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Billy has an unlimited supply of the following coins: pennie [#permalink]
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24 Oct 2011, 23:19
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75% (01:58) correct 25% (03:36) wrong based on 261 sessions
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Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and halfdollars (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
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Re: Algebra question [#permalink]
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24 Oct 2011, 23:40
gmatcracker24 wrote: Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and halfdollars (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 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.
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Re: Algebra question [#permalink]
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25 Oct 2011, 00:28
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
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Re: Algebra question [#permalink]
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25 Oct 2011, 21:31
gmatcracker24 wrote: Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and halfdollars (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 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.
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Re: Algebra question [#permalink]
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26 Oct 2011, 09:36
VeritasPrepKarishma wrote: gmatcracker24 wrote: Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and halfdollars (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 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. Thanks Karishma for providing a clear explanation.
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Re: Algebra question [#permalink]
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16 Nov 2011, 14:42
you can elimnate d and e because the cost of 5 candies at those prices are too high.



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Re: Algebra question [#permalink]
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05 Jan 2012, 02:30
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?
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Re: Algebra question [#permalink]
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05 Jan 2012, 21:31
siddharthmuzumdar wrote: 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? 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.
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Re: Algebra question [#permalink]
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05 Jan 2012, 21:38
Thanks you Karishma. This is really helpful.
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Re: Algebra question [#permalink]
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05 Jan 2012, 21:42
I did this by going the reverse methos and got the answer as 40. Dont know if this will work out in the GMAT.



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Re: Billy has an unlimited supply of the following coins: pennie [#permalink]
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02 Dec 2012, 00:25
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
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Re: Billy has an unlimited supply of the following coins: pennie [#permalink]
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29 Aug 2014, 07:11
Karishma,
How are you zeroing on the values 4 and 9 directly? Unable to understand it. Can we use LCM concept in this question?



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Re: Billy has an unlimited supply of the following coins: pennie [#permalink]
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11 Jan 2017, 06:56
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.
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Re: Billy has an unlimited supply of the following coins: pennie [#permalink]
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29 May 2017, 16:42
gmatcracker24 wrote: Billy has an unlimited supply of the following coins: pennies (1¢), nickels (5¢), dimes (10¢), quarters (25¢), and halfdollars (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 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.



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Re: Billy has an unlimited supply of the following coins: pennie [#permalink]
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20 Jan 2018, 17:51
sri30kanth wrote: Karishma,
How are you zeroing on the values 4 and 9 directly? Unable to understand it. Can we use LCM concept in this question? 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 zeroin 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.




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