Lesson Word problems on mixtures for antifreeze solutions
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<H2>Word problems on mixtures for antifreeze solutions</H2> <H3>Problem 1</H3>How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 40% antifreeze solution? <pre> Let "x" be the volume of pure antifreeze to be added, in gallons. Then the total volume of the solution will be 12+x gallons. The amount of the pure antifreeze in the new solution will be 0.1*12 + x gallons. The percentage concentration equation is {{{(0.1*12 +x)/(12+x)}}} = 0.4. Simplify it multiplying both sides by (12+x): 1.2 + x = 0.4*12 + 0.4x, 1.2 + x = 4.8 + 0.4x, x - 0.4x = 4.8 - 1.2 0.6x = 3.6, x = {{{3.6/0.6}}} = 6. <U>Check</U>: {{{(0.1*12 +x)/(12+x)}}} = {{{(1.2+6)/(12 + 6)}}} = {{{7.2/18}}} = 0.4. OK, check is done. <U>Answer</U>. 6 gallons of pure antifreeze is needed. </pre> <H3>Problem 2</H3>How many liters of water must be added to 26 liters of a 15% antifreeze solution to dilute it to a 12% solution? <B>Solution</B> <pre> Let "v" be the volume of water to be added, in liters. The amount of the antifreeze in 26 liters of the 15% solution is 0.15*26 = 3.9 liters. The amount of the antifreeze will not change after adding the water, but the volume of the solution will change and will get (26+v) liters. The percentage equation for the antifreeze content after adding v liters of water is {{{3.9/( v + 26)}}} = {{{0.12}}}. To solve it, multiply both sides by (v+26) and simplify: 3.9 = 0.12*(v + 26), 3.9 = 0.12v + 3.12, 0.12v = 0.78, v = {{{0.78/0.12}}} = 6.5. So, 6.5 liters of water must be added. <U>Check</U>: {{{3.9/(v + 26)}}} = {{{3.9/(6.5 + 26)}}} = {{{3.9/32.5}}} = 0.12. OK, check is done. <U>Answer</U>. 6.5 liters of water must be added. </pre> <H3>Problem 3</H3>How much of 50% antifreeze solution and 40% antifreeze solution should be combined to give 50 gallons of 46% antifreeze solution? <B>Solution</B> <pre> Let "v" be the amount (the volume) of the 50% antifreeze solution needed, in gallons. Then the amount (the volume) of the 40% antifreeze solution needed is (50-v) gallons. v gallons of the 50% antifreeze solution will contribute 0.5*v gallons of the pure antifreeze to the resulting solution. (50-v) gallons of the 40% antifreeze solution will contribute 0.4*(50-v) gallons of the pure antifreeze to the resulting solution. In all, the resulting solution of 50 gallons will contain 0.5*v + 0.4*(50-v) gallons of the pure antifreeze. The percentage equation is {{{(0.5*v + 0.4*(50-v))/50}}} = 0.46. (1) This is your equation to determine the unknown value of v. To solve (1), first multiply both sides by 50 to rid of the denominator. You will get (0.5*v + 0.4*(50-v) = 0.46*50 ---> 0.5v + 20 - 0.4v = 23 ---> 0.1v = 23 - 20 ---> 0.1.v = 3 ---> v = {{{3/0.1}}} = 30. So, 30 gallons of the 50% antifreeze solution must be mixed with 50-30 = 20 gallons of the 40% antifreeze solution. <U>Check</U>: {{{(0.5*v + 0.4*(50-v))/50}}} = {{{(0.5*30 + 0.4*(50-30))/50}}} = {{{(15 + 0.4*20)/50}}} = {{{(15 + 8)/50}}} = {{{23/50}}} = 0.46. OK, check is done. <U>Answer</U>. 30 gallons of the 50% antifreeze solution must be mixed with 20 gallons of the 40% antifreeze solution. </pre> <H3>Problem 4</H3>A 10-quart radiator contains a 30% antifreeze solution. How much of the solution needs to be drained out and replaced with pure antifreeze in order to raise the solution to 65% antifreeze? <B>Solution</B> <pre> Let "x" be (an unknown) volume of the 30% antifreeze solution which need to be drained out and replaced, in accordance with the condition. After draining, you will have (10-x) quarts of the antifreeze solution in the radiator (assuming that initially the radiator was full and contained 10 quarts of the solution). The amount of the pure antifreeze in the (10-x) quarts of the 30% solution is 0.3*(10-x). After replacing of "x" quarts by the pure antifreeze by "x" quarts of the pure antifreeze the amount of the pure antifreeze in the radiator is 0.3*(10-x) + x quarts. To get