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#1 2011-04-23 06:03:15

survival
Member
Registered: 2011-04-03
Posts: 655

Algebra word problems

wave
The school production of "Our Town" was a big success. For opening night, 364 tickets were sold. Students paid $250 each, while non-students paid $4.50 each. If a total of $1202.00 was collected, how many students and how many non-students attended?

The number of students was ____

The number of non-students was ____
Simplify

*
Finance
John Smith has $44000 invested in stocks paying 7%. How much additional money should he invest in certificates of deposit paying 4% so that the average return on the two investments is 5%?

He should invest $____ in certificates of deposit. Simplify.
Thank you!

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#2 2011-04-23 06:24:36

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Algebra word problems

hi survival,

Q1.  Let the number of students be x.

=> the number of non-students must be (364 - x)

So the money taken is

2.50x + 4.50 times ( 364-x ) = 1202.00

Multiply out the bracket, simplify and solve for x.


Q2.  44000 x 7% = 3080.

Let amount invested in certificates be x.

Then return from these is 4x/100

So total return is 3080 + x/25

Total investment is 44000 + x

So (3080 + x/25) / (44000 + x)  times 100 = 5

=>  3080 + x/25 = (44000 + x) /20

Again simplfy to find x

Post back if you need more help.

Bob

Last edited by Bob (2011-04-23 06:25:34)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2011-04-23 07:25:21

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

Thank you for your help. I will keep working on them.

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#4 2011-04-23 07:52:23

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi survival;

I see that BB unleashed just a bit of his algebraic mojo. You should have got something out of that.

You seemed to have left out a decimal point in your question. Bob covered for you and made the right choice.

survival wrote:

Students paid $250 each, while non-students paid $4.50 each.

$250 for a ticket is a tad high, is it not?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#5 2011-04-23 08:22:23

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

Sorry, I left a decimal out. It is 2.50. My mistake.

Still working on problem.
In the first problem. There is something I am not doing right since i ended up with a decimal in the number of students.

2.50 x 4.50 (364-x) =1202.00
elimination of decimals by multiplication with 100.
100 x2.50 + 100 x 4.50 (364)-x = 100 x 1202.00=120200
250x + 450 x 364 - 450x =120200
163800 - 200x =120200
163800-163800-200x=120200-16380
-200x = -103900

-200x  =103900
_____________
-200      -200
 
The 200's cancel each other out (cross) out and something is wrong with the way i worked it. And I get a decimal which is not correct of course. It is all messed up.

Last edited by survival (2011-04-23 11:19:15)

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#6 2011-04-23 16:02:17

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi;

2.50 x 4.50 (364-x) =1202.00

Where is the x after that 2.5?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#7 2011-04-23 16:36:05

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

I don't recall. I wrote that early this morning. So  how would you work it out? I knew it was wrong when the answer came out to 5. something and this is referring to students.
I didn't get the other problem right either.

Last edited by survival (2011-04-23 16:57:21)

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#8 2011-04-23 17:23:20

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi;


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#9 2011-04-23 17:27:24

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

Thank you!
How about that other one?
This one is about money too:
Othello Guaraz has a box of coins that he uses when playing poker with friends. The box currently contains 31 coins, consisting of pennies, dimes and quarters. The number of pennies is equal to the number of dimes, and the total value is $3.07. How many of each denomination of coin does Othello have?

The number of pennies, which is also equal to the number of dimes, is ____
The number of quarters is_______

Last edited by survival (2011-04-23 17:57:02)

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#10 2011-04-23 19:20:50

Bob
Administrator
Registered: 2010-06-20
Posts: 10,627

Re: Algebra word problems

hi survival,

Post your working for Q2.

Q3.

Same method.  Call the thing you want to find x.

In this case, number of pennies = number of dimes = x.

And so number of quarters = 31 - 2x.

Now form an equation to get the total amount of money.

I'm from the UK so I'm a bit uncertain how many pennies in a dime. * You'll have to complete the equation yourself; which will be a good exercise anyway.

x + ? times x + 25 times (31 -2x) = 307.

Bob

* I had a problem with the small coins when I came to the US because they don't have a value on them and the smaller one seemed to be worth more than the larger one.  I'm guessing quarter means quarter of a dollar = 25 cents.  Is a penny just the same as a cent?  Oh,  life will be so much simpler when we all use a universal currency.

