SOLUTION: A piggy bank contains Php 57 worth of coins in Php 1 and Php 5 denominations. The number of Php 1-coin is one more than twice the number of Php 5-coins. How many coins of each kind

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: A piggy bank contains Php 57 worth of coins in Php 1 and Php 5 denominations. The number of Php 1-coin is one more than twice the number of Php 5-coins. How many coins of each kind      Log On


   

Question 1150141: A piggy bank contains Php 57 worth of coins in Php 1 and Php 5 denominations. The number of Php 1-coin is one more than twice the number of Php 5-coins. How many coins of each kind does the piggy bank contain?
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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Let x be the number of Php 5-coins.

Then the number of Php 1-coins is  (2x+1), according to the condition.


The total money equation is


    5x + 1*(2x+1) = 57.


Simplify and solve for x.


ANSWER.  8 Php 5-coins and (2*8+1) = 17 Php 1-coins.


CHECK.   8*5 + 17*1 = 40 + 17 = 57 Php.    ! Correct !

Solved.