# Solving Investment Problems

**Solving Investment Problems**

**One of the most important applications of linear equations is found in solving investment problems. Investment problems use the Simple Interest formula I = Prt, where **

**P = Principal (Amount invested )**

**r = rate of simple interest**

**t = time in years**

**In this hub I present several problems with their solutions.**

**Problem Number One :**

**A man invested 35,000 pesos partly at 15% and the rest at 18%. The total income from these investments is 5,650 pesos. Find the sum invested at each rate.**

**Solution :**

**Let X = Amount invested at 15%**

**35,000 – X = Amount invested at 18%**

**Since Interest = Prt therefore**

**.15X = Interest earned at amount invested at 15%**

**.18(35,000 – X )= Interest earned at amount invested at 18%**

**Interest earned at 15% + Interest earned at 18% = 5,650**

**Working Equation :**

**.15X + .18(35,000 – X) = 5,650**

**Multiplying this equation by 100**

**15X + 18(35,000 – X ) = 565,000 **

**15X + 630,000 - 18X = 565,000**

**-3X = 565,000 – 630,000**

**(-3X = -65,000 ) -1/3**

**X = 21,667.67 ==****è Amount invested at 15%**

**35,000 – 21,666.67 = 13,3333.33 =****è Amount Invested at 18%**

**Problem Number Two :**

**A church congregation has 20,000 pesos to be invested in a fund , part at 3% and part at 7%. If the investment at 7% earns 500 pesos more per year than the other placement, how much is invested at each rate ?**

**Solution :**

**Let X = Amount invested at 3%**

**20,000 – X = Amount invested at 7%**

**Interest earned at 3% = .03X**

**Interest earned at 7% = .07( 20,000 – X )**

**Interest at 3% + 500 = Interest at 7%**

**Working equation :**

**.03X + 500 = .07(20,000 – X )**

**Multiplying this equation by 100**

**3X + 50,000 =7(20,000 –X )**

**3X + 50,000 = 140,000 – 7X**

**3X + 7X = 140,000 – 50,000**

**(10X = 90,000) 1/10**

**X = 9,000 =****è Amount invested at 3%**

**20,000 – 9,000 = 11,000 =****è Amount invested at 7%**

**Problem Number Three**

**Mr. Alfonso invested a sum of money at 4% simple interest and another sum which is 100,000 more than the first sum at 5%. The annual income from the two investments is 17,000 pesos. How much was Mr. Alfonso’s investment at each rate ?**

**Solution :**

**Let X = Amount invested at 4%**

**X + 100,000 = Amount invested at 5%**

**Interest at 4% = .04X**

**Interest at 5% = .05 (X + 100,000)**

**Interest at 4% + Interest at 5% = 17,000**

**Working equation :**

**.04X + .05(X + 100,000) = 17,000**

**Multiplying this equation by 100**

**4X + 5 ( X + 100,000) = 1,700,000**

**4X + 5X + 500,000 = 1,700,000**

**9X = 1,700,000 – 500,000**

**(9X = 1,200,000 ) 1/9**

**X = 133,333.33 =****è Amount invested at 4%**

**X + 100,000 = 133,333.33 + 100,000 = 233,333.33 ==****èAmount invested at 5%**

## More by this Author

- 6
SolvingWord Problems Involving Exponential Function Exponential function is one of the most important concept in Algebra. In this hub I present several problems involving exponential function with their...

- 4
Solving Geometry Problems Involving System of Equations In Two Variables Algebra has also found wide applications in Geometry. Among its common problems are problems involving perimeters and angles. In this hub I...

- 21
Solving Problems Involving Angular Velocity Among the challenging problems I encounter in Trigonometry are problems involving angular velocity. In this hub, I presented several problems...

## Comments 5 comments

sorry i will read this next timebut i will bookmark this

:) this helped me a lot! :)

it is very great aim crazy for this solving good luck

it so very great iam so crazy of this solving good luck

u can try this solving

5