Four Friends A, B, C and D are living in a hostel. One day A’s uncle came and gave him a box of laddoos, According to their rule whatever they get will be equally divided among them. So A divided the laddoos into four equal parts and he ate his part. Then B came to the room and he divided the laddoos into four equal parts and A and B ate their part. Then C came to the room he divided the laddoos into four equal parts and he ate his part. Then at the last D came to the room he also divided the laddoos into four equal parts and he ate his part. The number of laddoos D ate was 3. 1. How many laddoos did A eat?2. How many laddoos did B eat?3. How many laddoos did C eat?4. How many laddoos were there in the box altogether?

A eat 71 laddos B eat 39 laddos C eat 15 laddos Total laddos were 128

ApproachSince 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)

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)

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

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)

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.

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

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.

Let total no. of laddos be x Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â AÂ Â Â Â Â Â Â Â Â Â Â Â BÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â C Â Â Â Â Â Â Â D Â Â Â Â Â Â Â Â AÂ Â Â Â Â Â Â Â Â Â Â Â Â Â x/4Â Â Â Â Â Â Â Â (Â Â Â Â Â Â Â Â Â Â Â Â 3x/4Â Â Â Â Â Â Â Â Â Â Â Â )

Â Â BÂ Â Â Â Â Â Â Â Â Â ( 3x/16Â Â Â 3x/16 )Â Â Â Â Â Â (Â Â Â Â Â Â Â Â Â 6x/16Â Â Â )

Â Â CÂ Â Â Â Â Â Â Â Â 6x/64Â Â Â Â 6x/64 Â Â Â Â 6x/64Â Â Â Â Â 6x/64 Â Â

Â Â DÂ Â Â Â Â Â Â Â Â Â 6x/256Â Â Â 6x/256Â Â Â Â Â Â 6x/256Â Â Â Â Â Â 6x/256

Â Â nowÂ Â 6x/256=3 => x=128 Â so A's share = x/4+3x/16+6x/64+6x/256 =71 Â Â Â Â Â B's share = 3x/16+6x/64+6x/256=39 Â Â Â Â Â C's share= 6x/64+6x/256=15 Â Â Â Â Â D's share = 3 total laddos , x=128.

## Four Friends A, B, C and D are living in a hostel. One day A’s uncle came and gave him a box of laddoos, According to their rule whatever they get will be equally divided among them. So A divided the laddoos into four equal parts and he ate his part. Then B came to the room and he divided the laddoos into four equal parts and A and B ate their part. Then C came to the room he divided the laddoos into four equal parts and he ate his part. Then at the last D came to the room he also divided the laddoos into four equal parts and he ate his part. The number of laddoos D ate was 3. 1. How many laddoos did A eat?2. How many laddoos did B eat?3. How many laddoos did C eat?4. How many laddoos were there in the box altogether?

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