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Help with math word problem (intermediate algebra)
    #18562872 -

Ok, feel stupid asking this, but I am stumped on this:

Joe plans to invest $30,000, part at 3%, and part at 4% for one year. What is thee most that can be invested at 3% in order to make at least $1100 interest in one year?

I am sure the answer is really simple, and right infront of my eyes, but I am not seeing it. Please englighten me.


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18562897 -

(x*.03)+(30,000-x)*.04 = 1100

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Re: Help with math word problem (intermediate algebra) [Re: Patito]
    #18562915 -

What is X?


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18562920 -

nm think i got it


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18562928 -

scratch that. Don't get it.


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18562964 -

I would just do trial and error, tallying it up

30,000*.03 = X *.03 =Y*.03 until it goes higher then 31,100.


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18562969 -

Equation states: Amount of interest at 3% + Amount of interest at 4% = Total Interest.

x is the amount you're solving for. Both amounts need to add up to 30,000, so I used x and (30,000-x) for the two unknown amounts. That way you only need to have one equation with one unknown.

Solve for x and there's you're minimum. Spend any more at 3% and you won't be able to make 1100.

Someone can jump in if I'm wrong here. It's been awhile.

Edited by Patito (07/15/13 04:43 PM)

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Re: Help with math word problem (intermediate algebra) [Re: Konyap]
    #18562976 -

I hate having to turn word problems into equations. Especially with multiple variables.... I'm never sure which part I'm solving for. This is probably why I'm in retard math at my college.

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Re: Help with math word problem (intermediate algebra) [Re: Patito]
    #18563009 -

Patito said:
Equation states: Amount of interest at 3% + Amount of interest at 4% = Total Interest.

x is the amount you're solving for. Both amounts need to add up to 30,000, so I used x and (30,000-x) for the two unknown amounts. That way you only need to have one equation with one unknown.

Solve for x and there's you're minimum. Spend any more at 3% and you won't be able to make 1100.

Someone can jump in if I'm wrong here. It's been awhile.


I know what your saying, and have tried many variations on the same thing, but doing your way I don't think solving for X works.

x*.03 + (30,000 - x) * .04 =

.03x + (30,000 - x) * .04 =

.03x + 1200 - .04x <-- combine like terms

-.01x + 1200

Divide both terms by -.01x to isolate the variable and everything goes to shit.

Or maybe i am seeing it wrong?


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18563057 -

You have it right, except you're forgetting the other side of the equation.

-.01x + 1200 = 1100

100 = .01x

x = $10,000

Check your answer by choosing a higher number for x, say $11,000. The result will be $1,090 interest (on the right side of the equation). So 10,000 is the most that you can spend at 3% to get at least $1,100 interest.

e.g.

11000*(.03) + 19000*.04 = x
x = $1090

Note: it's a linear equation. As expected, if you spend less at 3% (and more at 4%), you will get a higher interest than $1,100.

Edited by Patito (07/15/13 05:09 PM)

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Re: Help with math word problem (intermediate algebra) [Re: Patito]
    #18563087 -

$30000 at 3% for a year, interest = $900
$30000 at 4% for a year, interest = $1200

Answer
$10,000 at 3% for a year, interest = $300
$20,000 at 4% for a year, interest = $800
Total = $1,100

The most that can be invested at 3% is $10,000


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Re: Help with math word problem (intermediate algebra) [Re: Patito]
    #18563165 -

Patito said:
You have it right, except you're forgetting the other side of the equation.

-.01x + 1200 = 1100

100 = .01x

x = $10,000

Check your answer by choosing a higher number for x, say $11,000. The result will be $1,090 interest (on the right side of the equation). So 10,000 is the most that you can spend at 3% to get at least $1,100 interest.

e.g.

11000*(.03) + 19000*.04 = x
x = $1090

Note: it's a linear equation. As expected, if you spend less at 3% (and more at 4%), you will get a higher interest than $1,100.




I was struggling to see "the other side of the equation." However, i just had the aha moment. It is quite simple, but these are always a bit mind bending when doing one for the first time.

I very much appreciate the time you took to explain that.


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18563254 -

wow i didn't even read that shit


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Re: Help with math word problem (intermediate algebra) [Re: Konyap]
    #18563315 -

aiyobro said:
wow i didn't even read that shit


The point was to solve it using algebra. Guess and check didn't help!


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Re: Help with math word problem (intermediate algebra) [Re: PDU]
    #18563360 -

I got a D the first time I took algebra cause I didn't buy the text book until like a month in then the next time I took it I never went to study center and got an A.


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Re: Help with math word problem (intermediate algebra) [Re: Konyap]
    #18565325 -

Hey, here is it in algebraz
x = 3% investment in $
y = 4% investment in $
We are given:
x + y = 30000
So:
x = 30000 - y
We are also given
x*3% + y*4% = 1100
This equals:
x*0.03 + y*0.04 = 1100
Multiply by 100
3x + 4y = 110000
Substitute x
3*(30000 – y) +4y = 110000
90000 -3y + 4y = 110000
-3y + 4y = 110000 – 90000
y = 20000
So:
x = 10000

Then at the bottom of your answer make sure u kick the maths teacher in the guts:
ps. 3% is a crappy investment, i've got no idea how Joe got $30k in cash if he's gonna invest at 3%. May as well give it to https://heml.is/ or mine bitcoins... or else he risks ending up coming a maths teacher


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