Lesson Coin problems
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<H2>Coin problems</H2> This lesson and the lesson next to this, <A HREF= http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson> More Coin problems</A>, show you some typical <B>Coin problems</B> and methods to solve them. The givens in this type of problems usually are a) an amount of money composed by the coins collection, and b) some information about the coins collection, for example, the number of coins. The collection of coins in these problems consists typically of two, rarely of three types of coins. The method to solve <B>coin problems</B> is to reduce them to the linear equation (if possible), or to the system of linear equations. This lesson and the lesson next to this, <A HREF= http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson> More Coin problems</A>, consider different sample problems to cover a variety of conditions that may be imposed to the coin collections. <H3>Problem 1</H3>Sue has $1.15 in nickels and dimes, totally 16 coins. How many nickels and how many dimes does Sue have? <B>Solution 1</B> <pre> Lets n be the number of nickels Sue has. Then the number of dimes is equal to (16-n). So, Sue has 5*n cents in nickels and 10*(16-n) cents in dimes. Since the total amount Sue has is $1.15, or 115 cents, it gives you an equation 5n + 10*(16-n) = 115. Simplify this equation: 5n +160-10n = 115 (after opening parentheses) -5n = 115-160 (after collecting like terms at the left side and moving 160 to the right side with the opposite sign) -5n = -45 (after collecting like terms at the right side) n = 9 (after dividing both sides by -5). So, the number of nickels is equal to 9. The number of dimes is equal to 16 minus the number of nickels, that is 16-n = 16-9 = 7. <B>Check</B>. The total number of coins is equal to 9 + 7 = 16. The total amount is equal to 9*5 + 7*10 = 45 + 70 = 115. The solution is correct. <B>Answer</B>. Sue has 9 nickels and 7 dimes. </pre> <B>Solution 2</B> <pre> This solution is reducing to the system of two linear equations in two unknowns. Let n be the number of nickels Sue has, and d be the number of dimes. Since the total number of coins is equal to 16, you can write the first equation d + n = 16. Sue has 5*n cents in nickels and 10*d cents in dimes. Since the total amount Sue has is $1.15 = 115 cents, you can write the second equation 5n + 10*d = 115. So, you have the system of two linear equations in two variables {{{system(d + n = 16, 5n + 10*d = 115 )}}} Solve it using the substitution method. Isolate the variable d from the first equation d = 16-n. Substitute it to the second equation. You get 5n + 10*(16-n) = 115. Simplify this equation (similar to that was done is the <B>Solution 1</B>): 5n +160-10n = 115 (after brackets opening) -5n = 115-160 (after collecting like terms at the left side and moving {{{160}}} to the right side with the opposite sign) -5n = -45 (after collecting like terms at the right side) n = 9 (after dividing both sides by {{{-5}}}). So, the number of nickels is equal to 9. The number of dimes is equal to 16 minus the number of nickels, that is 16-n = 16-9 = 7. <B>Check</B>. The total number of coins is equal to 9 + 7 = 16. The total amount is equal to 9*5 + 7*10 = 45 + 70 = 115. The solution is correct. <B>Answer</B>. Sue has 9 nickels and 7 dimes. </pre> <H3>Problem 2</H3>Michael has $1.95 in his collection, consisting of quarters and nickels. The number of nickels is in three more than the number of quarters. How many nickels and how many quarters does Michael have? <B>Solution 1</B> <pre> Let q be the number of quarters Michael has. Then the number of nickels is equal to (q+3). So, Michael has 25*q cents in quarters and 5*(q+3) cents in nickels. Since the total amount Michael has is $1.95, or 195 cents, it gives you an equation 25*q + 5*(q+3) = 195. Simplify this equation: 25q +5q + 15 = 195 (after brackets opening) 30q = 195 - 15 (after collecting like terms at the left side and moving {{{160}}} to the right side with the opposite sign) 30q = 180 (after collecting like terms at the right side) q = 6 (after dividing both sides by {{{-5}}}). So, the number of quarters is equal to 6. The number of nickels is in three more, that is 6+3 = 9. <B>Check</B>. The difference between the number of nickels and the number of quarters is equal to 9 - 3 = 6. The total amount is equal to 6*25 + 9*5 = 150 + 45 = 195. The solution is correct. <B>Answer</B>. Michael has 6 quarters and 9 nickels. </pre> <B>Solution 2</B> <pre> This solution is reducing to the system of two linear equations in two unknowns. Let q be the number of quarters Michael has, and n be the number of nickels. Since the number of nickels is in three more than the number of quarters, you can write the first equation n-q = 3. Michael has 5*n cents in nickels and 25*q cents in quarters. Since the total amount Michael has is equal to $1.95, or 195 cents, you can write the second equation 5*n + 25*q = 195. So, you have the system of two linear equations in two variables {{{system(n - q = 3, 5n + 25*q = 195 )}}} Solve it using the substitution method. Isolate the variable n from the first equation n = q+3. Substitute it to the second equation. You get 5(q+3) + 25*q = 195. Simplify this equation: 5q + 15 + 25q = 195 (after brackets opening) 30q = 195 - 15 (after collecting like terms at the left side and moving {{{160}}} to the right side with the opposite sign) 30q = 180 (after collecting like terms at the right side) q = 6 (after dividing both sides by {{{-5}}}). So, the number of quarters is equal to 6. The number of nickels is in three more, that is 6+3 = 9. <B>Check</B>. The difference between the number of nickels and the number of quarters is equal to 9 - 3 = 6. The total amount is equal to 6*25 + 9*5 = 150 + 45 = 195. The solution is correct. <B>Answer</B>. Michael has 6 quarters and 9 nickels. </pre> My other lessons on coin word problems in this site are - <A HREF=http://www.algebra.com/algebra/homework/word/coins/More-Coin-problems.lesson>More Coin problems</A> - <A HREF=http://www.algebra.com/algebra/homework/word/coins/Solving-coin-problem-without-equations.lesson>Solving coin problems without using equations</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Kevin-and-Randy-Muise-have-a-jar.lesson>Kevin and Randy Muise have a jar containing coins</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Typical-coin-problems-from-the-archive.lesson>Typical coin problems from the archive</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Three-methods-for-solving-standard-typical-coin-problem.lesson>Three methods for solving standard (typical) coin word problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/More-complicated-coin-problems.lesson>More complicated coin problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Advanced-coin-problems.lesson>Advanced coin problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Solving-coin-problems-mentally-by-grouping-without-using-equations.lesson>Solving coin problems mentally by grouping without using equations</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Non-typical-coin-problems.lesson>Non-typical coin problems</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Swapping-dimes-and-quarters.lesson>Swapping dimes and quarters</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/Santa-Claus-helps-solving-coin-problem.lesson>Santa Claus helps solving coin problem</A> - <A HREF=https://www.algebra.com/algebra/homework/word/coins/We-easily-deal-with-four-unknowns.lesson>We easily deal with four unknowns</A> - <A HREF=http://www.algebra.com/algebra/homework/word/coins/OVERVIEW-of-lesson-on-coin-word-problems.lesson>OVERVIEW of lessons on coin word problems</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.
