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Hint: Try breaking each cog size into prime factors.
Answer: 30 revolutions.
If we break each wheel into its prime factors, we get:
63 = 3 x 3 x 7
42 = 2 x 3 x 7
35 = 5 x 7
27 = 3 x 3 x 3
We now think of rotating the large wheel just once, and this is 3 x 3 x 7 teeth moved (3 x 21), and we can see that 42 tooth wheel also has a 3 x 7 (21 teeth) in it, with an extra 2. If we therefore rotate the 63 toothed wheel twice, the 42 will have rotated three times.
The answer involves cancelling any common factors from the large wheel. We can cancel 3, 3, 7 from any of the smaller ones to leave 2 (from the 42), 5 (from the 35) and 3 (from the 27). 2 x 5 x 3 = 30.
Puzzle 30
The local library decided to hand out some of its old books to its 1400 borrowers.
To each of the female borrowers they gave 6 books and to each male borrower they gave 4 books.
If only half of the females in town and three quarters of the males in town accepted the books, how many books were given away?
Hint: It didn't matter how many male or female borrowers there were.
Answer: 4200 books.
It doesn't matter how many female borrowers there were as 1/2 of the females accepted the books, and those that did accepted 6 books, which is 3 books per female on average.
Similarly it doesn't matter how many male borrowers there were as 3/4 of the males accepted 4 books each, which is again 3 books each.
So both the males and females accepted, on average, 3 books each. And 1400 * 3 = 4200.
Puzzle 31
At dawn on Monday a snail fell into a bucket that was 12 inches deep.
During the day it climbed up 3 inches, however, during the night it fell back 2 inches.
On what day did the snail finally manage to climb out of the bucket?
[Ref: ZQGM]
Hint: The answer isn't 12 days.
Answer: The following Wednesday, nine days later.
At dawn on Monday the snail was at the bottom. Each day it climbed 3 inches, and dropped 2 inches over night, so at the start of the next day it was 1 inch higher, so:
Hint: How many chickens did the last customer buy?
Answer: 15 chickens.
The first customer bought 15 ÷ 2 + ½ = 8 (leaving 7)
The second customer bought 7 ÷ 2 + ½ = 4 (leaving 3)
The first customer bought 3 ÷ 2 + ½ = 2 (leaving 1)
The first customer bought 1 ÷ 2 + ½ = 1
The answer can be worked out in the following way:
The fourth customer can only have bought one chicken, otherwise there would be some chickens left. The third customer bought all but 1, so total  (total ÷ 2 + ½) = 1. This can be solved algebraically or it's easy to see that the total must have been 3. This method keeps working until we reach the first customer.
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