The clock gains 12 minutes every hour, so in 60 real minutes it advances 72 minutes.→ Clock runs at a factor of 72/60=1.2 =1.2 times real time.
Let t be the number of real hours elapsed since midnight when the clock stopped.
The clock’s displayed hours at that moment = 1.2t hours (since it runs 1.2× real time).
The stopped clock shows 10:00 PM, i.e. 22 hours since midnight. So
1.2t=22
Solve for t:
t=22/1.2
18 hours 20 minutes → 6:20 PM (the real time when it stopped).
The clock stopped 4 hours ago, so current real time = stop time + 4 hours
6:20 PM + 4 hours = 10:20 PM.
Answer: the correct time now is 10:20 PM.
Given
Four cogs in constant mesh:
Largest cog: 168 teeth
Others: 49, 32, 15 teeth
When two gears mesh, their rotations are inversely proportional to their number of teeth.
If the largest cog makes N revolutions, then a cog with T teeth makes
revs= − (168/T)*N
(negative sign = opposite direction, but we only need them to make whole-number revolutions).
Requirement
Each cog must be back to its starting position, meaning each must rotate an integer number of whole revolutions.
Thus, for each small cog:
(168/T)*N must be an integer
So, 168N must be divisible by T.
1. Cog with 49 teeth
gcd(168,49)=7
(49/7)=7
→ N must be a multiple of 7
2. Cog with 32 teeth
gcd(168,32)=8
(32/8) = 4
→ N must be a multiple of 4
3. Cog with 15 teeth
gcd(168,15)=3
(15/3) = 5
→ N must be a multiple of 5
Final step: LCM
We need the smallest NNN that is a multiple of 7, 4, and 5.
LCM (7,4,5) = 140
The largest cog must make 140 revolutions for all cogs to return to their starting positions.
Answer: 888 + 88 + 8 + 8 + 8 = 1000
Fill the 3-gal jug. (5:0, 3:3)
Pour the 3 into the 5. (5:3, 3:0)
Fill the 3-gal jug again. (5:3, 3:3)
Pour from the 3 into the 5 until the 5 is full. This transfers 2, leaving 1 in the 3. (5:5, 3:1)
Empty the 5-gal jug. (5:0, 3:1)
Pour the 1 gallon from the 3 into the 5. (5:1, 3:0)
Fill the 3-gal jug. (5:1, 3:3)
Pour the 3 into the 5. Now 5 contains 4 and 3 is empty. (5:4, 3:0)
Fill the 3-gal jug one last time. (5:4, 3:3)
