# Questions on Word Problems: Mixtures answered by real tutors!

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Question 994765: If you wish to increase the percent of acid in 50 mL of a 15% acid solution in water to a 25% acid, how much pure acid must you add?
Thank you.

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If you wish to increase the percent of acid in 50 mL of a 15% acid solution in water to a 25% acid, how much pure acid must you add?
:
let x = amt of pure acid required
:
A typical mixture equation based on the amt of acid in the water
.15(50) + x = .25(x+50)
7.5 + x = .25x + 12.5
x - .25x = 12.5 - 7.5
.75x = 5
x = 5/.75
x = 6 mL of pure acid required

Question 994708: If sugar cost 0.75$for 250 ml how much would it cost for 355 ml of sugar ? Answer by fractalier(2141) (Show Source): You can put this solution on YOUR website! Set up a proportion... .75 is to 250 ml as x is to 355 ml...OR .75/250 = x/355 Now cross-multiply and solve for x: 250x = .75(355) x = .75(355) / 250 =$1.065

Question 994604: Mani Peanut Company wants to mix 20 kilos of peanuts worth P90 a kilo with cashew nuts worth P150 a kilo in order to make an experimental mix worth P110 a kilo.How many kilos of cashew nuts should be added to the peanuts?
Found 2 solutions by lwsshak3, josgarithmetic:
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Mani Peanut Company wants to mix 20 kilos of peanuts worth P90 a kilo with cashew nuts worth P150 a kilo in order to make an experimental mix worth P110 a kilo.How many kilos of
let x=amt of cashews to add to the peanuts
20*p90+x*p150=(x+20)p110
1800+p150x=1p110x+2200
40x=400
x=10
amt of cashews to add to the peanuts= 10 kilos

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y, the mass of the cashews to use.

Accounting for cost of the mixture, ;

Solve that for y.

Question 994590: A laboratory technician needs to make a 45-liter batch of a 40% acid solution. How can the laboratory technician combine a batch of an acid solution that is pure acid with another that is 10% to get the desired concentration?
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.
P=amount of pure acid; T=amount of 10% solution
P+T=45L
T=45L-P
.
(1.00)P+(0.10)T=(0.40)45L
P+(0.10)(45L-P)=18L
P+4.5L-0.10P=18L
0.90P=13.5L
P=15L
ANSWER 1: He should use 15 liters of pure acid.
T=45L-P=45L-15L=30L
ANSWER 2: He should use 30 liters of 10% acid solution.
.
CHECK:
(1.00)P+(0.10)T=(0.40)45L
15L+(0.10)30L=18L
15L+3L=18L
18L=18L

Question 994591: in a mixture of concrete there are three lb of cement mix for each pound of gravel .if the mixture contains a total of 76 lb of these two ingredients how many pounds of gravel are there?
Found 3 solutions by lwsshak3, macston, josgarithmetic:
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in a mixture of concrete there are three lb of cement mix for each pound of gravel .if the mixture contains a total of 76 lb of these two ingredients how many pounds of gravel are there?
let x=amt of gravel in mixture
76-x=amt of cement in mixture
(76-x)/x=3
3x=76-x
4x=76
x=19
76-x=57
amt of gravel in mixture=19 lbs

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.
G=amount of gravel; C=amount of cement=3G
.
G+C=76 lbs
G+3G=76 lbs
4G=76 lbs
G=19 lbs
.
ANSWER: There are 19 lbs of gravel in 76 lbs of mix.
.

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Fraction as cement: 3/4
Fraction as gravel: 1/4

Cut the 76 into four equal parts.

