Questions on Word Problems: Coins answered by real tutors!

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Question 1179758: there were 24 nickles, dimes, and quarters whose total value equaled $4.25. how many coins of each kind were there if the number of nickels equaled the number of dimes?
Answer by ikleyn(53748)

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.
there were 24 nickles, dimes, and quarters whose total value equaled $4.25.
how many coins of each kind were there if the number of nickels equaled the number of dimes?
~~~~~~~~~~~~~~~~~~~~~

Here is very simple way to solve this problem for 3 unknowns using only one equation.

Let x be the number of nickels.

Then the number of dimes also is x, according to the problem.


Then the number of quarters is  24-x-x = 24-2x.


Now write the total money equation

    5x + 10x + 25*(24-2x) = 425  cents.


Simplify and find x

    15x + 600 - 50x = 425

    600 - 425 = 50x - 15x

       175    =    35x

         x    =     175/35 = 5.


ANSWER.  5 nickels, 5 dimes and 24-5-5 = 14 quarters.


CHECK  for the total money  5*5 + 5*10 + 14*25 = 425 cents.   ! correct !

Solved completely, using only one equation.

Question 447885: a man goes to the bank with $6.00 and asks for change.he is given an equal number of nickels,dimes,and quarters.how many of each is he given
Answer by ikleyn(53748)

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A man goes to the bank with $6.00 and asks for change. He is given an equal number of nickels, dimes, and quarters.
How many of each is he given?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        It is a typical ARITHMETIC problem for 4th or 5th grade students to solve it MENTALLY by using
        the grouping method.

        I cannot miss this opportunity to present this solution here,
        because such problems are gems for developing thinking skills in elementary school students.

Group the coins, placing one nickel, one dime and one quarter in each group.


    +--------------------------------------------------------+
    |   According to the problem, it is possible to do so.   |
    +--------------------------------------------------------+


Then each such a group is worth  5 + 10 + 25 = 40 cents;

hence, the number of such groups is  600 cents divided by 40, which gives   = 15
for the number of these groups.


It gives the ANSWER:  there are 15 coins of each nomination.

Solved.

Question 434946: Bertha has 33 coins in her pocket totaling $5.25. If she only has nickles and quarters, how many of each type of coin does she have
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The formal algebraic solutions shown by the other tutors are fine.

The problem can be solved informally using logical reasoning and mental arithmetic. If your mental arithmetic is good, this can be faster than using formal algebra. That can be helpful in some case, such as a timed competitive exam.

Solving problems informally can also provide excellent brain exercise.

Here is an informal solution to the problem.

(1) If all 33 coins were nickels, the total would be 33(5) = 165 cents, or $1.65. The actual total, $5.25, or 525 cents, is 525-165 = 360 cents more.
(2) Each quarter is worth 25-5 = 20 cents more than each nickel.
(3) The number of quarters needed to make the additional 360 cents is 360/20 = 18.

So there are 18 quarters and 33-18 = 15 nickels.

ANSWER: 18 quarters, 15 nickels

Answer by josgarithmetic(39792)   (Show Source):
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COIN CENTS COUNT VALUES(cents) Nickel 5 33-q 5(33-q) Quarter 25 q 25q TOTALS 33 525

Accounting for all values

18 Quarters

Answer by ikleyn(53748)   (Show Source):
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Bertha has 33 coins in her pocket totaling $5.25. If she only has nickles and quarters,
how many of each type of coin does she have
~~~~~~~~~~~~~~~~~~~~~~~~~

        In the post by @mananth, his second equation in the 4th line is written incorrectly.
        It makes all the subsequent solution defective.
        So, I came to make the job accurately, in a way as it SHOULD be done.
        In my solution, I will use one unknown and one equation, since it provides a shorter way.

Let x be the number of quarters. Then the number of nickels is (33-x). Write the total value equation 25x + 5*(33-x) = 525 cents. Simplify and find x 25x + 165 - 5x = 525, 25x - 5x = 525 - 165, 20x = 360, x = 360/20 = 18. ANSWER. 18 quarters and 33-18 = 15 nickels. CHECK for the total 25*18 + 5*15 = 525 cents, precisely as it is given in the problem.

Solved correctly.

Question 419685: Hi tutor, i do not get this question what so ever. It his a problem solving, algebraic question. I really need help. It would be greatly appreciated, thank you so much, may god bless you. Q : 500 tickets for a football game were sold and the total receipts were $105. Some of the tickets sold for $15 and the rest sold for 25cents. Find the number of each sold. ( i am required too provide a equation/formula , write a table i already have that set up : i have it as ticket 1 & ticket two & value, number and amount. & i also would greatly appreciate if you show the work as too how you got the answers, so i can learn from it. Thank you so much.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
---------------------
500 tickets for a football game were sold and the total receipts were $105.
Some of the tickets sold for $15 and the rest sold for 25cents. Find the number of each sold.
---------------------

x at 0.25 dollars each
y at 15 dollars each

Subtract first equation from second equation.

