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A collection of 30 coins worth 5.50 consists of nickels dimes and quarters there are twice as many dimes as nickeles how many quarters are there?
Let the number of nickels, dimes and quarters be n, d, q respectively. Then n +d + q = 30 5n + 10d + 25q = 550 But d = 2n, so: n + 2n + q = 30 => 3n + q = 30 5n + 10(2n) + 25q = 550 => 25n + 25q = 550 => n + q = 22 Which gives two simultaneous equations to solve, resulting in: n = 4, q = 18 So there are 18 quarters (plus 4 nickels and 8 dimes).
10 coins consistinig of nickels and dimes equal 80 cents how many coins does she have?
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.
Bob has 60 coins consisting of quarters and dimes The coins combined value is 9.45 How many are quarters and how many are dimes?
there are 23 quarters 37 dimes to do this use simultaneous equations to make this simpler i will assign variables to amount of quarters and dimes x-quarters y-dimes you know you have 60 coins, so amount of quarters plus dimes is 60 x+y=60 you also know the amount is equal to $9.45 since quarters are $0.25 and dimes are $0.10 another equation can be made 0.25x+0.10y=9.45 this equation shows what you do when you count your change now solve the first equation for a variable using algebra x=60-y you can then plug this new value into your second equation 0.25(60-y)+0.1y=9.45 you can now solve for y, which is the number of dimes y=37 now plug this number back into x=60-y to get the amount of quarters x=60-(37) x=23 23 quarters 37 dimes
Rihanna has 5.25 in dimes and nickels for a total of 63 coins How many dimes does she have?
One equation will represent the number of coins, x + y = 63. The second equation will represent the value of the coins, 0.10x + 0.05y = 5.25. Solve the first equation for y (=63-x) and plug into the second equation. So, 0.1x + 0.05(63-x) = 5.25 Solve this and get x=42. your welcome mizz litta bay be aka anhellita richards
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?
You piggy bank has all nickels and quarters. You count up the coins and find out that there are 64 coins and you have a total of 7.40. How many of each type of coin do you have?
The idea is to write two equations, one for the number of coins, one for the amount of money. Then solve the equations.Assuming "n" is the number of nickels, and "q" the number of quarters, the equations for the coins, of course, is quite simply: n + q = 64 And the equation for the money (I'll use cents; you can just as well use dollars if you prefer): 5n + 25q = 740 You can solve the first equation for "n", then replace that in the second equation.
Kim has 40 coins with a total of 8.80 some of the coins are nickels and the rest are quarters how many of each kind of coin does Kim have?
Let's represent the number of nickels as (x) and the number of quarters as (y). We can create a system of equations to solve for (x) and (y): (x + y = 40) (total number of coins) (0.05x + 0.25y = 8.80) (total value of the coins) From the first equation, we can rewrite it as (x = 40 - y). Substituting this into the second equation gives us (0.05(40 - y) + 0.25y = 8.80). Solving this equation will give us the values of (x) (nickels) and (y) (quarters).
The coins in a cash register total 12.50 and there are only nickels dimes and quarters. There are twice as many dimes as nickels and there are twice as many quarters as dimes. How many many dimes are?
You need to define variables for the different types of coins, write the corresponding equations, then solve them. One equation for each fact. Here are the equations:5N + 10D + 25Q = 1250 D = 2N Q = 2D
A collection of 30 coins worth 5.50 consists of nickels dimes and quarters there are twice as many dimes as nickeles how many quarters are there?
Let the number of nickels, dimes and quarters be n, d, q respectively. Then n +d + q = 30 5n + 10d + 25q = 550 But d = 2n, so: n + 2n + q = 30 => 3n + q = 30 5n + 10(2n) + 25q = 550 => 25n + 25q = 550 => n + q = 22 Which gives two simultaneous equations to solve, resulting in: n = 4, q = 18 So there are 18 quarters (plus 4 nickels and 8 dimes).
