Answer / mayank

Let a gives x mangoes to D
Then x=x/4 + 1
==> x=4/3

So, 4/3 amount remaining after C transaction
,Let "m" no of mangoes remaining before C transaction then
m/3+1 no. of mangoes given to C and 4/3 remaining for D transaction
so, m=m/3 + 1 + 4/3
m=7/2


So, 7/2 amount remaining after B transaction
,Let "o" no of mangoes remaining before B transaction then
o/2+1 no. of mangoes given to B and 7/2 remaining for C transaction
so, o=o/2 + 1 + 7/2
o=9

so initially A was having 9 mangoes

Answer / zameer

9 mangoes. No Complete number is possible. Because at last
2/3rd is 1.

Answer / paul

The answer is 9 mangoes.

let y be the total/original no. of mangoes
b be the remaining mangoes after gave to B
c be the remaining mangoes after gave to C
d be the remaining mangoes after gave to D

equations:

b = y - 0.5y - 1
= 0.5y - 1

c = b - (1/3)b - 1
= (2/3)b - 1

d = c - (1/4)c - 1
= (3/4)c - 1

but d = 0, thus,

c = 4/3
b = 7/2
y = 9 mangoes -> answer

Answer / suraj

let no. of mangoes A has=x
Now, he gives 1/2 of x plus 1 to B.
So, A is left with (1/2ofx-1)=(x-2)/2
Now, he gives 1/3 of the remaining plus 1 to C.
Therefore C has (x-2)/2
------- + 1
3
So, A has 2*(x-2)/2 x-5
------- - 1 = ------
3 3
Now, he gives 1/4 of the remaining plus 1 to D.
Therefore A has 3*(x-5)/3 x - 9
-------- - 1 = --------
4 4
So, A is left with zero mangoes
=> (x-9)/4 = 0
=> x = 9
therefore A had 9 mangoes in the beginning.

Answer / nagicaj

Initially A has x
Now A gives B: x/2+1
So A has x/2-1 = (x-2)/2
Now A gives C: ((x-2)/2)/3+1
So A has ((x-2)/2)/3-1 = (x-8)/6
Now A gives D: ((x-8)/6)/4+1
So A has ((x-8)/6)/4-1=0
Solving the above x = 32 mangoes.

Answer / ashish

1 mangoe

Answer / vijaysankar

32 mangos

Answer / kalpana singh

let no. of mangoes A has=x
Now, he gives 1/2 of x plus 1 to B.
So, A is left with (1/2ofx-1)=(x-2)/2
Now, he gives 1/3 of the remaining plus 1 to C.
Therefore C has (x-2)/2
------- + 1
3
So, A has (x-2)/2 x-8
------- - 1 = ------
3 6
Now, he gives 1/4 of the remaining plus 1 to D.
Therefore D has (x-8)/6 x - 17
-------- + 1 = --------
4 24
So, A is left with zero mangoes
=> all the mangoes are with D
=> (x-17)/24 = 0
=> x = 17
therefore A had 17n the beginning.

Answer / venkatsundar

four mangos

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