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One Rupee to begger daily

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One person had some money. he gave one rupee to beggar and half of the remaining money was spent for buy something. same thing happened second day he gave one rupee to beggar from remaining money of first day. this continued on the six day he had 0 rupee, then how much money he had on first day?

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He was left with 0 rupee on sixth day, this means he had only 1 rupee on sixth day.

so he had 1*2 + 1 = 3 on 5th day

on 4th day start 3*2 +1 = 7

on 3rd day start 7*2 + 1 = 15

on 2nd day start 15*2 + 1 = 31

on first day start 31*2 + 1 = 63

When is Cheryl’s Birthday ?

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Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is. Cheryl gives them a list of 10 possible dates.

May 15 May 16 May 19

June 17 June 18

July 14 July 16

August 14 August 15 August 17

Cheryl then tells Albert and Bernard separately the month and the day of her birthday respectively.

Albert: I don’t know when Cheryl’s birthday is, but I know that Bernard does not know too.

Bernard: At first I didn’t know when Cheryl’s birthday is, but I know now.

Albert: Then I also know when Cheryl’s birthday is.

So when is Cheryl’s birthday?

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From the first statement, any of the day’s in that month should not be unique, else otherwise Albert can’t be sure of Bernard not knowing the birthday.

Lets take the possible answers

If it was May => Albert cant say the first statement, as if the day was 19, Bernard will know the birthday for sure.
If, it is was June => Same is the case with 18th.
If, it is in July => Albert can say the first statement.
If it is in August => Albert can say the first statement.

Now after hearing first statement, Bernard can clearly figure out that the month of birthday is either July or August.

So, if the day was 14 => It can be July 14 or August 14 =>as bernard know now, it is not possible
If the day was 15 => It can be Aug 15
If the day was 16 => it can be July 16
If the day was 17 => it can be Aug 17

Only three days possible are Aug 15, July 16 or Aug 17, bernard knows the day so he know the birthday.

From the third statement, Albert knows the birthday as well, so it can only be July 16, as if the month was Aug, Albert can not figure it out.

Thus the Cheryl’s birthday is on July 16.

Math Puzzle – solve this ?

by 1 Comment

solve this ?
a person sells book at a discount of 20% on the mark price and he gives 1 book free on very 11 books brought and still makes a profit of 10%.Find at what % above cost price is the marked price ?

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Lets say the mark price is X

then on selling 12 books he gets (0.8X)*11 = 8.8X

Lets say the cost price of each book is Y then the total cost of 11 books is 12Y

Now he made 10% profit thus 1.1*(12Y) = 8.8X

12Y = 8X

X = 3/2Y = 1.5Y

Thus X is 50% more than the Y, hence mark price is 50% more than the cost price.

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