Solving Fraction Word Problems in Algebra

In summary, the conversation was about two algebra word problems involving fractions and equations. The first problem asked for the number of mL of a 67% solution needed to be added to a 150mL solution of 75% alcohol to make a 72% alcohol solution. The equation set up for this problem was 0.75(150) + 0.67(x) = 0.72(150 + x). The second problem involved two pipes filling a tank in different times and the question asked for the time it would take for the larger pipe to fill the tank alone. The equation set up for this problem was t + (t - 7) = 12, where t represents the time it takes for the larger
  • #1
Pseudo Statistic
391
6
Hey,
I'm having some problems with some Fraction Word Problems in Algebra, can anyone assist me as to how I'm supposed to create my equations from the following two word problems in order to solve?
Thanks:

1) A 150mL solution of alcohol in water is 75% alcohol. How many mL of a 67% solution should be added to make the solution 72% alcohol? Let x - the number of mL of 67% solution to be added.

2) One pipe can fill a tank in 7 minutes less time than it takes a smaller pipe. Together they can fill the tank in 12 minutes. How long would it take the larger pipe to fill the tank? Let t = the number of minutes it takes the larger pipe to fill the tank alone.

For 2 I'm guessing it's t + (t - 7) = 12 ?

Thanks.
 
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  • #2
Pseudo Statistic said:
Hey,
I'm having some problems with some Fraction Word Problems in Algebra, can anyone assist me as to how I'm supposed to create my equations from the following two word problems in order to solve?
Thanks:

1) A 150mL solution of alcohol in water is 75% alcohol. How many mL of a 67% solution should be added to make the solution 72% alcohol? Let x - the number of mL of 67% solution to be added.

2) One pipe can fill a tank in 7 minutes less time than it takes a smaller pipe. Together they can fill the tank in 12 minutes. How long would it take the larger pipe to fill the tank? Let t = the number of minutes it takes the larger pipe to fill the tank alone.

For 2 I'm guessing it's t + (t - 7) = 12 ?

Thanks.

1. did you try to set this one up?

hint -- you are dealing with percent alchol times volume

2.

that method works, but look at what t is supposed to equal. is it what you've got it set equal to? should the time it takes the large pipe to fill up the tank be longer than the time it takes for the small pipe to fill up the tank?

what does t-7 symbolize? in that case what is the t in your equation?

hopefully those help
 
  • #3
Oh, sorry, for t I meant t + 7 where t is the larger tank and t + 7 is the smaller..
Is this a proper equation set up?
For 1, I have absolutely NO CLUE how to set it up, our teacher explained it terribly and I have a test tomorrow on this. :-\
Hope you could explain 1 for me (Just the formation of the equation..) and tell me if 2 was correct!
Thanks alot.
 
  • #4
Pseudo Statistic said:
Oh, sorry, for t I meant t + 7 where t is the larger tank and t + 7 is the smaller..
Is this a proper equation set up?
For 1, I have absolutely NO CLUE how to set it up, our teacher explained it terribly and I have a test tomorrow on this. :-\
Hope you could explain 1 for me (Just the formation of the equation..) and tell me if 2 was correct!
Thanks alot.

yes that is correct (numeber 2)

here's an idea for 1

amount of alcohol + amount of alcohol added = new % alcohol * total volume

how would you express the amount of alcohol knowing the solutions conentration and volume. you know the volume of one, your variable is the volume of the add. hopefully that helps?
 
  • #5
0.75(150) + 0.67x = 0.72(150) ?
Sorry if I'm getting this wrong.
Hopefully you can tell me if I'm thinking the right way?
Thanks.
 
  • #6
Pseudo Statistic said:
0.75(150) + 0.67x = 0.72(150) ?
Sorry if I'm getting this wrong.
Hopefully you can tell me if I'm thinking the right way?
Thanks.

look at your final volume -- is it 150 mL, or something else?

(hint: something else)
 
  • #7
...
I wrote 0.72(150)...
You said:
amount of alcohol + amount of alcohol added = new % alcohol * total volume
And the amount of Alcohol originally is 0.75(150) and we're adding 0.67 or 67% of x to get 72% of 150, or 0.75(150)
Isn't that how you described it?!
 
  • #8
Pseudo Statistic said:
...
I wrote 0.72(150)...
You said:
amount of alcohol + amount of alcohol added = new % alcohol * total volume
And the amount of Alcohol originally is 0.75(150) and we're adding 0.67 or 67% of x to get 72% of 150, or 0.75(150)
Isn't that how you described it?!

you're starting with 150 mL of the initial solution. when would the final volume equal the initial volume when you're adding something? (hint: only if you are adding 0 of the new solution).

so if you're diluting the initial sample, what way is your volume going to go? by how much?
 
  • #9
I'm going to guess that you meant 0.75(150) + 0.67(x) = 0.72(0.75(150))
Yes, I know, I am a hopeless case. :(
 
  • #10
Pseudo Statistic said:
I'm going to guess that you meant 0.75(150) + 0.67(x) = 0.72(0.75(150))
Yes, I know, I am a hopeless case. :(

look at what your right hand is saying. 72 % of 75 % of 150 mL

does that make sense?

your final conentration of solution is to be 72 %
your final volume is going to be the initial volume plus some amount of additional solution.
what is your variable representing?

in that case, what is your final volume?
 
  • #11
Ahhhhhhhhhhh...
Sorry, I finally get it.. (I need some sleep, and I hope I got it!)
0.75(150) + 0.67(x) = 0.72(150 + x)
This better be right. :-\
 
  • #12
looks good to me, test it out and see if the answer works and makes sense.
 
  • #13
x = 90
That makes sense to me..
It's 2AM and I need sleep, have a nice day! ;)
 

1. What is the best way to approach an algebra word problem?

The best way to approach an algebra word problem is to first carefully read and understand the problem. Then, identify the known and unknown variables and write them down. Next, create an equation or system of equations to represent the problem. Finally, use algebraic techniques such as substitution or elimination to solve for the unknown variable(s).

2. How do I know which operation(s) to use in an algebra word problem?

The operation(s) to use in an algebra word problem will depend on the relationship between the given variables. For example, if the problem involves finding the total of two quantities, addition would be the appropriate operation. If the problem involves finding the difference between two quantities, subtraction would be used. Multiplication and division are used for finding the product or quotient of two quantities, respectively.

3. How can I check if my algebra word problem solution is correct?

The best way to check if your solution to an algebra word problem is correct is to substitute your solution back into the original problem and see if it satisfies all of the given conditions. If it does, then your solution is likely correct. You can also use a calculator or online tool to verify your solution.

4. Can I use real-life examples to better understand algebra word problems?

Yes, using real-life examples can be a helpful way to better understand and solve algebra word problems. You can apply the given equations and variables to a real-life situation that you are familiar with, making the problem more relatable and easier to solve.

5. What are some common mistakes to avoid when solving algebra word problems?

Some common mistakes to avoid when solving algebra word problems include misinterpreting the given information, not carefully setting up the equations, and incorrectly performing operations or using the wrong operation(s). It is important to double-check your work and make sure it aligns with the given conditions of the problem.

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