Questions on Word Problems: Mixtures answered by real tutors!

Tutors Answer Your Questions about Mixture Word Problems (FREE)

Question 945776: The percentage of salt in 1L of water is 10%. If 500 mL of water is added to this mixture, what percentage of salt is there now?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39832)   (Show Source):
You can put this solution on YOUR website!
1L is 1000 ml.
Added 500 more ml. with no salt present.

percent salt

Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
The percentage of salt in 1L of water is 10%. If 500 mL of water is added to this mixture,
what percentage of salt is there now?
~~~~~~~~~~~~~~~~~~~~~~~~~~

        How the solution is worded in the post by @lwsshar3, it provokes homeric laughter.

        It is because the amount of salt in this problem is not measured in milliliters - it is measured in grams.

        See my correct solution below, written in right terms.

In this problem, the concentration of the mixture is measured in grams per milliliter. So, the fact that 1 liter of mixture is 10% salt MEANS that the amount of salt in the mixture is 0.1*1000 = 100 grams. When 500 mL of water is added to the mixture, the total volume of the mixture (of the liquid) becomes 1500 milliliters. It contains the same 100 grams of salt - hence, the concentration of salt in the mixture is now = = 0.066666..., or 0.0667 = 6.67% rounded. ANSWER. The final concentration of salt in the mixture is about 6.67%.

Solved and presented correctly, using adequate terminology.

/////////////////////////////////

What did shock me even more, is the fact that in Google Overview Artificial Intelligence,
the solution to this problem was presented in same terms and in same words as in the post by @lwsshar3,
i.e. in absolutely illiterate way.     <<<---===   May 20, 2026, 12:40 AM.

Question 961974: You have 0.35 liters of an acid solution whose acid concentration is 15%. How many liters of pure water should be added to lower the concentration to 7%?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39832)   (Show Source):
You can put this solution on YOUR website!
w, amount of water to add, assuming as volume (liters)

Solve!

Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
You have 0.35 liters of an acid solution whose acid concentration is 15%.
How many liters of pure water should be added to lower the concentration to 7%?
~~~~~~~~~~~~~~~~~~~~~~~~~~

        @lwsshar3 in his post gives an answer 40.
        40 of what ? Liters ?
        But this is an absurd.

        There is a rude error in calculations by @lwsshar3.
        I came to bring a correct solution.

let x=amt of pure water should be added (in liters).
0.15*0.35 = 0.07*(0.35+x)
0.0525 = 0.0245 + 0.07x
0.07x = 0.0525 - 0.0245
0.07x = 0.028
x = 0.028/0.07 = 0.4.
How many liters of pure water should be added to lower the concentration to 7%?     0.4 of a liter.

CHECK.     The final concentration is    = 0.07,   or   7%.     ! precisely correct !

Solved correctly.

The lesson to learn: when you make calculations, keep the dimensions consistent and be accurate.

Question 971955: A metal worker has a metal alloy that is 30% copper and another alloy that is 80% copper. How many kilograms of each alloy should the metalworker combine to creat 60 kilograms of a 51% copper alloy?
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13362)   (Show Source):
You can put this solution on YOUR website!

Two other tutors have provided essentially identical solutions using the standard formal algebraic method for solving 2-part mixture problems like this.

Here is a solution using a less formal method that can be used to solve any problem like this. The method can be especially fast and easy, especially if the numbers in the problem are "nice".

(1) Use a number line if it helps to observe/calculate that 51% is 21/50 of the way from 30% to 80%.
(2) That means 21/50 of the mixture should be the alloy that is 80% copper.

21/50 of 60 kg is (.42)(60) = 25.2 kg.

ANSWER: 25.2 kg of the 80% copper alloy and (60-25.2)=34.8 kg of the 30% copper alloy.

Answer by josgarithmetic(39832)   (Show Source):
You can put this solution on YOUR website!

ALLOY PERCENT QUANTITY PURE 'an' 30 60-x 30(60-x) 'another 80 x 80x Blend 51 60 51*60

. Solve this for x.

of the "another" or 80% copper alloy and
of the 30% copper alloy.

Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
A metal worker has a metal alloy that is 30% copper and another alloy that is 80% copper.
How many kilograms of each alloy should the metalworker combine to create 60 kilograms of a 51% copper alloy?
~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @lwsshak3 are incorrect due to arithmetic error.
        See my correct solution below.

let x=amt of 30% copper alloy to combine
60-x=amt of 80% copper alloy to combine
..
0.3x + 0.8*(60-x) = 0.51*60
0.3x + 48 - 0.8x = 30.6
48 - 30.6 = 0.8x - 0.3x
0.50x = 17.4
x = 34.8
How many kilograms of each alloy should the metalworker combine?
30% copper alloy: 34.8 kg
80% copper alloy: 60-34.8 = 25.2 kg

CHECK for the final concentration    = 0.51.         ! precisely correct !

