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Miguel has some dimes and some quarters. He has a minimum of 18 coins worth a maximum of $2.70 combined. If Miguel has 3 quarters, determine all possible values for the number of dimes that he could have. Answer:

Miguel has some dimes and some quarters. He has a minimum of 1818 coins worth a maximum of $2\$2.7070 combined. If Miguel has 33 quarters, determine all possible values for the number of dimes that he could have.\newlineAnswer:

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Full solution

Q. Miguel has some dimes and some quarters. He has a minimum of 1818 coins worth a maximum of $2\$2.7070 combined. If Miguel has 33 quarters, determine all possible values for the number of dimes that he could have.\newlineAnswer:
  1. Determine Total Value: Let's first determine the total value of the 33 quarters Miguel has.\newlineValue of 11 quarter = $0.25\$0.25\newlineValue of 33 quarters = 3×$0.253 \times \$0.25
  2. Calculate Value of Quarters: Now, calculate the value of the 33 quarters.\newlineValue of 33 quarters = 3×($0.25)=$0.753 \times (\$0.25) = \$0.75
  3. Find Maximum Value of Dimes: Subtract the value of the 33 quarters from the maximum total value to find the maximum value that the dimes can have.\newlineMaximum value of dimes = $2.70\$2.70 - $0.75\$0.75
  4. Calculate Maximum Value of Dimes: Calculate the maximum value of the dimes.\newlineMaximum value of dimes = $2.70\$2.70 - $0.75\$0.75 = $1.95\$1.95
  5. Find Maximum Number of Dimes: Now, let's find out the maximum number of dimes Miguel can have. Since each dime is worth $0.10\$0.10, we divide the maximum value of dimes by the value of one dime.\newlineMaximum number of dimes = $1.95$0.10\frac{\$1.95}{\$0.10}
  6. Calculate Maximum Number of Dimes: Calculate the maximum number of dimes.\newlineMaximum number of dimes = $1.95\$1.95 / $0.10\$0.10 = 19.519.5\newlineSince Miguel cannot have half a dime, the maximum number of whole dimes he can have is 1919.
  7. Find Minimum Number of Dimes: We know Miguel has at least 1818 coins in total. Since he already has 33 quarters, let's find the minimum number of dimes he must have to reach at least 1818 coins.\newlineMinimum number of dimes =18= 18 total coins −3- 3 quarters
  8. Calculate Minimum Number of Dimes: Calculate the minimum number of dimes.\newlineMinimum number of dimes = 18−3=1518 - 3 = 15
  9. Determine Range of Dimes: Now we have the range for the number of dimes Miguel can have. He must have at least 1515 dimes and at most 1919 dimes. However, we need to ensure that the total number of coins does not exceed 1818. If Miguel has 1919 dimes, the total number of coins would be 1919 dimes ++ 33 quarters == 2222 coins, which is more than 1818. Therefore, 1919 dimes is not a possible value.
  10. Check 1919 Dimes Possibility: We need to find the highest number of dimes Miguel can have without exceeding 1818 total coins. Since we already know that 1919 dimes is too many, let's check if 1818 dimes is a possibility.\newlineTotal coins with 1818 dimes = 1818 dimes + 33 quarters = 2121 coins, which is still more than 1818. Therefore, 1818 dimes is not a possible value either.
  11. Check 1818 Dimes Possibility: Next, let's check if 1717 dimes is a possibility.\newlineTotal coins with 1717 dimes = 1717 dimes + 33 quarters = 2020 coins, which is still more than 1818. Therefore, 1717 dimes is not a possible value either.
  12. Check 1717 Dimes Possibility: Finally, let's check if 1616 dimes is a possibility.\newlineTotal coins with 1616 dimes = 1616 dimes + 33 quarters = 1919 coins, which is still more than 1818. Therefore, 1616 dimes is not a possible value either.
  13. Check 1616 Dimes Possibility: Since we have already determined that the minimum number of dimes Miguel must have is 1515, and we have checked all values up to 1919 dimes and found them to exceed the total number of coins allowed, the only possible value for the number of dimes that Miguel could have is 1515.

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