**nmodi**
Answered On : Jul 17th, 2005

Ans. 48

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**sachin kumar rai**
Answered On : Aug 6th, 2005

according to my calc. the no. of breads = 64

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**ashish g**
Answered On : Oct 20th, 2005

I think answer is 63

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no. of breades are 63

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**Sheetal Salunkhe**
Answered On : Nov 21st, 2005

initially the breads should be 48. first thief took 24 of 48 second one took 12 of 24 third one have took 6 of 12 then remains 3 breads.

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**sunita**
Answered On : Dec 20th, 2005

ans:

it will be 63

procedure:

let initially der are x breads

1st thief takes=x/2 +1/2 ,so remaining=x-(x/2+1/2)=x/2-1/2

2nd thief takes=1/2+1/2*(x/2-1/2)=1/2+x/4-1/4=x/4+1/4 so remain=x/2-1/2-x/4-1/4=x/4-3/4

3rd thief takes=1/2 +1/2*(x/4-3/4)=1/2+x/8-3/8=x/8+1/8 so remain=x/4-3/4-x/8-1/8=x/8-7/8

4th thief takes=1/2+1/2*(x/8-7/8)=1/2+x/16-7/16=x/16+1/16 so remain=x/8-7/8-x/16-1/16=x/16-15/16

so x/16-15/16=3

x=48+15=63(ans)

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**jaspreet(jagz)**
Answered On : Feb 19th, 2006

63( ( ( ( X/2-B/2 )/2 -B/2 )/2 -B/2 ) /2 -B/2 ) =3B=>( X-15B) /16 =3B==>X=63B

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**HEMANTHKUMAR VARADHARAJAN**
Answered On : Jun 20th, 2006

Let the no. of breads initially present be x.Then1 st thief takes x/22 nd thief takes x/43 rd thief takes x/84th thief takes x/16.Remaining = 3 breadsEquating x/2 + x/4 + x/8 + x/16 + 3 = x(15x + 48)/16 = x15x + 48 = 16x=> x = 48The number of breads initially present were 48.

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**amateur**
Answered On : Aug 7th, 2006

Ans : 48

Let Total no of breads be X.

=> 1st Thief took X/2 remaining X/2

=> 2nd Thief took 1/2 * X/2 = X/4 remaining X/2-X/4= X/4

=> 3rd Thief took 1/2 * X/4 = X/8 remaining X/4-X/8=X/8

=> 4th Thief took 1/2 * X/8 = X/16 remaining X/8-X/16=X/16

Now its given that remaining bread = 3

Therefore X/16=3

=> X = 48

Ans = 48.

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**maximus**
Answered On : Oct 25th, 2006

63

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**Pooja**
Answered On : Dec 1st, 2006

The answer should be 48.

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**Jekin**
Answered On : Jun 30th, 2007

should be 63. the ones who think it's 48 didn't take half bread into consideration.

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**puspanjali**
Answered On : Aug 27th, 2007

if the total no of bread is=x

then the first thieves is stolen :(x/2+1/2)

2nd thieves is stolen half of the

present bread and half of the bread :(1/2(x/2+1/2)+1/2)

3rd :(1/2(1/2(x/2+1/2)+1/2)+1/2)

4th :(1/2(1/2(1/2(x/2+1/2)+1/2)+1/2)+1/2)

which is => 3(the remaining bread)=(1/2(1/2(1/2(x/2+1/2)+1/2)+1/2)+1/2)

=> x=36

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Last theif took 3 breads.

third one took 6,

second one took 12,

first one took 24 i.e. Half of the total breads

So, Answer is 48 breads..

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Answer: **48**

total=x

x=x/2+x/4+x/8+x/16+3

x=(15x+48)/16

x=48

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Definitely the ans is 63 because:

last time the bread left is 3 now we know that this comes when 4th theif took half of the remaining bread left by 3rd and 1/2 more, so reverse the process, add 12 to 3 that makes 72. and then multiply by 2. this will be 7 so 7 will be left by 3rd thief , now again reverse. 2(7+12) = 15 . do this 2 more times and the ans will come as 63 for last time,

NOTE: THis will be the minimum breads present originals

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Am sorry to say, none of you read the question fully,

"The thieves took half of what is present and also HALF A BREAD"

so 3*2*2*2*2 + halfbread*no.of thieves

= 48 + 2

= 50

correct me if i am wrong.

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The correct answer is 52.

Let x is the no. of initially breads present.

then

x-(x+1)/2-(x+3)/4-(x+7)/8-(x+15)/8=3

Hence x=52

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The correct answer is 52.

Let x is the no. of initially breads present.

then

x-(x+1)/2-(x+3)/4-(x+7)/8-(x+15)/16= 3

Hence x 52

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