Solved.

Question 971371: How long would it take to double your principle (Just any number) at an annual interest rate of 8% compounded continuously.
Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
How long would it take to double your principal (Just any number) at an annual interest rate
of 8% compounded continuously.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

In his post, @lwsshak3 gave the answer "approximately 9 years".

His calculations are correct, but the formulation of the answer is not adequate to the problem.

With the obtained value t = 8.66 years, the adequate answer is 8 years and 8 months.

This my correction is important, since it shows if a person, who solves the problem,
does understand correctly what he/she is doing.

/////////////////////////////

How the word "principal" is written in the original post, tells me
that the problem's creator does not know proper terminology in the area he writes about.

Question 971373: How much should you invest in a continuously compounded account at an annually interest rate of 6% if you want exactly $8000 after four years?
Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
How much should you invest in a continuously compounded account at an annually interest rate of 6%
if you want exactly $8000 after four years?
~~~~~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @lwsshak3 are incorrect.
        I came to provide an accurate solution.

Formula for a continuous compounding account: A=Pe^rt,
P = initial investment,
r = interest rate,
t=years,
A=amt after t-yrs

For given problem:
r = 0.06
t = 4
A = 8000

..

P = A/e^rt = 8000/e^(0.06*4) = 8000/e^(0.24)= 6293.02.
How much should you invest in the continuously compounded account? $6293.02.

Solved correctly.

Question 1013544: A Chemist has 100g of 25% acid solution. How much of these solution he needs to drain and replaced with 70% acid solution to obtain 100g of 60% acid solution
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13362)   (Show Source):
You can put this solution on YOUR website!

Here is a solution using a non-traditional method that can be used to solve any 2-part mixture problem like this. This method can be especially fast and easy if the numbers in the problem are "nice" (which in this problem they are not....)

(1) Use a number line if it helps to observe/calculate that 60% is 35/45 = 7/9 of the way from 25% to 70%.
(2) That means 7/9 of the mixture must be the 70% acid solution which is being added.

7/9 of 100g is 700/9 grams.

ANSWER: 700/9 grams

Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
A Chemist has 100g of 25% acid solution. How much of these solution he needs to drain and replace
with 70% acid solution to obtain 100g of 60% acid solution
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @lwsshak3 are incorrect and lead to wrong answer.
        I came to bring a correct solution.

Let x = the amount of 25% solution to drain and replace with 70% solution (in grams).
The balance equation for the acid is
0.25*(100-x) + 0.7x = 0.6*100
25 - 0.25x + 0.70x = 60
0.45x = 35
x = 35/0.45 = 77 grams,   or   77.7778 grams, approximately.         ANSWER

Solved correctly to teach you in a right way.

Question 295596: How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution to produce an 85% alcohol solution?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39832)   (Show Source):
You can put this solution on YOUR website!
v, amount of pure alcohol

Compute this v.

Answer by ikleyn(53906)   (Show Source):
You can put this solution on YOUR website!
.
How many ounces of pure alcohol must be added to 40 ounces of a 70% alcohol solution
to produce an 85% alcohol solution?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is incorrect.
        I came to bring a correct solution.

let quantity of pure alcohol to be added be x ounces
70% alcohol is 40 ounces
the total will be x+40 ounces
x + 0.7*40 = (x+40)*0.85
x + 28 = 0.85x + 34
0.15x = 34 - 28
0.15x = 6
x = 6/0.15 = 40.

ANSWER.   40 ounces of pure alcohol must be added.

CHECK   for the concentration:      = 0.85   ! Precisely correct !

Solved correctly.

        Notice that 85% concentration is precisely midpoint between 70% and 100%,
                so the answer is obvious and can be guessed MENTALLY.

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(13362)   (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(39832)   (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(53906)   (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(13362)   (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(53906)   (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(13362)   (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(39832)   (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(91)   (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(53906)   (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(39832)   (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(10845)   (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(39832)   (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(53906)   (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(39832)   (Show Source):
You can put this solution on YOUR website!
b, amount of soybeans
360-b, amount of cornmeal

Compute this,...

Answer by ikleyn(53906)   (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(53906)   (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(53906)   (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(39832)   (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(53906)   (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(39832)   (Show Source):
You can put this solution on YOUR website!
x amount of pure salt to add

-

.
.

Answer by MathTherapy(10845)   (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(39832)   (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(53906)   (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(53906)   (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(13362)   (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(53906)   (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(13362)   (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(53906)   (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(39832)   (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(53906)   (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(39832)   (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(53906)   (Show Source):
You can put this solution on YOUR website!
.

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