Merle's spare change jar has exactly 16 U.S. coins, each of which is a

Do we need also to prove that the remaining 225 value can be reached with a combination of seven 5-25-50 coins? What would happen if its not possible?

A quicker way to solve for option B->

There have to be at least 3 1 cent coins to make it to 288 (since 288/5 gives a remainder of 5). Hence 3 and 6 is one combination.
Now the next combination can only be 8 1 cent coins to make it to 288 (if you remove that 5), but that case is not possible, since 8 (1 cent coins) +16(10 cent coints)=24, which is more than 16(total number of coins)

1) Total values of these 3 types of coins : 60 + 125 + 100 = 285. Among the remaining (16- 13) = 3 coins, each has to be 1 cent coin, otherwise total value of coins in the jar will be more than 288 cents. Sufficient

2) We have to check whether any unique combo of coins is possible. (4)10 cent +(2)1 cent = 42. Remaining = 246 cents (10 coins). Not possible as any combo will result in a units digit of 5 or 0. Only possible no of 1 cents is either 3 or 8, but 8 is not possible as the total no of 10 cents cannot be 16. So the jar has 3 one cent coins. Sugfficient.

D is the answer

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.

The coins above account for 13 coins and its value is 285. We have 3 coins left and 3 cents remaining -- we can conclude the remaining 3 coins are 1-cent coins. SUFFICIENT.

(2) Among the 16 coins in the jar there are twice as many 10-cent coins as 1-cent coins.

We can have three 1-cent coins, eight 1-cent coins, 13 1-cent coins, etc. Why? Because if we had four 1-cent coins we would not arrive at 288 U.S. cents.

So we can have three 1-cent coins and six 10-cent coins.

Can we have eight 1-cent coins and 16 10-cent coins? No, because that would total 24 coins. SUFFICIENT.

Answer is D.

Hi, the question should be amended. Currently is not very clear. When we say that "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" we are saying that there is at least one coin of each denomination. The question should say each of which COULD be.

This is a tricky question. Do it only if you have sufficient time in the window. Otherwise, make a guess and move on.

I could be missing something here, but I am not sure why this is a 800+ level problem?

The first thing I noticed is the given denominations (which are not 1) end in 5 or 0. So, if we didn't have any ones, the sum of the values will be xx5 or xx0. This means the number of 1s should be 3 (or 13, or 23- not going to happen but you get the point) or 8 (or 18 ... so on).

Statement 1: if we add up the given values, we will end up with 13 coins worth 285. So, 3 coins remaining should be worth 3, i.e. 1 each. Pretty simple. A is sufficient.
Statement 2: number of 10s = 2x the number of 1s. As discussed above, number of 1s can be 3, 8, ... this means number of 10s can be 6, 16 ... hold on a minute. Can it be 16? Then the number of 1s and 10s alone would be 8 + 16 = 24 (which is way more than the limit 16). So the only possible value of number of 1s is 3. There you go, B is sufficient as well.

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.