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Tutors Answer Your Questions about Mixture Word Problems (FREE)

Question 275795: Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution. How much of each solution must she use?
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Tutor @josgarithmetic provides a response showing her favorite multiple-variable formula for solving 2-part mixture problems like this. Use that method if you love using formulas without having any understanding of how you are solving the problem.

Tutor @ikleyn provides a response showing a typical formal algebraic method for solving such problems. That is perfect if what you want is a formal algebraic solution method.

An informal (and usually faster) method for solving this kind of problem uses the logical fact that the ratio in which the two ingredients need to be mixed is exactly determined by where the target percentage lies between the two given percentages.

For this particular problem....
(1) Using a number line if it helps, observe/calculate that 45 is 23/58 of the distance from 22 to 80 (22 to 45 is a difference of 23; 22 to 80 is a difference of 58)
(2) That means that 23/58 of the mixture must be the higher percentage ingredient

ANSWER: The mixture should be made using 4.6 fl. oz. of the 80% acid solution and (11.6-4.6) = 7.0 fl. oz. of the 22% acid solution

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid
solution and a 80% acid solution. How much of each solution must she use?
----------------------------------------------------------------------
Make M fl. oz. of a T% acid solution by mixing together a L% acid
solution and a H% acid solution. How much of each solution to use?
----------------------------------------------------------------------

Use an unknown v fl. oz. of the H% solution and M-v fl. oz. of the L% solution.

Plug in the given values when you are ready and compute.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Beth wants to make 11.6 fl. oz. of a 45% acid solution by mixing together a 22% acid solution and a 80% acid solution.
How much of each solution must she use?
~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect, leading to wrong answer.
        I came to bring a correct solution.

Let x be the volume of the 80% solution, in oz. Then the volume of the 22% solution is (11.6-x) oz. The balance equation is 0.8x + 0.22(11.6-x) = 0.45*11.6. (1) It says that the combined mass of pure acid in ingredients (left side) is the same as the mass of pure acid in the mixture (right side). Simplify and find 'x' 0.8x + 0.22*11.6 - 0.22x = 0.45*11.6, 0.8x - 0.22x = 0.45*11.6 - 0.22*11.6 0.58x = 2.668 x = 2.668/0.58 = 4.6. ANSWER. 4.6 oz of the 80% solution is needed and (11.6-4.6) = 7 oz of the 22% solution. CHECK. Let's check the final solution for its concentration = 0.45, which is precisely correct.

Solved.

Question 276407: How many kgs of salt must be added to 24 kg of 20% salt solution in order to increase the concentration of salt to 40%?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Also ignoring the fact that a 40% solt solution is not possible....

Any 2-part mixture problem like this can be solved informally (and quickly, if the numbers are "nice") using the logical fact that the ratio in which the two ingredients must be mixed is exactly determined by where the target percentage lies between the percentages of the two ingredients.

For this problem...

(1) The target percentage of 40% is "3 times as close" to 20% as it is to 100% (the difference between 20 and 40 is 20; the difference between 40 and 100 is 60; 20 is one-third of 60)
(2) That means the amount of the 20% ingredient must be 3 times the amount of the 100% ingredient

The given amount of the 20% ingredient is 24 kg, so the amount of the added 100% salt must be one-third of 24 kg, which is 8 kg.

ANSWER: 8 kg

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
How many kgs of salt must be added to 24 kg of 20% salt solution in order to increase
the concentration of salt to 40%?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        When the problems are talking about salt solutions of high concentration, it is useful to remember
        that water solutions of NaCl can not have concentration higher than 27%.

        At 27%, the solution becomes saturated, and concentration can not go higher than this value.

        This is a standard fact, which students learn from the Science course in their 6-th grade.

        The second reason why I write these lines, is that the calculations in the post by @mananth
        are incorrect and inaccurate and lead to wrong answer.

        I came to bring a correct solution.

