Need help with some math word problems!

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<p>Hey, sorry these aren’t actually SAT questions but this place is usually pretty good with math.</p>

<li>Sally invested $12,000, a portion earning a simple interest rate of 4.5% per year and the rest earning a rate of 4% per year. After one year the total interest earned on these investments was $525. How much money did she invest at EACH rate?</li>
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<li>What annual rate of interest would you have to earn on an investment of $3500 to ensure receiving $262.50 interest after one year?</li>
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<p>I don’t need the answers really, I have them already. I would reall appreciate if someone could write out the work and explain to me how to go about solving them. Thanks!</p>

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<p>The equation is I (total interest) = amount invested x (rate #1) + amount invested x (rate #2).</p>

<p>You're given the two rates and the amount of money you started with, and I. Convert from percentages to decimals. Now you have 525 = .045x + .04y, where x and y are the amount of money he invested at each rate. But think, if he invested x dollars from the 12,000 he originally had, what does y equal? It equals (12,000-x), since x+y=12,000.</p>

<p>So now you have:
525 = .045x + .04(12,000-x)
525 = .045x + 480 - .04x
45 = .005x
x = 9000.
x + y = 12000
9000 + y =12000
y = 3000.</p>

<p>So he invested $9,000 at 4.5% and $3,000 at 4%.</p>

<p>The second one is easier.
Use the equation T (total amount, initial + interest) = I (initial) x (1 + r) to the t power.</p>

<p>You're given the initial amount, I, equals $3,500, the amount of interest, $262.5, and the amount of time, 1 year. You also know T, since T equals the sum of the interest and the initial amount, so T equals $3,762.50.</p>

<p>So now you have:
3762.5 = 3500(1+r)^1
3762.5 = 3500(1+r)
3762.5 = 3500 + 3500r
262.5 = 3500r
.075 = r.
Change the answer to a percentage to get 7.5%.</p>

<p>EDIT: Lol, I just realized there was an even easier way of doing this.
Interest = Amount x rate.
262.5 = 3500r
r=.075, or 7.5%</p>

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<p>This type of problem will not be tested.</p>