Question 1185540: Lawrence's wallet contains some ₱10.00, ₱5.00, ₱1.00, and ₱0.25 coins. There are 9 coins in his wallet. How many of each type does he have if the wallet has a total of ₱20.50?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13198)   (Show Source):
You can put this solution on YOUR website!
A pure mathematical solution to this problem would be very complex; however, there are many ways to solve this easily using logical reasoning.
You will learn nothing from this, and get no valuable mental exercise, if we simply show you the answer, or even if we show you the complete solution process.
I will instead suggest a few ways you can start on the problem and let you finish. To get the most benefit, try finishing the solution from each of the starts I show.
(1) A "greedy" approach -- starting with the highest value coins....
We can't have two ₱10 coins, because that would make ₱20, and we would have only ₱0.50 to make with the remaining 7 coins. So we can only have one ₱10 coin.
That leaves ₱10.50 to make with the other 8 coins.
A similar analysis then shows we can only have one ₱5 coin.
...then continue from there....
(2) Using the fact that only the ₱0.25 coins have values that are fractions....
With a total of ₱20.50, the number of ₱0.25 coins must be either 2 (to make ₱0.50) or 6 (to make ₱1.50). Quick analysis shows there can't be 6 ₱0.25 coins, because that would leave only 3 coins to make the remaining ₱19. So the number of ₱0.25 coins has to be 2.
... and again continue from there....
(3) Using the assumption that the statement of the problem implies that he has at least one of each coin....
1 of each coin make a total value of ₱16.25, leaving 5 more coins to make the remaining ₱4.25. Obviously one of the remaining coins has to be another ₱0.25 coin....
... and then continue on that path....
Answer by ikleyn(52778)  (Show Source):
You can put this solution on YOUR website! .
Lawrence's wallet contains some ₱10.00, ₱5.00, ₱1.00, and ₱0.25 coins. There are 9 coins in his wallet.
How many of each type does he have if the wallet has a total of ₱20.50?
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Definitely, there are at least two ₱0.25 coins.
Therefore, we can reformulate the problem equivalently by asking
How many of each type does he have if the total money is ₱20.00 and there are 7 coins, in all.
Next, there is at least one ₱10.00 coin, and it is clear that the number of ₱10.00 can not be two: it must be exactly one.
Therefore, we can reformulate the problem equivalently by asking
How many of each type does he have if the total money is ₱10.00 and there are 6 coins of ₱5.00 and/or ₱1.00.
Again, there is at least one ₱5.00 coin, and it is clear that the number of ₱5.00 can not be two: it must be exactly one.
So, finally, the answer is
one ₱10.00 coin; one ₱5.00 coin; five ₱1.00 coins, and two one ₱0.25 coins.
Solved.
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Notice, that from the linear algebra point of view, the problem allows only 2 equations for 4 unknowns;
but the requirement to have a solution in integer positive numbers leaves ONLY ONE possibility for the solution.
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