2 word problems

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michaelrlmt

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Any help getting these started would be appreciated, I am stumped and need to turn this in in the morning


How many liters of a 12% acid solution must be mixed with 4L of a 5% acid solution to make it an 8% solution?




AN investment advisor invested $10,000 in 2 accounts. One investment earned 9.8% annual simple interest, and the other investment earned 12.8% annual simple interest. The amount of interest earned in one year was $1085. How much was invested in the 9.8% account?


Thanks
 
The first exercise is a classic "mixture" problem. The second exercise is a classic "interest" problem. Do you have a textbook? There are also thousands of examples to look at on the Internet.

This web site is a cut above the usual answer mills; we like you to make a start.

On the first exercise, please tell me.

How many liters of acid are there in four liters of a 5% acid solution?

On the second exercise, start by assigning a symbol to represent the quantity for which the exericse is asking.

x = the amount of money invested at 9.8%.

Please tell me, what expression using x represents the amount of money invested at 12.8%?

If you answer these two questions of mine, then I'll move on from there based on your answers.

 
michaelrlmt said:
9.8 = x
12.8 = 10000-x
Click to expand...


I'm going forward with the second exercise first.

That's right, but let's write it out without equal signs, lest I get confused over the "equations".

x is the amount invested at 9.8%

10000 - x is the amount invested at 12.8%

Next, tell me how much interest is earned on the x dollars, and also tell me how much interest is earned on the (10000 - x) dollars.

 
lets try again on the first one? let 12% = X 5% = 4 and 8% = 4+X is that right?

so .12X+.2=.32+.o8X wich will = 4

on the second one i tried this
.098x+.128(10000-x)=1085
.098X+1280-.128X=1085
-.03X=-195
X=6500
 


Okay, on the first exercise, we're going to write an equation that equates the amount of acid coming from the two solutions getting mixed = the amount of acid in the new mixture.

Again, let's start by assigning a symbol to represent the value for which the exercise asks.

x = number of liters of 12% solution required in planned mixture

In other words, we are going to add x liters to 4 liters.

What is the volume of this new mixture?

How much acid does it contain if it turns out to be 8% acid?

You correctly told me that the four liters of 5% solution contains 4(0.05) = 0.2 liters.

How much acid is there in x liters of 12% solution?

If you've answered these questions, then you can write an equation that models the following:

(expression for amount of acid contributed by x liters of 12% solution) PLUS (0.2, because that's the contribution of acid from the four liters of 5% solution) EQUALS (expression for amount of acid in the new amount of 8% solution.)

Let me know if you need more help setting up this equation.

MY EDIT: fixed ambiguous language
 
michaelrlmt said:
lets try again on the first one? let 12% = X 5% = 4 and 8% = 4+X is that right?

It's certainly sloppy; I'm not sure if it's "right".

so .12X+.2=.32+.o8X wich will = 4
Click to expand...


Yes, that equation makes sense, to me.

0.12x + 0.2 = 0.08(x + 4)

0.12x + 0.2 = 0.08x + 0.32

However, I do not get x = 4 as the answer.

Check your algebra in solving the equation.

 
i guess my problem is addition
.12-.08 does not equal .03 it is .04 and that would make the answer 3 L of 12% solution
which makes a bit more sense
 
michaelrlmt said:
on the second one i tried this

.098x+.128(10000-x)=1085

.098X+1280-.128X=1085

-.03X=-195

X=6500
Click to expand...


You did good. Impress your teacher by writing the answer using a complete (and proper) sentence.

By the way, do you know how to check your answer?

 
yes just plug in the number for X and solve
I am just tired I think and I thank you very much for your help
the next year is gonna be ****
 
michaelrlmt said:
… the answer 3 L of 12% solution …
Click to expand...


0.12x + 0.2 = 0.08(x + 4)

0.12(3) + 0.2 = 0.08(3 + 4)

0.36 + 0.2 = 0.08(7)

0.56 = 0.56

It checks.

 
michaelrlmt said:
… next year is gonna be ****
Click to expand...


Not if you build yourself a very strong foundation throughout this year by ensuring that you're really learning this stuff, versus trying to get by with a minimum of effort.

My suggestion is to do lots of extra homework. The more you do, the better you get.

The same models, principles, concepts, and strategies repeat themselves throughout many different math topics. You want to have as many tools in your kit possible.

 
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