SOLUTION: a man had a a bag of sweets.he gave to his son one sweet and 1/7 of the remainning sweets.from what was left he gave to his daughter two sweets and 1/7 of the remainnig sweets.the

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Question 739840: a man had a a bag of sweets.he gave to his son one sweet and 1/7 of the remainning sweets.from what was left he gave to his daughter two sweets and 1/7 of the remainnig sweets.the two children found that they had the same numbers of sweets how many sweets were there in the orignal bag?
Answer by josgarithmetic(39616) About Me  (Show Source):
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N = Original number of sweets in the bag.

Son: highlight%281%2B%281%2F7%29N%29 received.


Number of sweets remaining in bag is R=+N-%281%2B%281%2F7%29N%29. The variable R is assigned as the number of sweets now REMAINING in the bag after the son received his allotment of sweets.

Daughter: 2%2B%281%2F7%29%28R%29 which is substituting for R,
2%2B%281%2F7%29%28+N-%281%2B%281%2F7%29N%29%29=2%2B%281%2F7%29%28N-1-N%281%2F7%29%5E2%29
=2+N/7-1/7-N/(7^2)
=highlight%282%2BN%2F7-1%2F7-N%2F%287%5E2%29%29 received.

Each child has the same number of sweets:
2%2BN%2F7-1%2F7-N%2F%287%5E2%29=1%2B%281%2F7%29N
Multiply left and right members by 7%5E2,
98%2B7N-7-N=49%2B7N
98-7-N=49
-N=49-98%2B7
N=98-49-7
highlight%28N=42%29
Forty-Two sweets originally in the bag.