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Alex, Bernard and Cindy had a total of 840 sweets. The ratio of the number of sweets Bernard had to the number of sweets Cindy had was 1:4. After Alex and Bernard each gave away 1/2 of their sweets, the 3 children had 620 sweets left. How many sweets did Bernard have at first?
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Answered by
Anonymous
Let "x" be the number of sweets Bernard had at first.
Then the number of sweets Cindy had initially was 4x;
and the number of sweets Alex had initially was the rest, (840-x-4x) = 840-5x.
After Alex and Bernard each gave away 1/2 of their sweets, the 3 children had 620 sweets left.
Mathematical translation is THIS equation
0.5 * ((840-5x) + x) + 4x = 620 for the remaining sweets.
Simplify
0.5*(840-4x) + 4x = 620
420 - 2x + 4x = 620
420 + 2x = 620
2x = 620 - 420 = 200
x = 200/2 = 100.
Bernard had 100 sweets at first.
Then the number of sweets Cindy had initially was 4x;
and the number of sweets Alex had initially was the rest, (840-x-4x) = 840-5x.
After Alex and Bernard each gave away 1/2 of their sweets, the 3 children had 620 sweets left.
Mathematical translation is THIS equation
0.5 * ((840-5x) + x) + 4x = 620 for the remaining sweets.
Simplify
0.5*(840-4x) + 4x = 620
420 - 2x + 4x = 620
420 + 2x = 620
2x = 620 - 420 = 200
x = 200/2 = 100.
Bernard had 100 sweets at first.
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