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A man had a bag of sweets. He gave his son one sweet and 1/7 of the remaining sweets. From what was left, he gave his daughter two sweets and 1/7 of the remainder. The two children found that they had the same number of sweets. How many sweets were there initially in the bag?

1 Answer

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Answer:

number of sweet in the bag = 36

Explanation:

Let

x = number of sweet in the bag

He gave 1 to his son

Remaining sweet = x - 1 sweets

1/7 of remaining sweets = 1/7 * (x-1)

= (x-1) / 7

Total sweets given to son = 1 + (x-1) / 7

= (7+x-1) / 7

= (x+6) / 7 (1)

Sweets remaining in the bag after giving to son

= x - {(x+6)/7}

= {7x-(x+6)} / 7

= (7x-x-6) / 7

= (6x-6)/7

He gave 2 sweets to daughter

Remaining sweets

= {(6x-6)/7}-2

= {(6x-6)-14} / 7

= (6x-6-14) / 7

= (6x-20)/7

He gave 1/7 from remaining to his daughter

= (6x-20)/7) * (1/7)

=(6x-20)/49

Total sweets given to daughter

= (6x-20)/49) + 2

= {(6x-20)+98} / 49

= (6x-20+98) / 49

=(6x+78)/49 (2)

Both children have the same number of sweets

Equate equation (1) and (2)

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

Cross product

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

49x + 294 = 42x + 546

49x - 42x = 546 - 294

7x = 252

x = 252/7

x = 36

number of sweet in the bag = 36

User Robin Winslow
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