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Tutors Answer Your Questions about Mixture Word Problems (FREE)
Question 241446: The manager of a bulk foods establishment sells a trail mix for $6 per pound and premium cashews for $12 per pound. The manager wishes to make a 135-pound trail mix-cashew mixture that will sell for $10 per pound. How many pounds of each should be used?
Answer by checkley77(7072)   (Show Source):
You can put this solution on YOUR website!12X+6(135-X)=135*10
12X+810-6X=1350
6X=1350-810
6X=540
X=540/6
X=90 LBS. OF $12 CASHEWS IS USED.
135-90=45 LBS. OF $6 TRAIL MIX IS USED.
PROOF:
12*90+6*45=135*10
1080+270=1350
1350=1350
Question 241234: Could someone please help me with this mixture problem? I have been pouring over it for two days (total of over 4 hours), and have tried various methods none of them yielding the correct answer.
A mechanic has 363 gallons of gasoline and 17 gallons of oil to make gas/oil mixtures. He wants one mixture to be 6% oil and the other mixture to be 2.5% oil. If he wants to use all of the gas and oil, how many gallons of gas and oil are in each of the resulting mixtures?
Answer by edjones(3299)  (Show Source):
You can put this solution on YOUR website!Let x=gals of gas required to make a .025 mixture.
Let y=gals of oil required to make a .025 mixture.
.
A) y/(x+y)=.025
B) (17-y)/(363-x+17-y)=.06
.
A)
y=.025x+.025y
.975y=.025x
y=.0256x
.
B)
(17-.0256x)/(380-x-.0256x)=.06
17-.0256x=22.8-.0615x
.0359x=5.8
x=161.56 gal
.
A)
y/(161.56+y)=.025
y=4.039+.025y
.975y=4.039
y=4.142 gal
It wont be exact because of rounding errors.
.
Ed
Question 241134: To estimate the size of the bear
population on the Keweenaw Peninsula, conservationists
captured, tagged, and released 50 bears. One year later, a
random sample of 100 bears included only 2 tagged bears.
What is the conservationist’s estimate of the size of the
bear population?
Answer by ankor@dixie-net.com(6693)  (Show Source):
You can put this solution on YOUR website! To estimate the size of the bear population on the Keweenaw Peninsula,
conservationists captured, tagged, and released 50 bears.
One year later, a random sample of 100 bears included only 2 tagged bears.
What is the conservationist’s estimate of the size of the bear population?
:
let x = estimation of no. of bears
;
 = 
cross multiply
2x = 100 * 50
2x = 5000
x = 2500 bears
Question 241109: Joseph has a 55% silver alloy. Ethan has a 75% silver alloy. They melted their alloys to create 160 grams of a 64% silver alloy. How many grams of each alloy?
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!1): Joseph has a 55% silver alloy. Ethan has a 75% silver alloy.
They melted their alloys to create 160 grams of a 64% silver alloy.
How many grams of each alloy?
:
Let x = amt of 75% alloy
then
(160-x) = amt of 55% alloy
:
Write an amt of silver equation:
.75x + .55(160-x) = .64(160)
:
You should be able to solve this now, if not, let me know via the comment.
Question 241085: If you have 200 ounces of 50% copper how many ounces of 20% copper would you add to get a 30% alloy?
Answer by nyc_function(260)  (Show Source):
Question 241000: If a pharmacist needs 70L of a 50% alchohol solution but only has a 30% alchohol solution and a 80% alchohol solution how many L of each would make a 50% solution?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!If a pharmacist needs 70L of a 50% alchohol solution but only has a 30% alchohol solution and a 80% alchohol solution how many L of each would make a 50% solution?
--------------------------------------
Equation:
alcohol + alcohol = alcohol
0.30x + 0.80(70-x) = 0.50*70
Multiply thru by 100 to get:
30x + 80*70 - 80x = 50*70
-50x = -30*70
x = 42 L (amt. of 30% solution needed in the mixture)
---
70-42 = 28 L (amt. of 80% solution needed in the mixture)
============================================================
Cheers,
Stan H.
Question 240531: How many millilitres of water much be added to 100mL of a 50% solution of sulphuric acid to make a 10% solution?
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!How many millilitres of water much be added to 100mL of a 50% solution of sulphuric acid to make a 10% solution?
