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22 Jul 2024, 01:30
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B
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Difficulty:
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82% (01:53) correct
18%
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Lucy stored 1-cent, 5-cent, 10-cent, and 25-cent coins in a piggy bank, amounting to a total of $12.21. If each type of coin was collected at least once, what is the fewest number of coins the piggy bank could contain?
A. 48
B. 52
C. 54
D. 56
E. 1221
A. 48
B. 52
C. 54
D. 56
E. 1221
03 Aug 2024, 09:45
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Official Solution:
Lucy stored 1-cent, 5-cent, 10-cent, and 25-cent coins in a piggy bank, amounting to a total of $12.21. If each type of coin was collected at least once, what is the fewest number of coins the piggy bank could contain?
A. 48
B. 52
C. 54
D. 56
E. 1221
Assume \(a\) 25-cent coins, \(b\) 10-cent coins, \(c\) 5-cent coins, and \(d\) 1-cent coins were collected. Then, we'd have:
\(25a + 10b + 5c + d = 1221\)
Our goal is to minimize \(a+b+c+d\). To achieve this, we need to maximize the number of the largest denomination coin, \(a\). Maximizing \(a\) is based on efficiency. Since \(a\) represents the number of the highest denomination coin, collecting more of these reduces the total number of other coins needed to reach the sum of $12.21. Each higher denomination coin contributes significantly more to the total, meaning fewer total coins are needed.
Since the value of \(a\) cannot exceed 48, we try \(a=48\). In this case, we have:
\(25*48 + 10b + 5c + d= 1221\)
\(10b + 5c + d = 21\)
Next, let's maximize the number of the next largest denomination coin, \(b\). The maximum value of \(b\) is 2. However, in this case, \(c\) would be 0, so \(b\) must be 1:
\(10*1 + 5c + d = 21\)
\(5c + d = 11\)
Here, \(c\) must be 2, making \(d\) equal to 1.
Therefore, \(a + b + c + d = 48 + 1 + 2 + 1 = 52\).
Answer: B
Lucy stored 1-cent, 5-cent, 10-cent, and 25-cent coins in a piggy bank, amounting to a total of $12.21. If each type of coin was collected at least once, what is the fewest number of coins the piggy bank could contain?
A. 48
B. 52
C. 54
D. 56
E. 1221
Assume \(a\) 25-cent coins, \(b\) 10-cent coins, \(c\) 5-cent coins, and \(d\) 1-cent coins were collected. Then, we'd have:
\(25a + 10b + 5c + d = 1221\)
Since the value of \(a\) cannot exceed 48, we try \(a=48\). In this case, we have:
\(25*48 + 10b + 5c + d= 1221\)
\(10b + 5c + d = 21\)
\(10*1 + 5c + d = 21\)
\(5c + d = 11\)
Therefore, \(a + b + c + d = 48 + 1 + 2 + 1 = 52\).
Answer: B
General Discussion
22 Jul 2024, 07:14
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$12.21=1221 cents
Each type of coin was collected at least once. Hence less value coins should be minimum
1*1 cent+1*5 cent+ 1*10 cent=16 cent
1*1 cent+2*5 cent+ 1*10 cent=21 cent
Remaining 1200 cents. 48 coins of 25 cents needed for 1200 cents
\(48+1+2+1=52\)
Answer is B
Each type of coin was collected at least once. Hence less value coins should be minimum
1*1 cent+1*5 cent+ 1*10 cent=16 cent
1*1 cent+2*5 cent+ 1*10 cent=21 cent
Remaining 1200 cents. 48 coins of 25 cents needed for 1200 cents
\(48+1+2+1=52\)
Answer is B
