Q: 150 total coins of dimes and nickels 70 more nickels than dimes?

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Helen has twice as many dimes as nickels and five more quarters than nickels the value of her coins is 4.75 how many dimes does she have?

Eighteen

4 Qs 12 Dimes 20 Nickles

7 nickels, 4 dimes, 3 quarters

there r 40 nikels

7 nickels, 4 dimes, and 3 quarters.

If there are 150 nickles and dimes and 70 more are nickles than dimes how many of each are there?

it would have to be pennies, 60 quarters is more than $7, same with dimes so pennies is the only one left.

13 nickels and 78 dime

Jack would have 11 Nickles && 7 Dimes. Promise you thattt, (:

You do. 6 dimes = 60 cents 9 nickels = 45 cents 6 dimes = 12 nickels

they were made that way

The dimes.The dimes.The dimes.The dimes.

3 dimes is 30 cents and 5 nickels is 25 cents so 3 dimes is worth more.

This question cannot be answered.Assume there is1 nickel. There must be 3 more than that in quarters - that makes4 quarters and there are13 dimes---- that adds up to18

30 pennies, 4 nickels and 5 dimes.

There are 11 combinations. Beginning with the largest coin, they are:1) 3 quarters, 1 dime, and 2 nickels2) 2 quarters, 4 dimes, 1 nickel3) 2 quarters, 3 dimes, 3 nickels4) 2 quarters, 2 dimes, 5 nickels5) 2 quarters, 1 dime, 7 nickels6) 1 quarter, 6 dimes, 2 nickels7) 1 quarter, 5 dimes, 4 nickels8) 1 quarter, 4 dimes, 6 nickels9) 1 quarter, 3 dimes, 8 nickels10) 1 quarter, 2 dimes, 10 nickels11) 1 quarter, 1 dime, 12 nickelsIf you do not need to use all three coins, there are 15 more combinations:12) 3 quarters, 2 dimes13) 3 quarters, 4 nickels14) 2 quarters, 9 nickels15) 1 quarter, 7 dimes16) 1 quarter, 14 nickels17) 9 dimes, 1 nickel18) 8 dimes, 3 nickels19) 7 dimes, 5 nickels20) 6 dimes, 7 nickels21) 5 dimes, 9 nickels22) 4 dimes, 11 nickels23) 3 dimes, 13 nickels24) 2 dimes, 15 nickels25) 1 dime, 17 nickels26) 19 nickels

80

You only need to solve for the number of either nickels or dimes. You can subtract that number from 86 to find the other. Since dimes are worth more, let's solve for dimes; we'll use the variable "d". First, make a chart of what we know:# dimes = "d"; 0.10 each; total cents in dimes = 10d# nickels = 86 - d, 0.05 each; total cents in nickels = 5(86-d)Total number/value of coins: total number is 86; total value $6.90, or 690 centsNow set up the equation:10d + 5(86-d) = 69010d + (5 . 86) + (5 . - d) = 69010d + 430 - 5d = 6905d + 430 = 6905d + 430 - 430 = 690 - 4305d = 260(1/5)5d = 260 (1/5)d = 52If there are 52 dimes, there must be 34 nickels (86 - 52 = 34).Let's check our work:52 dimes = $5.2034 nickels = $1.70$5.20 + 1.70 = $6.90

Nickels = 6 Dimes = 2 Quarters = 9 (1/3N) + N + (3+N) = 17 2 1/3N = 14 N = 6

(dimes X 10) + (nickels X 5) = 145 and dimes - nickels = 4 14 dimes + 1 nickel =145 13 dimes + 3 nickel =145 12 dimes + 5 nickel =145 11 dimes + 7 nickel =145 10 dimes + 9 nickel =145 since 11-7 = 4 11 dimes + 7 nickels =145

1) 10 dimes 2) 3 quarters, 2 dimes, 5 pennies 3) 2 quarters, 2 dimes, 6 nickels 4) 1 quarter, 6 dimes, 3 nickels As far as I can reason, there cannot be more than 4 ways.

65 pennies 0 dimes and 13 nickels 1 dime and 10 nickels 2 dimes and 9 nickels 3 dimes and 7 nickels 4 dimes and 5 nickels 5 dimes and 3 nickels 6 dimes and 1 nickel 2 quarters and 1 dime and 1 nickel That's all I can think of, but there are much more!

one dime = $0.10 plus one nickel = $0.15 plus 6 pennies = $0.21 *Used a total of 8 coins to make a sum of $0.21.*

Most InterestinglyYou have fifty US two-cent coins, a very interesting historical piece of currency.(Check out a great history at the related link)More Prosaically1 quarter, 2 dimes, 2 nickels, and 45 pennies.2 dimes, 8 nickels, and 40 pennies.