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How many dimes nickels and pennies are there in 1.10?
To find the number of dimes, nickels, and pennies in $1.10, we can set up a system of equations. Let x represent the number of dimes, y represent the number of nickels, and z represent the number of pennies. The total value equation is 10x + 5y + z = 110 cents. Since there are no constraints on the number of coins, there are multiple solutions to this problem. One possible solution is 8 dimes, 0 nickels, and 10 pennies.
If you receive 12 coins after inserting a dollar bill how many nickels did you receive?
One solution is 10 nickels, because if you put in a dollar it can be broken into: 2 quarters (50 cents) so 50 cents are left, and you have 10 coins left to use If these 10 coins are all nickels, (total: 50 cents), the problem is solved! .5 +.5 =1 dollar
150 total coins of dimes and nickels 70 more nickels than dimes?
110 nickels and 40 dimes my little brother had the same problem
How can you solve a ratio problem?
Here is one example of one way to do it.You have a ratio of five to one ( 5:1 or 5/1). You need to apply it to a specific situation. You must have five pennies for each nickel in two piles. Pile one has 60 pennies and pile two has 14 nickels.To find the correct number of coins to complete each pile, you can solve it in this manner if you have an aversion to other ways of doing this. I hope you understand what a ratio is. In this example as in real life, there are five pennies for every nickel.Pile one, the equation is: 5/1 = 60/N, where N is the number of nickels. This can be read out loud as "Five is to one as 60 is to N". To solve this, take 5 and divide it by 1. The answer is 5. Next take 60 and DIVIDE it by 5. This is 12, your answer: 60 pennies is to 12 nickels as 5 pennies is to 1 nickel. Here we divided because the unknown quantity was on the bottom of the fraction.Pile Two, the equation is: 5/1 = P/14, where P is the number of pennies. Read this out loud as "Five is to one as P is to 14. To solve this, take 5 and divide by 1. The answer is 5. Next take 14 and MULTIPLY by 5. This is 70, your answer: 70 pennies is to 14 nickels as 5 pennies is to 1 nickle. Here we multiplied because the unknown quantity was on the top of the fraction.
How many ways to make 43 cents?
To determine the number of ways to make 43 cents using U.S. coins (pennies, nickels, dimes, and quarters), we can use a combinatorial approach or a dynamic programming method. However, without specific calculations, the number of combinations can vary based on the denominations and their limits. Generally, the exact number of ways to make 43 cents can be computed programmatically or through extensive enumeration of combinations, but it's known to be a non-trivial problem. Would you like a specific breakdown or method to solve it?
13 coins that equal 1 dollar 5 ways?
2 quarters, 3 dimes 3 nickles, 5 pennies 7 dimes, 6 nickles 1 quarter, 7 dimes, 5 pennies 1 quarter, 3 dimes, 9 nickles It;s only four but maybe you van think of the fith
How many dimes nickels and pennies are there in 1.10?
To find the number of dimes, nickels, and pennies in $1.10, we can set up a system of equations. Let x represent the number of dimes, y represent the number of nickels, and z represent the number of pennies. The total value equation is 10x + 5y + z = 110 cents. Since there are no constraints on the number of coins, there are multiple solutions to this problem. One possible solution is 8 dimes, 0 nickels, and 10 pennies.
How many combinations of coins can be made into 36 cents?
Oh, isn't that a lovely question! Let's see, to make 36 cents, you can use different combinations of coins like quarters, dimes, nickels, and pennies. There are several ways to do this, and it's like creating a beautiful painting with different colors and textures. Just remember, there's no right or wrong way to make 36 cents with coins, so have fun exploring all the possibilities!
How many ways can you make 50 cents from pennies dimes nickels and quarters?
My third period amazing math class did this as their problem of the day today, and we discovered that there are 50 ways to make 50 cents. We included a 50-cent piece as well!
How many nickels would you have if you have 32.40 of only quarters and nickels and you have 78 more quarters?
Call the number of nickels n and the number of quarters q. From the problem statement, q = n + 78 and 25q + 5n = 3240. Substituting the first equation into the second yields 25(n + 78) + 5n = 3240, or 30n = 3240 - 1950 or n = 1290/30 = 43 nickels.
If she has five more dimes than quarters but twice as many nickels as dimes?
The question is incomplete. Please post a new version with the rest of the problem.
If you have fifty coins which have a total value of one dollar what are the coins and how many of each do you have?
There are two possible solutions: You could have one quarter, two dimes, two nickels and forty-five pennies, or you could have two dimes, eight nickels and forty pennies. An easy way of approaching this problem is to start by imagining that you have fifty pennies. You have the right number of coins, but are fifty cents short. Instead of adding other coins, you replace pennies with them: replacing a penny with a nickel gains four cents, a dime gains nine, and a quarter gains twenty-four. You can't possibly use more than two quarters, so there are few cases to consider there: If you replace two pennies with quarters, you've gained forty-eight cents, so you only need two more; but any further replacement will give you too much. If you use one quarter, you need to make up twenty-six more cents in steps of four or nine; it's easy to see that two of each works. Finally, with no quarters, you need to gain fifty cents using increments of four or nine; this yields the second solution.
If you receive 12 coins after inserting a dollar bill how many nickels did you receive?
One solution is 10 nickels, because if you put in a dollar it can be broken into: 2 quarters (50 cents) so 50 cents are left, and you have 10 coins left to use If these 10 coins are all nickels, (total: 50 cents), the problem is solved! .5 +.5 =1 dollar
What is 8 quarters greater than 25 nickels?
Call the unknown number of cents c. From the problem statement, c = (8 X 25) + (25 X 5) = 325 cents, which equals $3.25.
How many different ways can you make 26 cents from pennies nickels dimes and quarters?
Well honey, you can make 26 cents with 26 pennies, 5 nickels and 1 penny, 2 dimes and 6 pennies, or just 1 quarter and 1 penny. That's a total of 4 different ways to make 26 cents using those coins. Math can be a real hoot sometimes, can't it?
150 total coins of dimes and nickels 70 more nickels than dimes?
110 nickels and 40 dimes my little brother had the same problem
James has 33 coins in his pocket all of them nickels and quarters If he has a total of 2.65 how many quarters does he have?
This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.
