CAN YOU SOLVE THIS FOR ME PLEEEEAAAAASEEEEEEEE (even just one will be helpful) :D\n\n\n#1 The nickels and dimes had a value of $5.75. If there were 70 coins in all, how many were nickels?\n\n#2Karamuga had 60 nickels and dimes whose value was $5. How many of each kind of coin did he have.\n\n#3 The fishmonger sold codfish for 7 pence each and mussels for 1 pence each. If Harriet bought a total of 26 items and spent 86 pence, how many codfish did she buy?\n\nTHANKS :D
133082
They're all solved the same way.  Here's the first one.\n\n#1:  n is # of nickels; d is # of dimes\nn + d = 70;  n = 70 - d\n5n + 10d = 575\n5(70 - d) + 10d =575\n350 + 5d = 575\n5d = 225\nd = 45; n = 25\n\n#2:  Set it up and solve it just like the first one\n\n#3:  In the first two, the value of a nickel was 5 and the value of a dime was 10.  In this one, the value of a cod is 7 and the value of a mussel is 1.  Solve it the same way as the other two.\n\nRemember, anybody can make a mistake at this.  Always check your answer.  If you do, you can't be wrong.
364056
20 pennies\n15 pennies and a nickel\n10 pennies, two nickels\n10 pennies, one dime\n5 pennies, three nickels\n5 pennies, one nickel, one dime\n4 nickels\n2 nickels, one dime\n2 dimes
669740
she took the deal at 196,000.  \n\nthen Howie asked her which case would she have opened next, she picked 21, which had 400.00, so the money would have went up to 296,000.  \n\nthen she picked 13, which had 1.00 so the money would have went up to 419,000.  \n\nthen that left the last case and hers.  Hers had the 750,000. and the last one, on stage had 1 penny. \n\nBut, she walked away with a lot more money than she had at the start. \n\nWhat a good game!!  \n\n:-)
458274
1. 285000 X 7.481 = 2132085 gal\n\n2.  I'm almost positive that this one is written incorrectly.\n\n3.  Use FOIL\n5x X 5x = 25x^2\n5x X 3 = 15x\n5x X 3 = 15x\n3 X 3 = 9\n25x^2 + 30x + 9\n\n4. w = 12/5x\nw = 900/y^2
409147
Let n be the number of each coin (she has the same number of each coin).  Let's simplify things and do everything in cents (rather than dollars).\n\nThe value of the quarters would be 25n\nThe value of the dimes would be 10n\nThe value of the nickels would be 5n\nAll together she has 320 cents ($3.20).\n\nPutting this into an equation:\n25n + 10n + 5n = 320\n\nAdd the like terms:\n40n = 320\n\nDivide both sides by 40:\nn = 8\n\nTherefore Pat has 8 of each coin.\n\nHer $3.20 is broken down as follows:\n8 quarters = $2.00\n8 dimes = $0.80\n8 nickels = $0.40
236237
i assume you want a demonstration of the use of numbers and math in daily life... try these\n\n1. suppose brand a of candy sells for 2.37 for an 8 oz bag and brand b sells for 3.05 for a 11 oz bag ask audience which is a better deal ?, why ?, how can we compare the 2 brands to get a definite answer ? Now you introduce the idea of unit price which is $amount/weight do this for each brand. The smaller value for the two brands is the better unit price or the better deal.\n\nfind three different size circles. measure for each one the distance around the circle and the distance across the center of it write these values down for each circle.  Now for each pair of measurements divide the around value by the across value. ask the audience to compare the three quotients and note anything unusual.  what will happen is that no matter the size of the circle distance around/distance across is the same value about 3.14159. now state that pi is the ratio of dist. around / dist. across for a circle and is the same regardless of size of circle
426655
Think of this as an algebra problem.  Let's say N represents the number of quarters.  Therefore, the number of pennies = N+1 and the number of nickels = N+5.  From this you can get the equation\n\nN(.25) + (N+1)(0.01) + (N+5)(0.05) = 3.08\n\nMultiplying it out you get\n\n.25N + .01N + .01 + .05N + .25 = 3.08\n\nCombining the common values gives you\n\n.31N + .29 = 3.08\n\nFrom there you find that N is equal to 9 and the rest is easy.
