SOLUTION: The population of a town increased by 16 2/3 % in 1990 and by 10% in 1991. If the population of the town at the beginning of 1990 was 30,000, what was it at the end of 1991?

Question 1101180: The population of a town increased by 16 2/3 % in 1990 and by 10% in 1991. If the population of the town at the beginning of 1990 was 30,000, what was it at the end of 1991?
Found 2 solutions by richwmiller, greenestamps:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
corrected
16 2/3=50/3
16 2/3%=50/300
350/300=116 2/3%=1.16 2/3

(350/300)*1.1*30000=38500
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

The other tutor's answer is not right; you should be able to see where he made his arithmetic error.

16 2/3% as a fraction is 1/6; so the population of the town increased by 1/6 in 1990. Since the population at the beginning of 1990 was 30,000, it increase by 5,000 during 1990, making the population at the beginning of 1991 35,000.

10% as a fraction is 1/10; so in 1991 the population increased by 1/10 of 35,000, or 3,500, making the population at the end of 1991 38,500.
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