Coin Word Problems

The clandestine to success in solving mint bible problems is to be able to set up the decline systems of equations and precisely solve it using the substitution method acting or at times, the elimination method acting. It ’ sulfur besides worth pointing out that the presentation of the algebraic expressions in mint give voice problems are a bit different and not thus straightforward compared to what we ’ re used to. For exemplar, alternatively of saying “ the count of nickels is 2 more than the number of dimes “, you ’ ll often see this expressed in coin password problems as “ there are 2 more nickels than dimes “. Both algebraic expressions can be written in an equation as n=d+2 but just expressed differently. consequently, it is crucial to have a keen eye and understand of the information given in a coin word problem so we can translate the algebraic expressions correctly into equations. More importantly, you must be familiar with the value of each kind of US mint . Note: In solving our problems below, we will use the value of the coins in dollars which are listed under the third column in the mesa above. As you can see, the penny has a rate of 0.01, the nickel has a value of 0.05, the dime has a prize of 0.10, and the quarter has a value of 0.25.

Example 1: Tamara has 35 coins in nickels and quarters. In all, she has $ 4.15. How many of each kind of mint does she have ? right off the cricket bat, the problem gives us two important pieces of information. First, it tells us that there is a total number of 35 coins consist of nickels and quarters. second, the total value of the coins is $4.15. We need to translate these statements into algebraic equations to find how many nickels and how many quarters she has. Coin word problems normally involve two types of equations. One equality describes the sum count of coins while the other describes the full sum of money. But before we start writing our equations, let ’ s blue-ribbon variables that will stand for the unknown values in our problem .

• Let normality = number of nickels
• Let q = number of quarters

now that we have our variables, let ’ s write the two statements into equations. 1 ) Tamara has 35 coins in nickels and quarters. Equation #1: normality + q = 35 2 ) In all, she has $4.15. Equation #2: 0.05n + 0.25q = 4.15 note that in Equation # 2, we multiplied the respect of each coin ( in dollars ) by how many of that particular coin we have. therefore, since nickel is worth 5 cents, we multiplied 0.05 by the count of nickels ( north ) and multiplied 0.25 by the count of quarters ( q ) since a one-fourth is deserving 25 cents. We added both algebraic expressions and set it equal to the full value of the coins. This setup, however, results in Equation # 2 having two variables. so to solve for the total measure of money, we first need to define one of the coins in terms of the other. In other words, we need to express the issue of nickels in terms of the number of quarters or frailty versa. From Equation # 1, if we subtract the number of quarters from 35 ( which is the total number of coins ), what we ’ ll make is the act of nickels. We can write this algebraically as, newton = 35 – q By doing this, we nowadays have our number of nickels ( newton ) defined in terms of the act of quarters ( q ). Our following pace is to substitute the construction for newton into Equation # 2 and solve for the count of quarters ( q ). By defining normality in terms of q, we now have an equation with only one variable star. Solution: perfect ! This tells us that the number of quarters is 12. Since the problem is asking us to find how many of each coin Tamara has, we will simply subtract 12 from 35 to get the number of nickels . So we have ,

• Number of Nickels: 23
• Number of Quarters: 12

Answer: Tamara has 23 nickels and 12 quarters. Check: When solving word problems, it ’ s crucial to always verify if you got the correct answers. To check, we can plug in the values that we got for newton and q into Equation # 2 and see if both sides of the equation equal each early . eminence that you may besides use Equation # 1 to check your answers. Like what we precisely did, you merely have to substitute normality and q with 23 and 12, respectively, and see if they equal 35 which is the sum number of coins. Example 2: My brother has been putting merely nickels and dimes in his hoggish bank. He is able to save up to $ 3.65. If he has 4 more nickels than dimes, how many of each kind of mint does my brother have ? Let ’ s begin by picking our variables that will stand for the count of each type of coin .

• Let n = number of nickels
• Let five hundred = number of dimes

In this exemplar, we don ’ metric ton know how many coins there are in sum. however, we know that the coins inside the hoggish bank alone dwell of nickels and dimes which amount to $3.65. Translating this into an algebraic equation, we can write this as Equation #1: 0.05n + 0.10d = 3.65 We are besides told that there are 4 more nickels than dimes which means that the number of nickels ( north ) is peer to the count of dimes ( d ) plus four more. Equation #2: normality = d+4 As you can see in Equation # 2, we already have the nickels ( normality ) defined in terms of the dimes ( d ). In this case, we can go ahead and substitute newton with five hundred + 4 in Equation # 1, then solve for five hundred. Solution: therefore, this means that there are 23 dimes. Since there are 4 more nickels than dimes, then there must be 27 nickels. We besides arrive at the same solution when solving for north using Equation # 2 . Going second to our original problem, how many of each kind of mint does my brother have ? Answer: My brother has 27 nickels and 23 dimes. Check: Let ’ s plug in the values that we got for newton and five hundred into Equation # 1 to see if they amount to the sum value of the coins which is $ 3.65 . Example 3: A jar of coins has one-third as many dimes as quarters. If the sum amount of coins is $ 5.10, how many dimes and quarters are in the jar ? Before we translate the authoritative statements that are provided in the trouble, let ’ s first select the variables for the nameless values .