188323
This question is old. If the bill is $25 and they each paid $10 but the clerk returns $1 to each and the bellboy gets the $2 tip there is no lost dollar. It's all in the way the question is presented. \n\n$25/3 = $8.33 1/3 Or 2 people paid $8.33 and the final guy paid $8.34\n\nIn the question the cents get dropped making it seem as though they each paid an even $8 with an extra $1 back.\n\n$8 + $1 = $9. You can see the cents were dropped.\n\nThe other dollar is in the till where it belongs. The bill was $25 NOT $24\n\nEDIT: This is all a question of laziness. The clerk knew 3 guys split the bill. He should have sent up $1.67 cents each for 2 guys and $1.66 for the last. \n\n$1.67 + $1.67 + $1.66 = $5.00\n\nInstead he sent up 5 $1 bills. The guys were too lazy to work out who got what out of the final $2 after taking back a $1 bill each so they let the bellboy keep the $2. \n\n0.67 + 0.67 + 0.66 = $2.00 \n\nA cheap tip for the bellboy when you look at what each one tipped.\n\nTo sum this up\n\nGuy 1 paid\n\n$8.33 (room) + 0.67 (tip) + $1 (returned) = $10.00 \n\nGuy 2 paid\n\n$8.33 (room) + 0.67 (tip) + $1 (returned) = $10.00 \n\nGuy 3 paid\n\n$8.34 (room) + 0.66 (tip) + $1 (returned) = $10.00
455317
1.  Let x = numer of nickels and y = number of quarters\n(1) x + y = 20\n(2) 5x + 25y =400 (multiplied by 100 to get rid of decimal)\n-5x - 5y = -100 ((1) multiplied by -5)\n20y = 300\ny = 15 quarters\nx = 20 - 15 = 5 nickels\n\n2.  Same method as 1.\nlet x = no. of $12 seats, y = no.of $6 eats\nx + y = 30,000\n12x + 6y = 228,000\n. 6x + 6y = 180,000\n6x = 48,000\nx = 8,000\ny = 30,000 - 8,000\ny = 22,000
552987
1a)  \nIf you already drew a two, then there are 3 left.  And there are 51 cards left, so the probability is 3/51, which reduces to:\n= 1/17\n1b)\nIf you didn't already draw a two, then there are 4 left.  And there are 51 cards left, so the probability is:\n= 4/51\n\n2)\nDiameter = 108 miles (exactly twice the radius)\nCircumference = pi * d = 108 pi ≈339.2 miles\n\n3) \n37°C is hot; it is equivalent to normal body temperature (98.6°F).  You can prove this by plugging 37 in the following formula:\n\nF = (9/5)C + 32\nF = (9/5)37 + 32\nF = 333/5 + 32\nF = 66.6 + 32\nF = 98.6\n\nI suspect you'd be at the beach in a bathing suit, perhaps.\n\n4a)\nb = 24c / 4d  c=5, d = 3\nJust plug in the values:\n24(5) / 4(3) = 120/12 \n= 10\n4b)\nb = 24c / 4d  c=10, d = 15\n24(10) / 4(15) = 240/60 \n= 4\n\n5a)\ntow truck = 8 *tons*\n5b)\npostcard = 1 *oz*\n\n6a)\n18 + 4 - 16 + 3 - 9 \n= 0\n6b)\n55 - 32 + 6 - 7 + 1 \n= 23
54910
36 dimes ($3.60), 14 nickels (70¢)\n\nI did it via guess and check. =)
38050
Maximize the nickels (since they have the smallest value, it takes the most of them to make up the value you want) and use 1 each of nickels and dimes.\n\neg. $1.00 = .25 + .10 + .65\n         1 quarter, 1 dime, 13 nickels\n\n\n$15,348.30= .25 + .10 + 15,347.95\n         1 quarter, 1 dime, 306,959 nickels :)\n\nOr, for a formula, subtract $0.35 (the value of 1 quarter and 1 nickel) from the amount, then divide by $0.05.  This is the number of nickels you need, add 2 to get the total number of coins.