Something is wrong in the problem's description.
You should need positive whole numbers results for x and y.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
Hi tutor, i do not get this question what so ever. It his a problem solving, algebraic question. I really need help.
It would be greatly appreciated, thank you so much, may god bless you.
Q : 500 tickets for a football game were sold and the total receipts were $105.
Some of the tickets sold for $15 and the rest sold for 25cents.
Find the number of each sold.
( i am required too provide a equation/formula , write a table i already have that set up :
i have it as ticket 1 & ticket two & value, number and amount. & i also would greatly appreciate
if you show the work as too how you got the answers, so i can learn from it. Thank you so much.
~~~~~~~~~~~~~~~~~~~~~~~~~

Regarding this problem, I'd like to make two notices.

First, the solution and the answer in the post by @mananth (300 tickets at $0.25 and 200 tickets by $15)
both are incorrect.

They do not withstand the check: 300*0.25 + 200*15 = 3075, not 105 dollars.

Second notice is that the problem is posed incorrectly. It is SELF-CONTRADICTORY and describes
a situation which NEVER may happen.

Indeed, 500 $0.25 tickets cost 500*0.25 = 5*25 = 125 dollars, which is just MORE than $105.

So, this problem, as it is presented in your post, is FATALLY DEFECTIVE and is only good
to throw it to a garbage bin.

Question 729333: If you take 1 penny and double it each day for 31 days. How much money will you have? Express in $.
Answer by ikleyn(53748)   (Show Source):
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If you take 1 penny and double it each day for 31 days.
How much money will you have? Express in $.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The other person gave the answer 2147483648 in his post.

        Without making any calculations, I see and I am convinced that his answer is INCORRECT.

        How do I know it ?

Very simple. The answer is the sum 1 + 2 + 4 + 8 + . . . + = - 1. It is an odd number, while the other person provides an even number. The answer is cents, or 21,474,836.47 dollars.

Solved correctly.

Question 733000: Please help me solve this,
a coin collector has 2.45 in nickels and dimes. She has 5 times as many nickels as dimes. How many does she have of each coin?
123408@houndmail.macc.edu
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
d dimes
5d nickels

35 nickels and 7 dimes

check: 5*35+10*7=175+70=245

Answer by ikleyn(53748)   (Show Source):
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.
a coin collector has 2.45 in nickels and dimes. She has 5 times as many nickels as dimes.
How many does she have of each coin?
~~~~~~~~~~~~~~~~~~~~~~~~~

This problem is ideal to solve it mentally, without using equations.

Indeed, according to the problem, we can group the coins in sets, each containing 5 nickels and one dime. Each such set is worth 5*5 + 10 = 35 cents. Hence, the number of sets is = 7. So, the coin collector has 7 dimes and 5*7 = 35 nickels. CHECK. 7*10 + 35*5 = 245 cents = $2.45. ! correct !

Solved.

Question 733094: Kendra has $4.25 in dimes and quarters. If she has 3 times as many quarters as dimes, how many of each coin does she have?
Answer by ikleyn(53748)   (Show Source):
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Kendra has $4.25 in dimes and quarters. If she has 3 times as many quarters as dimes,
how many of each coin does she have?
~~~~~~~~~~~~~~~~~~~~~~~~~~~

This problem is ideal to solve it mentally, without using equations. Indeed, according to the problem, we can group the coins in sets, each containing 3 quarters and one dime. Each such set is worth 3*25 + 10 = 85 cents. Hence, the number of sets is = 5. So, Kendra has 5 dimes and 15 quarters. CHECK. 5*10 + 15*25 = 425 cents = $4.25. ! correct !

Solved.

The solution and the answer in the post by @lynnlo both are incorrect.

Question 743784: Michael's bank contains only nickels, dimes, and quarters. There are 50 coins is all, value at $4.15. The number of nickels is 10 short of being three times the sum of the number of dimes and quarters together. How many dimes are in the bank?
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
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Michael's bank contains only nickels, dimes, and quarters. There are 50 coins is all, value at $4.15.
The number of nickels is 10 short of being three times the sum of the number of dimes and quarters together.
How many dimes are in the bank?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let' solve it mentally. Let's add 10 short nickels mentally. We will get 50+10 = 60 coins, with the number of nickels being three times of the sum of the number of dimes and quarters together. It means that four times the number of dimes and quarters is 60/4 = 15 in the original collection, with the number of nickels 3*15 - 10 = 35 in the original collection. So, the original collection is 35 nickels plus 15 dimes and quarters (dimes and quarters are counted together). Hence, 15 dimes and quarters in the original collection are worth 415 - 35*5 = 240 cents. Had all 15 coins be dimes, it would be 150 cents, but we actually have 240 cents, or 90 cents more. Hence, some dimes should be replaced by quarters. Each such replacement increases the value of dimes and quarters by 15 cents, and the number of such replacements should be, therefore, 90/15 = 6. Thus, the original collection has 35 nickels, 6 quarters and 50 - 35 - 6 = 9 dimes. CHECK. Total coins = 35 + 6 + 9 = 50. Total money = 35*5 + 6*25 + 9*10 = 415 cents, or $4.15. The number of nickels is 3*(6+9) -10 = 3*15 - 10 = 45 - 10 = 35. Everything is correct. ANSWER. 35 nickels, 9 dimes and 6 quarters.

Solved.

Question 742399: Between 2 coins how many coins are there
Answer by ikleyn(53748)   (Show Source):
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.
Between 2 coins how many coins are there
~~~~~~~~~~~~~~~~~~~~~~~~~

This reminds me the question: how many stars are there in the sky?

Question 736645: john has twice the number of quarters as dimes. the amount of money in$1.80. find the number of dimes
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39792)   (Show Source):
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d dimes
q quarters

Only one variable needed.