• Let five hundred = number of dimes
• Let q = number of quarters

1 ) A jar of coins has one-third as many dimes as quarters Equation #1: \large { d = { 1 \over 3 } q } 2 ) The total amount of coins is $5.10 Equation #2: 0.10d + 0.25q = 5.10 This trouble involves a fraction but the apparatus is the same as our previous example. Again, we are not given the sum number of coins, but we are told that the total value of the dimes and quarters is $ 5.10. We were already able to define the number of dimes ( five hundred ) in terms of the number of quarters ( q ) in Equation # 1. thus in solving for q using Equation # 2, we will merely substitute five hundred with the expression \large { 1 \over 3 } q Solution: so there are 18 quarters. Let ’ s now use this value to find the number of dimes . therefore, we have

• Number of Dimes: 6
• Number of Quarters: 18

Answer: There are 6 dimes and 18 quarters in the jar. Check: Example 4: Damian had two-thirds as many pennies as nickels. The sum value of his coins was $ 2.38. Find the number he had of each kind of coin. This trouble is quite exchangeable to Example 3 so you should be familiar already with the steps needed to answer this bible trouble .

• Let phosphorus = number of pennies
• Let newton = number of nickels

What pieces of information are given to us ? 1 ) Damian had two-thirds as many pennies as nickels. Equation #1: \large { phosphorus = { 2 \over 3 } north } 2 ) The total value of his coins was $2.38. Equation #2: 0.01p + 0.05n = 2.38 Let ’ s now substitute the construction of phosphorus in Equation # 2 and solve for normality. Solution: now that we know how many nickels there are, get ’ s move on and find the number of pennies as well . We got the following values :

• Number of Pennies: 28
• Number of Nickels: 42

Answer: Damian had 28 pennies and 42 nickels.

Check: This time, I ’ ll leave it up to you to check if both of the values we got for the total of pennies and the number of nickels are right. You may substitute p and normality in Equation # 2 with the values and verify if indeed the sum of both coins when added, equals to $ 2.38. Example 5: Aunt Sheila has $ 2.50 in nickels and dimes in her wallet. How many of each kind of coin does she have, if the number of dimes exceeds doubly the number of nickels by 5 ? We ’ ll first pick our variables to stand for the nickels and dimes, then examine the crucial details given to us in the problem .

• Let n = number of nickels
• Let five hundred = number of dimes

Let ’ s translate each affirmation algebraically into an equation. 1 ) Aunt Sheila has $2.50 in nickels and dimes in her wallet. Equation #1: 0.05n + 0.10d = 2.50 2 ) The number of dimes exceeds twice the number of nickels by 5. Equation #2: d=2n+5 Since the phone number of dimes ( five hundred ) is already expressed in terms of the total of nickels, we can proceed in solving for normality using Equation # 1. Solution: This tells us that there are 8 nickels. Since it says that the number of dimes is 5 more than twice the act of nickels, then let ’ s use Equation # 2 to find how many dimes Aunt Shiela has . Answer: Aunt Shiela has 8 nickels and 21 dimes. Check: Example 6: I found coins worth $ 6.97 in my Grandpa ’ s drawer. surprisingly, the number of pennies, nickels, dimes, and quarters are all the like. How many of each kind of mint did I find ? This trouble is quite singular as it involves not merely two but four kinds of coins. Since the quantity for each mint is the lapp as the other coins, we will select only one variable star to stand for each kind. Let ’ s nibble “ deoxycytidine monophosphate “ .

• Let c = number of pennies
• Let c = number of nickels
• Let c = number of dimes
• Let c = number of quarters

