7 quarters and 11 nickles
This problem can be solved by solving the system of equation. Total worth of coins: $2.65 Total number of coins: 33 n= number of nickels q= number of quarters since we know that there are 33 coins total, we can set the equation like this: number of nickels + number of quarters = total number of coins => n+q=33 We also know that the worth of these coins is $2.65. each nickel is worth of $0.05 each quarter is worth of $0.25 therefore we can set the equation: 0.05 x number of nickels + 0.25 x number of quarters = total worth of coins. 0.05n+0.25q=2.65 However, for convienience, we should multiply the equation above by 100 to get rid of decimals. Thus it is 5n+25q=265 you will now have a following set of 2 equations: n+q=33 5n+25q=265 Use the SUBSTITUTION METHOD to solve either n or q for solving n: (replace q with n if you're willing to solve q instead) n+q=33 => n=33-q (since n is equal to 33-q, we can -q -q substitue n in the other equation.) 5(33-q)+25q=265 => 165-5q+25q=265 => 20q=100 => q=5 -165 -165 /20 /20 There are 5 quareters as a result.(or 28 nickels) since you know that q=5 you can substitute q in the first equation. n+(5)=33 => n=28 - 5 -5 therefore, there are 5 quarters and 28 nickels. ELIMINATION METHOD: n x -5 + q x -5 = 33 x -5 => -5n-5q=-165 5n+25q=265 + 5n+25q=265 ------------- 20q=100 => q=5 /20 /20 Or simply we can say: if we have x quarters, we have .25x value of them. So the value of nickels will be 2.65 - .25x. Since we have 33 coins, and x quarters, then the number of nickels will be 33 - x. So the value of all nickels would be also .05(33 - x). Thus, we have:2.65 - .25x = .05(33 - x)2.65 - .25x = 1.65 - .05x2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x1 = .20x1/.20 = .20x/.20x = 5 the number of quarters 33 - x= 33 - 5= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.
If Keoki has 14 quarters and 8 dimes (for a total of 22 coins), she has $3.50 and $0.80 or $4.30 in coins. If Keoki has 15 quarters and 7 dimes (for a total of 22 coins), she has $3.75 and $0.70 or $4.45 in coins. If Keoki has 22 coins that are all dimes and quarters and their value in total is $4.35 as asked, there isn't a combination of coins that will permit her to have both 22 coins and $4.35 worth of coins.
George saves nickels and dimes for tolls. If he has 8 coins worth $2.60,how many are nickels and how many are dimes? Answer this question by using system of equation.
A quarter
If there are 14 rows in the triangle, then there are 14*(14-1)/2 coins in the arrangement, i.e. 91 dimes. To find the value of the collection, simply multiply by the value of a dime. I believe this is 10 cents, so the coins are worth 910 cents, or $9.10.
7 quarters = 1.7511 nickels = 0.551.75 + 0.55 = 2.30
Ten (10) nickels and Three (3) quarters.
Two quarters, a dime, two nickels, and a penny
7 quarters 2 dimes 4 nickels
8 quarters and 12 nickels
19 quarters and 4 nickels.
14 Quarters = $3.50 28 nickels = $1.40 To get this answer you simple add 2 nickles to one quarter which = 35 cents divide 4.90 by 35 which equals 14 14 will be the number of quarters and double that, 28 will be the number of nickels
He has jillian.
You would have to know the exact number of coins per denomination. Modern U.S. quarters weigh 5.67 grams. Dimes 2.27 grams. Nickels 5 grams.
The value of the coins is going to be highly dependent upon the ratio of coins. In general: Dimes, quarters and half dollars are roughly $20/lb Nickels are $5/lb And there are 181 pennies to a pound ($1.81) So we're talking any where from $1600 to $150.
One bank box contains $100 worth of nickels, which is 2,000 coins.
Only dollars, half dollars an quarters dated 1776-1976 are "Bicentennial" coins. Dimes, nickels and cents are face value.