-------three dimes

Answer by ikleyn(53748)   (Show Source):
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.
john has twice the number of quarters as dimes. the amount of money in$1.80. find the number of dimes
~~~~~~~~~~~~~~~~~~~~~~~~~~~

        This problem is easier to solve mentally than using equation/equations.

According to the problem, we can group coins in sets, placing two quarters and one dime in each set. Then each set is worth 2*25 + 10 = 50 + 10 = 60 cents, and there are 180/60 = 3 such sets. Hence, the number of dimes John has is 3. ANSWER

Solved.

Question 869744: Polly has 20 coins in quarters and nickels. The total value is 2.20. Set up a system of equations and solve to determine the number of quarters, q, and the number of nickels ,n, she has
Answer by timofer(155)   (Show Source):
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n nickesl and q quarters

$2.20

multiply left and right by .

If a system is wanted then it is .

If you want to solve the system, ... your choice of method or, you might do some steps in your head:
and ;
you can try to recheck.

Question 670863: I am not sure how to figure out a coin problem without how much it's worth for example: A bank contains 44 coins (nickles, dimes, quarters). There are twice as many dimes as nickles and 8 fewer nickles than quarters. How much money is in the bank?
Found 4 solutions by n2, greenestamps, timofer, ikleyn:
Answer by n2(79)   (Show Source):
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A bank contains 44 coins (nickels, dimes, quarters). There are twice as many dimes as nickels
and 8 fewer nickels than quarters. How much money is in the bank?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let x be the number of nickels. Then the number of dimes is 2x, according to the problem, and the number of quarters is (x+8). Having this, write equation for the total number of coins x + 2x + (x+8) = 44. (here 44 is the given total number of coins). Simplify this equation and find x 4x + 8 = 44, 4x = 44 - 8 = 36, x =36/4 = 9. So, there are 9 nickels in the bank, 2*9 = 18 dimes and 9+8 = 17 quarters. Now you can easily find the value of the bank 5*9 + 10*18 + 17*25 = 650 cents, or 6.50 dollars. ANSWER. The value of the bank is $6.50.

Answer by greenestamps(13327)   (Show Source):
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A comment about the two responses you have received to this point, one using three variables and one using a single variable.

The easiest way to set the problem up is directly from the given information, which means using three variables and three equations; but then solving the problem means solving a system of three equations in three variables.

It takes a bit of effort (not much, really) to set the problem up using a single variable and a single equation; but then the effort needed to solve the problem is much less.

So the most efficient way to solve the problem using formal algebra is using a single variable and a single equation.

The given information compares the number of dimes to the number of nickels, and it compares the number of nickels to the number of quarters. The common item there is the number of nickels, so the clear logical choice for the single variable is the number of nickels.

So proceed as shown in the response from the tutor who uses a single variable:

let x = # of nickels
then 2x = # of dimes (there are twice as many dimes as nickels)
and x+8 = # of quarters (... and 8 fewer nickels than quarters -- i.e., 8 more quarters than nickels)

The equation then says the total number of coins is 44:

(x) + (2x) + (x+8) = 44

The solution from there is easy....

Answer by timofer(155)   (Show Source):
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x y and z for how many nickels dimes and quarters

"A bank contains 44 coins"

"twice as many dimes as nickles"

"8 fewer nickles than quarters. "

You have enough to write the total coin equation as one single variable.

-------------------With this variable (for the nickels) you can now find the dime and quarter counts, and find how much money.

Answer by ikleyn(53748)   (Show Source):
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.
I am not sure how to figure out a coin problem without how much it's worth for example:
A bank contains 44 coins (nickles, dimes, quarters). There are twice as many dimes as nickles
and 8 fewer nickles than quarters. How much money is in the bank?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        I will give a simple, straightforward, short and instructive solution,
        using only one single equation in one single unknown.

Let x be the number of nickels. Then the number of dimes is 2x, according to the problem, and the number of quarters is (x+8). Having this, write equation for the total number of coins x + 2x + (x+8) = 44. (here 44 is the given total number of coins). Simplify this equation and find x 4x + 8 = 44, 4x = 44 - 8 = 36, x =36/4 = 9. So, there are 9 nickels in the bank, 2*9 = 18 dimes and 9+8 = 17 quarters. Now you can easily find the value of the bank 5*9 + 10*18 + 17*25 = 650 cents, or 6.50 dollars. ANSWER. The value of the bank is $6.50.

Solved and explained in a simple and instructive way, using only one unknown,
and without introducing system of equation, which is not necessary here.

/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/

I wrote this my solution as an alternative to the long, wordy solution by @Theo.

We live now at the time when every our solution, produced at this forum,
goes directly to the database for Artificial Intelligence.

So not tens, but tens millions students/readers will read them.

