SOLUTION: roxanne has P4.65 in 5 cents and 25 cents. The number of 25 cents coins is 1 more than twice the number of nickels (5 cents). How many quarters (25 cents) she has?

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Question 1121179: roxanne has P4.65 in 5 cents and 25 cents. The number of 25 cents coins is 1 more than twice the number of nickels (5 cents). How many quarters (25 cents) she has?
Found 3 solutions by ikleyn, MathTherapy, greenestamps:
Answer by ikleyn(52788)   (Show Source):
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From the condition, you have this system of equation x - 2y = 1 (1) (counting coins) 25x + 5y = 465 (2) (counting cents). where x = the number of quarters and y = the number of nickels. From equation (1), express x = 1 + 2y and then substitute it into equation (2), replacing x. You will get a single equation for one unknown y: 25*(1+2y) + 5y = 465 Simplify and solve for y: 25 + 50y + 5y = 465 55y = 465 - 25 = 440 y = = 8. Then x = 1 + 2y = 1 + 2*8 = 17. Answer. 8 nickels and 17 quarters. Check. 8*5 + 17*25 = 465 cents. ! Correct !

Solved.     //   The method I used to solve the system is called the "Substitution method".

So, on the way, you learned how to solve the system of equations using the Substitution method.

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For coin problems and their detailed solutions see the lessons in this site:
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - Three methods for solving standard (typical) coin word problems
    - More complicated coin problems
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem

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and for all methods used - from one equation to two equations and even without equations.

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The referred lessons are the part of this online textbook under the topic "Coin problems".

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Answer by MathTherapy(10552)   (Show Source):
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roxanne has P4.65 in 5 cents and 25 cents. The number of 25 cents coins is 1 more than twice the number of nickels (5 cents). How many quarters (25 cents) she has?

Let number of nickels and quarters be N and Q, respectively
We then get: .05N + .25Q = 4.65
N + 5Q = 93 ------- Multiplying by 20
N = 93 - 5Q ------- eq (i)
Also, Q = 2N + 1 ------ eq (ii)
Q = 2(93 - 5Q) + 1 ---- Substituting 93 - 5Q for N in eq (ii)
Q = 186 - 10Q + 1
Q + 10Q = 187
Q, or number of quarters =

Answer by greenestamps(13200)   (Show Source):
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If a solution using formal algebra is not required, it can be good mental exercise solving this by logical analysis and mental arithmetic.

(1) Take away the one "extra" 25 cent coin; that leaves a total of 4.40, with the number of quarters now exactly twice the number of nickels.

(2) Since the number of quarters is now twice the number of nickels, group the coins in groups each containing two quarters and one nickel. The value of each group will be 0.55.

(3) The number of groups of those coins required to make the total of 4.40 is 4.40/0.55 = 8. So all together in the groups there are 16 quarters and 8 nickels.

(4) Now add back in the other quarter, to get the final answer of 17 quarters and 8 nickels.