document.write( "Question 1152784: Roger has 27 standard U.S. coins worth a total of $3.65. No coin is worth more than 30¢ or less
\n" ); document.write( "than 3¢. How many dimes could Roger have? (NOTE: There are several possible answers, but
\n" ); document.write( "you only need to give TWO correct answers.) \n" ); document.write( "
Algebra.Com's Answer #774900 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
Roger has 27 standard U.S. coins worth a total of $3.65. No coin is worth more than 30¢ or less
\n" ); document.write( "than 3¢. How many dimes could Roger have? (NOTE: There are several possible answers, but
\n" ); document.write( "you only need to give TWO correct answers.)
\n" ); document.write( "
The least number of VARIABLES you have, the easier these types of problems will be to correctly answer.\r\n" );
document.write( "\r\n" );
document.write( "As these are standard US coins and none is > 30c (.30) and < 3c (.03), the coins MUST consist of quarters, dimes, and nickels.\r\n" );
document.write( "\r\n" );
document.write( "By letting the number of quarters and dimes be Q and D, respectively, the NUMBER of nickels = 27 - Q - D, and the VALUE of the nickels would be: 3.65 - .25Q - .1D\r\n" );
document.write( "We then get: 3.65 - .25Q - .1D = .05(27  -  Q  -  D)\r\n" );
document.write( "3.65 - .25Q - .1D = 1.35 - .05Q - .05D\r\n" );
document.write( "3.65 - 1.35 = - .05Q + .25Q - .05D + .1D\r\n" );
document.write( "2.3 = .2Q + .05D\r\n" );
document.write( "46 = 4Q + D ------ Multiplying the above by 20 to get rid of decimals\r\n" );
document.write( "46  -  4Q = D\r\n" );
document.write( "\r\n" );
document.write( "From the above equation, Q CANNOT be 12, so Q MUST be ≤ 11, or ≥ 0 ====> 0 ≤ Q ≤ 11.\r\n" );
document.write( "\r\n" );
document.write( "By letting Q, or number of quarters be 11, we get the \"highlight%28matrix%281%2C3%2C+number%2C+of%2C+dimes%29%29\" and nickels as \"highlight%282%29\", and 14, respectively.\r\n" );
document.write( "\r\n" );
document.write( "By letting Q, or number of quarters be 10, we get the \"highlight%28matrix%281%2C3%2C+number%2C+of%2C+dimes%29%29\" and nickels as \"highlight%286%29\", and 11, respectively.\r\n" );
document.write( "\r\n" );
document.write( "Therefore, it can be deduced that as the number of quarters reduces by 1, the number of nickels reduces by 3, while the number of dimes increases by 4.\r\n" );
document.write( "\r\n" );
document.write( "From above, you already have the 2 possible numbers of dimes that he could have, but altogether, there are a total of 5 different scenarios and if you go\r\n" );
document.write( "by the statement above that I made about the alteration in the number of coins, you can easily find the other 3, if you so wish!\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "BE AWARE that you CANNOT have - 1 (< 0) nickels, which will produce 28 coins (CONTRADICTORY to the given info.)
\n" ); document.write( "
\n" );