gmatt1476 wrote:
Merle's spare change jar has exactly 16 U.S. coins, each of which is a 1-cent coin, a 5-cent coin, a 10-cent coin, a 25-cent coin, or a 50-cent coin. If the total value of the coins in the jar is 288 U.S. cents, how many 1-cent coins are in the jar?
(1) The exact numbers of 10-cent, 25-cent, and 50-cent coins among the 16 coins in the jar are, respectively, 6, 5, and 2.
(2) Among the 16 coins in the jar there are twice as many 10-cent coins as 1-cent coins.
Statement 1:
Amount rendered by the given coins = 6*10 + 5*25 + 2*50 = 285 cents.
To bring the total to 288 cents, the number of 1-cent coins = 3.
SUFFICIENT.
Statement 2:
Almost no work is needed here.
The combination of coins yielded in Statement 1 --
3 1-cent coins,
0 5-cent coins,
6 10-cent coins,
5 25-cents coins, and 2 50-cent coins -- satisfies the condition that there are twice as many 10-cent coins as 1-cent coins.
The only question is whether this combination can be altered such that the number of coins remains 16 and the total value remains 288 cents.
If we change the value yielded by the 9 red coins, we must increase or decrease by the same amount the value yielded by the 7 blue coins.
Since the blue coins are all multiples of 5, their value can change only by a multiple of 5.
For the red coins to change by a multiple of 5 while remaining in a ratio of 1 to 2, we would need to add or remove at least 5 1-cent coins and 10 10-cent coins, with the result that at least 15 red coins are either removed or added.
Neither option is viable.
Implication:
Only the combination yielded by Statement 1 will satisfy the condition in Statement 2 that there are twice as many 10-cent coins as 1-cent coins.
Thus, the number of 1-cent coins = 3.
SUFFICIENT.