Question 994592: A laboratory technician needs to make a 42-liter batch of a 40% acid solution. How can the laboratory technician combine a batch of an acid solution that is pure acid with another that is 10% to get the desired concentration?
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.
P=amount of pure acid; T=amount of 10% solution
P+T=42L
T=42L-P
.
(1.00)P+(0.10)T=(0.40)42L
P+(0.10)(42L-P)=16.8L
P+4.2L-0.10P=16.8L
0.90P=12.6L
P=14
ANSWER 1: He should use 14 liters of pure acid.
T=42L-P=42L-14L=28L
ANSWER 2: He should use 28 liters of 10% acid solution.
.
CHECK:
(1.00)P+(0.10)T=(0.40)42L
14L+(0.10)28L=16.8L
14L+2.8L=16.8L
16.8L=16.8L

Question 994589: A laboratory technician needs to make a 36-liter batch of a 40% acid solution. How can the laboratory technician combine a batch of an acid solution that is pure acid with another that is 20% to get the desired concentration?
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Those are all the same question, but just slightly different examples.

x, volume of the 20% acid
36-x, volume of the 100% acid

Question 994526: How many pounds of coffee that is 40% Java beans must be mixed with 80lbs of coffee that is 30% Java beans to make coffee that is 32% Java beans?
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Mixture: one material quantity unknown
http://www.algebra.com/my/Two-Part-Mixture-with-one-material-quantity-unknown.lesson?content_action=show_dev

Question 994496: How many grams of pure gold must be mixed with 40% gold alloy to produce 100 grams of 52% alloy?
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Let = grams of pure gold needed
= grams of gold in 40% alloy
--------------------

20 grams of pure gold are needed
-------------
check:

OK

Question 994415: joe and sam use a metal alloy that is 19% copper to make jewelry. How many ounce of an allow that is 12% copper must be mixed with an alloy that is 25% copper to form 55 ounces of desired alloy?
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Here is a video example that can work the same way:
two-part mixture using just one variable

x, how much 12% silver
55-x, how much 25% alloy

Question 994425: Jimmy mixed 4 oz. of peanuts with 6 oz. of
mixed nuts containing 40% peanuts. What
percent of the new mixture is peanuts?

4 ml of a 5% saline solution was mixed with 3
ml of a 89% saline solution. Find the
concentration of the new mixture.

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4 ml of 5%
3 ml of 89%
.05(4)+.89*3=7*unknown %
.20+2.67=7*unknown=2.87
divide by 7, both sides
unknown %=41 or 0.41

Question 994364: how many liters of a 20% solution of a chemical should be mixed with 10 liters of a 50% solution to obtain a mixture that is 40% chemical
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how many liters of a 20% solution of a chemical should be mixed with 10 liters of a 50% solution to obtain a mixture that is 40% chemical
-----
Equation:
active + active = active
0.20x + 0.50*10 = 0.40(x+10)
-------
20x + 50*10 = 40x + 40*10
-------
-20x = -10*10
x = 5 liters (amt. of 20% solution needed)
-------------
Cheers,
Stan H.
-----------

Question 994283: One type of chocolate mixture contains 30% of nuts, by mass. A second type of chocolate mixture contains 20% of nuts, by mass. What mass of each type of chocolate mixture needs to be mixed to make a 500-g chocolate mixture that will have 24% nuts, by mass?
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y, mass to use of the 30% mixture
500-y, mass to use of the 20% nuts mixture

Want mixture result to be 24%.

Question 994279: confused on this word problem..need helpppp
20% alcohol solution and 70% alcohol solution available. how many liters of each must the pharmacy use to make a mixture containing 50 liters of 60% alcohol solution?

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= liters of 20% alcohol solution to be used
= liters of 70% alcohol solution to be used
= liters of alcohol in liters of 20% alcohol solution
= liters of alcohol in liters of 70% alcohol solution
= liters of alcohol in liters of 60% alcohol solution
When you mix the two solutions you want:
a total volume of 50 liters, so , and
a total alcohol amount of 30 liters, so .
So you have to solve
.

Solving:
-->-->
-->-->
-->-->-->
-->-->
-->-->
The pharmacy should mix 10 liters of 20% alcohol solution and 40 liters of 70% alcohol solution.

