Question 480799
Let there be "x" sweets in original bag

Gave 1 to son so remaining are x-1 sweets
1/7 of remaining sweets = (x-1)*(1/7)=> (x-1)/7
 
Hence total sweets given to son
1+((x-1)/7)
(x+6)/7 ...............................(1)

Sweets remaining in the bag after giving to son
x-((x+6)/7)
(6x-6)/7

Out of this he gave 2 sweets to daughter so remaining sweets are
((6x-6)/7)-2
=(6x-20)/7

Further, he gave 1/7 from remaining lot to his daughter...
((6x-20)/7)*(1/7)
=(6x-20)/49

So total sweets with daughter
((6x-20)/49)+2
=(6x+78)/49 ..............................(2)

Since the problem specifies that the two children found that they have same number of sweets, so expressions (1) and (2) are equal.
Accordingly

(x+6)/7 = (6x+78)/49 

Solving for x gives value of 36

Ans :  36 sweets there in original bag