:
let x = amt of water required to accomplish this
:
Write an amt of acid equation
.50(100) = .10(100+x)
:
50 = 10 + .1x
:
50 - 10 = .1x
:
x = 
x = 400 ml of water required
:
:
Check solution in the amt of acid equation
.50(100) = .10(100 + 400)
50 = .10(500)
Question 240881: please help me....
How many grams of a 25% silver alloy is to be melted with 30 grams of a 55% silver alloy to obtain a 32% silver alloy?
Answer by solver91311(5072)  (Show Source):
Question 240619: a nurse mix 15 liters of a 10% solution of s drug with som 60% solution to get a 20% MIXTURE. how many liters of the 60% solution will be needed?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.60x+.10*15=.20(15+x)
.60x+1.5=3+.20x
.60x-.20x=3-1.5
.40x=1.5
x=1.5/.4
x=3.75 liters of 60% solution is used.
Proof:
.60*3.75+.10*15=.20(15+3.75)
2.25+1.5=.20*18.75
3.75=3.75
Question 240401: A chemist plans to mix 40 ml of a 5% alcohol solution with some 11% alcohol solution to make a 7% solution. How much of the 11% solution should be mixed with the 5% alcohol solution to obtain the 7% alcohol solution?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.11X+.05*40=.07(40+X)
.11X+2=2.8+.07X
.11X-.07X=2.8-2
.04X=.8
X=.8/.04
X=20 ML. OF 11% ALCOHOL IS USED.
PROOF:
.11*20+2=.07(40+20)
2.2+2=.07*60
4.2=4.2
Question 239567: 40 grams of a 50% acid solution. How much water must be added to make a 10% solution
Answer by checkley77(7072) (Show Source):
Question 239139: Mixture A is 7.5% acid. To raise the concentration of acid to 10%, some pure acid will be added to the mixture. How many liters of mixture A and how many liters of pure acid are needed to end up with 200 liters of 10% solution?
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!Mixture A is 7.5% acid. To raise the concentration of acid to 10%, some pure acid will be added to the mixture.
How many liters of mixture A and how many liters of pure acid are needed to end up with 200 liters of 10% solution?
:
Let x = amt of pure acid required,
It says the total has to be 200 liters, therefore
(200-x) = 7.5% stuff (A)
:
1.0x + .075(200-x) = .10(200)
:
x + 15 - .075x = 20
:
x - .075x = 20 - 15
:
.925x = 5
x = 
x = 5.4 liters of pure acid require
then
200 - 5.4 = 194.6 liters of 7.5% stuff (A)
;
:
Check solution in original equaiton
5.4 + .075(194.6) = .10(200)
5.4 + 14.6 = 20
Question 238768: I getting stumped on this problem, please help. A merchant has coffee worth $20 a pound that she wishes to mix with 80 pounds of coffee worth $90 a pound to get a mixture that can be sold for $60 a pound. How many pounds of the $20 coffee should be used?
Answer by Earlsdon(4900) (Show Source):
You can put this solution on YOUR website!Let x = the number of pounds of the $20-coffee. After mixing this amount with the 80 pounds of $90-coffee, the merchant will have (80+x) pounds of coffee worth $60 per pound.
Let's set up the algebraic equation to solve for x.
x($20)+80($90) = (80+x)($60) Simplify and solve for x.
20x+720 = 480+60x Subtract 20x from both sides.
720 = 480+40x Subtract 480 from both sides.
240 = 40x Finally, divide both sides by 40.
60 = x
The merchant will need to mix 60 pounds of the $20-coffee with 80 pounds of the $90-coffee to obtain (80+x = 140) pounds of $60-coffee.
Question 238771: I'm having some problems and need help. In a chemistry class, 3 liters of a 4% silver iodide solution must be mixed with a 10% solutin to get a 6% solution. How many liters of the 10% solution are needed?
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!In a chemistry class, 3 liters of a 4% silver iodide solution must be mixed with a 10% solution to get a 6% solution.
How many liters of the 10% solution are needed?
;
Let x = amt of 10% solution required
:
Write a decimal equiv equation:
.04(3) + .10x = .06(x + 3)
:
.12 + .1x = .06x + .18
"
.1x - .06x = .18 - .12
:
.04x = .06
x = 
x = 1.5 liters of 10% solution required
;
;
Check in the original equation
.04(3) + .1(1.5) = .06(1.5+3)
.12 + .15 = .06(4.5)
.27 = .27; confirms our solution
Question 238584: I need help setting up the following word problem.