While 10 years ago so wordy solutions as in the posts by @Theo
did not cause protest, now they become INAPPROPRIATE
for teaching millions of readers of AI.

It is why I work here now checking and re-developing many @Theo' solutions.

Because they SHOULD BE re-developed.

Question 659025: you have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have
Found 2 solutions by n2, ikleyn:
Answer by n2(79)   (Show Source):
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You have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let x be the number of pennies, y be the number of nickels, z be the number of dimes, w be the number of quarters. We have two equations x + y + z + w = 25, (1) for the total number of coins x + 5y + 10z + 25w = 100 (2) for the total value. From equation (2), subtract equation (1). You will get 4y + 9z + 24w = 75, 4y + 24w = 75 - 9z. (3) In the last equation, left side value is always a multiple of 4. Hence, right side must be a multiple of 4. z = 0 gives for the right side of (3) the value 75 - 9*0 = 75, which is not a multiple of 4. z = 1 gives for the right side of (3) the value 75 - 9*1 = 66, which is not a multiple of 4. z = 2 gives for the right side of (3) the value 75 - 9*2 = 57, which is not a multiple of 4. z = 3 gives for the right side of (3) the value 75 - 9*3 = 48, which is a multiple of 4. At z = 3, from (3), we have 4y + 24w = 48. (3) One possible solution for (3) is (y,w) = (0,2). Then x + y + z + w = x + 0 + 3 + 2 = 25 implies x = 20. So, one possible solution for the problem is 20 pennies, 0 nickels, 3 dimes and 2 quarters. Second possible solution for (3) is (y,w) = (6,1). Then x + y + z + w = x + 6 + 3 + 1 = 25 implies x = 15. So, second possible solution for the problem is 15 pennies, 6 nickels, 3 dimes and 1 quarters. Third possible solution for (3) is (y,w) = (12,0). Then x + y + z + w = x + 12 + 3 + 0 = 25 implies x = 10. So, third possible solution for the problem is 10 pennies, 12 nickels, 3 dimes and 0 quarters. Thus the problem has three possible solutions: (1) 20 pennies, 0 nickels, 3 dimes and 2 quarters, (2) 15 pennies, 6 nickels, 3 dimes and 1 quarters, (3) 10 pennies, 12 nickels, 3 dimes and 0 quarters. It is clear that there no other solutions.

Solved completely.

Answer by ikleyn(53748)   (Show Source):
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You have 25 coins, which have a total value of $1. what are the coins, and how many of each do you have ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        In his post, tutor @Theo found out one solution " 10 pennies, 12 nickels, 3 dimes and 0 quarters".
        He used the trial and error guessing.
        But the problem has two other solutions that eluded Theo.
        Below I give a complete analysis to catch all possible solutions.

Let x be the number of pennies, y be the number of nickels, z be the number of dimes, w be the number of quarters. We have two equations x + y + z + w = 25, (1) for the total number of coins x + 5y + 10z + 25w = 100 (2) for the total value. From equation (2), subtract equation (1). You will get 4y + 9z + 24w = 75, 4y + 24w = 75 - 9z. (3) In the last equation, left side value is always a multiple of 4. Hence, right side must be a multiple of 4. z = 0 gives for the right side of (3) the value 75 - 9*0 = 75, which is not a multiple of 4. z = 1 gives for the right side of (3) the value 75 - 9*1 = 66, which is not a multiple of 4. z = 2 gives for the right side of (3) the value 75 - 9*2 = 57, which is not a multiple of 4. z = 3 gives for the right side of (3) the value 75 - 9*3 = 48, which is a multiple of 4. At z = 3, from (3), we have 4y + 24w = 48. (3) One possible solution for (3) is (y,w) = (0,2). Then x + y + z + w = x + 0 + 3 + 2 = 25 implies x = 20. So, one possible solution for the problem is 20 pennies, 0 nickels, 3 dimes and 2 quarters. Second possible solution for (3) is (y,w) = (6,1). Then x + y + z + w = x + 6 + 3 + 1 = 25 implies x = 15. So, second possible solution for the problem is 15 pennies, 6 nickels, 3 dimes and 1 quarters. Third possible solution for (3) is (y,w) = (12,0). Then x + y + z + w = x + 12 + 3 + 0 = 25 implies x = 10. So, third possible solution for the problem is 10 pennies, 12 nickels, 3 dimes and 0 quarters. Thus the problem has three possible solutions: (1) 20 pennies, 0 nickels, 3 dimes and 2 quarters, (2) 15 pennies, 6 nickels, 3 dimes and 1 quarters, (3) 10 pennies, 12 nickels, 3 dimes and 0 quarters. It is clear that there no other solutions.

Solved completely.

-----------------------------

        Interesting fact:

Today, November 10, 2025, after completing my solution, I submitted this problem to two artificial intelligence sites.
I was guided by my curiosity: it was interesting to me, how they will treat this problem ?

One AI was Google AI Overview. The second AI was math-gpt.org

Both produced incomplete answers.

Google AI Overview referred to the incomplete @Theo solution.

Math-gpt.org did not share his links with me.

But I made my conclusion: they both can not think (in the usual meaning of this word)
- they both re-write (compile) from existing solutions in their databases.

Surely, as usual, I informed the Google AI Overview about their mistake through their
feedback system with the reference to the current link.

Question 201379: A coin bank contains 25 coins in nickles, dimes and quarters. There are four times as many dimes as quarters. The value of the coins is $2.05. How many dimes are in the bank?
Found 3 solutions by greenestamps, timofer, ikleyn:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The original solution from tutor @ikleyn was not correct, as it produced a total value of the coins that is greater than $2.05. But I see that she saw the error in her setup and has corrected her response.

