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Puzzle 17
During a recent BrainBashers race meeting, the fastest driver managed to complete a lap with an average speed of 150 mph.
They managed to complete the first two fifths of the lap at a speed of 123 mph and the second two fifths of the lap at a speed of 164 mph.
At what speed was the final one fifth of the lap length covered?
Reasoning
If we take the number of miles in a lap to be L miles and use Time = Distance ÷ Speed.
The first 2⁄5 are covered at 123 mph. Therefore, the time taken was [(2⁄5) x L] ÷ 123 hours.
The next 2⁄5 are covered at 164 mph. Therefore, the time taken was [(2⁄5) x L] ÷ 164 hours.
The last 1⁄5 is covered at y mph. Therefore, the time taken was [(1⁄5) x L] ÷ y hours.
The entire trip is covered at an average of 150 mph, and we can conclude that:
2L 2L L L
————— + ————— + ——— = ———
5x123 5x164 5y 150
Rearranging and simplifying gives:
1 1 2 2
——— = —— - ——— - ———
y 30 123 164
That means that y = 1 ÷ (1⁄30 − 2⁄123 − 2⁄164)
Using a calculator gives y = 205.
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Puzzle 18
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
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Puzzle 19
Which of the shapes are identical to the first one?
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.
