Solving Mixture Problems

Solving Mixture Problems

One of important and interesting applications of Algebraic equations is found in solving mixture problems. I find solving mixture problems challenging but a little bit confusing. In this hub I presented several mixture problems with their solutions. Hope you will enjoy reading this hub.


Problem Number One :

How many gallons of a liquid that is 74 percent alcohol must be combined with 5 gallons of another liquid that is 90 percent alcohol to obtain a mixture that is 84 percent alcohol ?

Solution :

Let X = The number in gallons of the first liquid

First Liquid :

Number of gallons : X

Percent of Alcohol : 74%

Number of gallons of alcohol : 0 .74X

Second Liquid :

Number of gallons : 5 gallons

Percent of Alcohol : 90%

Number of gallons of Alcohol : (.90)(5) = 4.5 gallons

Mixture:

Number of gallons : X + 5

Percent of Alcohol : 84%

Nunber of gallons of alcohol : 0.84(X + 5)

Working Equation :

The number of gallons of alcohol in the first liquid + The number of gallons of alcohol in the second liquid = The number of gallons of alcohol in the mixture.

0.74 X + 4.5 = 0.84 (X + 5)

0.74X + 4.5 = 0.84X + 4.2

0.74X - 0.84X = 4.2 - 4.5

-0.10 X = -0.3

10X = 30

1/10(10X) = 30 (1/10)

X = 3

Hence the required amount of first liquid is 3 gallons.


Problem Number Two :

A contractor mixed two batches of concrete that were 9.3 percent and 11.3 percent cement to obtain 4,500 lbs of concrete that was 10.8 percent cement. How many pounds of each type of concrete was used ?

Solution :

Let X = pounds of concrete with with 9.3% cement

4,500 – X = pounds of concrete with 11.3 % cement

First Concrete :

Amount in pounds : X

Percent of Cement : 9.3%

Number of pounds cement : 0.093X

Second Concrete:

Amount in pounds : 4,500 – X

Percent of Cement : 11.3%

Number of pound cement : (0.113)(4,500 – X)

Mixture

Amount in pounds: 4,500

Percent of Cement : 10.8%

Amount of cement in pounds : 0.108(4,500) = 486 pounds

Working Equation :

The amount in pounds of cement in concrete one + The amount in pound of cement in concrete two = Amount in pounds of cement in the mixture.

0.093X + 0.113(4,500 – X ) = 486

0.093X + 508.5 – 0.113X = 486

0.093X - 0.113X = 486 - 508.5

- 0.02 X = -22.5

2X = 2250

½(2X) = 2250(1/2)

X = 1,125 lbs

4500 - X = 3,375 lbs.

There were 1125 lbs of concrete with 9.3 % cement while 3375 lbs with 11.3% cement used.


Problem Number Three:

British Sterling is 7.5 percent copper by weight. How many grams of silver must be mixed with 159 grams of an alloy that is 10 percent copper in order to make sterling?

Solution :

Let X = Number of grams of silver that must be mixed with the alloy

Alloy :

Number of grams : 150 grams

Percent Copper : 10%

Amount of Copper in grams : (0.10) (150) = 15 grams

Silver :

Number of grams : X

Percent Copper : 0%

Amount of Copper : 0

Sterling :

Number of grams : 150 + X

Percent Coppet: 7.5%

Amount of Copper : 0.075 ( 150 + X )

Working Equations :

Grams of Copper in Alloy = Grams of Copper in Sterling

15 = 0.075(150 + X )

15 = 11.25 + 0.075 X

15 - 11.25 = 0.075 X

3.75 = 0.075 X

375 = 7.5 X

(1/7.5) 375 = 7.5 X ( 1/7.5)

X = 50 grams

50 grams of silver must be mixed with 150 grams of alloy which is 10% copper in order to make a sterling.

SOURCE :

COLLEGE ALGEBRA By

Rees

Sparks

Rees

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Comments 12 comments

lifegate profile image

lifegate 5 years ago from Pleasant Gap, PA

You amaze me. My math is terrible. Thanks for sharing.


RevLady profile image

RevLady 5 years ago from Lantana, Florida

Like LG said, you have an amazing mind for mathematics. I do not, but thanks for sharing.

Forever His,


cristina327 profile image

cristina327 5 years ago from Manila Author

Hi lifegate thank you for dropping by. Your visit and comment is much appreciated. Blessings and regards.


cristina327 profile image

cristina327 5 years ago from Manila Author

Hi RevLady thank you for appreciating this hub. Blessings and regards.


CMerritt profile image

CMerritt 5 years ago from Pendleton, Indiana

Okay, my brain hurts!!! but, I am amazed at those who can accomplish these mathamaical tasks with ease. I CAN, but only with some Tylenol in hand.

:)


frogyfish profile image

frogyfish 5 years ago from Central United States of America

OK, I skimmed through this - knowing I could come back and use your 'formula' when needed. I struggle with those kinds of problems' and always have but SOMETIMES I come up with a right answer - I think! :-) Wow, and thank you!


cristina327 profile image

cristina327 5 years ago from Manila Author

Hi Cmerritt it is great to hear from you. Thank you for dropping by and appreciating this hub. Your visit and comment is much appreciated. Blessings to you and regards.


Dave Mathews profile image

Dave Mathews 5 years ago from NORTH YORK,ONTARIO,CANADA

Ate Cristina: even with a Bsc i computers I never could solve one if thos problems. I always got lost with the value "X" in roman numerals "X" = 10

Brother Dave.


cristina327 profile image

cristina327 5 years ago from Manila Author

HI frogfish thank you for dropping by. Your visit is much appreciated. Blessings to you and regards.


cristina327 profile image

cristina327 5 years ago from Manila Author

HI Dave Mathews thanks for gracing this hub. your visit is much appreciated. Blessings and regards.


JHEM 4 years ago

AMAZING


susie 4 years ago

thanks a lot

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