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Puzzle 17
At the local mid-summers fayre, four ladies are planning to delight the visitors with their speciality cakes.
Mrs Archer will make a chocolate cake.
Neither Mrs Bowness, Louise nor Kay Cook propose to make a vanilla cake.
Julie, who is not Mrs Dancer, will make coffee.
Mary will not make a plain one.
You need to find their full names and who plans to make which cake.
Answer
Louise Archer will make the chocolate cake.
Julie Bowness will make the coffee cake.
Kay Cook will make the plain cake.
Mary Dancer will make the vanilla cake.
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Puzzle 18
At last week's Girl Guide's picnic, four girls were tucking into their packed lunches.
Jane had muffins, but Denise didn't have orange juice.
The girl who had blackcurrant juice also had scones and Frankie had blueberry juice.
The girl who had cake also had strawberry juice, whereas Alice didn't have the strawberry juice nor toast.
Can you work out who had which drink and which dessert?
Answer
Alice had blackcurrant juice and scones.
Jane had orange juice and muffins.
Denise had strawberry juice and cake.
Frankie had blueberry juice and toast.
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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.
Hint
Cockroaches have 6 feet, mice have 4, and birds have 2. They all 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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Puzzle 20
For being well-behaved at the garden fayre, four children were each given two sweets.
Jesse had an orange sweet. One child who had a red sweet also had a blue one. No child had two sweets of the same colour. A child who had a green sweet also had a red one. Jamie didn't have a red sweet, and Jo had a green one. Jordan didn't have an orange one, and Jesse had no blue sweets.
Knowing that there were two sweets of each colour, can you tell the colours of the sweets each child had?
Answer
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Reasoning
The children were: Jamie, Jesse, Jo, Jordan.
The colours were: Blue, Green, Orange, Red.
There were two sweets of each colour, and by (3) no child had two sweets of the same colour.
By (1), Jesse had an Orange sweet …
… by (6), didn't have Blue …
… by (2) and (4), Red was paired with Blue and Green …
… therefore, Jesse's other sweet can't have been Red, so was Green.
Jamie
Jesse Orange Green
Jo
Jordan
By (4) the other Green sweet was paired with Red, by (5) this must have been Jo.
Jamie
Jesse Orange Green
Jo Green Red
Jordan
By (2) the other Red sweet was paired with Blue, but by (5) this wasn't Jamie, so must have been Jordan.
Jamie
Jesse Orange Green
Jo Green Red
Jordan Red Blue
Leaving Jamie with Orange and Blue.
Jamie Orange Blue
Jesse Orange Green
Jo Green Red
Jordan Red Blue
It is now four years since I first asked for support, and I would like to sincerely thank everyone who has helped me. BrainBashers doesn't show adverts, and I continue to depend on the ongoing support from everyone who enjoys the site.
