Coin Counting
Jones gives Smith as many coins as Smith has. Smith then gives Jones as many coins as Jones has remaining. Now Jones has 36 coins, and Smith has 42 coins. How many coins did each of them have to begin with?
These word problems are easiest to solve by setting up math equations. First, we need to define a couple of variables: let j = the number of coins Jones started with, and let s = the number of coins Smith started with.
Now, we can begin to build equations. Consider how many coins Jones has at the end: 36 coins. Two things happen to get him from j coins to 36 coins. (1) Jones gives Smith as many coins as Smith has. How many coins is that? We've defined Smith's starting amount as s coins. After this transaction, Jones has (j - s) coins remaining. (2) Smith gives Jones as many coins as Jones has remaining (after giving Smith s coins). We've determined Jones has (j - s) coins left. Our final Jones equation looks like this:
Next, we need to write the Smith Equation. Smith winds up with 42 coins. Two things happen to get him from s coins to 42 coins. (1) Jones gives Smith as many coins as Smith has. Smith has s coins. Jones gives him s coins. Therefore, Smith now has (s + s) coins. (2) Smith gives Jones as many coins as Jones has remaining. We already determined that Jones has (j - s) coins remaining after giving Smith s coins. Therefore, Smith gives Jones (j - s) coins. Now Smith has (s + s) - (j - s) coins. Our final Smith equation looks like this:
Finally, we need to solve this system of equations. I like the elimination method.
A Corny Question
A merchant has 21 sacks of grain: seven full, seven half-full, and seven empty. He wants to divide them equally among his three sons. How can he do this--without transferring any portion of grain from sack to sack--so that each son will have not only an equal quantity of grain, but also an equal number of sacks?
This one stumped me as I tried to consider the full, half-full, and empty sacks of grain. I finally made progress by defining a weight for a full sack of grain. I chose an arbitrary 10 pounds. Next, I tried using equations to solve this puzzle, but didn't get very far--I struggled with defining variables, having too many variables, and then going 'round in circles with equations that weren't working. Finally, I used trial and error, and I was able to come up with the correct answers.
Using my arbitrary 10 lbs for a full sack of grain, we see that the merchant has a total of 10(7) + 5(7) + 0(7) = 105 lbs of grain among the 21 sacks. Therefore, each son needs to receive 105 / 3 = 35 lbs of grain and 21 / 3 = 7 sacks. I set up a table to begin giving the sons sacks of grain.
First, I started by giving the first son (boxed in green) the maximum number of full sacks he could get. Dividing 35 by 10 gives 3 and 1/2. So Son1 can get 3 full sacks, which is 30 lbs of grain. In order to give him 35 lbs of grain, he needs only 1 half-full sack. Then, in order to give him 7 sacks, he needs 7 - 3 - 1 = 3 empty sacks. His totals are perfect: 35 lbs of grain and 7 sacks. There are enough sacks of grain for Son2 to get the same as Son1: 3 full, 1 half-full, and 3 empty.
For Son3 (boxed in orange above), there is only 1 full sack of grain left (7 - 3 - 3 = 1). That gives him 10 lbs of grain. In order to make 35 lbs of grain, he needs 25 more pounds (35 - 10 = 25) or 5 half-full sacks (25/5 = 5). That is perfect because there are exactly 5 half-full sacks left (7 - 1 - 1 = 5). Now he has 1 full sack and 5 half-full sacks. In order to have 7 sacks total, he needs 1 empty sack. It's perfect again! There is exactly 1 empty sack left (7 - 3 - 3 = 1). We can check the first row by totaling the number of each type of sack and the total weight.
Therefore, solution 1 is as follows: Son1 and Son2 each get 3 full, 1 half-full, and 3 empty sacks; Son3 gets 1 full, 5 half-full, and 1 empty sacks.
We need to check for other possible solutions. Let's try giving Son1 (boxed in purple) one less than the maximum number of full sacks, that is only 2 full sacks. Then, in order to have 35 lbs of grain, he needs 15 more lbs or 3 half-full sacks (15/5 = 3). In order to have 7 sacks total, he needs 2 empty sacks (7 - 2 - 3 = 2). Let's give this same set-up to Son2.
For Son3 (boxed in blue), there are 3 full sacks left (7 - 2 - 2 = 3), giving him 30 lbs of grain. There is 1 half-full sack left (7 - 3 - 3 = 1), which gives him 35 lbs--perfect! Finally, there are 3 empty sacks remaining (7 - 2 - 2 = 3), which is the exact number he needs in order to have his 7 sacks total (7 - 3 - 1 = 3). Once again, we can check our solution by totaling the number of each type of sack and the total weight.
Solution 2, then, is as follows: Son1 and Son2 each get 2 full, 3 half-full, and 2 empty sacks; Son3 gets 3 full, 1 half-full, and 3 empty sacks of grain.
We can try for more solutions. In the third row (brown box), you see that I gave Son1 only 1 full sack this time. Then, he needed 5 half-full sacks to make 35 lbs and 1 empty sack to make 7 sacks. I can't give the same to Son2 because there are not enough half-full sacks (5 + 5 = 10 > 7). I can't give Son2 only 2 full sacks because then he would need 3 half-full sacks, and there are not enough for that either (5 + 3 = 8 > 7). If I give Son2 3 full sacks and continue on with the process, I find the solution is the same as Solution 1, just re-arranged a bit.
Lastly, I can try giving Son1 no full sacks (not shown in image). Then, I have to give him 7 half-full sacks in order for him to get 35 lbs of grain (35/5 = 7). Then he receives no empty sacks because he has his total of 7 sacks. After that, there are no more half-full sacks to give. There is no way to give any son 35 lbs of grain in only full and empty sacks because 10 does not evenly divide into 35. Therefore, there are no other solutions.
-------------------------------------
I hope you found these puzzles entertaining, as I did. Did you solve both correctly? Let me know in the comments! I would also love to read any alternate methods of solving them that you might have come up with--just post in the comments below!
Without equations, I tend to solve problems like coin counting by working backwards
ReplyDeleteJones ended with 36 and Smith with 42
The previous action doubled Jones coins (meaning he had 18 and got 18 more from Smith), which meant that Smith had 42 + 18 = 60 before that.
Now this 60 that Smith had was double his original number of 30 (after he go 30 from Jones), which means that Jones started with 18 + 30 = 48
To confirm:
Smith Jones
Start 30 48
Jones gives 30 to Smith
60 18
Smith gives 18 to Jones
42 36
That is an interesting way of approaching these problems. However, this can only be done if the final values are given. I can see how it is much faster, though!
ReplyDeleteMy solution is essentially the same as Victoria's, but I broke it up into little pieces.
ReplyDeleteLet j and s be the number of coins that Jones and Smith start out with, respectively. Then the story goes like this:
......................................... Jones.......................Smith........
Starting out.......................... j...............................s
Jones gives Smith
what Smith has................. j - s..........................2s
Smith gives Jones
what Jones has..............2(j - s)..................2s - (j - s)
or......................................2j - 2s.....................3s - j
which is..............................36..........................42
Now we have a system of two simultaneous
equations, namely
...................... 2j - 2s = 36
........................ 3s - j = 42
which gives us
..................... j = 48......and......s = 30
So Jones started out with 48 coins, and Smith started out with 30.
--Richard
Richard, That is a very good way to approach this problem. I like how you've laid things out into columns, with a row for each operation/step in the word problem. It provides a helpful visual. :-)
Delete