Problem Solving

Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents?

8
13
40
53
66

OA is C
Does anyone knw a short method to do this? MGMAT has two really complex methods...
thanks

vkb16 wrote:Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and half-dollars (50¢). On Monday, Billy bought one candy for less than a dollar and paid for it with exactly four coins (i.e., he received no change). On Tuesday, he bought two of the same candy and again paid with exactly four coins. On Wednesday, he bought three of the candies, on Thursday four of the candies, and on Friday five of the candies; each day he was able to pay with exactly four coins. Which of the following could be the price of one candy in cents?

8
13
40
53
66

OA is C
Does anyone knw a short method to do this? MGMAT has two really complex methods...
thanks

Before we do a lot of algebra, let's start with some common sense to narrow down the playing field.

On Friday he bought 5 candies with 4 coins. Since the biggest coin he has is 50 cents, 5 candies must be cheaper than 4*50 = 200 cents. Therefore, the maximum price per candy is 40 cents: eliminate D and E.

So, we have:

8
13
40

Now we have a couple of solid options: we could set up some algebraic equations; alternatively, we could work with the answer choices. Since we have a LOT of variables and a LOT of equations, I'm going for door #2.

Let's start with 8 cents:

1 candy (8 cents) for 4 coins: 5, 1, 1, 1
2 candies (16 cents) for 4 coins: 5, 5, 5, 1
3 candies (24 cents) for 4 coins: no way to get the "4" at the end... bzzt! Eliminate A.

13 cents:

1 candy (13 cents) for 4 coins: 10, 1, 1, 1
2 candies (26 cents) for 4 coins: 10, 10, 5, 1
3 candies (39 cents) for 4 coins: no way to get the "9" at the end... bzzt! Eliminate B.

C is the only choice left, so it must be correct.

Just for fun:

40 cents:

1 candy (40 cents) for 4 coins: 10, 10, 10, 10
2 candies (80 cents) for 4 coins: 50, 10, 10, 10
3 candies (120 cents) for 4 coins: 50, 50, 10, 10
4 candies (160 cents) for 4 coins: 50, 50, 50, 10
5 candies (200 cents) for 4 coins: 50, 50, 50, 50

Ding ding!

Now, if we had thought ahead a bit, we would have reasoned that it's going to be far easier to make our payments if the per candy price ends in a "5" or "0", since the only way to get a different units digit is to use pennies. If we had made that deduction, then we would have either just guessed C if we were in a rush or started by testing 40 cents before 8 or 13.

thanks a lot for the explanation!
btw Stuart, can you please explain this a bit>>?

''Before we do a lot of algebra, let's start with some common sense to narrow down the playing field.

On Friday he bought 5 candies with 4 coins. Since the biggest coin he has is 50 cents, 5 candies must be cheaper than 4*50 = 200 cents. Therefore, the maximum price per candy is 40 cents: eliminate D and E.''

why cheaper than 200 cents? we can have 4 coins of 50 cents each as well... Is it because of we had 3 candies @50 each, we cant set up 150 with 50,25,10, and 1 in 4 divisions?

vkb16 wrote:thanks a lot for the explanation!
btw Stuart, can you please explain this a bit>>?

''Before we do a lot of algebra, let's start with some common sense to narrow down the playing field.

On Friday he bought 5 candies with 4 coins. Since the biggest coin he has is 50 cents, 5 candies must be cheaper than 4*50 = 200 cents. Therefore, the maximum price per candy is 40 cents: eliminate D and E.''

why cheaper than 200 cents? we can have 4 coins of 50 cents each as well... Is it because of we had 3 candies @50 each, we cant set up 150 with 50,25,10, and 1 in 4 divisions?

Hi!

Sorry, small typo on my part - I meant to say:

"5 candies can cost at most 4*50 = 200 cents."

That's why I said the maximum price is 40 cents in the next sentence (as opposed to saying it had to be less than 40 cents) and why I didn't eliminate C.

thanks! that made things a lot clearer!

We can back solve this question by using the answer choices. Let's first check to make sure that each of the 5 possible prices for one candy can be paid using exactly 4 coins:
8 = 5+1+1+1
13 = 10+1+1+1
40 = 10+10+10+10
53 = 50+1+1+1
66 = 50+10+5+1

So far we can't make any eliminations. Now let's check two pieces of candy:

16 = 5 + 5 + 5 + 1
26 = 10 + 10 + 5 + 1
80 = 25 + 25 + 25 + 5
106 = 50 + 50 + 5 + 1
132 = 50 + 50 + 25 + 5 + 1 + 1

We can eliminate answer choice E here. Now three pieces of candy:

24 = 10 + 10 + 1 + 1 + 1 + 1
39 = 25 + 10 + 1 + 1 + 1 + 1
120 = 50 + 50 + 10 + 10
159 = 50 + 50 + 50 + 5 + 1 + 1 + 1 + 1.

We can eliminate answer choices A, B and D.
Notice that at a price of 40¢, Billy can buy four and five candies with exactly 4 coins as well:
160 = 50 + 50 + 50 + 10
200 = 50 + 50 + 50 + 50

hence answer is 40 cents.