Question

Sam invested $48,000 some at 6% interest and the rest at 10% . How much did he invest at each rate if he received $4,000 in interest in one year

Answer 1

I would start this problem by setting up a chart.

Chart is shown below.

This chart will be based on the interest formula which is below.

Amount Invested · Rate = Interest Earned

Down the left side of the chart, we have the

two different types of investments that Sam

makes, the 6% investment and the 10% investment.

For the amount invested column, we know that Sam invested

$48,000 but we don't know how much Sam invested at each rate.

So we use x and 48,000 - x to represent the

amount Sam invested at each rate.

For the rate column, we translate our two percentages

to decimal form so 6% becomes .06 and 10% becomes .10.

Remember that 6% means 6/100 or .06.

Based on our formula, we can multiply the first two

columns in our chart together to fill out the third column.

So we have x times .14 or which is .14x and

48000 - x times .10 or .10(48000 - x).

Reading through the last part of the problem, we know that Sam

earned a total of $4,000 in interest for the year.

So if we add up the two amounts of interest that Sam

earned at each rate, we will get a total of 4000.

Setting up our equation, we have .06 +.10(48000 - x) = 4000.

Solving from here, we first get rid of the decimals by

multiplying each side of the equation by 100.

This gives us 6x + 10(48000 - x) = 400000.

Eventually we find that x = 20000.

So Sam invested $20000 at 6% interest.

The amount he invested at 10% is 48000 - x

which is 48000 - 20000 or $28,000.

Answer 2

Answer with step-by-step explanation:

We are given that Sam invested $48,000 at 6% interest and the rest at 10%. Given that he received $4,000 interest in one year, we are to find how much he invest at each rate.

Assuming the amount invested at 6% interest rate as $x$, we can write it as:

$(\frac{6}{100} \times x)+(\frac{10}{100} \times [48000-x])=4000$

$0.06x+0.1(48000-x)=4000$

$0.06x+4800-0.1x=4000$

$0.1x-0.06x=4800-4000$

$0.04x=800$

$x=20000$

So the amount invested at 6% = $20,000 and amount invested at 10% = (48000 - 20000) = $28,000