You inherit one hundred thousand dollars. You invest it all in three accounts for one year. The first account pays 4% compounded
annually, the second account pays 3% compounded annually, and the third account pays 2% compounded annually. After one year, you earn $3,650 in interest. If you invest five times the money in the account that pays 4% compared to 3%, how much did you invest in the 4% account?
Let x = amount invested in the 1st account y = amount invested in the 2nd account z = amount invested in the 3rd account
Because the total investment is $100,000, therefore x + y + z = 100,000 (1) Interest earned in one year from the accounts is $3,650, therefore 0.04x + 0.03y + 0.02z = 3,650 or 4x + 3y + 2z = 365,000 (2)
Because x = 5y, therefore obtain these 2 equations: 5y +y +z = 100,000 or 6y + z = 100,000 (3) 4*(5y) +3y + 2z = 365,000 or 23y + 2z = 365,000 (4)
Substitute z=1000,000 - 6y from (3) into (4). 23y + 2(100,000 - 6y) = 365,000 23y + 200,000 - 12y = 365,000 11y = 165,000 y = $15,000 Therefore x = 5y = $75,000 z = 100,000 - 6y = $10,000
Answer: The amounts invested are 1st account: $75,000 2nd account: $15,000 3rd account: $10,000
Basically just algebra, just find out the cost for one water bottle, then find out the cost for ten. If 24 water bottles is 6 dollars, then they got four bottles of water for one dollar, meaning a bottle of water is 25 cents, or $0.25. Ten water bottles would be two dollars and fifty cents.
Y = n y = x^2 + n n > 0 since there are 2 equations equal to y, set them equal to each other and solve. n = x^2 + n Subtract n from both sides. 0 = x*2 Take the square root of both sides. x = 0
