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Comment on Mixture Questions
Hello Brent,
Could you please explain this question?
A mixture of 12 ounces of vinegar and oil is 40%n vinegar by weight. How many ounces of oil must be added to the mixture to produce a new mixture that is only 25% vinegar? Also is there an easier and faster approach than sketching the figures of solutions? There are many in other sources, but I do not want to confuse. I want a method from you.
INITIAL mixture: 12 ounces,
INITIAL mixture: 12 ounces, which is 40% vinegar
40% of 12 = 4.8
So, the INITIAL mixture contains 4.8 ounces of vinegar
Let x = number of ounces of oil that we must ADD to the initial mixture.
IMPORTANT: Once we add x ounces of oil to the initial mixture, the RESULTING mixture will have a total volume of 12+x ounces
Also, the RESULTING mixture will contain 4.8 ounces of vinegar
GOAL: We want the resulting mixture to be 25% vinegar.
In other words, we want the resulting mixture to be 1/4 vinegar
So, we get: 4.8/(x+12) = 1/4
Cross multiply to get: (1)(x+12) = (4)(4.8)
Expand and simplify: x + 12 = 19.2
Solve: x = 7.2
So, we must add 7.2 ounces of oil
Hello Sir,
For a mixture problem, what I normally do is, draw 3 boxes, and try to set up an algebraic equation.
for this problem, my equation was,
.40(12) + x = .25(12 + x)
but I see the, the correct equation is, .40(12) = .25(12 + x)
I don't know why, why the [ +x ] portion from the left side of the equation i wrote wasn't included into the equation.
is it because, the question is asking for a final mixture that is "only vinegar" instead of the phrase "resulting solution", that we normally see on other questions!
eagerly waiting for your reply.
Thanks :)
Notice that (0.40)(12)
Notice that (0.40)(12) represents the volume of VINEGAR in the initial mixture.
Likewise, (0.25)(12 + x) represents the volume of VINEGAR in the resulting mixture.
So, our equation is keeping track of the volume of VINEGAR throughout the process.
In the equation, x = number of ounces of OIL that we must add to the initial mixture.
Since there's no vinegar in the oil, we don't add x.
Alternatively, we COULD say that the oil we're adding is 0% vinegar.
So, we could write: 0.40(12) + (0.0)(x) = 0.25(12 + x)
Does that help?
100 percent clear Sir,
Thank you so much, you’re amazing :)
It is helpful
Thank you! The catch in this
Another way to approach this
300(30%) + 200(70%) = 500(c) --> 300 * .3 + 200 * .7 = 500*c
90 + 140 =230 = 500 * c
c = (230/500) *100 = 46%
Perfect!
Perfect!
A cup full of water when
grape solution.
Column A : volume of the cup
Column B: 400 cc
sir how to solve this question?
ASIDE: I'll use milliliters
ASIDE: I'll use milliliters (ml) instead of cubic centimeters (cc)
Start with 100 ml of 30% grape solution.
So, this initial solution contains 30 ml of grape solution, and 70 ml of water.
Now add x ml of pure water (i.e., x = the volume of the cup).
So, this part contains 0 ml of grape solution, and x ml of water.
When we combine the two amounts we get: 30 ml of grape solution and (70 + x) mls of water.
Also, note that the TOTAL volume of the resulting solution = (100 + x) mls
We want the resulting solution to be 10% grape solution.
In other words, we want: (volume of grape solution in resulting solution)/(TOTAL volume of resulting solution) = 10%
In other words: (30)/(100 + x) = 10/100
Simplify: (30)/(100 + x) = 1/10
Cross multiply to get: (1)(100 + x) = (30)(10)
Expand: 100 + x = 300
Solve: x = 200
So, the volume of the cup = 200 ml
We get:
QUANTITY A: 200
QUANTITY B: 400
Answer: B
Cheers,
Brent
During a college fun-fair,
Pls solve
Yikes! That's a poorly-worded
Yikes! That's a poorly-worded question!
The addition of "(and their family members)" makes this question unanswerable.