NOTE:If you did not study systems of equations, then use instead of , first find from ;
-->-->-->
-->-->-->--> .
Then you calculate

Question 994203: The perimeter of a rectangular farm is 10km.its length is one and half times its breadth.find the length and breadth of the farm
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The perimeter of a rectangular farm is 10km.its length is one and half times its breadth.find the length and breadth of the farm
-------
P = 2(length + width)
-----
10 km = 2((3/2)w + w)
----
5 = (5/2)w
---
length = (3/2)w = 3 km
----------
cheers,
Stan H.
----------

Question 994150: Suppose that one solution contains 50% alcohol and the other solution contains 80% alcohol. How many liters of each solution should be mixed to make 10.5 of a 70% solution?
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Suppose that one solution contains 50% alcohol and the other solution contains 80% alcohol. How many liters of each solution should be mixed to make 10.5 of a 70% solution?
----------------
Equation::
alcohol + alcohol = alcohol
0.50x + 0.80(10.5-x) = 0.70*10.5
----------
50x + 80*10.5 - 80x = 70*10.5
----------
-30x = -10*10.5
----------
x = (1/3)10.5
-----
x = 3.5 liters (amt. of 50% solution needed)
10.5-x = 7 liters (amt. of 80% solution needed)
-----------
Cheers,
Stan H.
-------------

Question 994088: the fastest temperature drop ever recorded was 49 deg F in 15 mins, recorded in Rapid City, South Dakota in 1911. how many degree C was this?
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Find the formula for converting degrees Fahrenheit to degrees Celsius. Substitute 49 for F in this formula. Do the arithmetic.

John

My calculator said it, I believe it, that settles it

Question 994006: A tank contains 2440 Liters of pure water. A solution that contains 0.03 kg of sugar per liter enters the tank at a rate of 5 liters per minute. The solution is mixed and drains at the same rate. Find the amount of sugar after 66 minutes.
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A tank contains 2440 Liters of pure water.
A solution that contains 0.03 kg of sugar per liter enters the tank at a rate of 5 liters per minute.
The solution is mixed and drains at the same rate.
Find the amount of sugar after 66 minutes.
:
Find the amt of sugar that entered the tank in 66 min
.03 * 5 * 66 = 9.9 kg of sugar entered the tank

Question 994064: A scientist needs 60 liters of a 60% acid solution. He currently has a 40% solution and a 70% solution. How many liters of each does he need to make the needed 60 liters of 60% acid solution?
Gabe neeeds ? liters of the 40% solution.
He also needs ? liters of the 70% solution.

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A scientist needs 60 liters of a 60% acid solution. He currently has a 40% solution and a 70% solution. How many liters of each does he need to make the needed 60 liters of 60% acid solution?
----------------
Equation:
acid + acid = acid
----------------------
0.40x + 0.70(60-x) = 0.60*60
-----
40x + 70*60 - 70x = 60*60
-----
-30x = -10*60
x = 20 liters
----
Gabe needs 20 liters of the 40% solution.
He also needs 40 liters of the 70% solution.
-----
Cheers,
Stan H.

Question 994028: How many 2 and 1/2 pound sugar jars can be filled out of 36 pound container?
Found 2 solutions by ikleyn, rfer:
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.
: = = .

14 jars can be filled.

15-th jar will be filled in of its capacity.

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36/2.5=14.4 jars
It says filled so it equals 14

Question 993897: Manha wants to make a 36 percent acid solution. She has already 3 leters of a 72 percent acid solution. How many litres of a 9 percent acid solution must she add.
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= liters of acid in 72% solution
Let = liters of 9% solution needed
= liters of acid in 9% solution
-----------------------------------

4 liters of 9% solution are needed
check:

OK

Question 993865: How many kg of wheat selling at Rs. 5.20 per kg should be mixed with wheat selling at 4.40 per kg to make a mixture of 30 kg selling at Rs. 144.80 ?
Found 2 solutions by MathTherapy, addingup:
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How many kg of wheat selling at Rs. 5.20 per kg should be mixed with wheat selling at 4.40 per kg to make a mixture of 30 kg selling at Rs. 144.80 ?
Let amount of Rs. 5.20 wheat to mix, be F
Then amount of Rs. 4.40 wheat to mix = 30 - F
We then get the equation: 5.2F + 4.4(30 - F) = 144.8
5.2F + 132 - 4.4F = 144.8
5.2F - 4.4F = 144.8 - 132
.8F = 12.8
F, or amount of Rs. 5.20-wheat to mix = , or  kg
Amount of Rs.-4.40 wheat to mix: 30 - 16, or  kg



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See the attached picture for the answer

Question 993840: write the equation of a line perpendicular to 9x+5y=4 that passes through the point (9,-5)
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.
The line

5x - 9y = n         (*)

is perpendicular to the given line.

