Question

Question #758c8

Answer 1

`750`

Explanation:

Let:

`x` be the number of euros invested in the `10%` certificate of deposit

and

`y` be the number of euros invested in the `6%` passbook account.

So now we can write a system of equations.

We know the total amount invested is `3000` so:

`x+y=3000`

The total annual interest on both accounts is `270` euros translates to:

`x(0.10)+y(0.06)=270`

Now, we just need to solve the system of equations:

`x(0.10)+y(0.06)=270`
`x+y=3000`

We could use elimination. Let's multiply the bottom equation by `0.10` and then subtract:

0.10x+0.06y=270
0.10x+0.10y=300

We get:

`-0.04y=-30`
`y=750`

So, the answer is 750 euros was invested at `6%`

We don't even need to solve for `x`.

Answer 2

Assumption: Combined interest is 270.
The wording in the question is a bit ambiguous.

Invested 750 at 6%

Explanation:

If you are able to combine the given information in such a way that you only have 1 unknown then you are 'home and dry'.

Let the sum invested at 10% be `x`
Then the sum invested at 6% is `(3000-x)`

Given that the total interest is 270 we have:

`10%x+6%(3000-x)=270`

`10%x-6%x+6%3000=270`

`4/100x+(6/(1cancel(00))xx30cancel(00))=270`

`4/100x+180=270`

`x=(270-180)xx100/4`

`x=2250` invested at `10%`

Thus amount invested at 6% is

`3000`
`ul(2250) larr" Subtract"`
`color(white)(0)750`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Check")`

`6%750+10%2250`
`color(white)("dd")45color(white)("dd")+color(white)("dd")225color(white)("dddd")=270` as required