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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