SOLUTION: Sam invested $2250, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $30 more than twice the income from the 6% investment. How much
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-> SOLUTION: Sam invested $2250, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $30 more than twice the income from the 6% investment. How much
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Question 1082375: Sam invested $2250, part of it at 6% and the rest at 8% yearly interest. The yearly income on the 8% investment was $30 more than twice the income from the 6% investment. How much did he invest at each rate?
the income from the amount invested at 6% is equal to .06 * x.
the income from the amount invested at 8% is equal to .08 * x.
the income from the amount invested at 8% is 30 more than 2 times the income invested at 6%.
.08 * y = 2 * .06 * x + 30
simplify this to get:
.08 * y = .12 * x + 30
solve for y to get:
y = (.12 * x + 30) / .08
replace y with this in the equation of x + y = 2250 to get:
x + (.12 * x + 30) / .08 = 2250
multiply both sides of this equation by .08 to get:
.08 * x + (.12 * x + 30) = 2250 * .08
simplify to get:
.2 * x + 30 = 180
subtract 30 from both sides of this equation to get:
.2 * x = 150
divide both sides of this equation by .2 to get:
x = 150 / .2 = 750
that's the amount invested at 6%.
the amount invested at 8% is equal to 2250 - 750 = 1500.
6% of 750 = 45
8% of 1500 = 120
the income from the amount invested at 6% is equal to 45.
the income from the amount invested at 8% is equal to 120.
120 is equal to 2 * 45 + 30 which is equal to 90 + 30 which is equal to 120.
the income from the amount invested at 8% is 30 more than 2 times the income from the amount invested at 6%.
solution looks good.
solution is:
amount invested at 6% is equal to 750.
amount invested at 8% is equal to 1500.
