SOLUTION: A jar has 190 coins consisiting of nickels, dimes and quarter the value of the coins totals $20. How many nickles, dimes and quarters are in the jar.
N + D + Q = 190
5N + 10D
Algebra.Com
Question 669304: A jar has 190 coins consisiting of nickels, dimes and quarter the value of the coins totals $20. How many nickles, dimes and quarters are in the jar.
N + D + Q = 190
5N + 10D + 25Q = 2000
-10N - 10D - 10Q = -1900
-5N + 15Q = 100
N - 3Q = -20
N = 3Q - 20
other answer is D = 210 - 4Q
Q can equal 52 and 7 therefore 7 <= Q <= 52
making the possibilities
52 Q = 13.00
2 D = .20
136 N = 6.80
20
similar for 52
My question is this, How do I do the same thing for
N + D + Q + H = 100 coins nickles, dime, quarters and half dollars
5N + 10D + 25Q + 50H = 1675
How do I solve for the possibilities
so far i have
D = 235 - 4Q - 9H
N = 3Q + 8H - 135
Answer by lynnlo(4176) (Show Source): You can put this solution on YOUR website!
20 nickels==================$1.00
10 dimes====================$1.00
4 quarters==================$1.00
you got =190 coins
total value==$20.00
dimes===========nickels====================quarters
50=$5.00 100=$5.00 40=$10.00
$5.00+$5.00+$10.00==========$20.00
50coins + 100coins +40coins=========190
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