Questions on Word Problems: Mixtures answered by real tutors!

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= liters of punch in 18% punch mix
= liters of punch in 30% punch mix
Let = liters of water to be added to get 25% punch mix
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4.4 liters of water should be added
check:




OK

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Interesting you ask, how much water.

If that is how the adjustment is to be done, then ignore the 18% punch and just use the 30% punch.
where w is how much liters of water to add. Solve for w.





liters water

Certainly there are other ways, such as to use some or (maybe) all(?) of the 18% punch along with the 30%.

A good alternative is say, let e be how much of the 18% punch to add to the 50 liter of 30% punch. Add no extra water.
... in fact, you would need to add ALL of the 20 liters of the 18% plus some water.
...

Solve for w, the liters of water to add to ALL of the available punch solutions.

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There are 3*0.4 = 1.2 gallons of pure antifreeze.

Let x = amount (in gallons) of pure antifreeze needed to add

If you add x gallons of pure antifreeze to 1.2 gallons of pure antifreeze, you'll have 1.2+x gallons of pure antifreeze

You are also adding x gallons of pure antifreeze to the 3 gallons of solution already there, so the total amount of solution is now 3+x gallons.

We can now say this

(1.2+x)/(3+x) = 0.5

since we want the percentage of pure antifreeze to total solution must be 50%

Now solve for x

(1.2+x)/(3+x) = 0.5

1.2+x = 0.5(3+x)

1.2+x = 1.5+0.5x

x = 1.5+0.5x-1.2

x-0.5x = 1.5-1.2

0.5x = 0.3

x = 0.3/0.5

x = 0.6

So you need to add 0.6 gallons of pure antifreeze to obtain 50% solution.

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All as weight per weight.

w = amount of water in grams
s = amount of salt in grams

and
Two variables unknown and two equations. Solve the system. Note, you do not really need the second equation. You essentially have just this:

, and the rest of the mass to make up to 100 grams would be the needed water. You may still use the second equation, mass balance.

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How many gallons of a 10% antifreeze solution: 0.10x
and a 25 % Antifreeze solution must be mixed: 0.25y
to make 3 gallons of a 20 % antifreeze solution: (3)(0.20)=0.6

but x+y=3 (so y=3-x)
and we write


0.1x+0.25(3-x)=0.6
which gives us x=1 gallon



:)

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how much of an alloy that is 20% copper: 0.20x
should be mixed with 500 ounces of an alloy that is 60% copper: 500(0.60)=300
in order to get: =
an alloy that is 30% copper: (x+500)(0.30)

putting them together:
0.2x+300=(x+500)(0.30)

which gives us x=1500 ounces


:)

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To produce 100 liters that is 17% acid, Liza took some amount from a 40% acid solution and a 60% acid solution then added 70 liters of pure water to the resulting mixture. How much of the 40% acid solution was used?
100 * .17 = 17 liters of acid
100 - 70 = 30 liters of the 60% and 40% solutions
let x be the amount of 40% solution, then 30-x is 60%solution, so we have
.40x + .60(30-x) = 17
.40x + 18 - .60x = 17
-.20x = -1
(1/5)x = 1
x = 5
so 5 liters of the 40% solution was used

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Let = cc's of 13% solution to be added
= cc's of salt in 13% solution
cc's of salt in 23% solution
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120 cc's of 13% solution need to be added
check:





OK

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you have 80 ml of a solution that is 60% acid

so you have 80*0.6 = 48 mL of pure acid


Let x = amount of pure acid

If you add x mL of pure acid to 48 mL of pure acid, then you'll have 48+x mL of pure acid total. This is out of 80 + x mL of new total solution.

This ratio (48+x)/(80+x) must be equal to 0.75 since we want this new percentage to be 75%, so

(48+x)/(80+x) = 0.75

now solve for x

(48+x)/(80+x) = 0.75

48+x = 0.75(80+x)

48+x = 60+0.75x

x = 60+0.75x - 48

x-0.75x = 60 - 48

0.25x = 12

x = 12/0.25

x = 48

So you need to add 48 mL of pure acid.

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How many gallons of 40% antifreeze should be mixed with 10 gallons of 85% antifreeze to obtain 65% antifreeze mixture?
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Equation:
alcohol + alcohol = alcohol
0.40x + 0.85*10 = 0.65(x+10)
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40x + 85*10 = 65x + 65*10
20*10 = 15x
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x = (2/3)20
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x = 13 1/3 gallons (amt. of 40% solution needed)
=======================
Cheers,
Stan H.
=======================

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(amount of acid)/(amount of mixture)=fraction as the acid or the concentration

Some parts are variable.

Let u = liters volume to add of the 7% acid.
Let v = liters of 27% acid.

Amount of acid = u*7%+v*27%
and
Although one of these is rational equation, they both are linear equations in two variables.
Just solve the system for u and v.

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A chocolate drink is 16% pure chocolate by volume. If 13 litres of pure milk are added to 22 litres of this drink, the percent of chocolate in the new drink is approximately.
***
let x= percent of chocolate in the new drink.
16%(22)=x%(13+22)
3.52=x(35)
x=3.52/35=0.10
percent of chocolate in the new drink≈10%

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We'll call the unknown quantity x
6x-total cost of 6 dollar coffee
(3)(4) -total cost of 4 dollar coffee

so our working equation is

6x+12=(x+3)(4.5)

which when solved gives us x=1 pound

:)

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c = how many cups of juice, which is 100%.

That was most of the solution. The rest is solve the equation for c.

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you have a parabola with a vertex at origin. From the graph I can see that if => and if => , then you have

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You can assume that this is a juice and water mixture.
Let = the cups of pure juice to be added
cups of pure juice in 5% mixture
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2.222 cups of pure juice are needed
check:




OK

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32litres42mililitre = litre?
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It's not clear what you want to do.

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how many % alcohol should a wine maker add to 20 gallons of wine that is 12%
alcohol to have 30 gallons of wine that is 14% alcohol?
:
We know that the amt to be added has to be 10 gal to get 30 gallons
Let x = the decimal equiv of the percent required
:
10x + .12(20) = .14(30)
10x + 2.4 = 4.2
10x = 4.2 - 2.4
10x = 1.8
x = 1.8/10
x = .18 * 100 = 18% required

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total pure water 150 ml
5% salt added =5/100 of 150 ml
=7.5 ml
salted water solution will be =150+7.50 =157.50 ml
answer 157.50 ml

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(285 tagged seals)/(1100 observed seals) found after the tagged ones during the tagging operation were mixed into the rest of the untagged seals.

Before the tagged seals were put into the collection of untagged seals,
(5758 tagged seals)/(U total seals)
U is the population of seals.

Proportion arranged conveniently:
U=22200 to the best of possible accuracy.