: 39%in 30m20 CALENDARTravel to HomeToday at 11:02 PMGreen Streetf... Lumen... Sign In orSchool..ohmRukayat Balogun오 | User Settings My ClassesLog Outork managerges Forums Calendar Gradebook1:30am Fall 2018> Assessmentpplications of Systems of EquationsDue in 11 hours, 28 minutes. Due Fri 12/21/2018 10:00 amMonico needs to mix a 10% alcohol solution with a 50% alcohol solution to create 100 milli!eters of a 26%solution. How many millileters of each solution must Monico use?Answer: Monico must mix-milli!eters of 10% solution andmilli!eters of 50% solution.Points possible: 1Unlimited attemptsLicenseSubmiton

Question
Asked Dec 21, 2018
: 39%
in 30m
20 CALENDAR
Travel to Home
Today at 11:02 PM
Green Street
f... Lumen... Sign In or
School..
ohm
Rukayat Balogun오 | User Settings My Classes
Log Out
ork manager
ges Forums Calendar Gradebook
1:30am Fall 2018> Assessment
pplications of Systems of Equations
Due in 11 hours, 28 minutes. Due Fri 12/21/2018 10:00 am
Monico needs to mix a 10% alcohol solution with a 50% alcohol solution to create 100 milli!eters of a 26%
solution. How many millileters of each solution must Monico use?
Answer: Monico must mix
-milli!eters of 10% solution and
milli!eters of 50% solution.
Points possible: 1
Unlimited attempts
License
Submit
on
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: 39% in 30m 20 CALENDAR Travel to Home Today at 11:02 PM Green Street f... Lumen... Sign In or School.. ohm Rukayat Balogun오 | User Settings My Classes Log Out ork manager ges Forums Calendar Gradebook 1:30am Fall 2018> Assessment pplications of Systems of Equations Due in 11 hours, 28 minutes. Due Fri 12/21/2018 10:00 am Monico needs to mix a 10% alcohol solution with a 50% alcohol solution to create 100 milli!eters of a 26% solution. How many millileters of each solution must Monico use? Answer: Monico must mix -milli!eters of 10% solution and milli!eters of 50% solution. Points possible: 1 Unlimited attempts License Submit on

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Expert Answer

Step 1

Lets assume required quantity of 10% alchol solution = x millileters

And required quantity of 50% alchol solution = y ml

Since quantity of resulting mixture = 100ml

x+y = 100           ...(1)

Quantity of alchol in x ml of 10% alchol solution = 10% of x = (x/10) ml

And quantity of alchol in y ml of 50% alchol solution = 50% of y = (5y/10) ml

Quantity of alchol in resulting mixture = 26% of 100 ml = 26 ml

 

 

Step 2
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