With 230 ants (1380 legs), 3 puppies (12 legs), and 1 teacher (2 legs).
If we call the number of puppies P, then there were (P + 20) children and 10 x (P + 20) ants, and a single teacher. If we now count the legs, we get:
Puppies:
P x 4 = 4P
Children:
(P + 20) x 2 = 2P + 40
Ants:
10 x (P + 20) x 6 = 60P + 1200
Teacher: 2
A total of 66P + 1242
Giving:
66P + 1242 = 1440
66P = 198
P = 3
So there were 3 puppies, hence 23 children.
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Puzzle 6
Starting with any C, using the move of a chess knight, can you find the letters of Christmas in the correct order?
Hint
How many minutes are there between 2:00 and 7:00?
Answer
65 minutes.
Reasoning
We need to break down the five hours in the following way:
3 x Answer minutes past two o'clock40 minutesAnswer minutes before seven o'clock
Why? It is now Answer minutes before seven o'clock, and then we have the 40 minutes before that, and the remaining time was 3 x Answer minutes past two o'clock.
So, five hours = 5 x 60 = 300 minutes. We already know about the 40 minutes, so we have to share the remaining 260 minutes in the ratio 3:1, which gives 195:65 minutes. Or using algebra 4 x Answer = 300 – 40, so Answer = 65 minutes.
Therefore, the five hours is made up of:
195 minutes past two o'clock40 minutes65 minutes before seven o'clock
So, it's currently 65 minutes before 7:00 = 5:55.
Double-Checking
It is now 65 minutes before seven o'clock = 5:55, so 40 minutes ago it was 5:15 – which is 195 minutes after 2:00, as required (as 65 x 3 = 195).
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Puzzle 8
In Farmer Stone's hay loft, there were several animals, in particular birds, mice, and cockroaches.
One day, feeling bored, I decided to count the animals. I found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice.
Hint
Cockroaches have 6 feet, mice have 4, and birds have 2. They each have one head.
Answer
35 birds, 5 mice, and 10 cockroaches.
Reasoning
Cockroaches have 6 feet, mice have 4, and birds have 2. They each have one head.
We can write down expressions for the heads and the feet – calling birds B, mice M, and cockroaches C.
Counting the heads: (1) B + M + C = 50
Counting the feet: (2) 2B + 4M + 6C = 150
We are told that for every mouse there are two cockroaches so, C = 2M. Update (1) and (2) to give: (3) B + 3M = 50 (4) 2B + 16M = 150
If we double (3) we get: (5) 2B + 6M = 100
We can now do (4) – (5) to give: 10M = 50 M = 5
So, we have 5 mice (and 10 cockroaches).
We can use M = 5 in (3) to give: B + 3 x 5 = 50 B + 15 = 50 B = 35
So, C = 10, M = 5 and B = 35.
Double-Checking
10 cockroaches have 10 heads and 60 feet.
5 mice have 5 heads and 20 feet.
35 birds have 35 heads and 70 feet.
The total number of heads = 10 + 5 + 35 = 50.
The total number of feet = 60 + 20 + 70 = 150.
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