However, if we IGNORE that proviso, and assume that all 75 attendees are either students or teachers, (no family members), then we can answer the question as follows:
Let t = # of teachers who went
Let x = # of students who went
So, we can write:
t + x = 75
30t + 12x = 1800
When we solve the system, we get: t = 50 and x = 25
Answer: 25 students.
Cheers,
Brent
I got the same answer but
https://gre.myprepclub.com/forum
Can you provide an explanation to the question posted above, I'm not quite understanding how the answer came to be.
Thank you!
Here's my full solution:
Here's my full solution: https://gre.myprepclub.com/forum/a-makes-up-8-percent-of-solution-r-and-...
Cheers,
Brent
https://gre.myprepclub.com/forum
When 6l pure acid is added isnt the resulting mixture total 26? since there is 8l of some other liquid?
Question link: https:/
Question link: https://gre.myprepclub.com/forum/a-container-holds-10-liters-of-a-soluti...
We start with 10 liters of ORIGINAL solution
Although this original solution consists of 2 liters of acid and 8 liters of some other liquid, the TOTAL volume is 10 liters.
So, if we add 6 more liters to the ORIGINAL 10 liters, the resulting volume = 10 + 6 = 16
Does that help?
Cheers,
Brent
This technique took some
Thanks, Kevin!!!
Thanks, Kevin!!!
Hello Sir,
Is there an algebraic solution to this question, I couldn't solve it the way I normally do, like drawing boxes and breaking it into parts!
"Suppose you have a 200-liter mixture that is 90% water and 10% bleach and ask how much water you would need to add to make it 5% bleach"
Thanks
We start with 200 liters of
We start with 200 liters of solution
10% of 200 = 20
So, the original solution contains 20 liters of bleach
Let x = the volume of water (in liters) we must add to get a solution that is 5% bleach
This means the NEW volume of the solution = 200 + x
The volume of bleach is still 20 liters (since we didn't add any bleach the original solution)
We want the new solution to be 5% bleach.
In other words, we want: (volume of bleach)/(volume of solution) = 5/100 (aka 5%)
Substitute to get: 20/(200 + x) = 5/100
Simplify to get: 20/(200 + x) = 1/20
Cross multiply: (1)(200 + x) = (20)(20)
Expand: 200 + x = 400
Solve: x = 200
So we must add 200 liters of water.
Got it, Sir, Thank you so
How many liters of 20%
Here's my full solution:
Here's my full solution: https://gre.myprepclub.com/forum/topic20540.html#p60664
To a sugar solution of 4
(A) 13 %
(B) 15%
(C) 20%
(D) 24%
(E) 30%
Your solution:
To a sugar solution of 4 liters containing 30% sugar, . . .
30% of 4 liters = 1.2 liters
So, the ORIGINAL solution contains 1.2 liters of sugar
. . . one liter of water is added.
So, the NEW volume = 5 liters
Included in the 5 liters is the original 1.2 liters of sugar
The percentage of sugar in the NEW solution is
1.2/5 = 2.4/10 = 24/100 = 24%
So, the NEW contains 24% sugar
What i did was i did 30% of 4ltr and found 1.2 till here i was good then i subtracted 1.2 (sugar) and got 2.8 water .....than as we are adding 1 ltr of water i added 1 to 2.8 so 3.8 and than subtracted 1.2 to get 0.2 as new sugar. multiply it by 100 and 20 % and got the answer as "c" which was wrong so where did i make a mistake can u please help.
Your mistake occurred when
Your mistake occurred when you wrote: "...and then subtracted 1.2 to get 0.2 as new sugar."
There is no "new sugar"
As you noted, there was 1.2 liters of sugar in the original solution.
When we added 1 liter of water to the solution, no new sugar is added to the solution.
So, once the 1 liter of water is added, the solution (as you noted) contains 1.2 liters of sugar and 3.8 liters of water.
Concentration of sugar in solution = (volume of sugar)/(total volume of solution)
= 1.2/(1.2 + 3.8)
= 1.2/5
= 24%
Does that help?
Hey thank u for the
That's correct.
That's correct.