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3 accessories with 100 rupees

June 3, 2015 by Shameer Sha 3 Comments

A man buy 3 accessories with 100 rupees. first one 5 rupees, second one 3 rupees, third one 0.5 rupees each. he bought total 100 pieces. How many pieces he bought for each accessory ?


First Accessory

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Second Accessory

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Third Accessory

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lets say he bought x,y and z pieces.
Then we can make these two equations
Eq1: x + y + z = 100
Eq2: 5x + 3y + 0.5z = 100

2*Eq2 => 10x + 6y + z = 200
2*Eq2-Eq1 => 9x + 5y = 100
from this equation x should be a multiple of 5, for x = 5 => y = 11
for x = 10 => y = 2
for x = 15 or more y becomes -ve

So two possible answers
x = 5, y = 11, z = 84
x = 10, y = 2, z = 88

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Vessel with water and syrup

May 23, 2015 by puzzler 1 Comment

vessel-with-water-and-syrup
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

A. 1/3
B. 1/4
C. 1/5
D. 1/7

Check your answer:-

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Lets assume that we take x part of the mixture out, and assuming uniform distribution of water and syrup, we can say out of the x part, 3x/8 was water and 5x/8 was syrup, now on replacing this x part by water, both water and syrup part should become equal, we can form the below equation

Water Part = Syrup part
3 – 3x/8 + x = 5 – 5x/8
x +2x/8 = 2
5x/4 = 2
x = 8/5

initially total quantity was 8(i.e. 3+5), and we should take out 1/5th part out and replace it with water to make both equal.

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How many children on school ground ?

May 3, 2015 by Dilip Lalwani Leave a Comment

There are some children on a school ground, a square has been drawn and all children are standing on a squares four lines, they are standing with same distance, four children are standing on a four corners. No 16 Is exactly opposite of No 6. How many children are there?

How many childer in school ground

Check your answer:-

Tried enough already?

Assuming there are x children in every side of the square, thus a total of 4x -4 children will be there(not counting corner child twice).

Now we know there are atleast 16 children, thus 4x-4 >= 16. x >= 5

Lets assume first child is sitting on bottom left corner, than assume three cases

Case 1: x = 5

then 6th child will be standing on second position in second line of square, thus the exactly opposite child will be sitting on x + x-2 + x + x-1 = 4x-3th position, i.e. 17th position.

Which is not the case as per the question, thus x can’t be 5

Case 2: x = 6

then 6th child is sitting on the corner(point B), thus exactly opposite child is on x + x-2 + x= 3x-2th position, i.e. 16th.

Thus it satisfies all conditions of the question.

Thus there are a total of 4x-4, i.e. 20 children in this case.

Case 3: x > 6

then 6th child is standing somewhere in the bottom line (point A), as per this exactly opposite child will be standing at position x + x-2 + x -6 = 3x – 8

for 3x-8 = 16 => x = 24/3 = 8

Thus there are a total of 4x-4 = 28 children in this case.

 

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The mystery of the missing euro

May 2, 2015 by Swapnil Guduru 1 Comment

two women were selling wrist bands in the close-by shops, one at three for an euro and other at two for an euro. one day both of them were obliged to return home when each had sold thirty bands unsold. They put together the two lots of marbles and handing them over to a friend and asked her to sell them at five for two euros.According to their calculation, after all, three for an euro and and two for an euro was exactly the same as five for two euros. Now they were expecting 25 euros as they would have normally got.Much to their surprise they got 24 euros for the whole lot. Now where did the one euro go? Can you explain the mystery?

Click here to See ExplanationHide

30 bands at the cost of 3 bands per euro were sold with 20 bands at the cost of 2 bands per euro, Thus they received 20 euros for this, while the other 10 bands were sold at 5 bands per two euro, thus received 4 euros additional.

Here there expectation were wrong due to the fact that 10 bands which were supposed to be sold originally at 4 bands per two euros, were instead sold at 5 bands per two euros.

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One Rupee to begger daily

April 21, 2015 by MITUL GOYANI 5 Comments

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

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When is Cheryl’s Birthday ?

April 19, 2015 by puzzler 6 Comments

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?

Check your answer:-

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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.

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Math Puzzle – solve this ?

April 18, 2015 by neo android 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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How many 0’s between 1 to 200

March 9, 2015 by Dilip Lalwani 8 Comments

How many 0’s are present in numbers from 1 to 200 ?

(including both)

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One zero after every 10 numbers, thus 20 + 10 extra (for 100’s series, 100, 101 to 109) + 1 extra for 200
Thus total number of zeros between 1 to 200 are 31.

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Solve this number puzzle

February 18, 2015 by dushyant singh 1 Comment

10408086_334897820029319_8703869613996047258_n

Click here to See SolutionHide
If we try to make equations

x1 + x2 + x3 = 9

x4 + x5 + 5 = 12

x7 + x8 + x9 = 7

x1 + x4 + x7 = 6

x2 + x5 + x8 = 9

x3 + 5 + x9 = 13

x1 + x5 + x9 = 5

x7 + x5 + x3 = 11

7 variables and 7 equations, its little tough and cumbersome to solve these equations so lets try some hit and trial.

you can see that x1 + x5 + x9 is only 5, thus it has less number of possibilities for x1, x5 and x9 as these cant be -ve.

lets assume x1 and x9 both are 0, in that case x5 will be 5.

17
0 1 8 9
2 5 5 12
4 3 0 7
——
6 9 13 5

diagonal equation does not satisfy, we can see x7 + x5 + x3 exceeds by 6.
so lets try to reduce it by 6,

lets try by x1 = 1, x9 = 1
11
1 1 7 9
4 3 5 12
1 5 1 7
6 9 13 5

it satisfies all the conditions
so, x1 = 1 , x2 = 1, x3 = 7, x4 = 4, x5 = 3, x7 = 1, x8 =5 and x9 = 1

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How many handshakes in African-US summit ?

February 11, 2015 by puzzler 1 Comment

During the African-US summit last year, 10 presidents left the conference room and greeted one another. Assuming no 2 presidents repeatedly greeted themselves, how many total handshakes were effected?

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Answer is 45 handshakes. The first president greeted 9 presidents, the second 8 and so on, so 9+8+7+6+5+4+3+2+1. This gives 45 handshakes.

in permutations and combinations way
A handshake is possible with two people, so it is similar to How many ways you can select two people out of 10.

i.e. 10C2 = 10!/(8!*2!) = 9*10/2 = 45

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