NOT LOOKING FOR THE ANSWER LOOKING FOR THE METHOD PLEASE.
How much pure acid should be mixed with 6 gallons of a 50% acid solution in order to get an 80% acid solution?
What I have is wrong I am sure:
.80=.50(x+6)+(x)
Any assistance would be greatly appreciated.
Found 2 solutions by ankor@dixie-net.com, Masaries7:
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!How much pure acid should be mixed with 6 gallons of a 50% acid solution in order to get an 80% acid solution?
:
using the decimal equiv of %; pure acid would be 1.0x
:
.50(6) + 1.0x = .80(x+6)
3 + 1x = .8x + 4.8
1x - .8x = 4.8 - 3
.2x = 1.8
x = 
x = 9
;
:
Check this in the amt of acid equation
.5(6) + 1(9) = .8(9 + 6)
3 + 9 = .8(15)
12 = 12
Answer by Masaries7(3)  (Show Source):
You can put this solution on YOUR website!Your question consists of ratios.
.5(6)/6= 3/6 3 is the gallons of acid. 6 is the total gallons
To find out how much more gallons of acid you need to obtain 80% acid levels we add our unknown amount to bottom and top of the ratio indicating for the top more acid gallons and for the bottom more total gallons overall.
3+x/6+x= .8
Solve for 'x'
The format for this type of problem is to figure out your initial amounts then set up a ratio equaling the new desired percentage. You must be sure of how the information is applied in the problem for instance had we been adding non-acid solution then there would be no 'x' variable on the top since the ratio we have is acid gallons/ total gallons. Adding non-acid does not apply to acid. Good tip is to visualize the problem in your mind (if it's possible don't get brain freeze however)
Question 238513: One week 10 copies of the novel The DaVinci Code by Dan Brown were sold for every 1.9 copies of John Grogan's Maryley and Me that were sold. If a total of 3570 copies of the two books were sold, how many copies of each were sold?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!One week 10 copies of the novel The DaVinci Code by Dan Brown were sold for every 1.9 copies of John Grogan's Marley and Me that were sold. If a total of 3570 copies of the two books were sold, how many copies of each were sold?
--------------
10:1.9 is the same as 10x:1.9x where 10 is the number of DaVinci sold.
---
Equation:
10x + 1.9x = 3570
11.9x = 3570
x = 3570/11.9
x = 300
---------------
DaVinci = 10x = 10*300 = 3000
Marly = 1.9x = 570
==============================
Cheers,
Stan H.
Question 238465: At a health food store, dried apple slices that cost $1.80 per pound are mixed with dried banana slices that cost $2.10 per pound. If the mixture weighs 5 pounds and costs $1.92 per pound, how many pounds of apple slices were used?
A) 1.5 pounds of apples B) 2 pounds of apples C) 3 pounds of apples
D) 3.5 pounds of apples E) none of these
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!2.10X+1.80(5-X)=1.92*5
2.10X+9-1.80X=9.6
.3X=9.6-9
.3X=.6
X=.6/.3
X=2 POUNDS OF BANANA SLICES IS USED.
5-2=3 POUNDS OF APPLE SLICES IS USED.
ANSWER C).
PROOF:
2.10*2+1.8*3=1.92*5
4.2+5.4=9.6
9.6=9.6
Question 238350: a 2004 Pontiac Montana has an engine coolant system capacity of 10.01 liter. If the system is currently filled with a mixture that is 30% antifreeze, how much of it should be drained and replaced with pure antifreeze so that they system is filled with a mixture that is 50% antifreeze?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Pontiac Montana has an engine coolant system capacity of 10.01 liter. If the system is currently filled with a mixture that is 30% antifreeze, how much of it should be drained and replaced with pure antifreeze so that they system is filled with a mixture that is 50% antifreeze?
-----------------------
Let "x" be the amount that should be drained and replaced.
------------------------------
Equation:
0.30*10.01 - 0.30x + 1.00x = 0.50*10.01
---
Multiply thru by 100 to get:
30*10.01 + 70x = 50*10.01
70x = 20*10.1
x = 2.89 liters (amount that should be drained and replaced)
================================================================
Cheers,
Stan H.
