Puzzle 1
My current age is the age of my favourite book (which is 14 years old) plus one third of my age.
How old will I be when my favourite book is twice its current age?
Puzzle Copyright © Kevin Stone
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Hint
A little algebra might help.
Answer
I will be 35 in 14 years.
Reasoning
I am currently 21.
If X is my current age, then:
X = 14 + X
———
3
so
2X = 14
———
3
and
2X = 42
So
X = 21
I am currently 21 and my favourite book is currently 14 years old, when it is 28 years old (double its age) I will be 35.
Puzzle 2
Starting in the bottom left corner and moving either up or right, adding up the numbers along the way, what is the largest sum that can be made?
| 2 |
3 |
5 |
4 |
1 |
| 3 |
2 |
4 |
3 |
4 |
| 5 |
1 |
3 |
5 |
2 |
| 3 |
3 |
2 |
3 |
1 |
| 1 |
4 |
2 |
4 |
4 |
Note: this puzzle is not interactive, and the squares cannot be clicked.
Puzzle Copyright © Kevin Stone
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Answer
The largest sum that can be made is 28.
Puzzle 3
Yesterday I went for a short bicycle ride around the local lakes. As the weather was very hot, I rode in shorter stages than I'd normally ride.
In Stage One, I rode half of the overall distance.
Stage Two saw half of the remaining distance plus 35 metres covered.
Stage Three covered three-quarters of the remaining distance.
Stage Four completed half of the remaining distance, plus 75 metres.
Stage Five completed the journey with a final burst of 150 metres.
How far did I cycle in total?
Puzzle Copyright © Kevin Stone
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Hint
Try working backwards.
Answer
7,340 metres.
Reasoning
Stage 5's clue tells us that 150m was left. So the distance left at the start of Stage 4 must have been:
Dist4 = Dist4 + 75 + 150
—————
2
Which simplifies to give Dist4 = 450m.
Stage 3's clue tells us that 450m was one quarter of the remaining distance, so Dist3 = 1,800m.
Stage 2's clue tells us that the distance at the start of Stage 2 was:
Dist2 = Dist2 + 35 + 1800
—————
2
Which simplifies to give Dist2 = 3,670m.
Stage 1's clue tells us that 3,670m was half the overall distance, which means the entire ride was 7,340m.
Double-Checking
Starting with 7,340m:
Stage 1: cycled 3,670m, leaving 3,670m
Stage 2: cycled 1,870m, leaving 1,800m
Stage 3: cycled 1,350m, leaving 450m
Stage 4: cycled 300m, leaving 150m
Stage 5: cycled 150m, leaving 0m
Puzzle 4
At midnight at the start of January 1st, Professor Stone set two old-fashioned clocks to the correct time.
One clock gains one minute every hour, and the other clock loses two minutes every hour.
When will the clocks next show the same time as each other?
When will the clocks both show the correct time?
Puzzle Copyright © Kevin Stone
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Hint
In order for the clocks to show the same time (e.g. 2 o'clock), what must the total time gained by one, and lost by the other, total?
Answers
Midnight, 10 days later, when they will both show 4 o'clock.
Midnight, 30 days later, when they will both show 12 o'clock.
Reasoning #1
In order for the clocks to show the same time, the total time gained by one, and lost by the other, must be 12 hours.
For example, if the first clock were to show 2 o'clock, it would have gained 2 hours. In order for the second clock to also show 2 o'clock, it would have had to have lost 10 hours. This is a total of 12 hours gained and lost.
We know that for every hour that passes, the first clock gains one minute, and the second clock loses 2 minutes, for a total time gained and lost of 1 + 2 = 3 minutes.
The total time gained and lost will equal 12 hours when 12 x 60 ÷ 3 = 240 hours have passed. 240 hours = 10 days.
The first clock will have gained 240 x 1 minutes = 240 minutes = 4 hours.
The second clock will have lost 240 x 2 minutes = 480 minutes = 8 hours.
So, they will both show 4 o'clock, 10 days later.
Reasoning #2
In the first answer, we can see that 10 days later, the clocks both show 4 o'clock.
If we move forward another 10 days, both clocks would show 8 o'clock.
If we move forward another 10 days, both clocks would show 12 o'clock.
This will now be the correct time.
So, they will both show 12 o'clock, 30 days later.
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