202269
Take it to your bank and desposit it.  They'll hate you, but their legally obligated to take it, I hear.  That's what "all debts, public and private" means!\n\nAccording to the US Mint, a nickel weighs 5 grams even and a dime weighs 2.268 g.\n\n(Yes, purists, I know grams ain't weight, but "masses" ain't a word!)\n\nKeep in mind your numbers have only one significant digit!  (i.e., if you want better accuracy, I need numbers more accurate than "20" and "30".  How about "21.3" or something?)  Playing with the numbers gives us:\n\nNickels:  ~2000, or $100\nDimes:  ~4000, or $400
122166
This problem was fun and took me a few minutes to realise the solution.  First I will restate the problem just for clarification.\n\nWe have 12 coins, one of which is fake (lighter or heavier).  We can only make 3 weight comparisons to discover which of the coins is the fake one.  My solution will tell you which coin is fake and if it is heavy or light in exactly 3 comparisons.\n\nI number the coins 1 - 12 and make the following 3 comparisons on a scale\n\nleft    vs   right\n-------------------\n1 2 3 10  vs  4 5 6 11\n1 2 3 11  vs  7 8 9 10\n1 4 7 10  vs  2 5 8 12\n\nnow I know which coin is false.  How do I do this? well I created a table which has 12 unique outcomes of 3 weighing comparisons on which side of the scale went down. for example left left left indicates that all 3 comparisons had the left side dropped.  Here is my table:\n\nfake coin  | correspondin weight comparisons\n--------------------------------------------\n1          |  left left left (heavy) or right right right (light)\n2          |  left left right (heavy) or right right left (light)\n3          |  left left balanced (heavy) or right right balanced (light)\n4          |  right balanced left (heavy) or left balanced right (light)\n5          |  right balanced right (heavy) or left balanced left (light)\n6          |  right balanced balanced (heavy) or left balanced balanced (light)\n7          |  balanced right left (heavy) or balanced left right (light)\n8          |  balanced right right (heavy) or balanced left left (light)\n9          |  balanced right balanced (heavy) or balanced left balanced (light)\n10         |  left right left (heavy) or right left right (light)\n11         |  right left balanced (heavy) or left right balanced (light)\n12         |  balanced balanced right (heavy) or balanced balanced left (light)\n\nnow I can find exactly which coin was the fake one by comparing the results of my 3 wieghs against my table. For example if I get two balances and then a right, I know the 12 coin is the fake and is heavy. Note that if I get balanced balanced left the the 12 coin is light.\n\nNotice that I first created this table of possible outcomes and then generated from it which coins would be compared against which. \n\nHeavy or Light:  I created this table in which no result on any row is the reverse of any result on any other row (as the reverse indicates whether it is light or heavy)\n\nThis is the simplest solution to this problem.  I hope this made sense.  Was really fun.
108465
First of all, you know that the total money collected was 57444 * 2, which is 114888.\n\nNext, you know that you have only 202 prizes.  Of those, three of them were valued: 2000, 1500 and 1000.  Keep in mind that after those three prizes, there are 199 prizes left, I'll get to that later.\n\nThe first three prizes will take a total of 4500 from the initial pot of 114888.\n\nNow, since each following prize is 5 less than the one before it, it is a summation of the next 199 prizes starting at an initial prize value of 1000 minus 5.  (Remember, each following was 5 less than the one before it.)\n\nSo, the summation is the sum of each value of the equation: (1000 - (5 * x)) where x starts at the value of 1 and increases to 199.  So this is the sum of: 995 + 990 + 985 + ... + 5.  The result of which is 99500.\n\nSo, 99500 + 4500 is the total amount paid out in prizes, being 104000.  Subtracting that from the initial intake of funds on $2 tickets leaves a total pure profit of 10888.\n\nNot a bad profit, it was almost like asking about one out of six of those people to just 'give' you one dollar. :)
526285
Do your own homework . \n\nI'll help you anyway since you probably haven't done your homework in ages.\n\n\n1.3\n2.$234,896\n3.$625\n4.$2.00\n5.$10.00
57031
Basically: There is an initial $30 charge.  It should have been $25, so $5 must be returned and accounted for.  $3 is given to the 3 friends, $2 is kept by the manager - there you have the $5.  \n\nThe trick to this riddle is that the addition and subtraction are done at the wrong times to misdirect your thinking - and quite successfully for most.  Each of the 3 friends did indeed pay $9, not $10, and as far as the friends are concerned, they paid $27 for the night.  But we know that the clerk will tell us that they were charged only $25 and when you add the $3 returned with the $2 kept by the bellhop, you come up with $30.