# 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

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

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

Forever His,

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.

:)

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!

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.

AMAZING

thanks a lot

12