Find the number of  n  from the condition that this line passes trough the point  (9,-5).  For it,  substitute  x=9  and y=-5  into the equation  (*).

Question 993766: This mixture problem has me stuck. The problem: A hospital needs a 50% dextrose solution. Only a 75% and 40% solution are in stock. How much of the 75% solution should be mixed with 500 oz of the 40% solution to get a 50% dextrose solution. My teacher showed us how to set up the chart with the percentages and how to multiply across and add down but the column where the variables is confusing me.

Found 2 solutions by MathTherapy, Theo:
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This mixture problem has me stuck. The problem: A hospital needs a 50% dextrose solution. Only a 75% and 40% solution are in stock. How much of the 75% solution should be mixed with 500 oz of the 40% solution to get a 50% dextrose solution. My teacher showed us how to set up the chart with the percentages and how to multiply across and add down but the column where the variables is confusing me.
Yes, the chart is good, but I’ll do the problem without it

Let the amount of 75% solution to be mixed, be S
Then amount of 75% solution in the resulting mixture = .75S
Amount of 40% solution to be mixed: 500 oz
Then amount of 40% solution in the resulting mixture = .4(500)
Equation for the resulting mixture is: .75S + .4(500) = .5(S + 500)
.75S + 200 = .5S + 250
.75S - .5S = 250 – 200
.25S = 50
S, or amount of 75% solution to mix = , or  oz



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i'm not sure how helpful they are here.

the basic problem is this.

you have two solutions.
we'll call them A and B.

A is a 40% solution
B is a 75% solution

C is the new solution that you want that is a 50% solution.

you have 500 ounces of the 40% solution.
you have x ounces of the 75% solution.
you will have 500 + x ounces of the 50% solution.

.40 * 500 + .75 * x = .50 * (500 + x)

you need to solve for x.

simplify the equation to get:

200 + .75 * x = 250 + .50 * x

subtract .50 * x from both sides of the equation and subtract 200 from both sides of the equation to get:

.75 * x - .50 * x = 250 - 200

combine like terms to get:

.25 * x = 50

divide both sides of this equation by .25 to get:

x = 50 / .25 = 200

go back to your original equation of .40 * 500 + .75 * x = .50 * (500 + x) and replace x with 200 to get:

.40 * 500 + .75 * 200 = .50 * (500 + 200)

simplify this equation to get:

200 + 150 = 250 + 100

combine like terms on each side of this equation to get:

350 = 350.

this confirms the value of x is equal to 200 to be good.

i had difficulty fitting a chart to this problem.

best i could come up with it this:

mix                      A             B            C
total ounces             500           x            x + 500
persent solution         40%           75%          50%
ounces solution          .4 * 500      .75 * x      .50 * (x + 500)


Question 993788: If there is 3 cups of sugar for every 2 cup of vinegar how much sugar for 1 cup of vinegar.
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ratio of sugar to vinegar is 3/2.

you want to keep that ratio the same.

3/2 = x/1

cross multiply to get 2x = 3

divide both sides of the equation by 2 to get x = 3/2.

you would need 3/2 cup of sugar for 1 cup of vinegar.

the ratio of 3/2 is the same as the ratio of (3/2) / 1

Question 993666: The store manager decides to make 40lbs of blend of coffees that he plans to sell for 10 dollars a lb. He mixes some coffee that sells for 9 dollars a lb with coffee that sells for 12 dollars a lb. How much of each should he use? (Hint- one coffee is x and the other coffee of the rest of the mixture so use 40 - x for the amount)
I already calculated that 40 lbs of coffee costing 10 dollars a lb would be 400 dollars. Now I need help on how to calculate how many pounds adding up to 40lbs. I would use of each that would bring me to 400 dollars total.
Thank you!