The balance equation for this problem is x + 0.2*24 = 0.4*(x+24), where 'x' is the salt mass to add. Simplify and find 'x' x + 4.8 = 0.4x + 9.6, x - 0.4x = 9.6 - 4.8, 0.6x = 4.8 x = 4.8/0.6 = 8. ANSWER. 8 kilograms of salt should be added.

Solved correctly and explained about the saturation limit.

Question 272631: How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion? (Round to the nearest tenth.)
Found 4 solutions by greenestamps, josgarithmetic, n2, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

While the problem was probably intended as one to be solved by formal algebra, here is quick and easy method for solving any 2-part mixture problem like this. The method is based on the fact that the ratio in which the two ingredients need to be mixed is exactly determined by where the target percentage lies between the two given percentages.

Without any more words than necessary, here is the solution to this problem using this method.

(1) The percentages of the two ingredients are 10% and 100%.
(2) The target percentage, 20%, is "8 times as close" to 10% as it is to 100% (the difference between 10% and 20% is 10%; the difference between 20% and 100% is 80%; 80% is 8 times as much as 10%).
(3) That means the amount of the 10% ingredient must be 8 times the amount of the 100% ingredient.
(4) The given amount of the 10% ingredient is 6 quarts, so the needed amount of the 100% ingredient is 1/8 of 6 quarts, which is 6/8 = 3/4 quarts.

ANSWER 3/4 of a quart, or 0.75 quarts

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!

