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$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 $10\%$?
$\endgroup$ 1 $\begingroup$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?
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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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