Found 2 solutions by MathTherapy, Theo:
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The store manager decides to make 40lbs of blend of coffees that he plans to sell for 10 dollars a lb. He mixes some coffee that sells for 9 dollars a lb with coffee that sells for 12 dollars a lb. How much of each should he use? (Hint- one coffee is x and the other coffee of the rest of the mixture so use 40 - x for the amount)
I already calculated that 40 lbs of coffee costing 10 dollars a lb would be 400 dollars. Now I need help on how to calculate how many pounds adding up to 40lbs. I would use of each that would bring me to 400 dollars total.
Thank you!
Since you wish to use x for the amount of one of the blends, then we'll make the amount of the $9/lb-blend to mix, be x Then amount of the$12/lb-blend to mix is: 40 - x
We then get the following equation:
9x + 12(40 - x) = 40(10)
9x + 480 - 12x = 400
- 3x = 400 - 480
- 3x = - 80
x, or amount of $9/lb-blend to mix = , or lbs Amount of$12/lb-blend to mix: , or  lbs



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x is the amount of coffee at 9 dollars a pound.
y is the amount of coffee at 12 dollars a pound.

you have two equations:

x + y = 40

9x + 12y = 400

solve these equations simultaneously, and you will get:

x = 26.67
y = 12.33

x + y = 40 becomes 26.67 + 13.33 = 40

9x + 12y = 400 becomes 9 * 26.67 + 12 * 13.33 which becomes 240.03 + 159.96 which becomes 299.99.

the difference from 400 is due to rounding.

26.67 is really 26.6666666666666........

13.33 is really 13.333333333333333........

to get this solution, you need to be able to solve the system of equations simultaneously.

here's a reference on how to do that.

http://www.regentsprep.org/regents/math/algebra/ae3/indexAE3.htm

i'll use elimination method.

x + y = 40
9x + 12y = 400

multiply both sides of the first equation by 9 to get:

9x + 9y = 360
9x + 12y = 400

subtract the first equation from the second equation to get:

3y = 40

divide both sides of this equation by 3 to get:

y = 13.33.....
this rounds to 13.33
use that value of y to solve for x in either equation to get:

x = 26.66.....
this rounds to 26.67

Question 993677: One solution contains 40% alcohol and another contains 72% alcohol. How much of each should be mixed together to obtain 16 ounces of a 62% alcohol solution?
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Let x be the amount of 40% alcohol. The 16-x must be the amount of 72% alcohol. The equation becomes, once you make percents into whole numbers...
x(40) + (16-x)(72) = 16(62)
40x + 1152 - 72x = 992
Simplifying gives
-32x = -160
and
x = 5
So 5 oz of 40% and 11 oz of 72% will do the trick.

Question 993597: How much of an alloy that is 20% copper should be mixed with 600 ounces of an alloy that is 70% copper in order to get an alloy that is 30% copper?
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Let x = the weight of the 20% alloy
Therefore the total weight of the mixture is x + 600
We can write the following equation which equates the amount of copper from each of the two alloys to the total amount contained in the mixture:
0.2x + 0.7*600 = 0.3(x+600)
Solve for x:
0.1x = = 240
x = 2400 oz
Check:
0.2*2400 + 0.7*600 = 480 + 420 = 0.3(3000) = 900

Question 993626: How many liters of a 25% salt solution must be added to 20 liters of a 12% solution to get a solution that is 20% salt?
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How many liters of a 25% salt solution must be added to 20 liters of a 12% solution to get a solution that is 20% salt?
-------
Equation:
salt + salt = salt
0.25x + + 0.12*20 = 0.20(x+20)
-----
25x + 12*20 = 20x + 20*20
-----
5x = 8*20
x = 8*4 = 32 liters (amt of 25% solution to add)
---------
Cheers,
Stan H.
------------

Question 993621: A jeweler has 5 rings, each weighing 18 grams, made of an alloy of 18% silver and 90% gold. She decides to melt down the rings and added enough silver to reduce the gold content to 75%. How much silver should she add?
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A jeweler has 5 rings, each weighing 18 grams, made of an alloy of 18% silver and 90% gold. She decides to melt down the rings and added enough silver to reduce the gold content to 75%. How much silver should she add?
-------
Weight of rings:: 90 grams
If the result is 75% gold, it must be 25% silver.
----
Equation:
silver + silver = silver
18%(weight) silver + 100%(weight) silver = 25%(weight) silver
-----------------------------------------
0.18*90 + 1*x = 0.25(90+x)
Multiply thru by 100 to get::
---
18*90 + 100x = 25*90 + 25x
Solve for "x", which is the amt. of 100% silver you add
----
100x-25x = 25*90-18*90
------
75x = 7*90
x = (19/15)*7
x = 8.8667 grams (amt. of 100% silver to add)
--------
Cheers,
Stan H.
----