PERCENT VOL. PURE 100 v 1.0v 10 6 0.1*6 20 v+6 v+0.1*6

Answer by n2(79)   (Show Source):
You can put this solution on YOUR website!
.
How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion?
~~~~~~~~~~~~~~~~~~~~~~~

Let x be the volume of pure antifreeze to add, in quarts. Then the governing equation is x + 0.1*6 = 0.2(x+6) <<<---=== the amount of the pure antifreeze in ingredients (left side) and in the mixture (right side) Simplify and find x x + 0.6 = 0.2x + 1.2, x - 0.2x = 1.2 - 0.6, 0.8x = 0.6, x = 0.6/0.8 = 6/8 = 3/4 = 0.75. ANSWER. 0.75 quarts of pure antifreeze should be added.

Solved correctly.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion?
(Round to the nearest tenth.)
~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is INCORRECT: its governing equation
        is written incorrectly.

        I came to provide a correct solution.

Let x be the volume of pure antifreeze to add, in quarts. Then the governing equation is x + 0.1*6 = 0.2(x+6) <<<---=== the amount of the pure antifreeze in ingredients (left side) and in the mixture (right side) Simplify and find x x + 0.6 = 0.2x + 1.2, x - 0.2x = 1.2 - 0.6, 0.8x = 0.6, x = 0.6/0.8 = 6/8 = 3/4 = 0.75. ANSWER. 0.75 quarts of pure antifreeze should be added.

Solved correctly.

Notice that the instruction for rounding is not appropriate to the problem.

It tells me that the problem's composer was not focused on the problem when created it.
Or simply does not understand what he/she composes.

Question 558198: Hello:
please help me fiqure out this word problem for future math problems with the step to perform: i do recall this problems but it has been a while since i done this word math problem
this is the problem: How much of an alloy that is 40% copper should be xixed with 500 ounce tha is 80% copper in order to get an alloy that is 60% copper.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!

PERCENT COPPER QUANTITY PURE COPPER 40 x 0.4x 80 500 oz. 0.8*500 60 x+500 0.4x+0.8*500

Compute this.

Answer by MathTherapy(10806)   (Show Source):
You can put this solution on YOUR website!

Hello: please help me fiqure out this word problem for future math problems with the step to perform: i do recall this problems but it has been a while since i done this word math problem this is the problem: How much of an alloy that is 40% copper should be xixed with 500 ounce tha is 80% copper in order to get an alloy that is 60% copper. ************************************* Adding a 40% substance to an existing 80% substance, to get a 60% substance results in an addition of the same amount of the INITIAL substance. In other words, Percent: , and AMOUNTS: . So, amount of 40% copper to be added will be the INITIAL amount, 500 ounces. The above is NOT STANDARD, so future mixture problems may definitley NOT be similar to this one. And, as you would like to "fiqure out this word problem for future math problems with the step to perform...", you can do this problem, as follows: Let the amount of 40% copper to be added, be C Percent of copper in INITIAL amount of alloy: 80, or .8 Amount of copper in INITIAL amount of alloy: .8(500) = 400 oz Percent of copper to be ADDED to INITIAL alloy: 40, or .4 Amount of copper to be ADDED: .4C Percent of copper in RESULTANT alloy: 60, or .6 Amount of copper in RESULTANT alloy: .6(500 + C) = 300 + .6C We then get: Initial amount of copper + ADDED amount of copper = RESULTANT amount uf copper, OR 400 + .4C = 300 + .6C .4C - .6C = 300 - 400 - .2C = - 100 Amount of copper to be ADDED to alloy, or Some folks tend to use a table, which could prove to be more organized and easier to follow, But, that's up to you.

Question 269702: The time it takes to do homework includes a fixed amount of time to prepare plus a constant amount of time per problem. If a student can do 5 homework problems in 40 minutes, and 10 problems in 70 minutes, how many minutes will 25 problems take?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
x, how many problems
y, how much time
points (x,y)

Points described are (5,40) and (10,70).

Slope for the linear relationship

Taking point (5,40) and using found slope 6

Finding y if x=25

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The time it takes to do homework includes a fixed amount of time to prepare plus a constant amount of time per problem.
If a student can do 5 homework problems in 40 minutes, and 10 problems in 70 minutes, how many minutes will 25 problems
take?
~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect.
        I came to bring a correct solution.

Let the fixed time taken be x
The time to solve a problem be y
x+5y=40------------1
x+10y=70-----------2
subtract equation 2 from 1
(x+5y) - (x+10y) = 40-70
x+5y -x-10y = -30
-5y = -30
y = -30/-5
y = 6. Time taken to solve each problem is 6 minutes
x + 5y = 40 substitute the value of y in the equation
x + 30 = 40
x = 40-30 = 10 minutes will be the fixed time
25 problems will take (10 minutes + 6 *25)
= 160 minutes or 2 hours and 40 minutes.         ANSWER

Question 263188: Soybean is 18%protein and cornmeal is 9% protein, how many lbs you mix for 360lbs at 16% protein?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
b, amount of soybeans
360-b, amount of cornmeal

Compute this,...

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Soybean is 18% protein and cornmeal is 9% protein, how many lbs you mix for 360lbs at 16% protein?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is incorrect due to his arithmetic error.
        See my correct solution below.

soyabean---------------- cornmeal-------------- mix
18---------------------- 9---------------------16

let quantity of soybean be x
quantity of cornmeal will be 360-x
0.18x + 0.09*(360-x) = 360*0.16
0.18x + 32.4 - 0.09x = 57.6
0.09x = 25.2
x = 25.2/0.09 = 280 lbs soybean, cornmeal 360-280 = 80 lbs.         <<<---===     ANSWER

CHECK.    = 0.16,   or 16%.     ! correct !

Solved correctly.

Question 262142: Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 40 miles per hour and train B is traveling at 60 miles per hour. Train A passes a station at 5:25 A.M. If train B passes the same station at 5:37 A.M, at what time will train B catch up to train A?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Trains A and B are traveling in the same direction on parallel tracks. Train A is traveling at 40 miles per hour
and train B is traveling at 60 miles per hour. Train A passes a station at 5:25 A.M. If train B passes the same
station at 5:37 A.M, at what time will train B catch up to train A?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is incorrect due to his errors in reasoning and in calculations.
        I came to bring a correct solution.

At 5:37 AM, train A is 40 mph x 12 minutes = 40 mph x 0.2 of an hour = 8 miles ahead train B, and the rate of approaching train B to train A is 60-40 = 20 mph. So, train B will catch train A in = = 24 minutes. So, the catching time is 5:37 AM plus 24 minutes, which is 6:01 AM. ANSWER. The catching time is 6:01 AM.

Solved correctly.

Question 1002017: Okay, so I'm working on Aleks, and they all of the sudden switched the problems I was working on and didn't give me an example on how to solve questions like these. I'm completely lost and stuck while trying to figure this out. Could someone please explain to me how I solve these types of questions?
Two factory plants are making TV panels. Yesterday, Plant A produced twice as many panels as Plant B. Four percent of the panels from Plant A and 3%of the panels from Plant B were defective. How many panels did Plant B produce, if the two plants together produced 660 defective panels?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
Two factory plants are making TV panels. Yesterday, Plant A produced twice as many panels as Plant B.
Four percent of the panels from Plant A and 3%of the panels from Plant B were defective.
How many panels did Plant B produce, if the two plants together produced 660 defective panels?
~~~~~~~~~~~~~~~~~~~~~~~~~~~

