Question

Una parte de 25 000 dólares se invierte al 10% de interés, otra parte al 12 % y el resto al 16%. El ingreso anual total de las tres inversiones es de 3200 dólares. Además, el ingreso de la inversión al 16% es igual al ingreso de las otras dos inversiones combinadas. ¿Cuánto se invirtió a cada tasa de interés?

Answer 1

• $5000 at 10%, $10000 at 12% and 10000 at 16%

Explanation:

• One part of $ 25,000 is invested at 10% interest, another part at 12%, and the rest at 16%. The total annual income from the three investments is $ 3,200. Also, the income from the investment at 16% is equal to the income from the other two investments combined. How much was invested at each interest rate?

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Let the parts be x, y and z

As per given we get below system of equations:

• x + y + z = 25000
• 0.1x + 0.12y + 0.16z = 3200
• 0.1x + 0.2y = 0.16z

Substitute 0.1x + 0.2y in the second equation:

• 0.16z + 0.16z = 3200
• 0.32z = 3200
• z = 3200/0.32
• z = 10000

Now we have:

• x + y + 10000 = 25000 ⇒ x + y = 15000

and

• 0.1x + 0.12y + 0.16*10000 = 3200 ⇒ 0.1x + 0.12y = 1600

Multiply the second equation and then subtract the first one:

• 10(0.1x + 0.12y) = 10(1600) ⇒ x + 1.2y = 16000
• x + 1.2y - (x + y) = 16000 - 15000
• 0.2y = 1000
• y = 10000

Then

• x = 15000 - 10000 = 5000

So the parts are:

• $5000 at 10%, $10000 at 12% and 10000 at 16%