I have 23 coins valued at $4.25.They consist of quarters and dimes.Findout how may of each kind of coin i have
546151
Do your own damn middle school homework.
833462
If he had NO quarters, then he would have 18 dimes and 22 nickels, worth $2.90.  Every time you replace 1 nickel and 1 dime with 2 quarters, it keeps the coin count right, and the extra nickels right, while the value goes up by (2 * 25 - 10 - 5) = 35 cents.  But to get from $2.90 to $4.05, the total value needs to go up by $1.15 which is NOT an even multiple of $0.35, so there is no solutuion in whole numbers.  One of the figures in the question must be wrong.
343331
What math test do you take?  I'm in Algebra 1 and I only remember 1 question.  It's impossible.\n\nSomeone has a total of 100 coins in dimes and quarters.  The total value adds up to $14.05. How many quarters does she have?\n\nA) 27 (I chose this-it's a dollar off)\nB) 40\nC) 56\nD) 79 (I forget-it's way too high)\n\nWell, that's all I have.  Bye bye!\n\nThanks to the guy who solved that.  I won't get a zero now!
199752
1 cent is the value of 1 penny.\n1 dollar is the value of 1 dollar.\n\n1 penny = 1 cent = 0.01 dollars\n1 nickel = 5 cents = 0.05 dollars\n1 dime = 10 cents = 0.10 dollars\n1 quarter = 25 cents = 0.25 dollars\n\nThere are a couple more rare or obscure coins such as the 50 cent piece (not made anymore), and the dollar coin (many unpopular and thus little used versions of this).
172025
This problem is easy because we can only have 2 coins.  Each penny is worth 1 cent, but also 1 coin.  What I did to solve this problem was use a guess and check method.  \n82 cents- 22 cents (all in pennies) = 60 cents.  Because there are 34 coins, we subtract 22 from 32 to get 12.  That means that 12 coins have to be nickels and to check that this is correct:\n12 * 5 = 60 (cents from nickels) + 22 (cents from pennies)= 82 cents.  12 nickels were found.
406839
collectible coin values at:
259226
here's one with coins that was on one of the worksheets i have:\n\nlet's say, for example, that you have $8.75 in dimes and quarters and there are 65 coins. how many dimes and quarters are there?\n\nso there would be 2 equations:\nd+q=65 (which is how many coins there are)\n\nand\n\n10d+25q=875 (how much money there is in dimes and quarters)\n\nfrom there, you'd solve it up and down like\n\n 10d+25q=875\n+  d+  q= 65 (Multiply by 10)\n---------------------------\n 10d+25q  =875\n+(-10d-10q=-650)\n-----------------------\n       15q= 225\n         q=15\n\nthen you plug in 15 for q for one of the equations.\n\n15+d=65\n65-15=50\nd=50\n\nyou'd have 50 dimes and 15 quarters. (i hope)\n\ni also hoped this helped ^^;
459515
2 quarters 3 dimes and 20 pennies
1085726
Check eBay.  Lots of coins bought/sold there.
1014242
i just checked ebay and they are going for about 20 buck. good luck. check out www.coinfacts.com
589542
Your best bet is to find a coin dealer in your area.  You can check out approximate values by looking at what the coin and silver certifates are selling for on ebay, or at one of the online price guides.  Good luck!
1128414
It looks like you have a 1932 Five Lati Crown slver coin.\n\nBelow is a link to a coin dealer that was offering the same coin for between $23 and $42 US.  I don't jknow if those are current prices.\n\nWith all coins - the value of the coin is related to the condition / wear of the coin.
550830
7 dimes and 15 quarters\n\nLet q = the number of quarters\nLet d = the number of dimes\n\nq = d + 8 (because there are 8 more quarters than dimes)\n4.45 = .25q + .10d (the sum of the number of quarters and dimes equals 4.45.\n\nBecause q = d + 8, substitute d+8 into the second equation for q.\n 4.45 = .25(d + 8) + .10d\n4.45 = .25d + 2 +.1d  (distribute .25)\n4.45 = .35d + 2    (Combine like terms on the right side)\n2.45 = .35d    (Subtract 2 from both sides)\n7 = d    (Divide both sides by .35)\n\nNow plug 7 in the first equation for d and solve for q:\n\nq = 7 + 8\nq = 15\n\nThus 7 dimes and 15 quarters\n\nCheck:\n\n7 dimes = .70 \n15 quarters = 3.75\n.70 + 3.75 = 4.45
199689
At the very least, they are worth there bullion value, that is, whatever silver is selling for today. If on the other hand, they are in excellent condition, they could be worth significantly more. Take them to a coin dealer and find out. It won't cost you anything.
877044
$1.19 - 3 quarters, 4 nickels, 4 pennies\n\nThe people above me who have changed there answers are cheats and should be reported as such !
737305
3 gold coins maybe 1 silver
789045
Let\nd = number of dimes\nq = number of quarters\nn = number of nickels\np = number of pennies\n.10d = value of dimes\n.25q = value of quarters\n.05n = value of nickels\n.01p = value of pennies\n\n\n\n\nWe have the following equations:\n\nKeinv has $6.45 in coins in his cash box. \n.10d + .25q + .05n + .01p = 6.45\n\nThe number of quarters is one less than twice the number of dimes. \nq = 2d - 1\n\nThe number of nickels is one less than twice the number of \nquarters. \nn = 2q - 1\n\nThe value of the pennies is the same as the value of \nthe nickels. \n.01p = .05n\n\n\n\nIn summary,\n(1) .10d + .25q + .05n + .01p = 6.45\n(2) q = 2d - 1\n(3) n = 2q - 1\n(4) .01p = .05n\n\nWe simplify these:\n(1) 10d + 25q + 5n + p = 645\n(2) q = 2d - 1\n(3) n = 2q - 1\n(4) p = 5n\n\nSubstitute (2) into (3)\nn = 2(2d - 1) - 1\nn = 4d - 2 - 1\nn = 4d - 3\n\nSubstitute this into (4)\np = 5(4d - 3)\np = 20d - 15\n\nSubstitute n and p into (1)\n10d + 25(2d - 1) + 5(4d - 3) + (20d - 15) = 645\n\nWe distribute\n10d + 50d - 25 + 20d - 15 + 20d - 15 = 645\n\nWe add the d's and the constants\n100d - 55 = 645\n\nWe transpose the constant to the right, and simplify\n100d = 700\n\nThus,\nd = 7\nq = 2d - 1 = 13\nn = 4d - 3 = 25\np = 20d - 15 = 125\n\nTherefore, Keinv has 7 dimes, 13 quarters, 25 nickels, and 125 pennies (having a total value of $6.45).\n^_^