Question 238300: how many liters of a 75% solution should be mixed with 55% solution to get 70 liters of a 63% solution?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.75X+.55(70-X)=.63*70
.75X+38.5-.55X=44.1
.20X=44.1-38.5
.20X=5.6
X=5.6/.20
X=28 LITERS OF 75% SOLUTION.
70-28=42 LITERS OF 55% SOLUTION.
PROOF:
.75*28+.55*42=.63*70
21+23.1=44.1
44.1=44.1
Question 238189: How many liters of 20% alcohol solution and 10% alcohol solution must be mixed to obtain 50 liters of a 12% alcohol solution?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.20x+.10(50-x)=.12*50
.20x+5-.10x=6
.10x=6-5
.10x=1
x=1/.1
x=10 liters of 20% alcohol is used.
50-10=40 liters of 10% alcohol is used.
Proof:
.20*10+.10*40=.12*50
2+4=6
6=6
Question 238262: how many liters of 20% salt solution must be added to 72 liters of 59% salt solution to get a 44% salt solution?
Answer by alicealc(98) (Show Source):
You can put this solution on YOUR website!volume of 20% salt solution = x
20% * x + 59% * 72 = 44% * (x + 72)
0.2x + 0.59 * 72 = 0.44 * (x + 72)
0.2x + 42.48 = 0.44x + 0.44 * 72
0.2x + 42.48 = 0.44x + 31.68
42.48 - 31.68 =0.44x - 0.2x
10.8 = 0.24x
x = 10.8/0.24 = 45
so, the volume of 20% salt solution to be added is 45 liters
Question 238210: how many liters of 20% alcohol solution and 10% alcohol solution must be mixed to obtain 50 liters of a 12% alcohol solution
Answer by nyc_function(260) (Show Source):
Question 238192: i need help with all kind of words problems. i just dont understand how to do them.
Answer by College Student(217)  (Show Source):
Question 238118: How much drug would you need to make two onces of a 20% Mixture.
Answer by solver91311(5072) (Show Source):
Question 237613: A 25 – ounce solution is 20% alcohol . If 50 ounces of water added to it , what
percent of the new solution is alcohol?
A. 5 B. 6 2/3 C. 8 D. 10 E. none of these
Answer by Edwin McCravy(2922) (Show Source):
You can put this solution on YOUR website!A 25 – ounce solution is 20% alcohol . If 50 ounces of water added to it , what
percent of the new solution is alcohol?
20% of the 25 ounces is alcohol and the rest of the
25 ounces is water.
20% of the 25 ounces is 5 ounces, so 5 ounces of the
25 ounces is alcohol and the other 20 ounces is water.
Now if we add 50 ounces of water to those 20 ounces,
there is 70 ounces of water, but still only 5 ounces
of alcohol, so  or about 7.1% alcohol.
Edwin
Question 237399: An advertisement for grape drink claims that the drink contains 10% grape juice. How much pure grape juice would have to be added to 5 quarts of the drink to obtain a mixture containing 40% grape juice?
Answer by josmiceli(3013) (Show Source):
You can put this solution on YOUR website!In words:
(quarts of pure grape juice at end)/(total quarts of drink at end) = 40%
given:
Let  = quarts of pure grape juice to be added
In  quarts of drink, I have
 quarts pure grape juice
-----------------------------------------
2.5 quarts of pure grape drink must be added
check:
OK
Question 237023: I'm not sure how to even set this up.
My son has $112 dollars made up of dimes, quarters and one dollar bills. The amount of one dollar bills is 52 dollars and the amount of quarters is two times the amount of dimes. How many does he have of each?
I got as far as 112-52= 70 (to be left with only the coins to resolve), but I didn't understand how to get any further. Please help!
Thanks!
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!My son has $112 dollars made up of dimes, quarters and one dollar bills. The amount of one dollar bills is 52 dollars and the amount of quarters is two times the amount of dimes. How many does he have of each?
--------------------------------------------------
Value Equation: 10d + 25q + 5200 = 11200 cents
10d + 25q = 6000 cents
q = 2d
10d + 50d = 6000
60d = 6000
d = 100 (# of dimes)
q = 2d = 200 (# of quarters)
b = 52 (# of dollars)
========================================
Cheers,
Stan H.
Question 236811: A chemist needs 160 milliliters of a 25% solution but has only 5% and 69% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
Answer by nerdybill(2448)  (Show Source):
You can put this solution on YOUR website! A chemist needs 160 milliliters of a 25% solution but has only 5% and 69% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
.
