Asked by Anonymous
Three boys, Jason, Keith and Lucas, had equal number of coins. Jason had only twenty-cent coins. Keith and Lucas each had a mix of twenty-cent and one-dollar coins. Keith had 16 one-dollar coins while Lucas had 7 one-dollar coins.
(a) Of the three boys, who had the least amount of money and who had the most?
(b) What was the difference in the total value of Jason and Lucas' coins?
(c) Keith used half of his one-dollar coins and some of his twenty-cent coins to buy a book that cost $10.40. Then, he had 7/10 of his twenty-cent coins left. How many twenty-cent coins did Keith have at first?
(a) Of the three boys, who had the least amount of money and who had the most?
(b) What was the difference in the total value of Jason and Lucas' coins?
(c) Keith used half of his one-dollar coins and some of his twenty-cent coins to buy a book that cost $10.40. Then, he had 7/10 of his twenty-cent coins left. How many twenty-cent coins did Keith have at first?
Answers
Answered by
oobleck
Suppose
Jason had x 20¢ coins. He had 20x ¢
Keith had 16 $1 coins and m 20¢ coins. His money was
1600+20m = 1600+20(x-16) = 1280+20x ¢
Lucas had 7 $1 coins and n 20¢ coins. His money was
700+20n = 700+20(x-7) = 560+20x ¢
(a) shoud be clear
(b) also clear
(c) 800 + 20a = 16040
a = 3/10 m
Now you can find m
Jason had x 20¢ coins. He had 20x ¢
Keith had 16 $1 coins and m 20¢ coins. His money was
1600+20m = 1600+20(x-16) = 1280+20x ¢
Lucas had 7 $1 coins and n 20¢ coins. His money was
700+20n = 700+20(x-7) = 560+20x ¢
(a) shoud be clear
(b) also clear
(c) 800 + 20a = 16040
a = 3/10 m
Now you can find m
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