        Calculations and the answer in the post by @mananth are incorrect
        His check is incorrect too - so, his check skipped the wrong answer.

        I came to bring a correct solution.

Let B produce x panels
Then A produces 2x panels
Four percent of the panels from Plant A and 3%of the panels from Plant B were defective.
0.04(2x) +0.03x = 660
0.08x +0.03x = 660
0.11x = 660
/0.11
x = 660/0.11
x= 6000
Panels produced by B are 6000.         ANSWER
CHECK
0.04(2*6000) + 0.03(6000) = 660.         ! correct !

Solved correctly.

Question 1003698: How much water must be mixed with pints of solution that is 60% developer to produce a mixture that is 18% developer?
14 pt
19 2/3 pt
20 pt
14/1/3 pt
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Mix with HOW MANY pints of solution? You find what that is. Until then, let it be p pints of
60% developer solution.

Want result of 18% developer solution, by adding some v pints of water.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
How much water must be mixed with pints of solution that is 60% developer to produce a mixture that is 18% developer?
14 pt
19 2/3 pt
20 pt
14/1/3 pt
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

It looks like a key numerical value, necessary for the solution, is inadvertently missed in the post.

Question 27219: How much pure salt must be added to to 25 grams of a 12% salt solution to make it a 20% solution. I am having a very hard time trying to figure this problem out. Thanks for your help.
Dennis
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
x amount of pure salt to add

-

.
.

Answer by MathTherapy(10806)   (Show Source):
You can put this solution on YOUR website!

How much pure salt must be added to to 25 grams of a 12% salt solution to make it a 20% solution. I am having a very hard time trying to figure this problem out. Thanks for your help. Dennis With amount of pure salt being S, equation is: .12(25) + S = .2(25 + S) Solving for S, amount of pure salt to be added = , or 2.5 grams. FYI: The answer is NOT 10 grams, as the other person who responded, states!

Question 452936: a 10% and a 20% salt solution are combined to make 12 gallons of 15% salt solution. How much of each strength solution is required?

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
6 gallons of each. Arithmetic steps should not be necessary, since 15% is exactly in the middle of 10 and 20.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

In his post, @mananth treats ingredients in lb, which unit represents pounds.

It is incorrect. The unit to use in this problem is gallons.

So, lb at each and every appearance here must be replaced by gallons.

