Billy has an unlimited supply of the following coins: pennie

Question

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?

a)8¢
b)13¢
c)40¢
d)53¢
e)66¢

I was not able to understand the question at first go.Please do share your understanding along with the approach.

Thanks

A. 8¢
B. 13¢
C. 40¢
D. 53¢
E. 66¢

I was not able to understand the question at first go.Please do share your understanding along with the approach.

Thanks

KarishmaB · Accepted Answer

Since we have coins of denomination 1, 5, 10, 25 and 50 and we are supposed to use only 4 coins, my first thought is that I cannot make a sum which ends in a 4 or a 9 (except 4 itself). To make a sum ending in 4 or 9, I would need four coins of 1¢ and some more coins to make whatever is left e.g. to make 9, we need 5+1+1+1+1
So we cannot make 8*3 = 24
13*3 = 39
53 and 66 are anyway out as discussed above. The only option left is 40.

g106 · Answer

Billy bought 5 candies with 4 coins on Friday . So maximum amount paid for 5 candies on Friday = 200¢

From this we can eliminate 53 ¢ and 66¢.

in the remaining options if we just count the possibilities 40¢ is the only possible answer.

LalaB · Answer

hm not sure that I am right , but still
i began with answer choices-
a)the price of one candy is 8¢
then

8*1 candy then 4 coins could be - 5 ; 1; 1; 1
8*2(16) candies then 4 coins could be -5;5;5;1
8*3(24) candies then no chances to divide coins properly 
so A is out
B) the price of one candy is 13 then 4 coins could not be divided properly among 1;5;10;25;50 so B is out

C) the price of one candy is 40 
then
40*1 candy then 4 coins could be 10;10;10;10
40*2 candies (80) then 4 coins could be 50;10;10;10
40*3 candies (120) then 4 coins could be 50;50;10;10
40*4 candies(160) then 4 coins could be 50;50;50;10
40*5 candies(200) then 4 coins could be 50;50;50;50

hope it helps and my way of thinking is ok hehe

gmatcracker24 · Answer

Thanks Karishma for providing a clear explanation.

Bowtie · Answer

you can elimnate d and e because the cost of 5 candies at those prices are too high.

siddharthmuzumdar · Answer

I have a doubt about this question.
It is clear that options D and E are out.
For options A, B and C, do we have to try out all possible combinations within 2 minutes? Unless I start with C, I could end up wasting a lot over 2 minutes. The last digit approach suggested by Karishma will come in handy but that will only be handy after I have already started enlisting the combinations and fitting in the possible values. Is there another way to handle this?

KarishmaB · Answer

No, you don't need to enlist the combinations or fit any values.

Coins available: 1, 5, 10, 25, 50
You have to use exactly 4 coins.
So you cannot make anything ending with 4 or 9 (except 4 itself)

Can you make 8, 16, 24, 32 and 40 (just the first 5 consecutive multiples of 8. You don't even need to write it down.) using 4 coins? One thing we know for sure is that we cannot make 24. So we don't need to check anything else.

Can you make 13, 26, 39, 52 and 65 using 4 coins? Again, we cannot make 39 since it ends with 9.

Once you work out that anything ending in 4 or 9 doesn't work for you, it will take you half a minute to think the solution through.

The only other approach I can think of is to check each option and each value which is extremely time consuming. I don't see a straight forward and quick algebraic approach to this problem.

siddharthmuzumdar · Answer

Thanks you Karishma. This is really helpful.

subhajeet · Answer

I did this by going the reverse methos and got the answer as 40. Dont know if this will work out in the GMAT.

mbaiseasy · Answer

I tested each value if it's possible to pay with 4 coins for 1,2,3,4, and 5 candies.

A. 8 cents

1 candy - 8 cents - 5,1,1,1
2 candies - 16 cents - 5,5,5,1
3 candies - 24 cents - 10,10,1,1,1,1 (No) - Going with a bigger coin is also not possible. A is OUT!

B. 13 cents

1 candy - 13 cents - 10,1,1,1
2 candies - 26 cents - 10,10,5,1
3 candies - 39 cents - 10,10,10,5,1,1,1,1 (No) - 25,10,1,1,1,1 (No)... B is OUT!

C. 40 cents

1 candy - 40 cents - 10,10,10,10
2 candies - 80 - 50,10,10,10
3 candies - 120 - 50,50,10,10
4 candies - 160 - 50,50,50,10
5 candies - 200 - 50,50,50,50

BINGO!

Answer: C

sri30kanth · Answer

Karishma,

How are you zeroing on the values 4 and 9 directly? Unable to understand it. Can we use LCM concept in this question?

anairamitch1804 · Answer

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.

mbaapp1234 · Answer

It's probably best to quickly test cases here.

A) 8 is possible with four coins, 16 with four coins, but 24 is not. Eliminate A.

B) 13 is possible with 4 coins, 26 with four coins, but 39 is not. Eliminate B.

C) 40 is possible with 4 coins, 80 is possible with 4 coins, and 120 is possible with four coins. At this point, given the time constraint, it's probably best to pick C and move on.

ravish844 · Answer

Pls note that when we have to exactly pay by 4 coins every time, and if we use 4 pennies (1¢), then only we get 4 in the end.
Now, if we use any 1 other coin, each of which is ending with 0 or 5, we can use max of 3 pennies (1¢), because we have already used 1 coin other than penny.
That means, we can get 1, or 2 or max we will get is 3, for 3 pennies used, depending on the number of pennies used for only 1 coin used which is ending with 0.
Similarly, we will get 6, or 7, or max we will get is 8, for 3 pennies used, depending on the number of pennies used for only 1 coin used which is ending with 5. 
This eliminates the use of a coin total ending with 4 or 9. Hence the quick elimination...

Hope this help and clears the doubt, how we zero-in on 4 and 9.
I am not sure how to use the LCM concept or if it is applicable in this particular question.

Thanks a lot Karishma Ma'm, for your wonderful insight, it got me thinking really and finally even I understood how.

Aditi10 · Answer

I've been hanging around this page for months to prepare for GMAT but VeritasPrepKarishma 's solution here just blew my mind. Usually, you read solutions and have an "ah" moment but it took me some time to just understand the logic here. Probably the most intelligent solution I've seen in GMATClub. Thanks! You're making some of us smarter than we were.

puneetfitness · Answer

Hi Karishma please advise me on can we use logic that as every time Billy is buying candy with exactly 4 coins the. The price of each coin must me divisible by four and then we can rule out 13,53,66

And then we are left with 8 and 40

These we can try this way

8*4= 32= 10+10+10+1+1 not valid

40*4=160= 50+50+50+10

Posted from my mobile device

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Billy has an unlimited supply of the following coins: pennie

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