Question 480799: a man had a bag of sweets he gave one to his son and 1/7 of the remaining and what he was left he gave to his daughter two sweet and 1/7 of the remaining sweets.the two children found that they have same number of sweets how many sweets are there in original bag
Answer by mathstutor494(120)   (Show Source):
You can put this solution on YOUR website! 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
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