Question 449124: The radiator in Natalie�s car contains 6.3 L of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of 50% antifreeze?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The radiator in Natalie's car contains 6.3 L of antifreeze and water. This mixture is 30% antifreeze.
How much of this mixture should she drain and replace with pure antifreeze so that there will be a mixture of 50% antifreeze?
~~~~~~~~~~~~~~~~~~~~~~~~

        In the post by @mananth, the solution is incorrect, because his starting equation is written incorrectly.
        I came to bring a correct solution.

Let x be the volume of the original mixture to drain and replace with pure antifreeze. Then the balance equation for pure antifreeze content is 0.3*(6.3-x) + x = 0.5*6.3. Simplify and find x 0.3*6.3 - 0.3x + x = 0.5*6.3, -0.3x + x = 0.5*6.3 - 0.3*6.3, 0.7x = (0.5*-0.3)*6.3 0.7x = 0.2*6.3 x = 0.2*9 = 1.8. ANSWER. 1.8 liters of the original mixture should be drained and replaced by the pure antifreeze.

Solved correctly.

Question 448309: A milk tank can be filled by pip A in 6 hours, by pipe B in 8 hours and by pipe C in 12 hours. how long will it take all three pipes working together to fill the tank?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Tutor @ikleyn has provided a response showing a solution by the method that is usually taught.

Here is a solution using an alternative method.

Consider the least common multiple of the three given times, which is 24 hours.

In 24 hours...
pipe A could fill the tank 24/6 = 4 times;
pipe B could fill the tank 24/8 = 3 times; and
pipe C could fill the tank 24/12 = 2 times

So in 24 hours the three pipes could fill the tank 4+3+2 = 9 times.

That means the number of hours required for the three pipes together to fill the tank one time is 24/9 = 8/3.

ANSWER: 8/3 hours, or 2 2/3 hours

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
A milk tank can be filled by pipe A in 6 hours, by pipe B in 8 hours and by pipe C in 12 hours.
how long will it take all three pipes working together to fill the tank?
~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are inaccurate; his solution and his answer are incorrect.
        I came to bring a correct/accurate solution in the form as it SHOULD be presented.

First pipe makes 1/6 of the job per hour. Second pipe makes 1/8 of the job per hour. Third pipe makes 1/12 of the job per hour. Working together, the three pipes make + + = + + = = = of the job per hour. Hence, it will take hours, or 2 hours, or 2 hours and 40 minutes to fill the tank, if all three pipes work together.

Solved.

/\/\/\/\/\/\/\/

And my separate notice in the margins of my solution
        I do not think that somebody will fill a milk tank during 12 hours.

Question 433981: 10 gallons of a 15.0% alcohol solution are to be mixed with a 28.0% alcohol solution to make a 20.0% alcohol solution. How many gallons of a 28.0% alcohol must be used? How many gallons of a 20.0% alcohol solution are made?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The ratio in which the two alcohol solutions must be mixed is exactly determined by where the 20% of the mixture lies between the 15% and 28% of the two ingredients.

The difference between 15 and 20 is 5; the difference between 20 and 28 is 8.

So the two ingredients must be mixed in the ratio 5:8. Note that since 20% is closer to 15% than it is to 28%, the larger portion must be the 15% alcohol solution.

Solve the problem using a proportion, given that we are using 10 gallons of the 15% alcohol.

ANSWERS:
(1) 6.25 gallons of the 28% alcohol should be used
(2) the mixture will be 10+6.25 = 16.25 gallons

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
10 gallons of a 15.0% alcohol solution are to be mixed with a 28.0% alcohol solution to make a 20.0% alcohol solution.
(a) How many gallons of a 28.0% alcohol must be used?
(b) How many gallons of a 20.0% alcohol solution are made?
~~~~~~~~~~~~~~~~~~~~~~~~~~~

       In the post by @mananth, his answer to question (a) is correct,
       but his answer to question (b) is not correct and should be fixed.
       So, I place below the solution with modified/corrected part for question (b).

