Posted by Sarah Mae B. Oquindo on Friday, February 12, 2010 at 10:15am.
I have already solved two problems like this for you. Set up the algebraic equations that describe what you know, and solve them. There will always be as many independent equations as there are unknowns.
number of $20's ---- x
number of $5's ----- y
number of $1's ----- z
but z = x+y - 1
x+y+z = 39
x + y + x+y-1 = 39
2x + 2y = 40
x + y = 20 (#1)
also
20x + 5y + z = 194
20x + 5y + x+y-1 = 194
21x + 6y = 195
7x + 2y = 65 (#2)
double #1 and subtract fron #2
5x = 25
x = 5
then in x+y = 20 , ----> y = 15
and finally
z = x+y-1 = 19
check: is 5+15+19 = 39 ? YES
is 20(5) + 5(15) + 19 = 195 ? YEA!!
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