Question 1152784
Roger	has	27	standard	U.S.	coins	worth	a	total	of	$3.65.	No	coin	is	worth	more	than	30¢	or	less	
than	3¢.	How	many	dimes	could	Roger	have?	(NOTE:	There	are	several	possible	answers,	but	
you	only	need	to	give	TWO	correct	answers.)
<pre>The least number of VARIABLES you have, the easier these types of problems will be to correctly answer.

As these are standard US coins and none is > 30c (.30) and < 3c (.03), the coins MUST consist of quarters, dimes, and nickels.

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
We then get: 3.65 - .25Q - .1D = .05(27  -  Q  -  D)
3.65 - .25Q - .1D = 1.35 - .05Q - .05D
3.65 - 1.35 = - .05Q + .25Q - .05D + .1D
2.3 = .2Q + .05D
46 = 4Q + D ------ Multiplying the above by 20 to get rid of decimals
46  -  4Q = D

From the above equation, Q CANNOT be 12, so Q MUST be ≤ 11, or ≥ 0 ====> 0 ≤ Q ≤ 11.

By letting Q, or number of quarters be 11, we get the {{{highlight(matrix(1,3, number, of, dimes))}}} and nickels as {{{highlight(2)}}}, and 14, respectively.

By letting Q, or number of quarters be 10, we get the {{{highlight(matrix(1,3, number, of, dimes))}}} and nickels as {{{highlight(6)}}}, and 11, respectively.

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.

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
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!


BE AWARE that you CANNOT have - 1 (< 0) nickels, which will produce 28 coins (CONTRADICTORY to the given info.)</pre>