SOLUTION: Hi Elaine had some sweets. When she ate 2/7 of them and gave 36 to her brother she had 2/7 left. How many did she have at first. Thanks

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Question 1167892: Hi
Elaine had some sweets. When she ate 2/7 of them and gave 36 to her brother she had 2/7 left. How many did she have at first.
Thanks
Found 4 solutions by josgarithmetic, ikleyn, MathTherapy, greenestamps:
Answer by josgarithmetic(39620)   (Show Source):
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Try again.
and then the same , but of what?

Answer by ikleyn(52803)   (Show Source):
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.

+ 36 + = x Multiply all the terms by 7 2x + 36*7 + 2x = 7x 36*7 = 7x - 2x - 2x 36*7 = 3x 12*7 = x x = 84. ANSWER. At first, she had 84 sweets.

Solved.

Answer by MathTherapy(10552)   (Show Source):
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Hi
Elaine had some sweets. When she ate 2/7 of them and gave 36 to her brother she had 2/7 left. How many did she have at first.
Thanks

With the original number of sweets being S, and after eating , she will have of sweets, remaining
We then get:
3S = 7(36) ------ Cross-multiplying
Original number of sweets, or

Answer by greenestamps(13200)   (Show Source):
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She ate 2/7 of the sweets and still had 2/7 of them left after she gave some to her brother.

That means the 36 sweets she gave her brother were 3/7 of the number she started with. (2/7 plus 2/7 plus 3/7 = 7/7 = all of them)

(3/7)x = 36
x = 36(7/3) = 12*7 = 84