**deepa**
Answered On : Nov 27th, 2007

calculation is as below:

D = 3*4 = 12

C= 12*4/3 = 16

B= (16*2)/4 = 8

A= 32*4/3= 42.6

Hence A's share is 10.6+8

B ate 8 Laddoos

C ate 4

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A eat 71 laddos

B eat 39 laddos

C eat 15 laddos

Total laddos were 128

Approach Since D eat 3 laddos that means when he came he saw there were 12 laddos (3*4)That means when C came in he must have seen 48 (12*4) laddos and divided then in four parts each which left 12 for D. Now important point to note here is that when C came he saw 48 laddos which were actually for C and D 24 each because if B has divided in 4 equal part he would have eat 24, give A as 24 and left 24 each for C and D. That gives us step that When B came in must have seen 96 (24*3) laddos. Which were actually for B, C and D. That gives us answer that when A got the box there should be 128 laddos which he divided in 4 equal parts of 32 each. So if we sum up laddos which A must have eat it will be (32+24+12+3)

B (24+12+3) and C (12+3) and D we know it’s 3

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d=3

3*4=12

a. b . c , d ate 3 sweet = 3

12*4=48/3=16

a, b , c ate = 16each

16*4=64/2=32

a,b=32each

32*4=128

a=128

a=128+32+16+3

a=179

b=32+16+3=51

c=16+3=19

d=3

total= 252

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Let number of laddoos be n

Now

n = A, 3A

3A = 3A/4(B), 9A/4

9A/4 = 9A/16(C), 27A/16

27A/16 = 27A/64(D), 81A/64

Given, D=3

=>A = 64/9

B = (3*64) / (9*4) = 16/3

C = (9*64) / (16*9) = 4

Total No of Laddoos = A+3A = 256/9

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Hi All,

My approach is quite simple:

Say there are x laddoes.

A shares x/4 each to himself, B,C,D. n eates it (A=x/4)

B comes and shares x/4*4 to each and eat (A=x/4+x/16, B=x/16)

C comes , shares x/4*4*4 to each and eat (A=x/4+x/16, B=x/16, c=x/64)

D comes, shares x/4*4*4*4 to each and eat (A=x/4+x/16, B=x/16, c=x/64, D=x/256)

as its given D eats 3: x/256 =3 => x=768 (Clear)

let us check it with c = x/64 => 768/64 = 12

B = x/16 =>768/16 = 48

A = x/4 + x/16 => 192+48 = 240

A=240, B=48,C=12 D=3.

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no: of laddoos A ate is 240

no: of laddoos B ate is 48

no: of laddoos c ate is 12

total no: of laddoos in the box is 768

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Total laddood is 24

d ate 3

so when D comee remaining laddoos 12

3/4 of x is 12

so x is 16

similarly

do it it comes to 24.

Am I right?

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total 20 laddos

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1. How many laddoos did A eat?

2. How many laddoos did B eat?

8

3. How many laddoos did C eat?

4

4. How many laddoos were there in the box altogether?

3

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1. How many laddoos did A eat?

18.66

2. How many laddoos did B eat?

8

3. How many laddoos did C eat?

4

4. How many laddoos were there in the box altogether?

42.66

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a

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1.a=27

2.B=10

3C=3

4TOTAL=43

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A eats 71 ladoos

B eats 39 ladoos

C eats 15 ladoos

total was 128 ladoos

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Suppose, uncle had brought x no. of ladoos.

now, they are divided in 4 parts equally:

a=x/4

b=x/4r

c=x/4

d=x/4

now, A ate his part,thus left number of ladoos are:

3x/4 which are again further divided into 4 equal parts.

thus:

a=3x/16

b=3x/16

c=3x/16

d=3x/16

now, a and b are people who eat thus remaining ladoos are:

3x/4-3x/8=3x/8

i.e. remaining ladoos are: 3x/8.

now, again C came to room and divided the remaining ladoos in 4 equal parts.

thus,

a=3x/32

b=3x/32

c=3x/32

d=3x/32

now , only C eats his part thus,

remaining ladoos r: 3x/32

Finally, D comes and he divided into 4 equal parts

a=3x/128

b=3x/128

c=3x/128

d=3x/128

thus, after eating his own part the remaining ladoos r: 3x/128

which is equal to given as 3:

=> 3x/128=3

so, total number of ladoos were 128, and find the others.

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It is not at all possible for D to have only 12 chocolates to divide among his four friends.

Let the number of Chocolate be X.

How Many Chocolates D had with him before he shared that with his friends?

Remember that D should have more than 1/4th of total number of chocolates before he shared with his friends.