Note that she recommends setting up the problem using a single variable; I concur. That requires a bit more effort than setting the problem up using two or even three variables; but it reduces significantly the amount of effort required to solve the problem.

Meanwhile, I will present an informal solution that can be obtained using mental arithmetic and logical reasoning. While of course the student should understand the solution using formal algebra, it should be noted that solving a problem like this informally can provide excellent brain exercise.

There are four times as many dimes as quarters. So consider groups of 5 coins consisting of one quarter and four dimes, with a total value of 65 cents. Then solve the problem by trying different numbers of those groups of 5 coins.

If there were one such group, with a total of 5 coins, the remaining value to be made with the nickels would be $2.05 - $0.65 = $1.40; but that requires 140/5 = 28 nickels, making a total of 5+28 = 33 coins, so that doesn't work.

If there are two such groups, with a total of 10 coins, the remaining value to be made with the nickels is $2.05 - $1.30 = $0.75. That requires 75/5 = 15 coins, which makes the correct total of 10+15 = 25 coins.

So there are two groups each containing one quarter and four dimes, making 8 the total number of dimes.

ANSWER: 8 dimes

Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!
n number of nickels
4q number of dimes
q number of quarters

For how many coins
same as

For the money value in cents
same as

This system can be made.

same as

solving for the quarters first

the number of quarters

Question asked is how many dimes.
dimes

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
A coin bank contains 25 coins in nickels, dimes and quarters.
There are four times as many dimes as quarters. The value of the coins is $2.05.
How many dimes are in the bank?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        At the first glance, this problem is for three unknowns.
        But it is so only for the first glance.

        Actually, this problem and many other similar problems admit elegant solution in one single unknown,
        and it is the reason why the problem is designed in this way and proposed to you to teach you to true Math.

Let 'x' be the number of quarters. Then the number of dimes is 4x, according to the problem, and the number of nickels is 25 - x - 4x = 25-5x. Write an equation for the total value 5*(25-5x) + 10*(4x) + 25x = 205 cents. Simplify and find x 125 - 25x + 40x + 25x = 205, 40x = 205 - 125, 40x = 80, x = 80/40 = 2. Thus we found that there are 2 quarters. Hence, the number of dimes is 4*2 = 8. ANSWER. There are 8 dimes in the bank.

Solved elegantly, as promised.

It is a good style as the problem should be solved,
and a good style as the solution should be presented.

----------------------------------------------------

A few words about why I consider this approach correct and attractive.

First, it reduces the problem to one unknown and one equation, making the solution much simpler
than dealing with a system of three equations.

Second, it allows students to solve such problems at an earlier age, when they are not yet familiar
with systems of three equations. Therefore, it allows to accelerate the students' mathematical
education/development, which is a great benefit.

Third, this approach trains the students minds and makes their minds more flexible
(comparing with the stupid setup using 3 unknowns),
which is the important goal of mathematical education in middle and high schools.

Fourth, it is beautiful, simple and powerful at the same time, and this teaches students
to see the beauty and power in simplicity, which makes life more harmonious.

Question 1209788: roberta has 4 dollars in dimes and quarters. She has 5 more dimes than quarters.In total, roberta has 25 coins

Answer by CPhill(2189)   (Show Source):
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Here's how we can solve this problem:
Let $q$ be the number of quarters and $d$ be the number of dimes.
Since Roberta has 5 more dimes than quarters, we have the equation:
$d = q + 5$
The total number of coins is 25, so we have:
$q + d = 25$
Substituting the first equation into the second equation, we get:
$q + (q + 5) = 25$
Combining like terms, we have:
$2q + 5 = 25$
Subtracting 5 from both sides, we get:
$2q = 20$
Dividing both sides by 2, we find:
$q = 10$
Substituting this value back into the equation $d = q + 5$, we find:
$d = 10 + 5 = 15$
So, Roberta has 10 quarters and 15 dimes.
We can check our answer by verifying that the total amount of money is $4:
$0.25(10) + $0.10(15) = $2.50 + $1.50 = $4.00
Therefore, our answer is correct.

Question 1209144: wilson has 8 coins of same value and seven paise. Chloe has 11 coins of same value as those that wilson has and she has five paise. Wilson says "If you give me one of your coins and four paise, will have same amount of your money". What is the value of each coin that wilson and chloe have?
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Let x be the value of the unknown coins. Then

8x+7 = paise Wilson has
11x+5 = paise Chloe has

If Chloe gives Wilson one coin and 4 paise, then

(8x+7) + (x+4) = 9x+11 will be the paise Wilson has
(11x+5) - (x+4) = 10x+1 will be the paise Chloe has

Those two amounts will be the same

9x+11 = 10x+1
x=10

The value of the coins that Wilson and Chloe have is

ANSWER: 10 paise

Question 1209004: Quinn and Charlotte have been saving their coins and notes. Quinn has only saved
$2 coins and Charlotte only $5 notes.
