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##### set up the system and solve by elimination method.

label Algebra
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schedule 0 Hours
account_balance_wallet \$5

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?

Oct 16th, 2017

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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Oct 16th, 2017
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