the nearest 1000 of 85654 is:_______
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Let the number of prizes of money 100 be X, and the number of prizes of cost 25 be
y;
x+y = 63
y = 63-x
--(1)
100x+25y = 300 4x+y = 120 y = 120-4x (2)
63-x = 120-4x
4x-x = 120-63
3x = 57
x = 19
y = 63-x = 63-19
y = 44
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Answer:
The nearest 1000 of 85654 is:
86000
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