the antifreeze concentration of 65%, the volume "x" should satisfy the equation {{{(0.3*(10-x) + x)/10}}} = {{{0.65}}}. (1) First step to solve this equation is to multiply both sides by 10 to rid of the denominator. You will get 0.3(10-x) + x = 6.5, or 3 - 0.3x + x = 6.5 ---> 0.7x = 6.5 - 3 ---> 0.7x = 3.5 ---> x = 5. <U>Answer</U>. The amount of the 65% antifreeze solution that needs to be drained and replaced by the pure antifreeze is 5 quarts. </pre> My other lessons on word problems for mixtures in this site are - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Mixture-problems.lesson>Mixture problems</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/More-Mixture-problems.lesson>More Mixture problems</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Solving-typical-mixture-problems.lesson>Solving typical word problems on mixtures for solutions</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Solving-typical-word-problems-on-mixtures-for-dry-substances-coffee-nuts.lesson>Word problems on mixtures for dry substances like coffee beans, nuts, cashew and peanuts</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-dry-substances-like-cookies-candies.lesson>Word problems on mixtures for dry substances like candies, dried fruits</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-dry-substances-like-soil-and-sand.lesson>Word problems on mixtures for dry substances like soil and sand</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Word-problems-on-mixtures-for-alloys.lesson>Word problems on mixtures for alloys</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Typical-word-problems-on-mixtures-from-the-archive.lesson>Typical word problems on mixtures from the archive</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Advanced-mixture-problems.lesson>Advanced mixture problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Advanced-mixture-problem-for-three-alloys.lesson>Advanced mixture problem for three alloys</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Unusual-word-problem-on-mixtures.lesson>Unusual word problem on mixtures</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Check-if-you-know-the-basics-of-mixtures-from-Science.lesson>Check if you know the basics of mixtures from Science</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Special-type-mixture-problems-on-DILUTION-adding-water.lesson>Special type mixture problems on DILUTION adding water</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Increasing-concentration-of-an-acid-solution-by-adding-pure-acid.lesson>Increasing concentration of an acid solution by adding pure acid</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Increasing-concentration-of-alcohol-in-solution-by-adding-pure-alcohol.lesson>Increasing concentration of alcohol solution by adding pure alcohol</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/How-many-kilograms-of-sand-must-be-added-to-a-mixture-of-sand-and-cement.lesson>How many kilograms of sand must be added to a mixture of sand and cement</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Draining-replacing-mixture-problems.lesson>Draining-replacing mixture problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/How-much-water-should-be-evaporated.lesson>How much water must be evaporated</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Advanced-problems-on-draining-and-replacing.lesson>Advanced problems on draining and replacing</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Using-effective-methodology-to-solve-many-steps-dilution-problems.lesson>Using effective methodology to solve many-steps dilution problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/mixtures/Entertainment-problems-on-mixtures.lesson>Entertainment problems on mixtures</A> - <A HREF=http://www.algebra.com/algebra/homework/word/mixtures/OVERVIEW-of-lesson-on-word-problems-for-mixtures.lesson>OVERVIEW of lessons on word problems for mixtures</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-I.