Last edited by Bob (2011-04-23 19:31:14)


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#11 2011-04-23 19:33:29

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

Thank you, but I have already tried the chart thing, where you fill it in and work the equation as you go, step by step etc. I posted after the first problems with my equation incorrect. I am math illiterate unless I am spending money.

About the money system/universal? It is the same with the metric system. I have never been to other countries, but my Geology professor is Canadian and said the US was the only country that does not use kilometers as it should be. He said one lady was on a trip overseas and saw a sign that reflected 100, on the highway and she was happy to drive faster. Geology is a wonderful part of the sciences, absolutely amazing.

Off topic, sorry, solving the word problems are thinkers, the student problem,  bobbym solved it and it reflects where I went wrong. The coins, quarters, dimes, etc. (last problem) is the same way. I am off. Thank you and the forum for your generosity in sharing your wisdom with others. It is a gift to learn and be taught.

Last edited by survival (2011-04-23 19:54:49)

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#12 2011-04-23 20:40:37

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi survival;

If you can not finish by yourself.

Using 2) you get:

cleaned up:

Multiply A by 25:

Subtract B from A.

pennies = 12, dimes = 12, quarters = 7


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#13 2011-04-23 21:45:54

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

Thank you! All are correct as they always are.
Edited to add that this is also a good review before taking a math quiz. It is akin to a math library with answers.

Last edited by survival (2011-04-23 21:52:47)

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#14 2011-04-23 21:53:52

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

I do not agree. Libraries have books and librarians. This place has no librarians or books.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#15 2011-04-24 06:42:05

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

You are correct.
Another money equation:
Florence has $14000 invested in stocks paying 8%. How much additional money should she invest in certificates of deposit paying 4% so that the average return on the two investments is 5%?

She should invest %_____ in certificates of deposit.
Thank you!

Last edited by survival (2011-04-24 06:43:20)

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#16 2011-04-24 08:32:58

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi survival;

$42000 is the answer I am getting.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#17 2011-04-24 09:39:27

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

You are exact. Thank you!

I hope this is the last finance problem:
A politician invested some money at 6% simple interest, and $32,000 more than four times the amount at 12%. The total annual interest earned from the investment was $36,240. How much did he invest at each amount?
Thank you again!

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#18 2011-04-24 09:58:50

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

$60000 at 6% and $272 000 at 12%.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#19 2011-04-24 10:16:04

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

That is exact.
What does the answer look like when you switch the numbers for this problem. Investing money at
6% interest, and 36,000 more than twice the amount at 9%. The total annual interest earned from the investment was $24,840.
How much did he invest at each amount?
He invested $____ at 6% and $_____at 9%.

Thank you!

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#20 2011-04-24 10:19:31

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi survival;

$90 000 at 6% and $216 000 at 9%

Taking a little break to eat.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#21 2011-04-24 11:56:22

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

It is exact as always. Thank you so much.
When I work a problem, it takes forever, because I write the steps down. I have spent hours on one problem before and wish I could solve them quickly.

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#22 2011-04-24 12:41:04

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

Look at this, I am back with figures. LOL

5x  6^ y 4^ + 35x 4^y3^ + 25xy

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#23 2011-04-24 13:27:21

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

Hi survival;

5x  6^ y 4^ + 35x 4^y3^ + 25xy

Where is the rest of that problem?


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#24 2011-04-24 13:44:55

survival
Member
Registered: 2011-04-03
Posts: 655

Re: Algebra word problems

I will look again at the problem to see what else is there.

Meanwhile, I am working on this one:About the girth and weight of a fish.
The weight of a fish varies jointly as its girth and the square of its length. One fish weighed in at 21.7 lb and measured 34 inches long with 20 in girth. How much would a fish 31 inches long with 19 in girth weigh?
The fish would weigh approximately_____lb round to six decimal places. Round the weight to the nearest tenth, if necessary.
I am at this point and it won't "take."
K = 21.7/20 (34)2^
I keyed it on the calculator and it comes out 15 something, which is wrong..
Thanks and I will look for the other problem now.

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#25 2011-04-24 14:41:30

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: Algebra word problems

The fishes weight is 17.137556 lbs rounded to 6 decimal places. To the nearest tenth it would weigh 17.1 lbs.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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