Let x = liters of 5% solution
then
160-x = liters of 69% solution
.
.05x + .69(160-x) = .25(160)
.05x + 110.4 - .69x = 15
110.4 - .64x = 15
-.64x = -95.4
x = 149.0625 milliliters (5% solution)
.
69% solution:
160-x = 160-149.0625 = 10.9375 milliliters (69% solution)
Question 236651: I need help, can you please help me?!?!?
Simplify each expression using the square root method.
sq root of -16 plus 2
Found 2 solutions by stanbon, checkley77:
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!Simplify each expression using the square root method.
sq root of -16 plus 2
------------------------------
sqrt(-16) + 2
= 4i + 2
= 2 + 4i
======================
Cheers,
Stan H.
Answer by checkley77(7072) (Show Source):
Question 236541: Soybean meal is 12% protein; Cornmeal is 6% protein. How many pounds of each should be mixed together to get a 240 pound mixture that is 7% protein?
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.12x+.06(240-x)=.07*240
.12x+14.4-.06x=16.8
.06x=16.8-14,4
.06x=2.4
x=2.4/.06
x=40 pounds of soybeans was used.
240-40=200 pounds of cornmeal was used.
Proof:
.12*40+.06*200=16.8
4.8+12=16.8
16.8=16.8
Question 236564: how much does an alloy that is 20% copper should be mixed with 200 onces of an alloy that is 90% copper in order to get an alloy that is 30% copper
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!how much does an alloy that is 20% copper should be mixed with 200 onces of an alloy that is 90% copper in order to get an alloy that is 30% copper
-----------------
Equation:
copper + copper = copper
0.20x + 0.90*200 = 0.30(x+200)
Multiply thru by 100 to get:
20x + 90*200 = 30x + 30*200
-10x = -60*200
x = 1200 onces (amt of 20% alloy needed in the mixture)
==========================================================
Cheers,
Stan H.
Question 236081: Suppose an exam has 28 questions on it and students need to answer 18 of the questions. How many ways can this be done if the first 5 and last 4 questions must be answered and question #19 must not be answered because of a typo?
Answer by Theo(675)  (Show Source):
You can put this solution on YOUR website!28 questions
answer 18
must answer first 5 and last 4
must not answer number 19
numbers are 1 through 28
take away first 5 to get 6 through 28
take away last 4 to get 6 through 24
take away 19 to get:
6 through 18 and 20 through 24
18 have to be answered.
4 + 5 = 9 slots that have to be picked in all cases.
that leaves 9 slots that can be chosen out of the ones that are still available.
slots available for those 9 are 6 through 18 and 20 through 24
that's 13 + 5 = 18 slots that are available for the 9 slots that still have to be filled.
since 9 have to be answered, we have number of possible combinations of:
9 out of 18.
formula for number of combinations equals n! / ((n-x)!*x!)
with n = 18 and x = 9, this formula becomes:
18! / ((18-9)!*9!)
that becomes 18! / (9!*9!) = 48620 different ways
your answer is:
48620 different ways that the questions can be answered ***********************
to see how this works, pick smaller numbers.
assume the following:
numbers are 1 through 10
first 2 and last 1 have to be picked.
slot number 6 is not available.
available slots become 3 through 5 and 7 through 9.
you have to answer 5 questions
number of available slots is 3 and 3 = 6
number of slots frozen are 3
total of 5 slots have to be filled leaving 2 slots with options.
we have 6! / ((6-2)! * 2!)
this becomes 6! / (4! * 2!)
this becomes 6*5*4!/(4!*2!) which becomes 6*5/2*1 which becomes 3*5 = 15
the possible ways the test can be answered is 15.
let's see how that works out.
slots 1 and 2 and 10 are required.
slot 6 can't be used.
slots available are:
345789
how many possible combinations of 2 can we make of this?
we can make.............
34
35
37
38
39
45
47
48
49
57
58
59
78
79
89
that's a total of 15.
stick 12 in front of each of these and add 10 at the end of each of these and we have the full set including the frozen slots that had to be filled.
example substituting hexadecimal digit a for 10:
total of 5 slots have to be filled.
slots 1 and 2 and a are frozen for all possible combinations.
1234a
1235a
1237a
1238a
1239a
1245a
1247a
1248a
1249a
etc.....
