GMAT Number Properties Explained — United States Mint produces coins in 1 -cent, 5 -cent, 10 -cent,

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

Options:

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

simultaneous equationsnumber theory

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