Give it a try and see if your brain hurts. 😁
[Spoiler Alert]: Scroll down to find the answers:.
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Problem 1: The Missing Dollar
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Image credit: Canvas
Answer: The three guests each spent $9 at the end. That makes it $9 x 3 = $27. The hotel owner kept $25 and the bellhop kept $2. That makes it $25 + $2 = $27.
The common mistake people tend to make is that they use $27 + $2 = $29, instead of realizing it is really $27 the 3 guests spend and the $3 the guests got back: $27 + $3 = $30. The $2 here really should be -$2, which makes it $27 - $2 = $25, where $25 is the real cost of the hotel stay for the 3 guests.
The official name of the problem is Missing Dollar Riddle.
Problem 2: Cut the Loss
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Image credit: iStockPhoto
Answer:
This problem seems very confusing due to the many moving parts and money (real or counterfeit) exchanging hands. The easiest way to think through this is from the perspective of each person involved and just check the difference at the end.
If we treat the counterfeit bill as worth $0.
Customer: $0 out and $20 worth of merchandize plus $30 cash in. Total gain: $50
Tom: $20 worth of merchandize, $30 cash to customer out. Nothing in ($50 from Charlie was returned to Charlie at the end, so just a wash). Total loss: $50
Charlie: $50 out to Tom and then $50 in from Tom at the end. Total loss: $0
Everything balances out. So the answer is $50 total loss for Tom. There are many variations of this Counterfeit Money Problem out there you can read about.
Problem 3: Green Onion Vendor
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Image credit: The Virginian Pilot
Answer: The original price of the green onions is $1/lb. So you spend $1 you get 1lb. of green onions. Once separating into stems and leaves, the prices become $0.70/lb. for stems and $0.30/lb. for leaves. So spending $1 now you can get 1lb. of stems and 1lb. of leaves. That is actually a total of 2lb. of green onions. That's why the customer was able to spend $50 and get 100lb. of green onions.
Problem 4: Find the Odd Ping Pong Ball
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Image credit: 123RF.com
Answer:
The key challenge in this problem is that we don't know if this odd ball is lighter or heavier.
There are many solutions to this problem. Here's one:
If you divide 12 balls to 3 groups of 4 each. You can compare two groups with the first weighting. Let's name them group 1, 2, and 3. If group 1 and 2 are identical (case 1), then the odd ball has to be in group 3. If group 1 and 2 are not equal (case 2), then group 3 must have all normal balls.
For case 1, take 3 balls out of group 3 and compare to 3 normal balls (just pick them from group 1 and 2).
Case 1.1: if the weights are identical, the 3 balls from group 3 must be normal. The remaining ball from group 3 is the odd ball. We just have to compare it against a normal ball in the third weighting to see if it is lighter or heavier.
Case 1.2: if the 3 balls from group 3 are heavier, now we know the odd ball is heavier. Pick 2 out of these 3 and compare them. If the weights are identical, then the remaining ball is the odd ball. If the weights are not equal, then the heavier one is the odd ball.
Case 1.3: if the 3 balls from group 3 are lighter, now we know the odd ball is lighter. Then perform same steps as above, but looking for lighter ball instead.
For case 2, let's just say group 1 is the heavier side (we can swap group names if not), remove 3 balls from group 1, take 3 balls from group 2 and add to group 1, and take 3 balls from group 3 (normal balls) and add to group 2. Remember the balls originally in group 1 and 2.
Case 2.1: If group 1 is still heavier, that means either that original ball in group 1 is the odd ball and is a heavier ball, or the original ball in group 2 is the odd ball and is a lighter ball. Just compare one of them to a normal ball, we would know which one out of the two original balls is the odd ball and if it is lighter or heavier.
Case 2.2: If group 1 is now lighter after the swap, that means one of the three balls moved from group 2 to to group 1 is the odd ball and it is lighter. Now you just have to find the light odd ball out of the 3 by comparing 2 of them similar to case 1.2.
Case 2.3: If group 1 and 2 have same weights now, that means the one of the three balls we removed from group 1 is the odd ball and it is heavier than a normal ball. Now just find the heavier ball out of those 3 using methods similar to case 1.2.
The official name of this problem is the Balance Puzzle. You can read the Wiki page if you want to lean more.
Problem 5: Silver Utensils Problem
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Image credit: Shutterstock
Answer:
The key of solving this problem is to realize that there are many possible combinations of how much the fork, spoon and knife each cost. But there is only one answer on how much money Bill has once the prices for the fork, spoon and knife are set.
The easiest way to solve this problem is to just pick some easy numbers for the prices of the utensils that meet the requirements, and then figure out how many sets Bill can buy.
For example, let's just say a fork is $1 and a spoon is also $1. That means 21 forks and 21 spoons would cost $42 and that's how much money Bill has. If $42 can buy 28 knives, each knife must cost $1.50, and a set of fork, spoon and knife would cost $3.50. Now divide $42 by $3.50, we get 12. So Bill can buy 12 sets of utensils with all his money.
If we double the prices of each utensils, so fork is $2, spoon is $2 and knife is $3, the only thing changes is how much money Bill has. He now has $84, and he can still buy 12 sets of utensils.
What if the fork is $1, spoon is $2, and knife ends up been $2.25, and with the $63 Bill has, he can still buy 12 sets of utensils. You just have to make sure each utensil ends up in whole cents.
You can of course solve this problem using system of linear equations. You will end up with 12 set and you also don't need to know what price each utensil has.
Hope you enjoyed the reading. Do you feel smarter now?
BTW: The easiest way to remember my blog address is http://lanny.lannyland.com
