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Anna35
18 hours ago
6

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 inte

rest in one year

Mathematics
2 answers:
lawyer [4K]18 hours ago
5 0

To tackle this issue, I recommend creating a chart.

The chart is illustrated below.

We will base this chart on the interest formula provided below.

Amount Invested · Rate = Interest Accrued

On the left side of the chart, there will be two investment types made by Sam, specifically the 6% and 10% investments.

While we know Sam's total investment is

$48,000, the individual amounts at each interest rate remain unknown.

Thus, we will use x and 48,000 - x to symbolize the respective amounts he invested at the two rates.

The rate column will convert our percentages into decimal form: 6% becomes 0.06 and 10% becomes 0.10.

This translates 6% into 6/100 or 0.06.

Now utilizing our formula, we can compute the products of the first two columns to fill in the third column.

This means we have x times 0.06 equal to 0.06x and also 48,000 - x times 0.10, resulting in 0.10(48000 - x).

Referring to the last segment of the problem, we know Sam generated $4,000 in interest for the year.

By adding the interest amounts he earned from both rates, we equate that to 4000.

The equation can be structured as 0.06x + 0.10(48000 - x) = 4000.

We can eliminate the decimals by multiplying every part of the equation by 100.

This results in 6x + 10(48000 - x) = 400000.

We ultimately determine that x = 20000.

Therefore, Sam invested $20000 at an interest rate of 6%.

The investment amount at 10% is 48000 - x, resulting in 48000 - 20000, or $28,000.

Zina [3.9K]18 hours ago
3 0

Step-by-step explanation of the answer:

Sam allocated $48,000 with 6% interest and the remainder at 10%. He earned a total of $4,000 in interest over the course of one year, and now we need to determine the amount invested at each rate.

Let us denote the amount invested at 6% as x.

(\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

This results in the 6% investment being $20,000 and the 10% investment totaling = (48000 - 20000) = $28,000

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