sherry invested a sum of money at 6% and invested a second sum that was $1,500 greater than the first sum, at 5%. if the total annual income was $570, how much was invested at each rate? Thank you for your assistance.
.06 x + .05 (x + 1500) = 570
x is the 1st sum ... x + 1500 is the 2nd
The answer is 4500 and 6000. I can't make this work. 6x+5(x+1500) 570
11x + 7500 = 5700 - 7500= 1800. 11x won't divide equally in to that. Any other suggestions? I tried to solve it initially by the same method you show, but couldn't get to work. What am I missing? Thanks again.
.06 x + .05 x + 75 = 570 ... .11 x = 495 ... x = 4500
don't like decimals??
6x + 5(x + 1500) = 57000 ... eliminating decimals means adjusting factors
11x + 7500 = 57000 ... 11x = 49500 ... x = 4500
yea! thank you.
To find out how much money was invested at each rate, we can set up a system of equations based on the given information.
Let's assume that Sherry invested an amount x at 6% interest rate. According to the problem, the second sum, which is invested at 5% interest rate, is $1,500 greater than the first sum. So, the second sum can be expressed as (x + $1,500).
Now, let's calculate the annual income from each investment. The income from the first investment can be calculated using the formula: income = principal * rate. So, the income from the first investment, at 6% interest rate, is 0.06x.
Similarly, the income from the second investment, at 5% interest rate, is 0.05(x + $1,500).
According to the problem, the total annual income from both investments is $570. So, we can set up the following equation:
0.06x + 0.05(x + $1,500) = $570
Now, let's solve this equation to find the value of x, which represents the amount invested at 6% interest rate.
0.06x + 0.05x + $75 = $570
0.11x = $495
x = $495 / 0.11
x = $4,500
Therefore, Sherry invested $4,500 at 6% interest rate. The second sum, which is $1,500 greater than the first sum, can be calculated as $4,500 + $1,500 = $6,000.
So, Sherry invested $4,500 at 6% interest rate and $6,000 at 5% interest rate.