These puzzles are the second batch I'm taking from Math-E-Magic by Raymond Blum, Adam Hart-Davis, Bob Longe and Derrick Niedermann. The first was a geometric question; these are based on algebra.
These puzzles are entitled Cookie Jar and Fleabags, but they are very similar to a wide range of puzzles (typically related to the relationships between people's ages).
Cookie Jar
Joe and Ken each held a cookie jar and had a look inside them to see how many cookies were left.
Joe said, "If you gave me one of yours, we'd both have the same number of cookies."
Ken replied, "Yes, but you've eaten all of yours - you have none left!"
How many cookies does Ken have?
This is a relatively straightforward puzzle, helped by the fact that Joe has zero cookies, and there's only one other constraint - if Ken gives Joe a cookie, they'll have the same number (one). So, if Joe will have one cookie after the transaction, then so will Ken.
But that isn't the answer. We have to remember that Ken has one cookie after the transaction, but that he also had the one he would give to Joe - so he has two.
Fleabags
Two shaggy old dogs were walking down the street.
Captain sits down and says to Champ, "If one of your fleas jumped onto me, we'd have the same number."
Champ replies, "But if one of yours jumped onto me, I'd have five times as many as you!"
How many fleas are there on Champ?
This one is going to take a little more work - and we can use algebra to help solve it.
Let's have the number of fleas on Captain as A, and the number of fleas on Champ as H (taking the second letter of the two dogs' names).
If one flea jumps onto Captain, he will have A+1. And if that flea has come from Champ, then he will have H-1. And these numbers are the same, so A+1 = H-1 (1)
Now, if one flea jumps from Captain, he will have A-1. And this number is five times greater than Champ's new total H+1. So 5(A-1) = H+1 (2)
If A+1 = H-1 then A+2 = H (from 1)
And we can use this new value of H in (2), to give us 5(A-1) = (A+2) + 1
Expanding and simplifying:
5A - 5 = A + 3
4A = 8
A = 2
Captain has two fleas.
And since A+2 = H, Champ has four fleas.
These puzzles are entitled Cookie Jar and Fleabags, but they are very similar to a wide range of puzzles (typically related to the relationships between people's ages).
Cookie Jar
Joe and Ken each held a cookie jar and had a look inside them to see how many cookies were left.
Joe said, "If you gave me one of yours, we'd both have the same number of cookies."
Ken replied, "Yes, but you've eaten all of yours - you have none left!"
How many cookies does Ken have?
This is a relatively straightforward puzzle, helped by the fact that Joe has zero cookies, and there's only one other constraint - if Ken gives Joe a cookie, they'll have the same number (one). So, if Joe will have one cookie after the transaction, then so will Ken.
But that isn't the answer. We have to remember that Ken has one cookie after the transaction, but that he also had the one he would give to Joe - so he has two.
Fleabags
Two shaggy old dogs were walking down the street.
Captain sits down and says to Champ, "If one of your fleas jumped onto me, we'd have the same number."
Champ replies, "But if one of yours jumped onto me, I'd have five times as many as you!"
How many fleas are there on Champ?
This one is going to take a little more work - and we can use algebra to help solve it.
Let's have the number of fleas on Captain as A, and the number of fleas on Champ as H (taking the second letter of the two dogs' names).
If one flea jumps onto Captain, he will have A+1. And if that flea has come from Champ, then he will have H-1. And these numbers are the same, so A+1 = H-1 (1)
Now, if one flea jumps from Captain, he will have A-1. And this number is five times greater than Champ's new total H+1. So 5(A-1) = H+1 (2)
If A+1 = H-1 then A+2 = H (from 1)
And we can use this new value of H in (2), to give us 5(A-1) = (A+2) + 1
Expanding and simplifying:
5A - 5 = A + 3
4A = 8
A = 2
Captain has two fleas.
And since A+2 = H, Champ has four fleas.