Amy has 63 coins with a total value of $5.05,Some are dimes some nickles, how many dimes does he have?
960458
Let x be the number of dimes and y be the number of nickels. From the question we get\n\nx + y = 63 --> (1)\n10x + 5y = 505 --> (2) (this is in cents)\n\nrearranging (1), we get\n\nx = 63 - y\n\nSubstituting into (2), we get\n\n10(63 - y) + 5y = 505\n630 - 10y + 5y = 505\n-5y = -125\ny = 25\n\nhence, we get\n\nx = 63 - 25\nx = 38\n\ntherefore she has 25 nickels and 38 dimes
982570
13 dimes, 5 quarters.  You know that there has to be an odd number of quarters because the amount ends in a 5 and there are no nickels.  Since the number is only $2.55, you know there's probably only 3 or 5 quarters, and 5 works.
788328
For your first, |x - 3| > 10\n\nFor your second, if two cans of soda cost $1.40, how much would 7 cans cost?
459515
2 quarters 3 dimes and 20 pennies
767799
Give her a break. Everyone is here coz they didn't really listen in class or need help. That's why we are here to answer them.\n\n1) 11% (275/2500 x 100 = Answer)\n\n2) 963 (900/100 x 107 = Answer)\n\n3) 750 s (300/40 x 100 = Answer)\n\n4) 7550 (8882/100 x 85 = Answer)\n\n5) $20442.50 (18500/10 x 110.5 = Answer)\n\n6) $5875.20 (6400/100 x 8.2 = Answer) \n\nI've done it for u, u better buck up in ur maths. These are the basics.
882485
1 Nickel = 5 c\n1 Dime = 10 c\n1 Quarter = 25 c\n5.50$ = 550 c\n\nAssume number of nickels = X\n\nThen number of quarters = X + 4\n\nThen number of dimes = total number of coins – (number of nickels + number of quarters) = 37 – (X + X + 4) = 33 – 2X\n\nnumber of dimes * dime value + number of nickels * nickel value + number of quarters * quarter value = Total value\n\n(33 – 2X) * 10 + X * 5 + (X + 4) * 25 = 550\n330 – 20X + 5X + 25X + 100 = 550\n10X + 430 = 550\n10X = 550 – 430\n10X = 120\nX = 120 / 10\nX = 12\n\nNumber of nickels = X = 12\nNumber of dimes = 33 – 2X = 33 – 2 * 12 = 9\nNumber of quarters = X + 4 = 16\n\n12 * 5c + 9 * 10c + 16 * 25c = 60c + 90c + 400c = 550 c = 5.50 $\n\n12 + 9 + 16 = 37 coins\n\nAnswer: Jo has 9 dimes :-)
956295
We need a certain amount of each kind of candy.\n\nCall the amount (in pounds) of the 50 cent candy X. Call the amount (in pounds) of the 75 cent candy Y.\n\nTo get 80 pounds of the mixture, X + Y = 80.\n\nThe candy mixture has a total value of:\n\n80 pounds * (55 cents per pound) = $44.\n\nThe value of the 50 cent candy used is 0.50X. The value of the 75 cent candy used is 0.75Y.\n\nTo have the values add up, 0.50X + 0.75Y = 44.\n\nHere is the system of equations:\n\nX + Y = 80\n0.50X + 0.75Y = 44\n\nThere are many ways to solve systems. I'll use substitution here:\n\nX + Y = 80\nY = 80 - X\n\n0.50X + 0.75Y = 44\n0.50X + 0.75(80 - X) = 44\n0.50X + 60 - 0.75X = 44\n-0.25X = -16\nX = 64\nY = 80 - X\nY = 16\n\nUse 64 pounds of the 50 cent candy, and 16 pounds of the 75 cent candy.\n\nCheck:\n\nClearly, 64 + 16 = 80 pounds.\n\nThe 64 pounds of 50 cent candy costs 32 dollars. The 16 pounds of 75 cent candy costs 12 dollars.\n\nThe total value of the 80 pounds of candy is 44 dollars.\n\n44 dollars/80 pounds = 55 cents per pound\n\nIt works!
861025
Take your own hint.  If one just gives you the answer, you don't learn anything.\n\nStart with $3.\nConsider her and the 4 friends (5 total).\nLast consider the teacher.\n\nEasy as pie.
1025123
100/2=50\n 50    50\n-1      -1\n49      49\n49+49= 98\nyou gave them each 1 dollar back so 100-2=98.\nYou aren't missing a dollar.
138815
36 dimes ($3.60), 14 nickels (70¢)\n\nI did it via guess and check. =)
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.
636056
one coin is not a nickel. the other one is. the answer is a quarter and a nickel
121136
need 1 more coin to show heads...\n\n4 possibilities. \n-Both coins have heads (25%)\n-1st coin has a head (25%)\n-2nd coin has a head (25%)\n- both coin has a tail (25%)\n\nadd em up.. 50%
927660
Hi, \n\nThey way to do this problem is to be careful in the way you set up your equations. \n\nD= dimes\nQ= Quarters\nN = Nickels\n\nSince there are twice as many dimes as quarters, you'd have to multiply the number of quarters by 2 to get it to equal the number of dimes. \n\n2Q = D \n\nSimilarly, if there are two more nickels than quarters, you'd have to add 2 to the number of quarters to get the number of Nickels\n\nQ + 2 = N\n\nNow, The total value (the number of each coin times the value of the coin.) is $7.60\n\n(0.25)Q + (0.1)D + (0.05)N = $7.60\n\nNow you have three equations with three unknowns. Now all you have to do is isolate variables, and substitute. \n\nHope that helps, \nMatt
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...
1004932
whats a nickel? if one isn't a nickel that means the other one could be and the first coin is the remaining amount, is it a dime? i dunno, am rubbish at cents dimes dollars nickels and quarters, being English and all.
404031
End result they paid 24 coins each 3 x 24 = 72\nDisbursed as follows: Cashier 70 coins + waiter 2 coins = 72 coins all accounted for.