Question 236320: a motor boat travels 36km down stream in 2 hours . in coming back up streams, the
trip takes 3 hours. find the rate of the boat in still water and the rate of the current.(solve using linear system)
Answer by Theo(675) (Show Source):
You can put this solution on YOUR website!rate * time = distance.
the rate is a combination of the boat and the stream.
with the stream, the rate is (b+s)
against the stream, the rate is (b-s)
b is the rate of the boat
s is the rate of the stream.
h is the time in hours.
d = distance
going downstream, the formula for rate * time = distance becomes:
(b+s) * 2 = 36 (equation 1)
coming back up stream, the formula for rate * time = distance becomes:
(b-s) * 3 = 36 (equation 2)
since both these formulas are equal to 36, then they both must be equal to each other, so we have:
(b+s) * 2 = (b-s) * 3
remove parentheses to get:
2b + 2s = 3b - 3s
subtract 2b and add 3s to both sides of this equation to get:
3s + 2s = 3b - 2b which becomes:
5s = b
replace b with 5s in equation 1 to get:
(b+s) * 2 = 36 (equation 1) becomes:
(5s+s) * 2 = 36 which becomes
6s * 2 = 36 which becomes:
12s = 36 which becomes
s = 3
you have:
s = 3
b = 5s = 5*3 = 15
replace s and b in both original equations to confirm the answers are good for both equations.
both original equations are:
(b+s) * 2 = 36 (equation 1)
(b-s) * 3 = 36 (equation 2)
replace b with 15 and s with 3 to get:
(15+3) * 2 = 36 (equation 1)
(15-3 * 3 = 36 (equation 2)
these equations bgecome:
18*2 = 36
12*3 = 36
since both equations are true, the answer is confirmed.
your answer is:
rate of the boat is 15 miles per hour.
rate of the stream is 3 miles per hour.
Question 236154: Carol has 2 rose bushes; one with red flowers and one with yellow flowers. The red bush has 3 times as many flowers as her yellow floers,Together there are 124 flowers. How many of each color flower does she have? What iis the equation(or system equation)
Answer by rfer(2688) (Show Source):
Question 236048: how many kilograms of water must be added to 10 kg of a 75% antifreeze solution in order to produce a 50% solution?
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website! how many kilograms of water must be added to 10 kg of a 75% antifreeze solution in order to produce a 50% solution?
----------------------------------
Let "x" be the amount of water added.
Equation:
alcohol + alcohol = alcohol
0.75(10) + 0(x) = 0.50(10+x)
Multiply thru by 100 to get:
75*10 = 50*10 + 50x
50x = 25*10
x = 5 kg (amount of water needed in the mixture)
====================================================
Cheers,
Stan H.
Question 236010: how many gallons of 70% antifreeze solution must be mixed with 60 gallons of 10% antifreeze to get a mixture that is 60% antifreeze
Answer by checkley77(7072) (Show Source):
You can put this solution on YOUR website!.70x+.10*60=.60(x+60)
.70x+6=.60x+36
.70x-.60x=36-6
.10x=30
x=30/.10
x=300 gal. of 70% anti freeze is used.
Proof;
.70*300+.10*60=.60*(300+60)
210+6=.60*360
216=216
Question 235976: A chemist wants to create 60 grams of an nickel alloy that is 65% pure. She has one mixture of an nickel alloy that is 45% pure and another mixture that is 75% pure. How much of each mixture must she use?
Answer by josmiceli(3013) (Show Source):
You can put this solution on YOUR website!Let  = grams of 45% mixture to be used
Let  = grams of 75% mixture to be used
given:
(1)  grams
(2)  = pure nickel in final mixture
------------------------------
In words:
( grams of pure nickel in final mixture )/(total grams) = 65%
(3) 
Multiply both sides of (1) by  and subtract
(1) from (3)
(3)
(1) 
And, since
20 grams of the 45% mixture and 40 grams of the 75% mixture are needed
check:
OK
Question 235884: Two ships leave port at the same time. One sails south at 15 mi/h and the other sails east at 20 mi/h. Find a function that models the distance D between the ships in terms of the time t (in hours) elapsed since their departure.
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(3013) (Show Source):
Answer by ankor@dixie-net.com(6693) (Show Source):
You can put this solution on YOUR website!Two ships leave port at the same time.
One sails south at 15 mi/h and the other sails east at 20 mi/h.
