Time remaining:
set up the system and solve by elimination method.

Algebra
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A collection of dimes and quarters has a value of $3.55. if there are a total of 25 coins, how many dimes and quarters are there?

Apr 26th, 2015

Let the 
number of quarters = x
number of dimes = y 
Since total has 25 coins, so
x+y = 25 ...........Equation (1) 
1 quarter = 25 cents...so x quarters = 25*x cents
1 dime = 10 cents.....so y dimes = 10*y cents
$3.55 = 3.55 * 100 = 355 cents 
Since the total value is $3.55 or 355 cents
Therefore, 
25x+10y = 355.........Equation (2) 
Solving equation (2)
5*5x + 5*2y = 5*71
=> 5(5x+2y) = 5(71)
=> 5x+2y=71.........Equation (3) 
From Equation(1)
y = 25-x 
Substituting the value of y in Equation(3)
5x + 2y = 71
=> 5x + 2(25-x) = 71
=> 5x + 50 - 2x = 71
=> 3x = 21
=> x = 7
Therefore, No. of Quarters = 24 
As, y = 25 -x
=> y = 25-7=18
Number of dimes = 18, quarters=24 
Hence solved


Apr 26th, 2015

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