per hectare on 1.5 hectares and sells at ₱20 per kilogram. Find the gross revenue.

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  Answer: 126000 ₱

  1. Total yield: $4.2 \times 1.5 = 6.3$ tonnes $= 6{,}300$ kg.
  2. Revenue: $6{,}300 \times 20 = 126{,}000$ pesos.
• 8. Standard Functions and Their Graphs a.farm-data.3

  Producing one hectare of palay costs ₱62,000 and yields 4,000 kg. What farmgate price per kilogram lets the farmer break even?

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  Answer: 15.5 ₱

  1. Break-even when revenue equals cost: $4{,}000p = 62{,}000$.
  2. $p = 62{,}000 \div 4{,}000 = 15.50$ pesos per kilogram.
  3. Any price above ₱15.50 gives profit; the revenue function $R(p)=4000p$ is linear in price.
• 9. Standard Data and Descriptive Statistics a.farm-data.4

  Average palay yields (tonnes per hectare) in five provinces: 4.2, 3.8, 4.5, 3.6, 4.9. Find the mean yield.

  Your answer (t/ha)

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  Answer: 4.2 t/ha

  1. Sum: $4.2+3.8+4.5+3.6+4.9 = 21.0$.
  2. Mean $= 21.0 \div 5 = 4.2$ tonnes per hectare.
• 10. Challenge Non-routine Earning Money and Percentages a.farm-data.5

  Urea fertiliser rose from ₱1,400 to ₱1,750 per bag. A farmer uses 8 bags per hectare on 3 hectares. By how many pesos did the fertiliser cost increase, and what is the percentage increase? Enter the peso increase.

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  Answer: 8400 ₱

  1. Increase per bag: $1{,}750 - 1{,}400 = 350$; percentage $350/1{,}400 = 25\%$.
  2. Bags used: $8 \times 3 = 24$.
  3. Extra cost: $24 \times 350 = 8{,}400$ pesos.