Question 993573: How many liters of 20% acid solution should be mixed with 50% acid solution to obtain 12 liters of a 30% acid solution?
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Same method as here, but just different values:
Only one variable needed for two-part mixture problem

Question 993498: How much fresh water must be added to 3 L of 10% salt water to make the solution 5% salt water?
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3 l *0.10=0.3
add x liters with no salt
(3+x) *0.05=0.15+0.05x
0.3=0.05x+0.05 x
0.10x=0.3
x=3 liters
====
Easier way
You are diluting the solution in half from 10% salt to 5% salt.
This will mean you are doubling the amount of solution, so you add 3 liters of water.

Question 993456: a factory uses 15 pounds of every steel 18 pounds of copper. how much copper will the factoru use for 2,700 pounds of steel?
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.
x=pounds of copper
.

ANSWER: The factory will use 3240 lbs
of copper for every 2700 lbs of steel.

Question 993445: Clifford is making a really big white cake for his sister's birthday. The recipe calls for 1.5 times as much flour as sugar. It also calls for 1/2 as much butter as sugar. All the butter, flour, and sugar together total 7.5 cups. How much butter, sugar, and flour does clifford need? write and solve an equation.
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What part don't you understand; writing the equation or solving the equation once you have written it?

[Student reply]: "I don't understand how to write the equation."

Let represent the number of cups of sugar. I chose sugar to be represented by the basic variable because everything else is expressed by a multiplier of the amount of sugar. The amount of flour is cups and the amount of butter is cups. The total is 7.5 cups so:

Solve for , then calculate and .

John

My calculator said it, I believe it, that settles it

Question 993316: Some warehouse stores charges members an annual fee to shop there. On his first trip to a warehouse store, Mr. Marsh pays $50 membership fee. Cases of bottled water cost$4 at the warehouse store. Write and solve an equation to find the total amount mr. Marsh spent on his first trip before tax if he bought 8 cases of water.
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a = b + c * d
a is the total cost
b is the cost of membership
c is the cost per case
d is the number of cases.

the actual letters used don't really mean anything.
it's what they represent that's important.
sometime you can make them meaningful, but other times you can't because the same letter would have to be used more than once, or you might have to resort to variable names that are more than one letter long.

i'll show you another solution where i used more meaningufl names for the variables.

you know that:

b = 50
c = 4
d = 8

the actual letters used don't really mean anything.
it's what they represent that's important.

formula becomes:

a = 50 + 4 * 8 which becomes:

a = 50 + 32 which becomes:

a = 82.

total cost is equal to 82.

now here's the same solution with more meaninful names to the variables involved.

tc = total cost
mf = membership fee
cpc = cost per case
nc = number of cases.

formula is:

tc = mf + cpc * nc

you have:
mf = 50
cpc = 4
nc = 8

formula becomes:

tc = 50 + 4 * 8

you can even just use the names in english as the variable names.

example:

total cost = membership fee + cost per case * number of cases.

membership fee = 50
cost per case = 4
number of cases = 8

formula becomes:

total cost = 50 + 4 * 8

Question 993323: Adding 25 gallons of acid to a mixture, makes it 80% Acid solution, while adding 25 gallons of water makes it 60% Acid solution. What is the original acid percentage of the mixture?
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You are not given the amount of original acid mixture, so give this a variable for its volume.
v, the initial volume in gallons of the starting mixture of unknown percent concentration.

Let p be the percent concentration of your original acid mixture.
Additionally, you are forced to assume that your acid to add in pure form, the 100%, is a liquid.

The system of equations slightly reworked is

STEPS SOLVING THE SYSTEM

-
Next use the simpler, dilution equation.

Question 993223: A silversmith mixed 35 g of an 80% silver alloy with 40 g of a 25% silver alloy. What is the percent concentration of the resulting alloy? (Round your answer to one decimal place.)