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 - .05x
2.65 - 1.65 - .25x + .25x = 1.65 - 1.65 - .05x + .25x
1 = .20x
1/.20 = .20x/.20
x = 5 the number of quarters 33 - x
= 33 - 5
= 28 the number of nickels. Thus, we have 5 quarters and 28 nickels.
7 quarters and 11 nickles
She has 11 nickels.
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
The 8 coins are: 3 quarters, 2 dimes, 1 nickel and 2 pennies.
The idea is to write two equations, one for the number of coins, one for the amount of money. Then solve the equations.Assuming "n" is the number of nickels, and "q" the number of quarters, the equations for the coins, of course, is quite simply: n + q = 64 And the equation for the money (I'll use cents; you can just as well use dollars if you prefer): 5n + 25q = 740 You can solve the first equation for "n", then replace that in the second equation.
4 nickels and 5 quarters
47 Quarters 83 Nickels
7 nickels, 4 dimes, and 3 quarters.
7 quarters and 11 nickles
7 quarters = 1.7511 nickels = 0.551.75 + 0.55 = 2.30
10 quarters and 5 nickels
7 nickels, 3 quarters
3 quarters & 2 nickels
3 quarters and 10 nickels.
8 of them.
She has 11 nickels.
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?