How many ways can you make 75 cents only using quarters dimes and nickels?
There are 18 possible combinations of quarters, dimes and nickels that total 75 cents. They are: 1) -- 3 Quarters 2) -- 2 Quarters, 2 Dimes, 1 Nickel 3) -- 2 Quarters, 1 Dime, 3 Nickels 4) -- 2 Quarters, 5 Nickels 5) -- 1 Quarter, 5 Dimes 6) -- 1 Quarter, 4 Dimes, 2 Nickels 7) -- 1 Quarter, 3 Dimes, 4 Nickels 8) -- 1 Quarter, 2 Dimes, 6 Nickels 9) -- 1 Quarter, 1 Dime, 8 Nickels 10) -- 1 Quarter, 10 Nickels 11) -- 7 Dimes, 1 Nickel 12) -- 6 Dimes, 3 Nickels 13) -- 5 Dimes, 5 Nickels 14) -- 4 Dimes, 7 Nickels 15) -- 3 Dimes, 9 Nickels 16) -- 2 Dimes, 11 Nickels 17) -- 1 Dime, 13 Nickels 18) -- 15 Nickels
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.
How do you solve by substitution y equals x-1?
George saves nickels and dimes for tolls. If he has 8 coins worth $2.60,how many are nickels and how many are dimes? Answer this question by using system of equation.
Karen has 20 coins worth 1.35 the coins are all nickels and dimes how many of each coin does she have?
Oh, what a lovely little math puzzle we have here! Let's paint a happy little picture to solve it. Since Karen has 20 coins total worth $1.35, we can set up a system of equations to find out she has 9 nickels and 11 dimes. Keep on painting those beautiful math problems, my friend!
10 coins consistinig of nickels and dimes equal 80 cents how many coins does she have?
You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.You could (a) use trial and error; (b) write two equations; or (c) call the number of nickels "n", and the number of dimes "10-n" (since there are 10 coins in total). I will use the latter approach.Number of nickels: nNumber of dimes: 10-nFor the main equation, multiply the number of coins by the value of each coin:value of nickels + value of dimes = 805n + 10(10-n) = 80Now solve the equation:5n + 100 - 10n = 80-5n = -205n = 20n = 4So, you have 4 nickels, and 6 dimes.
The coins in the store's cash register total 4.00 The cash register contains only nickels dimes and quarters There is an equal number of each coin How many dimes are in the cash register?
There are 10 of each coin in the cash register. 10 quarters (2.50) plus 10 dimes (1.00) plus 10 nickels (.50) = 4.00 The formula to solve this problem is: (X x .25) + (X x .10) + (X x .05) = 4.00 X (.25+.10+.05) = 4.00 X = 4.00/.4 = 10Woah that is really hard math
A cash register contains 12.50 and there are only nickels dimes and quarters There are twice as many dimes as nickels and twice as many quarters as dimes How many quarters are in the cash register?
Nickles - 10 $0.50dimes - 20 $2.00quarters - 40 $10.000.50+2.00+10.00 = $12.50 containing 40 quarters---Here's how to solve this with Algebra :Let N be the number of nickels, so that2N is the number of dimes, and2(2N) or 4N is the number of quarters.A nickel is 5 cents, a dime is 10 cents, and a quarter is 25 cents,and the total in the cash register is $12.50Multiplying by their cents values, we have5(N) plus 10(2N) plus 25 (4N) = 12505N + 20N + 100N = 1250125 N = 1250N= 10So the number of nickels is 10, dimes 20, and quarters 40.
How many coins are worth 1.45 if the coins are only in dimes and nickels?
To find out how many coins are worth $1.45 using only dimes and nickels, we can set up a system of equations. Let x be the number of dimes and y be the number of nickels. The value of x dimes is 10x cents, and the value of y nickels is 5y cents. We can write the equation 10x + 5y = 145 (since $1.45 is equivalent to 145 cents). To solve for the number of coins, we need to find a combination of x and y that satisfies this equation.