Charlotte has 27 $5 notes and Quinn 76 $2 coins.
How much do they have together
Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.

Together, they have   27*5 + 76*2 = 287   dollars.

Question 1208876: In Stephen’s piggy bank there are as many nickels as pennies three more dimes the. Nickels and five fewer quarters than pennies. If he has 3.97 in his bank how many of each kind of coin does he have
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

The formal algebraic solution from the other tutor is fine. Note, however, that most students at any level will find it easier to work the problem in cents instead of dollars so that all the calculations are in whole numbers instead of decimals.

But you can get good problem-solving practice by solving a relatively simple problem like this using logical reasoning and simple arithmetic. Here are two such solutions.

(1) Changing the numbers of each kind of coin

Take away the three "extra" dimes so that the numbers of pennies, nickels, and dimes are all the same; that makes the total value of the coins $3.97 - $0.30 = $3.67.

Similarly, add five more quarters so that the numbers of all denominations of coins are the same. The new total value is $3.67 + $1.25 = $4.92. Let's change that to 492 cents to continue with the problem.

There are now equal numbers of pennies, nickels, dimes, and quarters. The value of one of each of those coins is 1+5+10+25 = 41 cents.

Divide the new total of 492 cents by 41 cents to find that there are 12 of each coin.

ANSWERS:
pennies: 12
nickels: 12
dimes (3 more than that): 15
quarters (5 fewer than that): 7

CHECK: 12(1) + 12(5) + 15(10) + 7(25) = 12+60+150+175 = 397 cents = $3.97

(2) Using logical reasoning, along with the given total value of $3.97 and the fact that the total value of the nickels, dimes, and quarters is a multiple of 5 cents, to solve the problem using "smart" trial and error.

With a total value of 397 cents, the number of pennies can only be 2, or 7, or 12, or 17, or....

There can't be only 2 pennies, because the number of quarters is 5 less than the number of pennies.

If there were 7 pennies, then the number of quarters would be only 7-5 = 2, and you aren't going to get a total of 397 cents with just 2 quarters.

Trying 12 pennies next, we find that gives us the correct total value of 397 cents.

From there the solution is the same as above.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
Written more clearly, As many nickels as pennies, three more dimes than nickels, and five fewer quarters than pennies. He has $3.97.

COIN QUANTITY AMOUNT penny x 0.01x nickel x 0.05x dime x+3 0.1(x+3) quarter x-5 0.25(x-5) TOTAL 3.97

, and from this you can find the others.

Question 1208422: if i had 1 $2, 1 $1, 4 50c, 5 20c and 5 10c coins what are the different ways I can make $2.00
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20077)   (Show Source):
You can put this solution on YOUR website!

Use Excel or whatever computer program you have access to: $2 $1 $0.50 $0.20 $0.10 1 0 0 0 0 0 1 2 0 0 0 1 1 2 1 0 1 1 1 3 0 1 1 0 5 0 1 0 5 0 0 1 0 4 2 0 1 0 3 4 0 0 4 0 0 0 0 3 2 1 0 0 3 1 3 0 0 3 0 5 0 0 2 5 0 0 0 2 4 2 0 0 2 3 4 0 0 1 5 5 Edwin

Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Make an organized list using a "greedy" algorithm, in which you start your list with the largest possible numbers of the largest value coins.

(1) If you have the $2 coin, you are done -- you can't have any more coins.
(2) If you don't have the $2 coin, then you can have 2, 1, or 0 $1 coins.
(3) If you don't have the $2 or $1 coin, you can have 4, 3, 2, 1, or 0 50-cent coins.
(4) if you don't have any of the $2, $1, or 50-cent coins, you don't have enough other coins to make $2.

Make your list using that outline. I will do part of the work for you; you don't learn anything by having me do the whole problem for you.
$2 $1 50c 20c 10c case # from above --------------------------------------------- 1 0 0 0 0 (1) 0 2 0 0 0 (2) 0 1 2 0 0 0 1 1 2 1 0 1 1 1 3 0 1 1 0 5 0 1 0 5 0 0 1 0 4 2 0 1 0 3 4 0 0 4 0 0 (3) 0 0 3 ....... (3) (there are 3 solutions here) 0 0 2 ....... (3) (there are 3 solutions here) 0 0 1 ....... (3) (there is 1 solution here)

So there is a total of 17 solutions....

Question 1100355: Sam has $3.15 in quarters nickels and dimes he has half as many quarters as he has dimes and nickels together he also has two more nickels than dimes how many of each coin are there
Found 5 solutions by MathTherapy, Edwin McCravy, mccravyedwin, greenestamps, timofer:
Answer by MathTherapy(10806)   (Show Source):
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Sam has $3.15 in quarters nickels and dimes he has half as many quarters as he has dimes and nickels together he also has two more nickels than dimes how many of each coin are there Appears as though this should be written as: "Sam has $3.15 in quarters, nickels, and dimes. He has half as many quarters as he has dimes and nickels, together. He also has two more nickels than dimes. How many of each coin are there?" Let number of dimes be D Then number of nickels is D + 2 (He has 2 more nickels than dimes) Number of quarters: (He has half as many quarters as he has dimes and nickels, together) Amount in DIMES: .1D Amount in NICKELS: .05(D + 2) = .05D + .1 Amount in QUARTERS: .25(D + 1) = .25D + .25 Total amount: $3.15 We then get: .1D + .05D + .1 + .25D + .25 = 3.15 .4D + .35 = 3.15 .4D = 2.8 Number of dimes, or Number of nickels: D + 2 = 7 + 2 = 9 Number of quarters: D + 1 = 7 + 1 = 8

Answer by Edwin McCravy(20077)   (Show Source):
You can put this solution on YOUR website!

