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A college student Earned $6500 during summer vacation working as a waiter. The student invested Part of the money at 10% and the rest at 9%. If the student received a total of $607 in interest at the end of the year, how much was invested at 10%?

Respuesta :

Answer:

The amount invested at 10% rate is $ 2200

Step-by-step explanation:

Given as :

The total amount earn by student = $6500

The total interest received at the end of year = $607

Let the amount invested at 10% rate = $x

So, The amount invested at 9% rate = ($6500 - $x)

Let the time for which money invested = 1 year

Now from Simple Interest method :

SI = [tex]\frac{principal\times Rate\times Time}{100}[/tex]

Or, SI_1 + SI_2 = [tex]\frac{x\times 10\times 1}{100}[/tex] + [tex]\frac{(6500-x)l\times 9\times 1}{100}[/tex]

Or, SI_1 + SI_2 = [tex](\frac{10x}{100})+\frac{(6500-x)\times 9}{100}[/tex]

Or, $607 × 100 = 10x + (6500-x) × 9

Or, $60700 = 10x- 9x + 58500

Or, $60700 - $58500 = x

Or , 2200 = x

∴ x = 2200

So, The amount invested at 10% rate = $x = $ 2200

Hence The amount invested at 10% rate is $ 2200   Answer

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