Puzzle 17
My first is in bridge, but not in ridge.
My second is in awake, and in mistake.
My third is in danger, but not in ranger.
My fourth is in strange, and in orange.
My fifth is in spline, and in nine.
My last is in river, and in diver.
My whole likes the darkness.
What am I?
Puzzle Copyright © Kevin Stone
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Hint
The first letter can only be B.
Answer
A badger.
Reasoning
My first is in bridge, but not in ridge = B
My second is in awake, and in mistake = A K E
My third is in danger, but not in ranger = D
My fourth is in strange, and in orange = R A N G E
My fifth is in spline, and in nine = I N E
My last is in river, and in diver = R I V E
Puzzle 18
Each of these words can be split in half, and two letters can be placed in the middle to create a new six-letter word.
The four added pairs of letters, in order, spell an 8-letter word. What is that word?
lend
liar
pale
coin
For example:
door + ct = do + ct + or = doctor
suit + nl = su + nl + it = sunlit
Puzzle Copyright © Kevin Stone
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Hint
The 8-letter word begins with the letter G.
Answer
Generous.
lend + ge = le + ge + nd = legend
liar + ne = li + ne + ar = linear
pale + ro = pa + ro + le = parole
coin + us = co + us + in = cousin
Puzzle 19
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.
How many of each animal were there?
Puzzle Copyright © Kevin Stone
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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.
Puzzle 20
Alex can eat 27 chocolates in an hour, Billie can eat 2 chocolates in 10 minutes, and Charlie can eat 7 chocolates in 20 minutes.
How long will it take them to share and eat a large box of 120 chocolates while watching a movie?
Puzzle Copyright © Kevin Stone
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Hint
How many chocolates can they eat in one hour?
Answer
2 hours.
Reasoning
In one hour, Alex eats 27 chocolates, Billie eats 12, and Charlie eats 21.
A total of 60 chocolates.
Therefore, 120 chocolates would take 2 hours.
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