------ Percent ---------------- quantity
Alcohol 15 ----------------10
Alcohol II 28----------------x
Total 20 ----------------10+x

15*10+28*x=20(10+x)
150+28x =200+20 x
28 x - 20 x = 200 - 150
8x=50
/8

x=6.25 gallons 28% Alcohol II must be used.                 <<<---=== ANSWER to q.  (a)

10+x = 16.25 gallons of the 20% alcohol are made.         <<<---=== ANSWER to q.  (b)

Question 419707: A 10-quart radiator contains a 30% concentration of antifreeze. how much of the sloution must be drained and replaced by 100% antifreeze to bring the solution up to 50%
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
v, to remove and replace.
Total finished contents is to be unchanged.

basic setup

When this makes sense, just compute v.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

The answer "x=2.86 liters of 100% antifreeze" in the post by @matanth is incorrect.

The correct answer is "x=2.86 quarts of 100% antifreeze".

Question 428537: How many liters of distilled water must be added to 140 liters of an 80 percent alcohol solution to obtain a 70 percent solution?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Starts as liters of pure alcohol.

Add x liters of distilled water to make 70% alcohol.

Simple algebra.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

The correct answer is 20 liters - - - not 20 gallons.

Question 428518: How many liters of distilled water must be added to 140 liters of an 80% alcohol solution to obtain a 70% solution? i think the answer is 2.3 liters.
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

The correct answer is 20 liters - - - not 20 gallons.

Question 729631: How many liters each of 8%, 10%, and 20% hydrogen peroxide should be mixed together to get 8 liters of 12.5% solution, if the amount of 8% solution used must be 2 liters more than the amount of 20% solution used?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!

Concentration, % Volume, liters pure hydrogen peroxide 8 v+2 0.08(v+2) 10 8-(v+2)-v=6-2v 0.1(6-2v) 20 v 0.2v TOTAL 12.5 8

, to represent the final concentration.
Simplify and solve....

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
How many liters each of 8%, 10%, and 20% hydrogen peroxide should be mixed together to get 8 liters
of 12.5% solution, if the amount of 8% solution used must be 2 liters more than the amount of 20% solution used?
~~~~~~~~~~~~~~~~~~~~~~~

Ingredients: x liters of the 20% hydrogen peroxide; (x+2) liters of the 8% hydrogen peroxide; and (8 - x - (x+2)) = 6-2x liters of the 10% hydrogen peroxide. Write equation for the content of hydrogen peroxide 0.2*x + 0.08*(x+2) + 0.1*(6-2x) = 0.125*8. Simplify and find x 0.2x + 0.08x + 0.16 + 0.6 - 0.2x = 1, 0.2x + 0.08x - 0.2x = 1 - 0.16 - 0.6 0.08x = 0.24 x = 0.24/0.08 = 3. ANSWER. Use 3 liters of the 20% hydrogen solution PLUS 3+2 = 5 liters of the 8% hydrogen solution and do not use the 10% hydrogen solution, at all. CHECK. The final concentration will be (0.2*3 + 0.08*5)/8 = 1/8 = 0.125 = 12.5%, correct.

Solved.

Question 729572: A company contracts to paint 3 houses. Mr.Orange can paint a house in 6 days while Mr.Grey would take 8 days and Mr.Red 12 days. After 8 days Mr.Orange goes on vacation and Mr.Grey begins to work for a period of 6 days. How many days will it take Mr.Red to complete the contract?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
A company contracts to paint 3 houses. Mr.Orange can paint a house in 6 days while Mr.Grey would take 8 days
and Mr.Red 12 days. After 8 days Mr.Orange goes on vacation and Mr.Grey begins to work for a period of 6 days.
How many days will it take Mr.Red to complete the contract?
~~~~~~~~~~~~~~~~~~~~~~~

The whole job is to paint 3 houses. Mr. Orange' rate of work is of the house per day. Working 8 days, Mr Orange painted = houses, and 3 - = houses remained to paint. Mr. Gray' rate of work is of the house per day. Working 6 days, Mr. Gray painted = of a house, and - = - = of a house remained to paint. Mr. Red' rate of work is of the house per day. Hence, he needs = 11 days to complete the contract. ANSWER. Mr. Red needs 11 days to complete the job.

Solved.

