For three consecutive years, Sam invested some money at the start of the year. The first year, he invested x dollars. The second

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1 month ago

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For three consecutive years, Sam invested some money at the start of the year. The first year, he invested x dollars. The second

year, he invested $2,000 less than 5/2 times the amount he invested the first year. The third year, he invested $1,000 more than 1/5 of the amount he invested the first year.
During the same three years, Sally also invested some money at the start of every year. The first year, she invested $1,000 less than 3/2 times the amount Sam invested the first year. The second year, she invested $1,500 less than 2 times the amount Sam invested the first year. The third year, she invested $1,400 more than 1/4 of the amount Sam invested the first year.

If Sam and Sally invested the same total amount at the end of three years, the amount Sam invested the first year is $ and the amount Sally invested the last year is $ .

Mathematics

1 answer:

AnnZ [12.3K] · Answer 1 · 2026-03-31T04:01:26+03:00

Let’s define x as the amount invested by Sam in the first year.
Here are the corresponding expressions derived from the provided descriptions for Sam's investments.

For Sam:
2nd year: investment = 5x/2 - 2000
3rd year: investment = x/5 + 1000

The total Sam invested is:
x + (5x/2 - 2000) + (x/5 + 1000)

Next, we can form the expressions for Sally’s investments.
For Sally
1st year: investment = 3x/2 - 1000
2nd year: investment = 2x - 1500
3rd year: investment = x/4 + 1400

Thus, Sally's total investment is,
total = (3x/2 - 1000) + (2x - 1500) + (x/4 + 1400)

Setting both totals equal gives us:
(x) + (5x/2 - 2000) + (x/5 + 1000) = (3x/2 - 1000) + (2x - 1500) + (x/4 + 1400)

Solving for x,
x = 2000
For Sally's investment for the third year:
investment = x/4 + 1400 = (2000/4 + 1400) = 1900

RESULTS:
Sam's first year = $2000
Sally's third year = $1900