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I've been having trouble with these two questions. The first is simple interest, the second is rate. I'm sure they're easy but I can't focus on getting the solution because I'm terrible at focusing on word problems.

1) Pat invested a total of 3,000 dollars. Part of the money was invested in a money market account that paid 10% simple annual interest, and the remainder of the money was invested in a fund that paid 8% simple annual interest. If the interest earned at the end of the first year from these investments was $256, how much did Pat invest at 10% and how much at 8%?

2) Two cars started from the same point and traveled on a straight course in opposite directions for exactly 2 hours, at which time they were 208 miles apart. If one car traveled, on average, 8 miles per hour faster than the other car, what was the average speed of each car for the 2-hour trip?

Thank you

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    $\begingroup$ 1) let x be the money he invested in 1st money market and (3000-x) the money he invested in the 2nd one. 2) let x be the avg speed of first car and (x+8) the avg speed of 2nd car. Can you continue? $\endgroup$ Dec 3, 2013 at 3:06
  • $\begingroup$ The first isn't compound interest, it's simple interest, you say so yourself. $\endgroup$
    – Tyler
    Dec 3, 2013 at 3:10
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    $\begingroup$ One problem at a time please $\endgroup$
    – user28877
    Dec 3, 2013 at 3:18
  • $\begingroup$ @ThanosDarkadakis I was able to solve the first one (simple interest) quickly with that help.. but for some reason still stuck on the 2nd one. $\endgroup$
    – user3871
    Dec 3, 2013 at 3:32

4 Answers 4

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Hint for 1): If Pat had invested it all at $10\%$, what would it have earned? How much less is earned for each dollar invested at $8\%$ rather than $10\%$?

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  • $\begingroup$ Yeah, I struggle coming up with things like designating x for a value, then the remainder as total - x (in this case, 3000-x). Idk why I struggle to think of that... but I do. $\endgroup$
    – user3871
    Dec 3, 2013 at 3:54
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enter image description here

Car heading west gives $y=x*2$, car heading east gives $(208-y)=(x+8)*2$

NOTE: keep time in hours as speed is given in miles per hours.

substitute $y=2x$ in second equation to get $x$.

Do you find the term "average speed" confusing?

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For part 2): Sometimes it helps to break out the dimensions of the problem when you form your algebraic expression. For example,

$$2 [\operatorname{hours}]\left( x\frac{[\operatorname{miles}]}{[\operatorname{hours}]} +(x+8)\frac{[\operatorname{miles}]}{[\operatorname{hours}]}\right)=208[\operatorname{miles}].$$

This way you can see that you have the right dimensions. After solving, you can of course check to see that your solution makes sense.

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Pat invested a total of 3,000 dollars. Part of the money was invested in a money market account that paid $10\text{%}$ simple annual interest, and the remainder of the money was invested in a fund that paid $8\text{%}$ simple annual interest. If the interest earned at the end of the first year from these investments was 256 dollars, how much did Pat invest at 10% and how much at $8\text{%}$?

Solution: let $X$ be invested amount for $10\text{%}$

      remaining $(3000-X)$ be invested amount for $8\text{%}$.

       which earned him a interest of 256dollars

            therefore simple substitution  

$\frac{(X\times 10\times 1)}{100} + \frac{(3000-X)\times 8\times 1}{100} = 256$

solve for $X$

$X=800\ @ 10\text{%}$ ; $2200\ @ 8\text{%}$

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