Find a function that models the distance D between the ships in terms of the time t (in hours) elapsed since their departure.
:
This is right triangle, legs formed by the path of the two ships
distance between the ships is the hypotnenus
a^2 + b^2 = c^2
;
a = 15t
b = 20t
c = d
:
d(t) = 
:
d(t) = 
:
d(t) = 
d(t) = 25t
Question 235817: Complete the factorization.
16x^7 = 2x^2( )
Answer by checkley77(7072) (Show Source):
Question 235629: Chemistry. A chemist has two solutions: one containing 40% alcohol and another containing 70% alcohol. How much of each should be used to obtain 80 liters of a 49% solution?
Im trying to find a way to solve this using matrices. I'm really unsure about how to go around solving this problem. Thanks
Answer by Theo(675) (Show Source):
You can put this solution on YOUR website!The solution that you want will equal 80 liters.
The amount of alcohol that you want in this solution would therefore be .49*80 = 39.2 liters of alcohol.
Let x = the number of liters in the first solution that you have.
Let y = the number of liters in the second solution that you have.
The amount of alcohol in the first solution that you have is therefore .4*x
The amount of alcohol in the second solution that you have is therefore .7*y
You will mix these together to get 39.2 liters of alcohol which represents .49 of your new solution.
We have an equation that says:
.4*x + .7*y = 39.2
We have another equation that says that:
x + y = 80
The first equation tells us how much alcohol is in the new solution.
The second equation tells us how much total solution is in the new solution.
This has to be since the 80 liters are coming from x and y only.
These equations need to be solved simultaneously.
The equations are:
.4*x + .7*y = 39.2
x + y = 80
We can solve by substitution or by elimination.
You can also solve using matrix algebra if you wish.
The equations should be in the form you can use to do that.
I'll use substitution.
Since x + y = 80, I'll solve for y to get:
y = 80 - x
I'll substitute in the equation:
.4*x + .7*y = 39.2 to get:
.4*x + .7*(80-x) = 39.2
Expand parentheses to get:
.4*x + .7*80 - .7*x = 39.2
Simplify to get:
.4*x + 56 - .7*x = 39.2
Combine like terms to get:
-.3*x + 56 = 39.2
Subtract 56 from both sides of the equation to get:
-.3*x = 39.2 - 56 = -16.8
Divide both sides by -.3 to get:
x = 56
Since x = 56 and x + y = 80, this means that:
y + 56 = 80 which means that y = 24
We have:
x = 56
y = 24
Plug these values into the original equations and see if they work out.
The original equations are:
.4*x + .7*y = 39.2
x + y = 80
replace x with 56 and y with 24 to get:
.4*56 + .7*24 = 39.2
56 + 24 = 80
The second equation becomes 80 = 80 so that one is good.
The first equation becomes:
22.4 + 16.8 = 39.2 which becomes:
39.2 = 39.2 so the first part is good two.
The answer to your probem is:
You need 56 liters of the .4 solution and 24 liters of the .7 solution to make 80 liters of a .49 solution.
Question 235504: outline the steps you would use to prepare a 10% salt solution from a 25% salt solution for a 100 ml.
Answer by stanbon(26297) (Show Source):
You can put this solution on YOUR website!outline the steps you would use to prepare a 10% salt solution from a 25% salt solution for a 100 ml.
------------------
Equation:
salt + salt = salt
0.25x + 0(100-x) = 0.10(100 ml)
Multiply both sides by 100 to get:
25x = 10(100)
x = 1000/25 = 40 ml (amount of 10% solution you need in the mix)
100-x = 60 ml (amount of water you need in the mix)
=======================================================
Cheers,
Stan H.
Question 235297: A chemist has two solutions: one containing 40% alcohol and another containing 70% alcohol. How much of each should be used to obtain 80 liters of 49% solution.
Answer by nerdybill(2448) (Show Source):
You can put this solution on YOUR website! A chemist has two solutions: one containing 40% alcohol and another containing 70% alcohol. How much of each should be used to obtain 80 liters of 49% solution.
.
Let x= liters of 40% alcohol
then
80-x = liters of 70% alcohol
.
.40x + .70(80-x) = .49(80)
.40x + 56 - .70x = 39.2
56 - .30x = 39.2
-.30x = -16.8
x = 56 liters of 40% alcohol
.
70% alcohol:
80-x = 80-56 = 24 liters of 70% alcohol
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