Toy has 4.45 in quarters and dimes.She has 8 more quarters than dimes,how many coins of each type does she has
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
305502
a quarter and a nickel because if one of them is not a nickel then try the other. Hah ha!!
310299
since two nickels equal a dime multiply 8 by 2 and you get 16 nickels and since five nickels equals one quarter multiply 5*12 to get 60 nickels so the ratio is 4:15
789401
Let's see.\n\nTwo dimes and one nickel\nOne dime and three nickels\nFive nickels.\n\nYou can have five pennies in place of one nickel in any of the above sitiuations.\n\nSo you could have:\n\nTwo dimes and five pennies\nOne dime and two nickels and five pennies\nFour nickels and five pennies.\n\nBut once again, you can have five pennies in place of one nickel in any of the above sitiuations. (Now, D = dime, N = nickel, P = penny)\n\n1D, 1N, 10 P\n3N, 10P\n\nOne more time\n\n1D, 15P\n2N, 15P\n\nAnd again:\n\n1N, 20P\n\nOne last time\n\n25P\n\nI count 12 ways.
466300
The waitress calculation is wrong...\n\nYou me and Sally gave the waitress $10 each for a total of $30.\n\nThe waitress gave back $5\n\nWe gave the waitress back $2.\n\nTherefore, the waitress has $30 - $5 + $2 = $27
957112
The answer was A..       dime, nickel and quarter or something like that.
726629
Here's one...\n\nI have 30 cent in my hand.  One coin is not a quarter.  What are the two coins?\n\nAnswer: a quarter and a nickel.  JUST one coin in not a quarter.  NOT BOTH coins!\n\nI hope that works for you!\n\nIf not, here's another one...\n\nWhich one is correct:\nThe yolk are white.  OR The yolk is white.\n\nAnswer: NEITHER.  The yolk is YELLOW!\n\nI hope I was of some help!
345803
D is # of dimes (each worth 10 cents)\nQ is # of quarters (worth 25 cents)\n\n10D + 25Q = 255 (equation about value)\n\nD = Q + 8 (equation about number of each coin)\n\nSubstitute Q + 8 for D in the first equation and solve for Q. Use that answer to solve for D.
878065
.28 = 1 quarter and 3 pennies\n.28 = 2 dimes 1 nickel and 3 pennies\n.28 = 1 dime 3 nickels and 3 pennies\n.28 = 5 nickels and 3 pennies\n.28 = 2 dimes and 8 pennies\n.28 = 1 dime 2 nickels and 8 pennies\n.28 = 1 dime 1 nickel and 13 pennies\n.28 = 1 dimes and 18 pennies\n.28 = 4 nickels and 8 pennies\n.28 = 3 nickels and 13 pennies\n.28 = 2 nickels and 18 pennies\n.28 = 1 nickel and 23 pennies\n.28 = 28 pennies\n\ntotal 13 ways to mack change for .28.
876730
A quarter and a nickel.\n\nONE of them is not a nickel...its a quarter!
501856
a penny weighs 2.5 g.  there are 454 g in a lb.\n\n454/2.5 = 181.6, so there are 182 pennies, or thereabouts.\n\nROFL @ the other answer...
834677
Since Delano gives Nadine four dimes so that they have the same amount of money, the following equations are true. \n\na=number of dimes that Delano had to start\n\n5(7+a)+40=10a-40\n\n35+5a+40=10a-40\n\n75+5a=10a-40\n\n115=5a\n\na=23\n\nTherefore, Delano with 23 dimes, and Nadine has 23+7, or 30 nickels. Nadine has $1.50, while Delano has $2.30. When Delano gave Nadine 4 dimes, he lost 40 cents while she gained 40, so both have $1.90. Therefore, they have $3.80 combined.
809813
Seen this riddle at least 2 times already... getting old.... \n\nanswer is the other coin is the nickle and the first one is a quarter...
604316
One coin is not a ten-cent piece, but the other one is.  You have a 5- and a 10-cent piece.
405715
x+y=50\n.02x + .05y=1.87\n\nfrom the first equation, subtract y from both sides:\n\nx=50-y\n\nsubstitute this into x in the second equation:\n.02(50-y) + .05y = 1.87\n\nand solve for y\n1-.02y+.05y=1.87\n.03y=.87\ny=29\n\nso there are 29 5 cent coins and 21 2 cent coins.
896257
1. \nLet n be the number of nickels\nLet d be the number of dimes\nLet q be the number of quarters\n\nNumber of coins is 30:\n#1) n + d + q = 30\n\nThere are twice as many nickels as quarters:\n#2) n = 2q\n\nTotal value of the coins is $3.35.  (Use the value in cents, rather than dollars).\n#3) 5n + 10d + 25q = 335\n\nThree equations and three unknowns...\n\nSubstitute #2 into #1:\nd + 3q = 30\nAnd into #3:\n10d + 35q = 335\n\nNow solve for d:\nd = 30 - 3q\n\nSubstitute into the last one:\n10(30 - 3q) + 35q = 335\n300 - 30q + 35q = 335\n5q = 35\nq = 7\n\nPlug q back into d = 30 - 3q:\nd = 30 - 3(7)\nd = 9\n\nPlug q back into n = 2q\nn = 2(7)\nn = 14\n\nSo you have 14 nickels, 9 dimes, 7 quarters\n\n(As a double check --> 0.70 + 0.90 + 1.75 = 3.35)\n\n2. \nKaren is Anthony (statement #3)\nSince Karen is Anthony, John must be Juliet (statement #1)\nBeth is not Cleopatra, so she must be Romeo (statement #4)\nThat leaves Dan to be Cleopatra.\n\nFinal result:\nKaren = Anthony\nJohn = Juliet\nBeth = Romeo\nDan = Cleopatra\n\nThose cross-dressing friends...
459515
2 quarters 3 dimes and 20 pennies