Also the chocolates are shared only among the four friends

From A 1/4 of X, multiple of 4 (as B divide into 4 parts)

From B 1/16 of X, multiple of 4

From C 1/16 X + 1/64 X, multiple of 4 by 4

As 1/16 X is divisible by 4, 1/64 X also multiple of 4 (D divides his chocolates into four parts)

Therefore the minimum value of X is 256

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My Calculations are

Total laddoos division was = x/4+x/4+x/4+x/4

Now A eats x/4 + x/16 laddoos

B eats x/16 laddoos

C eats x/16 laddoos

D eats x/16 laddoos which is 3 as given in question

So total laddoos were 48,

A -> 15

B -> 3

C -> 3

D -> 3. Note in the question it is explicitly said in C and D's case that **HE** ate his part and in both the cases A and B

do not eat.

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let there are x laddoos.

A share is x/4;

remaining is 3x/4;

now B share is 3x/16;

and A share will be x/4+3x/16;

now the remaining no. of laddoos will be 3x/8 becoz a and b ate thier part;

now C will share 3x/32;

remaining will be 3x/8-3x/32=9x/32;

and finally D share will be 9x/128;

9x/128=3

x=128/3=42.6

A ate 128/12+(3/16)*128/3=10.6+8=18.6;

B ate 3*128/3*16=8;

C ate 4

total no . of laddoos will be 42.6

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A -->12 +3laddoos

b---> 9 laddoos

c --> laddues

total 46 laddoos

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Well answer is a fraction and ladoos can't be in fraction.

So though the answer is not logically correct; stilltotal number of ladoos is 128/3

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1. A had 6 ladoos

2. B had 2 ladoos

3. C had 4 ladoos

4. There were 15 ladoos in the box

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I have seen the answers but could find only 1 correct answer i.e. vivekjain17's ans.

Many ppl have come to the conclusion dat there r 240 total Laddoos.. where as they missed the fact when A ate 1/4th of the total, he divided the REMAINING 3/4th laddoos into 4 parts again.

Check out this Solution:-

Let total Laddoos = x

1. A ate x/4 Laddoos, so Remaining Laddoos = 3x/4

2. Then A & B ate 1/4[3x/4] Laddoos each, so Remaining Laddoos = 1/2[3x/4] = 3x/8

3. Then A, B & C ate 1/4[3x/8] Laddoos each, so Remaining Laddoos = 1/4[3x/8] = 3x/32

4. Then A, B, C & D ate 1/4[3x/32] each and finished the Laddoos.

Date 3 Laddoos i.e. 1/4[3x/32]=3 ; x=128.

~ A ate 4 times i.e. x/4 + 1/4[3x/4] + 1/4[3x/8] + 1/4[3x/32] = 71

~ B ate 3 times i.e. 1/4[3x/4] + 1/4[3x/8] + 1/4[3x/32] = 39

~ C ate 2 times i.e. 1/4[3x/8] + 1/4[3x/32] = 15

~ D ate 1 time i.e. 1/4[3x/32] = 3

Cross Checking (71+39+15+3 =128)

Very Simple but nt Time efficient method. I guess vivekjain17's method is time efficient but the explanation was little confusing 4 me.

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Let 'l' b number of laddoos

A divides it into 4 parts, so each wil get l/4...

now B has l/4 ladoos which he again divides into 4 parts...l/16

ate A(l/4+l/16) B(l/16)

Now C has l/4+l/16=5l/16 ladoos...when divides into 4 equal parts 5l/64

A(l/4+l/16+5l/64) B(l/16+5l/64) C(5l/64)

Now D has l/4+l/16+5l/64 ladoos..when didivded into 4 equal parts 25l/256

A(l/4+l/16+5l/64+25l/256) B(l/16+5l/64+25l/256) C(5l/64+25l/256) D(25l/256)

given 25l/256=3, therefore l=30.72,

C=5l/64+25l/256=5.4,

B=l/16+5l/64+25l/256=7.32

A=l/4+l/16+5l/64+25l/256=15

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A...7

B...5

C...4

D...3

Total=28

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Let total no. of laddos be x

A B C D

A x/4 ( 3x/4 )

B ( 3x/16 3x/16 ) ( 6x/16 )

C 6x/64 6x/64 6x/64 6x/64

D 6x/256 6x/256 6x/256 6x/256

now 6x/256=3 => x=128

so A's share = x/4+3x/16+6x/64+6x/256 =71

B's share = 3x/16+6x/64+6x/256=39

C's share= 6x/64+6x/256=15

D's share = 3

total laddos , x=128.

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4. There were total 128 ladoos.

3. C ate 15 ladoos

2. B ate 39 ladoos

1. Finally a ate 71 ladoos

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