Three countrymen met at a cattle market. "Look here," said Hodge to Jakes, "I'll give you six of my pigs for one of your horses, and then you'll have twice as many animals here as I've got."
"If that's your way of doing business," said Durrant to Hodge, "I'll give you fourteen of my sheep for a horse, and then you'll have three times as many animals as I."
"Well, I'll go better than that," said Jakes to Durrant; "I'll give you four cows for a horse, and then you'll have six times as many animals as I've got here."
No doubt this was a very primitive way of bartering animals, but it is an interesting little puzzle to discover just how many animals Jakes, Hodge, and Durrant must have taken to the cattle market.
[Ref: ZUCK] At A Cattle Market. Amusements In Mathematics by Henry Ernest Dudeney (1917).
Hint: A little algebra might help.
Jakes must have taken 7 animals to market, Hodge must have taken 11, and Durrant must have taken 21. There were thus 39 animals altogether.
This can be solved with a little algebra, where J - Jakes, H = Hodge and D = Durrant:
From the clues:
2 x (H - 6 + 1) = J + 6 - 1 
3 x (D - 14 + 1) = H + 14 - 1 
6 x (J -4 + 1) = D + 4 - 1 
These can be rearranged to give:
2H - J = 15 
3D - H = 52 
6J - D = 21 
We can now use 2 x  +  to give:
6D - J = 119 
We can now use  + 6 x  to give:
35J = 245
J = 7
We can then use J = 7 in  and  to give H = 11, and D = 21. QED.
Can you find a five digit number which has no zeros nor ones in it and no digit is repeated, where:
The fourth digit is a quarter of the total of all of the digits.
The second digit is twice the first digit.
The third digit is the largest.
The last digit is the sum of the first two digits.
[Ref: ZPXQ] © Kevin Stone
Hint: The number contains no zeroes.
Remember the question stated that the number contained no zeros nor ones.
Alan's son Daniel, is exactly one fifth of Alan's age.
In 21 years time, Alan will be exactly twice Daniel's age.
Betty's is exactly seven times the age of her daughter, Jessica.
In 8 years time, Betty will be three times the age of Jessica.
How old are Daniel and Jessica now?
[Ref: ZQZZ] © Kevin Stone
Hint: A little bit of algebra might help.
Answer: Daniel is 7 years old and Jessica is 4 years old.
The first half of the question involves Daniel (D) and Alan (A).
Daniel is currently one fifth of Alan's age, so:
A = D x 5 (1)
In 21 years time, Alan will be twice his age, so:
A + 21 = (D + 21) x 2 (2)
Using (1) in (2) gives:
A + 21 = (D + 21) x 2
5D + 21 = (D + 21) x 2
5D + 21 = 2D + 42
3D = 21
D = 7
So Daniel is 7 (and Alan is 35).
The second half of the question involves Jessica (J) and Betty (B).
Betty's is exactly seven times the age of Jessica:
B = J x 7 (3)
In 8 years time, Betty will be three times the age of Jessica, so:
B + 8 = (J + 8) x 3 (4)
Using (3) in (4) gives:
B + 8 = (J + 8) x 3
7J + 8 = (J + 8) x 3
7J + 8 = 3J + 24
4J = 16
J = 4
So Jessica is 4 (and Betty is 28). QED.
Within the BrainBashers school, the science department has three disciplines.
In total, 280 students study chemistry, 254 students study physics and 280 students study biology.
97 students study both chemistry and physics, 138 students study both physics and biology, 152 students study both chemistry and biology.
73 students study all three disciplines.
Can you determine how many students there are in the science department? The answer is well below 814.
[Ref: ZZYU] © Kevin Stone
Hint: How many students study chemistry and physics, but not biology?
The answer is most easily seen if three intersecting circles are drawn, and the numbers inside each section worked out.