Nico is saving money for his college education. He invests some money at 6​%, and ​$900 less than that amount at 4 %. The investments produced a total of ​$154 interest in 1 yr. How much did he invest at each​ rate?

1 answer

Let \( x \) be the amount of money Nico invests at 6%. Then, the amount he invests at 4% would be \( x - 900 \).

The interest earned from the 6% investment in one year is:

\[ 0.06x \]

The interest earned from the 4% investment in one year is:

\[ 0.04(x - 900) \]

According to the problem, the total interest from both investments is $154. Therefore, we can set up the following equation:

\[ 0.06x + 0.04(x - 900) = 154 \]

Now, let's distribute and simplify the equation:

\[ 0.06x + 0.04x - 0.04 \times 900 = 154 \]

Calculating \( 0.04 \times 900 \):

\[ 0.04 \times 900 = 36 \]

So, the equation becomes:

\[ 0.06x + 0.04x - 36 = 154 \]

Combining like terms:

\[ 0.1x - 36 = 154 \]

Next, add 36 to both sides of the equation:

\[ 0.1x = 154 + 36 \]

\[ 0.1x = 190 \]

Now divide by 0.1:

\[ x = \frac{190}{0.1} = 1900 \]

So, Nico invests \( x = 1900 \) at 6%. Now we will find the amount he invests at 4%:

\[ x - 900 = 1900 - 900 = 1000 \]

Thus, Nico invests:

  • $1900 at 6%
  • $1000 at 4%

To verify, we can calculate the interest from both investments:

Interest from 6% investment:

\[ 0.06 \times 1900 = 114 \]

Interest from 4% investment:

\[ 0.04 \times 1000 = 40 \]

Total interest:

\[ 114 + 40 = 154 \]

The calculations check out, confirming the amounts invested:

  • $1900 at 6%
  • $1000 at 4%