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Question

The United States Mint produces coins in 1 -cent, 5 -cent, 10 -cent, 25 -cent, and 50 -cent denominations. If a jar contains exactly 100 cents worth of these coins, which of the following could be the total number of coins in the jar? I. 91 II. 81 III. 76

Answer Choices

A. $I only$
B. $II only$
C. $III only$
D. $I and III only$
E. $I, II, and III$

Steps

Explanation	Calculations	Help
$We name variables for the number of coins in each denomination: "1-cent, 5-cent, 10-cent, 25-cent, and 50-cent denominations" and write one equation for the total count and one for the total value.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = n c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100$
$We test option I by setting the total number of coins to 91 and subtracting the count equation from the value equation using Simultaneous Equations - Combination . This leads to an extra cents equation that implies there must be exactly one dime, so option I is possible.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = 91 c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100 4 c_{5} + 9 c_{10} + 24 c_{25} + 49 c_{50} = 9 c_{10} = 1$	Theory & Tactics Method Card EQUA3-B Click to view full details
$We test option II by setting the total number of coins to 81 and performing the same subtraction using Simultaneous Equations - Combination . This yields an extra cents equation with no integer solution for the other coins, so option II is not possible.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = 81 c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100 4 c_{5} + 9 c_{10} + 24 c_{25} + 49 c_{50} = 19$	Theory & Tactics Method Card EQUA3-B Click to view full details
$We test option III by setting the total number of coins to 76 and subtracting the count equation from the value equation using Simultaneous Equations - Combination . This gives an extra cents equation implying exactly one quarter, so option III is possible.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = 76 c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100 4 c_{5} + 9 c_{10} + 24 c_{25} + 49 c_{50} = 24 c_{25} = 1$	Theory & Tactics Method Card EQUA3-B Click to view full details

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Final Answer

I and III only

Question

The United States Mint produces coins in 1 -cent, 5 -cent, 10 -cent, 25 -cent, and 50 -cent denominations. If a jar contains exactly 100 cents worth of these coins, which of the following could be the total number of coins in the jar? I. 91 II. 81 III. 76

Answer Choices

A. $I only$
B. $II only$
C. $III only$
D. $I and III only$
E. $I, II, and III$

Steps

Explanation	Calculations	Help
$We name variables for the number of coins in each denomination: "1-cent, 5-cent, 10-cent, 25-cent, and 50-cent denominations" and write one equation for the total count and one for the total value.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = n c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100$
$We test option I by setting the total number of coins to 91 and subtracting the count equation from the value equation using Simultaneous Equations - Combination . This leads to an extra cents equation that implies there must be exactly one dime, so option I is possible.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = 91 c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100 4 c_{5} + 9 c_{10} + 24 c_{25} + 49 c_{50} = 9 c_{10} = 1$	Theory & Tactics Method Card EQUA3-B Click to view full details
$We test option II by setting the total number of coins to 81 and performing the same subtraction using Simultaneous Equations - Combination . This yields an extra cents equation with no integer solution for the other coins, so option II is not possible.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = 81 c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100 4 c_{5} + 9 c_{10} + 24 c_{25} + 49 c_{50} = 19$	Theory & Tactics Method Card EQUA3-B Click to view full details
$We test option III by setting the total number of coins to 76 and subtracting the count equation from the value equation using Simultaneous Equations - Combination . This gives an extra cents equation implying exactly one quarter, so option III is possible.$	$c_{1} + c_{5} + c_{10} + c_{25} + c_{50} = 76 c_{1} + 5 c_{5} + 10 c_{10} + 25 c_{25} + 50 c_{50} = 100 4 c_{5} + 9 c_{10} + 24 c_{25} + 49 c_{50} = 24 c_{25} = 1$	Theory & Tactics Method Card EQUA3-B Click to view full details

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Final Answer

I and III only