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

total US coins = 16
types of coins = 1-cent,5-cent,10-cent,25-cent, and 50-cent
total value ofr coins = 288
how many 1-cent coins are in the jar?

STATEMENT (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.
value of 6- 10-cent coin = 10*6 = 60
value of 5- 25-cent coin = 25*5 = 125
value of 2- 50-cent coin = 50*2 = 100
remaining value = total value of all coins - value of these 3 types of coins = 288-285 = 3
now we are left with 1-cent and 5-cent coins
5-cent coins are not possible because value of remaining coins = 3

so, there are 3 -- 1-cent coins in the jar
SUFFICIENT

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

let's write all the possible combinations of 10-cent and 1-cent coins

1-cent coin = 1 then 10-cent coin = 2
value of these coins = 20+1 = 21
value of remaining coins = 288-21 = 267 (not the multiple of 5 or 10)
we cant get 267 from coins--5-cent,25-cent, and 50-cent (not possible)

1-cent coin = 2 then 10-cent coin = 4
value of these coins = 40+2 = 42
value of remaining coins = 288-42 = 246 (not the multiple of 5 or 10)
we cant get 246 from coins--5-cent,25-cent, and 50-cent (not possible)

1-cent coin = 3 then 10-cent coin = 6 (only possible value)
value of these coins = 60+3 = 63
value of remaining coins = 288-63 = 225 (multiple of 5)
we can get 225 from coins--5-cent,25-cent, and 50-cent

1-cent coin = 4 then 10-cent coin = 8
value of these coins = 80+4 = 84
value of remaining coins = 288-84 = 204 (not the multiple of 5 or 10)
we cant get 204 from coins--5-cent,25-cent, and 50-cent (not possible)

1-cent coin = 5 then 10-cent coin = 10
value of these coins = 100+5 = 105
value of remaining coins = 288-105 = 183 (not the multiple of 5 or 10)
we cant get 183 from coins--5-cent,25-cent, and 50-cent (not possible)

1-cent coin = 6 then 10-cent coin = 12 (not possible because total coin is 16)

how many 1-cent coins are in the jar?
--3
SUFFICIENT

D is the answer

Statement 1-
Value of the coins of 10-cent, 25-cent, and 50-cent coins present in the jar= 6*10+5*25*+2*50=285

Value of the coins of 1-cent and 5-cent coins present in the jar= 288-285=3 cents

Only 1 case is possible, that is there are 3 coins of 1-cent and 0 coins of 5 cents in the jar

Sufficient

Statement 2- the total value of the coins in the jar is 288

Value of 5-cents, 10-cent, 25-cent, and 50-cent coins must be a multiple of 5

Hence value of 1-cent
= 288-5k
= 285-5k +3
= 5(57-k) +3
= 5X+3

Hence number of 1-cent can be 3, 8, 13.....so on

Case 1
If number of 1-cent=3, number of 10-cent= 6
This case is possible...(Look at case in statement 1)

Case 2
If number of 1-cent=8, number of 10-cent= 16
Total Number of coins>8+16 (Not possible)

We can clearly see that there is no other case possible except case 1

Sufficient

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.
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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.

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