The mystery to success in resolving coin word problems is to it is in able to set up the correct systems of equations and also precisely solve it using the substitution method or at times, the elimination method.

It’s also worth mentioning that the presentation of the algebraic expressions in coin word difficulties are a little different and also not for this reason straightforward contrasted to what we’re used to. Because that example, rather of saying “the number of nickels is 2 more than the variety of dimes“, you’ll regularly see this to express in coin word troubles as “there room 2 more nickels 보다 dimes“. Both algebraic expressions have the right to be written in one equation together n=d+2 but just expressed differently.

Therefore, the is an essential to have a crawl eye and also understanding the the information given in a coin word problem so we have the right to translate the algebraic expression correctly right into equations. Much more importantly, you should be acquainted with the worth of each kind of united state coin.

You are watching: Nickels dimes and quarters word problems


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Note: In resolving our problems below, us will use the value of the coins in dollars i beg your pardon are detailed under the 3rd column in the table above. As you have the right to see, the penny has a worth of 0.01, the nickel has a value of 0.05, the dime has actually a value of 0.10, and the quarter has a value of 0.25.

Example 1: Tamara has actually 35 coins in nickels and also quarters. In all, she has $4.15. How countless of each sort of coin go she have?

Right off the bat, the difficulty gives us two important pieces of information. First, that tells us that over there is a total variety of 35 coins consisting that nickels and also quarters. Secondly, the total worth of the coins is $4.15. We require to analyze these statements into algebraic equations to uncover how many nickels and also how numerous quarters she has.

Coin word troubles usually indicate two varieties of equations. One equation explains the total number of coins while the other defines the full amount of money.

But prior to we start writing ours equations, let’s select variables that will stand because that the unknown worths in our problem.

Let n = variety of nickelsLet q = variety of quarters

Now the we have our variables, let’s compose the two statements into equations.

1) Tamara has 35 coins in nickels and also quarters.

Equation #1: n + q = 35

2) In all, she has actually $4.15.

Equation #2: 0.05n + 0.25q = 4.15

Note the in Equation #2, we multiplied the worth of each coin (in dollars) through how countless of that particular coin we have. So, since nickel is worth 5 cents, we multiplied 0.05 by the number of nickels (n) and also multiplied 0.25 by the variety of quarters (q) since a quarter is precious 25 cents. We added both algebraic expression and set it equal to the full value that the coins.

This setup, however, results in Equation #2 having actually two variables. Therefore to resolve for the total amount of money, we first need to define one that the coins in regards to the other. In other words, we should express the variety of nickels in terms of the number of quarters or evil versa.

From Equation #1, if we subtract the variety of quarters from 35 (which is the total variety of coins), what we’ll acquire is the variety of nickels. We deserve to write this algebraically as,

n = 35 - q

By act this, us now have our number of nickels (n) defined in regards to the number of quarters (q).

Our following step is to substitute the expression because that n into Equation #2 and solve for the number of quarters (q). By specifying n in regards to q, we now have an equation with just one variable.

Solution:


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Perfect! This tells us that the number of quarters is 12. Due to the fact that the difficulty is asking united state to discover how many of each coin Tamara has, we will just subtract 12 native 35 to acquire the number of nickels.


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So us have,

Number the Nickels: 23Number of Quarters: 12

Answer: Tamara has 23 nickels and 12 quarters.

Check:

When resolving word problems, it’s essential to constantly verify if you obtained the exactly answers. Come check, we deserve to plug in the worths that we acquired for n and also q right into Equation #2 and also see if both sides of the equation equal each other.


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Note that you may additionally use Equation #1 to examine your answers. Like what we just did, you simply need to substitute n and q v 23 and 12, respectively, and see if they equal 35 which is the total variety of coins.

Example 2: mine brother has actually been putting only nickels and dimes in his piggy bank. That is able to conserve up to $3.65. If he has 4 more nickels than dimes, how plenty of of each type of coin go my brothers have?

Let’s begin by choose our variables that will certainly stand because that the number of each type of coin.

Let n = number of nickelsLet d = variety of dimes

In this example, we don’t understand how plenty of coins there space in total. However, we recognize that the coins inside the piggy financial institution only covers nickels and dimes i beg your pardon amount to $3.65. Translating this right into an algebraic equation, we deserve to write this as

Equation #1: 0.05n + 0.10d = 3.65

We are additionally told that there are 4 an ext nickels 보다 dimes which method that the variety of nickels (n) is equal to the number of dimes (d) plus 4 more.

Equation #2: n = d+4

As you have the right to see in Equation #2, we currently have the nickels (n) defined in regards to the dimes (d). In this case, we deserve to go ahead and substitute n through d + 4 in Equation #1, then fix for d.

Solution:


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So, this way that there space 23 dimes. Due to the fact that there space 4 more nickels 보다 dimes, climate there must be 27 nickels. We also arrive in ~ the same answer when solving because that n making use of Equation #2.


