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Yuliya22 [10]
3 years ago
9

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:
wariber [46]3 years ago
3 0
To answer we let x be the amount of money that Sam invested during the first year. 

Below are the expressions translated from the given word forms for the amount invested.

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

The sum of money invested by Sam is:
  x + (5x/2 - 2000) + (x/5 + 1000)

Similarly, we derive the expressions that we use for the amount that Sally invested.
Sally
1st year  :    amount  = 3x/2 - 1000
2nd year :    amount = 2x - 1500
3rd year :     amount = x/4 + 1400

The total amount that Sally invested is, 
        total = (3x/2 - 1000) + (2x - 1500) + (x/4 + 1400)

Equating the two equations:
 (x) + (5x/2 - 2000) + (x/5 + 1000) = (3x/2 - 1000) + (2x - 1500) + (x/4 + 1400)

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

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