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Question 1187485: father borrowed money from mother. mother lent him 530 consisting of 20-peso, 50-peso, 100-peso bills. mother gave him 10 bills. how many pieces of each bill did mother give father?
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! father borrowed money from mother. mother lent him 530 consisting of 20-peso, 50-peso, 100-peso bills. mother gave him 10 bills. how many pieces of each bill did mother give father?
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4 20's
3 50's
3 100's
Answer by greenestamps(13206)  (Show Source):
You can put this solution on YOUR website!
Finding a solution by pure trial and error is a good way to solve the problem; but it doesn't teach you any mathematical methods.
There are at least a couple of ways to solve this using logical reasoning and simple mental arithmetic. Let's look at those methods first.
One informal method is using a "greedy" algorithm, in which you start by trying the largest possible number of the largest bills.
There is a total of 530 pesos, so the maximum number of 100-peso bills is 5. But we can't make the remaining30 pesos using five 20- or 50-peso bills.
If there are 4 100-peso bills, that leaves 130 pesos to be made using six 20- and 50-peso bills. But trial and error (or formal calculations) show we can't do that.
If there are 3 100-peso bills, that leaves 230 pesos to be made using seven 20- and 50-peso bills. Trial and error shows we CAN do that -- using 4 20-peso bills and 3 50-peso bills.
ANSWER: 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills = 300+150+80 = 530 pesos
A more sophisticated but still informal solution uses logical reasoning to quickly cut down the number of possible combinations.
Any combination of 50- and 100-peso bills will make a total that is a multiple of 50 pesos; the total is 530 pesos. Subtracting 20-peso bills to reduce the remaining total to a multiple of 50 pesos, we find there must be 4 20-peso bills to make 80 pesos, leaving 450 pesos to be made with the 100- and 50-peso bills.
Then trial and error shows the way to make the remaining 450 pesos using 6 100- and 50-peso bills is with 3 of each.
And again the answer is 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills.
And here is a formal algebraic solution....
x = # of 100-peso bills
y = # of 50-peso bills
z = # of 20-peso bills
[1] x+y+z = 10 (the total number of bills is 10)
[2] 100x+50y+20z = 530 (the total value is 530 pesos)
Simplify [2]: 10x+5y+2z=53
Eliminate z:
2x+2y+2z=20
10x+5y+2z=53
8x+3y=33
Solve that equation for one variable in terms of the other, and use the fact that x, y, and z are positive integers less than 10 to deduce the solution.
3y=33-8x
y=11-(8/3)x
y is an integer, and 11 is an integer, so (8/3)x must be an integer. That means x must be either 3 or 6. x=6 would make y negative, so x must be 3.
Then y=11-(8/3)3=11-8=3; and x=3 and y=3 means z=4.
And once again, using formal mathematics, the solution is 3 100-peso bills, 3 50-peso bills, and 4 20-peso bills.
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