now that we have our variable, let ’ s examine the information given to us closely. We are told that the coins amount to $6.97. Since the measure for each type of mint ( coulomb ) is the same as the others, we will multiply the rate of each mint in dollars with c, add all values of the coins together then set it equal to the total rate of all the coins which is $ 6.97. therefore, we have Equation: 0.01c + 0.05c + 0.10c + 0.25c = 6.97 We have covered all the crucial details from our problem in the equation above so we can go ahead and proceed to solve for degree centigrade. Solution: great ! We now know that there are 17 coins for each type of coin. Answer: There are 17 pennies, 17 nickels, 17 dimes, and 17 quarters in my Grandpa ’ s drawer . Check: The death thing that we need to do is to verify that all the coins indeed amount to $ 6.97 . Example 7: Mrs. Potter received $ 3.44 of deepen after paying for her groceries. The cashier gave her a total of 47 coins in pennies, nickels, and dimes. If she received the same act of pennies and nickels, how many of each coin did she receive ? here we have a coin give voice trouble that is packed with information – which is good ! The more details are provided to us, the better. 👍 As you can see, we are given the sum value of the coins, the total number of coins, and the kinds of coins. Since Mrs. Potter received the lapp measure of pennies and nickels but a different measure of dimes ; we ’ ll use the variable speed of light to stand for the numeral of pennies and nickels, and the variable d to represent the count of dimes .

• Let c = number of pennies
• Let c = number of nickels
• Let five hundred = number of dimes

If you haven ’ deoxythymidine monophosphate realized however, this trouble is quite unique barely like our former exemplar. It involves three types of coins but we ’ re only using two variables to represent them due to the same number of pennies and nickels. first base, we ’ ra told that Mrs. Potter received $3.44 in pennies, nickels, and dimes. To translate this instruction into an algebraic equality, it just makes sense to add the values for each type of mint together then set it peer to $ 3.44. Equation #1: 0.01c + 0.05c + 0.10d = 3.44 next, it says that the teller gave her a total of 47 coins in pennies, nickels, and dimes. This one is easier to translate into an equation since we equitable have to add the total of pennies, nickels, and dimes on the left-hand side of the equation then set it equal to 47. Equation #2: cytosine + c + five hundred = 47 At this point, Equation # 1 still has two variables. so before we can proceed to find how many of each type of coin she received, we need to express the number of dimes ( vitamin d ) in terms of the count of pennies and nickels ( coke ) foremost. We can do this by solving for ( five hundred ) using Equation # 2 . We ’ ll substitute the rate of vitamin d with 47 – 2c in Equation # 1 then solve for c. In doing thus, we will alone be working with one variable star. Solution: perfect ! This tells us that there were 9 pennies and 9 nickels. so immediately, all we have left to do is to find how many dimes there were. We already had the equation for d which is five hundred = 47 – 2c. Using the value we got for degree centigrade, we have Let ’ s recapitulation :

• Number of Pennies: 9
• Number of Nickels: 9
• Number of Dimes: 29

Answer: Mrs. Potter received 9 pennies, 9 nickels, and 29 dimes. Check: Let ’ s verify our answers promptly by plugging the values of coulomb and five hundred, into Equation # 2 : Another option besides is to do your check using Equation # 1. Example 8: Mom gave the pizza manner of speaking guy $ 7.88 in pennies, dimes, and quarters as a tip. She gave him 12 more quarters than pennies and 8 more dimes than quarters. Find the act of each kind of coin that she gave as a tip. This last example is besides an interest one. Right off, we can see that we are given the entire rate of the coins and that the gratuity consists of three types of coins. The quarters are besides defined in terms of the pennies while the dimes are described in terms of the quarters. fair like in our former examples, let ’ s begin by selecting variables to stand for each type of coin .

• Let p = number of pennies
• Let vitamin d = number of dimes
• Let q = number of quarters

immediately let ’ s examine and translate each instruction algebraically into an equality that we can solve. 1 ) ma gave the pizza pitch guy $7.88 in pennies, dimes, and quarters as a tip. Equation #1: 0.01p + 0.10d + 0.25q = 7.88 2 ) She gave him 12 more quarters than pennies and 8 more dimes than quarters. We can translate this instruction into two equations to show the total of quarters defined in terms of the act of pennies and the number of dimes defined in terms of the total of quarters. Equation #2: q = phosphorus + 12 Equation #3: vitamin d = q + 8 At this time, the pennies are not defined in terms of any of the two coins. But since the dimes are already described in terms of the quarters and the quarters are described in terms of the pennies, it just makes feel for us to describe the count of pennies in terms of the issue of quarters arsenic well. We can do this using Equation # 2. q = p + 12 → \color { crimson } p = q – 12 In doing so, both the number of dimes and pennies are defined in terms of q making it potential for us to solve Equation # 1 with only one variable. In other words, we will substitute phosphorus and five hundred in Equation # 1 with their respective algebraic expressions then solve for q. Solution:

excellent ! This tells us that there were 20 quarters. But since our master problem asks us to find the number for each type of coin, let ’ s besides find out how many pennies and dimes were there, by plugging in the rate of the issue of quarters into Equation # 3 and in the equation for the number of pennies ( phosphorus ) .

• Number of Pennies:

• Number of Dimes:

Answer: Mom gave 8 pennies, 28 dimes, and 20 quarters as a gratuity. Check:

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Category : Economy