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30 Nov 2020, 09:46
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$1 bills: 1
$20 bills: 16
Percival has a total of 26 bills, consisting of a combination of $1, $5, $10, and $20 bills.
If the total value of these bills is $386, and Percival has at least one bill of each denomination, select for $1 bills the possible number of $1 bills Percival could have, and for $20 bills the possible number of $20 bills Percival could have that are jointly consistent with the information provided. Make only two selections, one in each column
If the total value of these bills is $386, and Percival has at least one bill of each denomination, select for $1 bills the possible number of $1 bills Percival could have, and for $20 bills the possible number of $20 bills Percival could have that are jointly consistent with the information provided. Make only two selections, one in each column
| $1 bills | $20 bills | |
| 1 | ||
| 6 | ||
| 10 | ||
| 12 | ||
| 16 | ||
| 18 |
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Official Answer
$1 bills: 1
$20 bills: 16
16 Jan 2025, 14:41
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I like the solution - it’s helpful. i get the solution but doing this during the exam under 3min is it realistically possible to do all the permutation and combination ? what mindset should i have approaching such question.
Thank you
Thank you
General Discussion
30 Nov 2020, 09:46
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Official Solution:
Let's analyze the provided options and determine how many $20 bills Percival could have.
• Percival cannot have 18 $20 bills, as that would amount to $360, leaving only $26 to be distributed among the remaining 8 bills consisting of $1, $5, and $10 denominations. Since the sum of $5 and $10 bills will always end with a units digit of 0 or 5, there can only be 1 or 6 $1 bills to make the total units digit equal 6. If there were 1 $1 bill, the remaining 7 bills of $5 and $10 denominations would exceed $25. Conversely, if there were 6 $1 bills, the remaining 2 bills of $5 and $10 denominations would total less than $20. Thus, 18 $20 bills is not possible.
• He also cannot have 12 $20 bills because, in that case, the total from the $20 bills would be $240. The remaining 14 bills, even if all were $10 bills, would amount to only $140, resulting in a total of $380, which is less than $386.
• Therefore, from the options, Percival must have 16 $20 bills, which amounts to $320.
This leaves $66 to be distributed among the remaining 10 bills consisting of $1, $5, and $10 denominations. Since the sum of $5 and $10 bills will always end with a units digit of 0 or 5, there can only be 1 or 6 $1 bills to make the total units digit equal 6.
• If there were 6 $1 bills, they would amount to $6, leaving $60 to be accounted for by the remaining 4 bills. However, even if all 4 bills were $10 bills, this would only amount to $40, which is less than $60.
• Therefore, there must be exactly 1 $1 bill.
Correct answer:
$1 bills "1"
$20 bills "16"
Bunuel
Let's analyze the provided options and determine how many $20 bills Percival could have.
• Percival cannot have 18 $20 bills, as that would amount to $360, leaving only $26 to be distributed among the remaining 8 bills consisting of $1, $5, and $10 denominations. Since the sum of $5 and $10 bills will always end with a units digit of 0 or 5, there can only be 1 or 6 $1 bills to make the total units digit equal 6. If there were 1 $1 bill, the remaining 7 bills of $5 and $10 denominations would exceed $25. Conversely, if there were 6 $1 bills, the remaining 2 bills of $5 and $10 denominations would total less than $20. Thus, 18 $20 bills is not possible.
• He also cannot have 12 $20 bills because, in that case, the total from the $20 bills would be $240. The remaining 14 bills, even if all were $10 bills, would amount to only $140, resulting in a total of $380, which is less than $386.
• Therefore, from the options, Percival must have 16 $20 bills, which amounts to $320.
This leaves $66 to be distributed among the remaining 10 bills consisting of $1, $5, and $10 denominations. Since the sum of $5 and $10 bills will always end with a units digit of 0 or 5, there can only be 1 or 6 $1 bills to make the total units digit equal 6.
• If there were 6 $1 bills, they would amount to $6, leaving $60 to be accounted for by the remaining 4 bills. However, even if all 4 bills were $10 bills, this would only amount to $40, which is less than $60.
• Therefore, there must be exactly 1 $1 bill.
Correct answer:
$1 bills "1"
$20 bills "16"