Question 45662: Hi again,
I need some assistance with this mixture problem:
It takes a pipe 10 minutes less than another one to fill a tank of water. Both pipes together can fill the tank in 12 minutes. How long will it take each one to fill the tank separately?
I am having trouble on how to start setting the equation up.
Thanks again,
Lou
Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

Hello, in your post you call this problem as a "mixture problem".

Actually, it is not a "mixture" problem and never was.

The problems of this kind have their traditional name, which is "joint work" problems.

Do not create your own names for subjects in the area, where they are well established
just for hundred years and with which you are unfamiliar with.

Making disorder in a well ordered space is not a good style and is disrespect to traditions.

Besides of it, it shows that you don't know a standard terminology,
which is not in your interests to demonstrate around.

Answer by MathTherapy(10806)   (Show Source):
You can put this solution on YOUR website!

Hi again, I need some assistance with this mixture problem: It takes a pipe 10 minutes less than another one to fill a tank of water. Both pipes together can fill the tank in 12 minutes. How long will it take each one to fill the tank separately? I am having trouble on how to start setting the equation up. Thanks again, Lou Let time it takes FASTER pipe, be F Then time it takes the SLOWER pipe = F + 10 Fraction of tank filled by FASTER pipe in 1 munute: Fraction of tank filled by SLOWER pipe in 1 munute: With the time taken by both pipes to fill the tank being 12 minutes, we get the following equation: 12(F + 10) + 12F = F(F + 10) ------ Multiplying by LCD, 12F(F + 10) (F - 20)(F + 6) = 0 F - 20 = 0 OR F + 6 = 0____F = - 6 (ignore) Time faster pipe takes to fill tank, by itself, or F = 20 minutes Time slower pipe takes to fill tank, by itself, or F + 10 = 20 + 10 = 30 minutes The answer can also be seen HERE ====> Question 742195: One pipe can fill a tank in 20 minutes, while another takes 30 minutes to fill the same tank. How long would it take the two pipes together to fill the tank?

Question 740885: The coffee merchant wants to blend a mildly strong coffee called Nervous Nellie with an extremely strong coffee called Nervous Breakdown to create a medium strength coffee he plans to call Nervous Wreck. If the Nellie flavor costs $1.20 per pound and the Breakdown flavor costs $1.40 per pound, how many pounds of the Nellie is needed to create a 30 pound blend which costs $1.35 per pound?
Please include how you got the solution
Found 3 solutions by timofer, greenestamps, ikleyn:
Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!

--------------------------------------------------------------- price pounds costs --------------------------------------------------------------- nellie 1.2 30-c 1.2(30-c) breakdown 1.4 c 1.4c wreck(blend) 1.35 30 1.2(30-c)+1.4 ----------------------------------------------------------------

To account for cost

, the pounds of Breakdown
which means , the pounds of Nellie

Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The response from the other tutor shows a solution using the standard formal algebraic method for solving this mixture problem.

For any 2-part mixture problem like this one, the problem can be solved informally by a different method; in this problem the solution by this alternative method is very quick and easy because the numbers are "nice".

The short, quick, and easy solution....

Because $1.35 is three-fourths of the way from $1.20 to $1.40, 3/4 of the mixture must be the more expensive flavor. 3/4 of 30 pounds is 22.5 pounds.

ANSWER: 22.5 pounds of the Breakdown flavor, so 30-22.5 = 7.5 pounds of the Nellie flavor

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
The coffee merchant wants to blend a mildly strong coffee called Nervous Nellie with an extremely strong coffee called Nervous Breakdown to create
a medium strength coffee he plans to call Nervous Wreck. If the Nellie flavor costs $1.20 per pound and the Breakdown flavor costs $1.40 per pound,
how many pounds of the Nellie is needed to create a 30 pound blend which costs $1.35 per pound?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let x be the number of pounds of the $1.20 per pound coffee (Nellie). Then the number of pounds of the $1.40 per pound coffee (Breakdown) is (30-x). Write the total money equation 1.20x + 1.40(30-x) = 30*1.35 dollars. Simplify and find x 1.20x + 1.40*30 - 1.40x = 30*1.35 -0.2x = 30*1.35 - 30*1.40 -0.2x = 30*(1.35-1.40) -0.2x = 30*(-0.05) 0.2x = 30*0.05 x = = = 7.5 pounds. ANSWER. 7.5 pounds of the Nellie coffee is needed. CHECK. = 1.35 dollars per pound, the price for the pound of the blend. ! correct !