Going ago to our original problem, how countless of each sort of coin walk my brothers have?

Answer: mine brother has actually 27 nickels and also 23 dimes.

Check:

Let’s plug in the values that we gained for n and also d right into Equation #1 to check out if they amount come the full value that the coins i m sorry is $3.65.


Example 3: A jar of coins has one-third as many dimes together quarters. If the complete amount the coins is $5.10, how countless dimes and also quarters space in the jar?

Before we analyze the vital statements that are noted in the problem, let’s an initial select the variables because that the unknown values.

Let d = number of dimesLet q = number of quarters

1) A jar of coins has actually one-third as many dimes as quarters

Equation #1: \larged = 1 \over 3q

2) The total amount that coins is $5.10

Equation #2: 0.10d + 0.25q = 5.10

This difficulty involves a portion but the setup is the exact same as our previous example. Again, we room not given the total number of coins, yet we room told that the full value that the dimes and quarters is $5.10.

We were currently able to define the number of dimes (d) in terms of the number of quarters (q) in Equation #1. So in fixing for q making use of Equation #2, us will just substitute d with the expression \large1 \over 3q

Solution:


Therefore, us have

Number of Dimes: 6Number of Quarters: 18

Answer: There room 6 dimes and 18 soldier in the jar.

Check:


Example 4: Damian had actually two-thirds as many pennies together nickels. The total value the his coins was $2.38. Uncover the number he had of each type of coin.

This trouble is quite comparable to instance 3 for this reason you should be familiar currently with the actions needed come answer this word problem.

Let p = number of penniesLet n = number of nickels

What pieces of details are given to us?

1) Damian had two-thirds as numerous pennies together nickels.

Equation #1: \largep = 2 \over 3n

2) The total value of his coins to be $2.38.

Equation #2: 0.01p + 0.05n = 2.38

Let’s currently substitute the expression of ns in Equation #2 and also solve for n.

Solution:


Now that we understand how countless nickels over there are, let’s relocate on and find the variety of pennies as well.


We acquired the following values:

Number of Pennies: 28Number the Nickels: 42

Answer: Damian had 28 pennies and 42 nickels.

Check:

This time, I’ll leave it as much as you to examine if both of the worths we obtained for the variety of pennies and the variety of nickels space correct. You may substitute p and n in Equation #2 through the values and also verify if without doubt the amount of both coins once added, amounts to to $2.38.

Example 5: Aunt Sheila has $2.50 in nickels and dimes in she wallet. How numerous of each type of coin walk she have, if the number of dimes exceeds twice the variety of nickels by 5?

We’ll an initial pick our variables to was standing for the nickels and also dimes, then examine the necessary details provided to united state in the problem.

Let n = number of nickelsLet d = number of dimes

Let’s interpret each declare algebraically into an equation.

1) Aunt Sheila has actually $2.50 in nickels and dimes in her wallet.

Equation #1: 0.05n + 0.10d = 2.50

2) The number the dimes exceeds double the number of nickels through 5.

Equation #2: d=2n+5

Since the number of dimes (d) is already expressed in regards to the variety of nickels, we can proceed in resolving for n using Equation #1.

Solution:


This tells united state that there space 8 nickels. Because it claims that the number of dimes is 5 much more than twice the variety of nickels, then let’s usage Equation #2 to uncover how numerous dimes Aunt Shiela has.


Example 6: I uncovered coins worth $6.97 in my Grandpa’s drawer. Surprisingly, the variety of pennies, nickels, dimes, and quarters are all the same. How plenty of of each type of coin did i find?

This problem is quite distinct as it involves not simply two however four kinds of coins. Since the quantity for every coin is the exact same as the various other coins, we will choose only one variable to stand for every kind. Let’s pick “c“.

Let c = variety of penniesLet c = number of nickelsLet c = number of dimesLet c = variety of quarters

Now the we have actually our variable, let’s research the information given to us closely. We space told that the coins amount to $6.97. Since the quantity for each kind of coin (c) is the exact same as the others, we will multiply the worth of each coin in dollars v c, include all values of the coins with each other then collection it equal to the complete value of all the coins i beg your pardon is $6.97. Therefore, us have

Equation: 0.01c + 0.05c + 0.10c + 0.25c = 6.97

We have actually covered every the essential details from our difficulty in the equation above so we have the right to go ahead and proceed to settle for c.

Solution:


Great! we now know that there room 17 coins because that each kind of coin.

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Answer: There are 17 pennies, 17 nickels, 17 dimes, and also 17 quarters in mine Grandpa’s drawer.

Check:

The last thing that we have to do is to verify the all the coins without doubt amount come $6.97.


Example 7: Mrs. Potter received $3.44 of adjust after paying for her groceries. The cashier offered her a complete of 47 coins in pennies, nickels, and dimes. If she received the same variety of pennies and also nickels, how countless of each coin go she receive?

Here we have actually a coin word difficulty that is pack with information – i beg your pardon is good! The much more details are noted to us, the better.