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Two women are selling apples. The first sells 30 apples at 2 for $1,
earning $15. The second sells 30 apples at 3 for $1, earning $10. So
between them they’ve sold 60 apples for $25.
The next day they set the same goal but work together. They sell 60
apples at 5 for $2, but they’re puzzled to find that they’ve made only
$24.
What became of the other dollar?
Here, could be that lost dollar!
What is the conundrum?
They sell 60 apples at 5 for $2, that means 12 such sets of 5 apples. Suppose, out of each such set, 1 woman takes out $1 for 2 apples and other takes $1 for 3 apples. So, first woman earns $12 by selling 24 apples and second woman sells 36 apples for $12.
In short, first woman gives away 6 apples (from her 30 apples) to second woman increasing her count to 36 reducing her own count to 24. First woman would have made $3 from those but second woman only made $2 from those 6 apples. And there is that lost dollar in earning.
So, 60 apples can't be divided equally to find the earning as they had sold apples at different rates on previous day.
Other way, if they wanted to sell apples together with 30 apples each, then they should have sold apples at average of (1/2 + 1/3)/2 = $5/12 per apple (i.e. 12 apples for $5) instead of $2/5 per apple.
The difference in price per apple (5/12 - 2/5) = (1/60).
So the difference in earning after selling 60 such apples = (1/60) x 60 = 1.
And there is that other dollar!
What's wrong gone here on next day? Instead of averaging dollars per apple, apples per dollar are added directly which resulted reduced cost of each apple.
Andrea’s only timepiece is a clock that’s fixed to the wall. One day she forgets to wind it and it stops.
She travels across town to have dinner with a friend whose own clock is always correct. When she returns home, she makes a simple calculation and sets her own clock accurately.
How does she manage this without knowing the travel time between her house and her friend’s?
That's how she manages to set it accurately!
How it was mistimed?
Andrea winds her clock & sets it to the arbitrary time. Then, she leaves her house and when she reaches her friend's house, she note down the correct time accurately. Now, after having dinner, she notes down the correct time once again before leaving her friend's house.
After returning to home, she finds her own clock acted as 'timer' for her entire trip. It has counted time that she needed to reach her friend's house + time that she spent at her friend's house + time she needed to return back to home.
Since, Andrea had noted timings at which she reached & left her friend's house, she can calculate the time she spent at her friend's house. After subtracting this time duration from her unique timer count she gets the time she needed to reach to & return from her friend's house.
She must have taken the same time to travel from her home to her friend's home and her friend's home to her home. So dividing the count after subtracting 'stay time' she can get how much time she needed to return back to home.
Since, she had noted correct time when she left her friend's home, now by adding time that she needed to return back to home to that, she sets her own clock accurately with correct time.
Let's try to understand it with example.
Suppose she sets her own clock at 12:00 o' clock and leave her house. Suppose she reaches her friend's home and note down the correct time as 3:00 PM. After having dinner she leaves friend's home at 4:00 PM.
After returning back to home she finds her own clock showing say 2:00 PM. That means, she spent 120 minutes outside her home with includes time of travel to and from friends home along with time for which she spent with her friend. If time of stay at her friend is subtracted from above count, then it's clear that she needed 60 minutes to travel to & return back from friend's home.
That is, she needed 30 minutes for travel the distance between 2 homes. Since, she had noted correct time as 3:00 PM when she left friend's home, she can set her own clock accurately at 3:30 PM.
I place four balls in a hat: a blue one, a white one, and two red ones.
Now I draw two balls, look at them, and announce that at least one of
them is red. What is the chance that the other is red?
Well, it's not 1/3!
What was the puzzle?
It's not 1/3. It would have been 1/3 if I had taken first ball out, announced it as red and then taken second ball out. But I have taken pair of ball out. So, there are 6 possible combinations.
red 1 - red 2
red 1 - white
red 2 - white
red 1 - blue
red 2 - blue
white - blue
Out of those 6, last is invalid as I already announced the first ball is red. That leaves only 5 valid combinations.
And out of 5 possible combinations only first has desired outcome i.e. both are red balls.
Hence, there is 1/5 the chance that the other is red
Okay, I’ll ask three questions, and if you miss one I get your house. Fair enough? Here we go:
1.A clock strikes six in 5 seconds. How long does it take to strike twelve?
2.A bottle and its cork together cost $1.10. The bottle costs a dollar more than the cork. How much does the bottle cost?
3.A train leaves New York for Chicago at 90 mph. At the same time, a bus leaves Chicago for New York at 50 mph. Which is farther from New York when they meet?
You need little common sense in answering above!
