If a shopkeeper can only place the weights on one side of the common balance. For example, if he has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names of the weights you will need to measure all weights from 1 to 1000? This is a fairly simple problem and very easy to prove also.
There is a shop where written:
Buy 1 for $1
10 for $2
100 for $3
I needed 999 and still only paid $3. How could this be financially viable for the shop-keeper?
I want to fill my bucket using both cold and hot water.
I have two taps for both cold and hot water. The hot water tap fills the bucket in exact 6 hours and the cold water tap fills the bucket in exact 4 hours.
I turn both of them simultaneously but I forgot to turn off another tap which removes the water out of the bucket. This tap can empty the bucket in 12 hours.
Dwayne Johnson was running away with the loot from a heist in his car along with Vin Diesel. One tire was punctured and he dropped down to replace it. While changing the wheel, he dropped the four nuts that were holding the wheel and they fell into a drain. Vin Diesel gave him an idea using which they were able to drive till the rendezvous point.
