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Mellyy

Guest

Solve using algebra

An investor has $50000 to invest in one year. She decides to put some in a secure deposit account and the rest in a more risky investment.
At the end of the year the deposit account pays interest of 8%, the more risky pays 18% and the investor receives a total of $5600 in interest

How much did she put in each account
Please show workings

=]

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Solve using algebra

This calls for a system of equations.

So this investor has $50,000 to invest in two different ways. The regular one and the more risky.

let x = regular invesment
let y = risky investment.

You know the total of both investments is $50,000 so:

x + y = 50000

You also know how much each investment yields.

0.08x + 0.18y = 5600

That means that the 8% yield on the regular investment plus the 18% yield on the risky investment is equal to $5600. This was given to us.

So now you have 2 equations with x and y. So now just solve both variables.

0.08x + 0.18y = 5600
x + y = 50000

y = 50000 - x

0.08x + 0.18(50000 - x) = 5600
0.08 + 9000 - 0.18x = 5600
-0.1x = -3400
x = 34000

y = 50000 - x
y = 50000 - 34000
y = 16000

So now you know that this investor invested $34000 in the regular investment and $16000 in the more risky investment.

Hope I helped.

Last edited by LuisRodg (2008-03-09 23:14:54)

Melllyy

Guest

Re: Solve using algebra

Thank you!