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.
4 nickels and 5 quarters
47 Quarters 83 Nickels
7 quarters and 11 nickles
7 nickels, 4 dimes, and 3 quarters.
10 quarters and 5 nickels
7 quarters = 1.7511 nickels = 0.551.75 + 0.55 = 2.30
7 nickels, 3 quarters
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?
3 quarters and 10 nickels.
10 quarters and 50 pennies
In the US, we use pennies, nickels, dimes and quarters.
4 quarters and 3 nickels
Ten (10) nickels and Three (3) quarters.
8 quarters and 12 nickels
3 quarters & 2 nickels
The coins in the store's cash register total $12.50. The cash register contains only nickels, dimes, and quarters. There are twice as many dimes as nickels. There are also twice as many quarters as dimes. How many quarters are in the cash register?
you have 3 quarters 31 dimes and 65 pennies
The question suggests that there are 24 coins. 13 of them are pennies, 14 are nickels, and 16 are dimes and the rest are quarters. To answer this question, One would add the number of pennies, nickels, and dimes and subtract the sum of those coins from 24. The difference of the two numbers would be the amount of quarters. However, 13+14+16=43. 24-43= -19 There can't be -19 quarters.
4 quarters. 100 pennies. 10 dimes. 20 nickels
10 nickels and 2 quarters
Two quarters, a dime, two nickels, and a penny