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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.

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โˆ™ 2009-07-23 01:24:59
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Q: James has 33 coins in his pocket all of them nickels and quarters If he has a total of 2.65 how many quarters does he have?
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