In fact, you can even use artificial intelligence. You cannot trust AI for advanced stuff, but it can solve simple stuff very easily. I copied and pasted exactly word-for word what you posted above (without punctuation marks) into: https://poe.com/ and here's what I got (copied and pasted) from that site: ------------------------------------- Let's define the variables for the coins: Let 𝑑 be the number of dimes. Let 𝑛 be the number of nickels. Let 𝑞 be the number of quarters. From the problem, we have the following relationships: The total value of the coins is $3.15, which is 315 cents: 10𝑑+5𝑛+25𝑞=315 Sam has half as many quarters as the total number of dimes and nickels: 𝑞=1/2(𝑑+𝑛) He has two more nickels than dimes: 𝑛=𝑑+2 Now, we can substitute 𝑛 in terms of 𝑑 into the other equations. From the third equation, we have: 𝑛=𝑑+2 Substituting 𝑛 into the second equation: 𝑞=(1/2)(𝑑+(𝑑+2))=(1/2)(2d+2)=d+1 Now substituting 𝑛 and 𝑞 into the first equation: 10𝑑+5(𝑑+2)+25(𝑑+1)=315 Expanding this gives: 10𝑑+5𝑑+10+25𝑑+25=315 40𝑑+35=315 Now, subtract 35 from both sides: 40𝑑=280 Dividing by 40: 𝑑=7 Now that we have d, we can find 𝑛 and 𝑞 𝑛=𝑑+2=7+2=9 𝑞=𝑑+1=7+1=8 Thus, the solution is: Dimes (d): 7 Nickels (𝑛): 9 Quarters (𝑞): 8 Final Answer: Dimes: 7 Nickels: 9 Quarters: 8 ------------------------- That was all copied and pasted exactly as poe.com printed instantly. But don't depend on it for more advanced math. It will sometimes make stupid mistakes. AI is not perfected yet. But someday it will be, and will take us all over, just as Stephen Hawking warned. LOL (Maybe I shouldn't be laughing). Edwin

Answer by mccravyedwin(421)   (Show Source):
You can put this solution on YOUR website!

Apparently the younger generation doesn't see any need for punctuation marks or capital letters to separate sentences. They just run them all together. And soon the correct way to write will include "u" for "you", "r" for "are", "before" as "b4", "8" for "ate", etc. In that light, there are plenty of sites online where you only have to type in a system of equations, point and click, and the answer pops out instantaneously. So why not use them? You can even do it on a TI-84 calculator, if you install the right program. Why hitch a horse to a buggy and get there in an hour, when you can drive a car and get there in 10 minutes? If you use a separate letter for each unknown, setting up the equations is a cinch. This system below is set up directly from the words: Go to one of those sites. Type that in. Point and click, and instantly you get q=8, d=7, n=9. Here is a link to such a site: https://www.emathhelp.net/en/calculators/algebra-2/system-of-equations-solver/? Edwin

Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

He has two more nickels than dimes:
let x = # of dimes
then x+2 = # of nickels

He has half as many quarters as nickels and dimes together:
# of nickels and dimes together = x + (x+2) = 2x+2
# of quarters = (1/2)(2x+2) = x+1

The total value of the coins is $3.15, or 315 cents:
10(x)+5(x+2)+25(x+1)=315
10x+5x+10+25x+25=315
40x=280
x=280/40=7

ANSWERS:
nickels: x+2 = 9
dimes: x = 7
quarters: +1 = 8

CHECK: 9(5)+7(10)+8(25) = 45+70+200 = 315

Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!
.

Go to

https://www.algebra.com/algebra/homework/word/coins/Word_Problems_With_Coins.faq.question.1100355.html

to see all the solutions.

Question 1207581: how do I find how many dimes he has?
Maurice has 280 coins - all quarters and dimes - won from playing poker. If his winnings
total $49.45, how many dimes does he have?"
and help for this please.
"The perimeter of a triangle is 64 metres. The longest side of the triangle is 3 times the
length of the shortest side, and the third side is 8 metres less than double the length of the
shortest side. Determine the dimensions of all three sides of the triangle."
Found 3 solutions by mananth, ikleyn, josgarithmetic:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website!
Maurice has 280 coins - all quarters and dimes - won from playing poker. If his winnings
total $49.45, how many dimes does he have?"
Let the number of dimes maurice has be d
and quarters be q
Maurice has 280 coins
d+q=280...........................................(1)
his winnings total $49.45
10d+25q=49.45*100.........................(2)
10d+25q=4945
multiply (1) by 10 and subtract from (2)
10d+10q= 2800
10d+25q=4945
15q= 2145
q= 2145/15 = 143
d+q= 280
Therefore d = 280-143=137
Maurice has 137 dimes.
Check
0.25×143+0.10×137=35.75+13.70=49.45

The perimeter of a triangle is 64 metres. The longest side of the triangle is 3 times the
length of the shortest side, and the third side is 8 metres less than double the length of the
shortest side. Determine the dimensions of all three sides of the triangle."

Let x be the longest side ,y be the shortest side and z the third side.
The longest side of the triangle is 3 times the
length of the shortest side,
x=3y ......................................(1)
The third side is 8 metres less than double the length of the
shortest side.
z= 2x-8 ..................................(2)

The perimeter of a triangle is 64 metres (given)
x+y+z= 64
x+(x/3)+(2x-8)=64
x+x/3 +2x-8=64
3x+x/3 = 72
multiply by 3
3x+3x = 216
6x=216
x= 36
x=3y
y=x/3
y = 36/3 = 12
x+y+z= 64
12+36+z= 64
z = 64-48
z= 16
The dimensions of the sides of the triangle are 12 meters, 36 meters, and 16 meters.

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
help for this please.
"The perimeter of a triangle is 64 metres. The longest side of the triangle is 3 times the
length of the shortest side, and the third side is 8 metres less than double the length of the
shortest side. Determine the dimensions of all three sides of the triangle."
~~~~~~~~~~~~~~~~~~~