Solved.

The answer 7 pounds in the post by @lynnlo is incorrect.

Question 740881: If the snack mix is made of sea monster scales, which cost $3.50 per pound , and baby whale barnacles, which cost $5 per pound, how many pounds of sea monster scales will she need to create s 45 pound blend thats worth $4.50 per pound
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

You have received two responses showing similar typical formal algebraic solutions to this 2-part mixture problem.

Here is a VERY short and quick solution using a method that can be used for any 2-part mixture problem like this. (The solution using this method is very quick for this example, because the numbers are "nice". But it can be used for any problem like this; the arithmetic just won't be so simple.)

Short solution: $4.50 is two-thirds of the way from $3.50 to $5.00, so 2/3 of the mixture must be the more expensive baby whale barnacles.

ANSWER: 30 pounds of baby whale barnacles and 15 pounds of sea monster scales.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!

The given described information: PRICE POUNDS COST s. m. scales 3.50 b. w. barnackles 5.00 mixture 4.50 45

PRICE POUNDS COST s. m. scales 3.50 45-b 3.5(45-b) b. w. barnackles 5.00 b 5b mixture 4.50 45 3.5(45-b)+5b

Relating cost per pound to the desired price

30 pounds of baby whale barnacles, 15 pounds sea monster scales.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
If the snack mix is made of sea monster scales, which cost $3.50 per pound, and baby whale barnacles,
which cost $5 per pound, how many pounds of sea monster scales will she need to create s 45 pound blend
that's worth $4.50 per pound ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Take x pounds of the $3.50 ingredient per pound and (45-x) pounds of the $5 ingredient per pound. The blend will cost 3.50x + 5*(45-x) dollars for 45 pounds. Equate it to the cost of 45 pounds of the blend at $4.50 per pound 3.50x + 5*(45-x) = 45*4.50 dollars. Simplify this equation and find x 3.50x + 5*45 - 5x = 45*4.50, 3.50x - 5x = 45*4.50 - 5*45, -1.50x = 45*(4.50-5) -1.50x = 45*(-0.5) x = = = 15. ANSWER. 15 pounds of the monster scale at $3.50 should be mixed with 45-15 = 30 pounds of the baby shale barnacles at $5 per pound. CHECK. We will check the price of the blend per pound = 4.50 dollars per pound of the blend. ! correct !

Solved and checked.

Question 740890: The coffee merchant wants to blend a mildly strong coffee called Nervous Nellie with an extremely strong coffee called Nervous Breakdown to create a medium strength coffee he plans to call Nervous Wreck. If the Nellie flavor costs $1.20 per pound and the Breakdown flavor costs $1.40 per pound, how many pounds of the Nellie is needed to create a 30 pound blend which costs $1.35 per pound?

Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The response from tutor @ikleyn shows a standard formal algebraic solution for mixture problems like this.

You can also solve this or any similar mixture problem using the idea of weighted averages; the problem can be solved quickly with simple mental arithmetic if the numbers in the problem are "nice".

In this problem the price per pound of the mixture is 3/4 of the way from the lower price to the higher price; that means 3/4 of the mixture must be the more expensive ingredient.

3/4 of 30 pounds is 22.5 pounds, so the mixture must contain 22.5 pounds of the Breakdown flavor, which means 30-22.5 = 7.5 pounds of the Nellie flavor.

ANSWER: 7.5 pounds