If you stop your analysis there, where @josgarithmetic stopped it, your solution
will be 100% incomplete and 100% incorrect.

As you read the problem, you may see that if x is the length of the shortest side of the triangle, then the longest side is 3x, and the third side is 2x-8. But such triangle can not exist, since the triangle inequality is violated. Indeed, the sum of sides x and (2x-8) is 3x-8, and it is SHORTER than the length of the third side, 3x, while, according to the triangle inequality, it should be LONGER than the third side. It means that the problem's condition is SELF-CONTRADICTORY and such a triangle CAN NOT exist.

Solved.

--------------------

The " solution " by @mananth is 100% incorrect.

It is " rotten to the core ", since a triangle with side lengths
12 meters, 16 meters and 36 meters does not exist.

Answer by josgarithmetic(39792)   (Show Source):
You can put this solution on YOUR website!
------------------------------------------------------
Maurice has 280 coins - all quarters and dimes - won from playing poker. If his winnings total $49.45, how many dimes does he have?"
---------------------------------------------------

dimes and the other coin type not dimes
Totaling of 280 of these coins:
--------Write this equation and solve; that is how.

As for the triangles side lengths exercise question,
This part is said very very badly: Determine the dimensions of all three sides of the triangle.

How about restating that exercise like this:
-------------------------------------------------------
The perimeter of a triangle is 64 metres. The longest side of the triangle is 3 times the length of the shortest side, and the third side is 8 metres less than double the length of the
shortest side. Determine the length of each side of the triangle.
-------------------------------------------------------

SIDES LENGTH shortest x medium 2x-8 longest 3x SUM 64

This allows a perimeter equation,
and the, ...,

-------and from this, find the length of the medium side and of the longest side.

Would this triangle work starting with the variable side lengths?
-
one of the Triangle Inequality inequalities:

NO. The described triangle will not work.

Question 1207580: how can i find the answer? Dave has eight more dimes than nickels in his pocket. In total, the coins in his pocket are
worth $3.05. How many of each type of coin does he have?

Answer by ikleyn(53748)   (Show Source):
You can put this solution on YOUR website!
.
how can i find the answer? Dave has eight more dimes than nickels in his pocket.
In total, the coins in his pocket are worth $3.05. How many of each type of coin does he have?
~~~~~~~~~~~~~

Let x be the number of nickels; then the number of dimes is (x+8). The value of nickels is 5x cents; the value of dimes is 10*(x+8) cents. The total value is $3.05, or 305 cents; so we write this equation 5x + 10*(x+8) = 305. Simplify and find x 5x + 10x + 80 = 305 15x = 305 - 80 15x = 225 x = 225/15 = 15. ANSWER. 15 nickels and 15+8 = 23 dimes. CHECK. The total value is 15*5 + 23*10 = 75 + 230 = 305 cents = $3.05. ! correct !

Solved.

Question 1205719: Katie has a collection of nickels, dimes and quarters with a total value of $6.80. There are 5 more dimes than nickels and 6 more quarters than nickels, how many of each coin does she have?
Found 2 solutions by MathTherapy, greenestamps:
Answer by MathTherapy(10806)   (Show Source):
You can put this solution on YOUR website!

Katie has a collection of nickels, dimes and quarters with a total value of $6.80. There are 5 more dimes than nickels and 6 more quarters than nickels, how many of each coin does she have? Let number of nickels be N Then, number of dimes and quarters = N + 5, and N + 6, respectively Value of nickels: .05N Value of dimes: .1(N + 5) = .1N + .5 Value of quarters:.25(N + 6) = .25N + 1.5 As total value is 6.80, we get: .05N + .1N + .5 + .25N + 1.5 = 6.8 .4N + 2 = 6.8 .4N = 4.8 Number of nickels, or Now, calculate number of dimes (N + 5), and number of quarters (N + 6).

Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

You can solve the problem informally, using steps that are nearly the same as the formal algebraic solution shown by the other tutor.

Probably this problem was supposed to be solved using formal algebra; but you can get good mental exercise solving the problem without it.

The value of the 5 "extra" dimes is 5($0.10) = $0.50; the value of the 6 "extra" quarters is 6($0.25) = $1.50. The total value of those "extra" coins is $0.50 $1.50 = $2.00.

So the value of the remaining coins is $6.80 - $2.00 = $4.80.

Those coins are equal numbers of nickels, dimes, and quarters. The total value of a group consisting of one nickel, one dime, and one quarter is $0.40. To make the remaining $4.80, the number of those groups must be $4.80/$0.40 = 12.

So the number of nickels is 12; the number of dimes is 12+5 = 17, and the number of quarters is 12+6 = 18.

ANSWERS: 12 nickels, 17 dimes, 18 quarters

Question 1205731: 50 coins in dimes and quarters have a total value of $11.00. How many dimes and quarters are there?
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13327)   (Show Source):
You can put this solution on YOUR website!

Informally....

If all 50 coins were dimes, the total value would be $5.00.
The actual total value is $11.00, which is $6.00 more.
Since each quarter is worth $0.15 more than each dime, the number of quarters must be $6.00/$0.15 = 600/15 = 40.

ANSWER: 40 quarters and 10 dimes

CHECK: 40($0.25)+10($0.10) = $10.00+$1.00 = $11.00

----------------------------------------------------------------------

And here is a really unorthodox solution, treating the problem as a mixture problem.

You are mixing quarters worth 25 cents each with dimes worth 10 cents each to get a mixture of coins worth an average of $11.00/50 = 22 cents each.
Since 22 cents is 4/5 of the way from 10 cents to 25 cents, 4/5 of the coins must be quarters. (From 10 to 25 is a difference of 15; from 10 to 22 is a difference of 12; 12/15 = 4/5.)
4/5 of 50 is 40.

ANSWER (again, of course): 40 quarters and 10 dimes

Answer by josgarithmetic(39792)   (Show Source):
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50-q of dimes
q quarters

There are still 50 dimes and quarters just as before.