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# At a particular store, candy bars are normally priced at $1.  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Intern Joined: 18 Dec 2012 Posts: 5 Followers: 0 Kudos [?]: 10 [1] , given: 11 At a particular store, candy bars are normally priced at$1. [#permalink]  13 Jan 2013, 02:16
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At a particular store, candy bars are normally priced at $1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for$0.50. A third candy bar would cost $1.00, a fourth would cost$0.50, and so on.

If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?

(1) D is prime.

(2) D is not divisible by 3.
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] 15 Jan 2013, 05:37 1 This post received KUDOS 1 This post was BOOKMARKED trex16864 wrote: At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on. If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd? (1) D is prime. (2) D is not divisible by 3. 1 candy cost 1 2 candies cost 1+.50=1.50 ( here D is not an integer, hence we cannot buy 2 candies . so we can reject all cases where D is non Integer) 3 candies cost 1.50 +1 =2.50 4 candies cost 2.50+.50= 3 5 candies cost 3+1= 4 6 candies cost 4+.50= 4.50 7 candies cost 4.50+1=5.50 8 candies cost 5.50.+.50= 6 9 candies cost 6+1= 7 ..... 13 candies cost =10 (i) D is prime D=3 and N=4 (N is even) D=7 N=9 (N is odd ) not sufficient (ii) D is not Divisible by 3 D=1 N=1 D=4 N =5 D=7 N=9 D=10 N=13 so we see if D is not divisible 3 then N is always odd. Hence B is sufficient Hope it's clear _________________ - Stne Manager Joined: 29 Mar 2010 Posts: 143 Location: United States Concentration: Finance, International Business GMAT 1: 590 Q28 V38 GPA: 2.54 WE: Accounting (Hospitality and Tourism) Followers: 1 Kudos [?]: 57 [0], given: 16 Re: At a particular store, candy bars are normally priced at$1. [#permalink]  27 Jun 2013, 20:26
Can this problem be turned into an algebraic expression?
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] 04 Jul 2013, 09:34 stne wrote: trex16864 wrote: At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on. If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd? (1) D is prime. (2) D is not divisible by 3. 1 candy cost 1 2 candies cost 1+.50=1.50 ( here D is not an integer, hence we cannot buy 2 candies . so we can reject all cases where D is non Integer) 3 candies cost 1.50 +1 =2.50 4 candies cost 2.50+.50= 3 5 candies cost 3+1= 4 6 candies cost 4+.50= 4.50 7 candies cost 4.50+1=5.50 8 candies cost 5.50.+.50= 6 9 candies cost 6+1= 7 ..... 13 candies cost =10 (i) D is prime D=3 and N=4 (N is even) D=7 N=9 (N is odd ) not sufficient (ii) D is not Divisible by 3 D=1 N=1 D=4 N =5 D=7 N=9 D=10 N=13 so we see if D is not divisible 3 then N is always odd. Hence B is sufficient Hope it's clear Is there any way to do this problem within 2 mins. Writing out all the values takes time and one is bound to make mistakes. It took almost 4 mins for me to complete Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 629 Followers: 64 Kudos [?]: 776 [1] , given: 135 Re: At a particular store, candy bars are normally priced at$1. [#permalink]  04 Jul 2013, 11:45
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avinashrao9 wrote:
Is there any way to do this problem within 2 mins.
Writing out all the values takes time and one is bound to make mistakes.

It took almost 4 mins for me to complete

trex16864 wrote:
At a particular store, candy bars are normally priced at $1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for$0.50. A third candy bar would cost $1.00, a fourth would cost$0.50, and so on.

If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd?

(1) D is prime.

(2) D is not divisible by 3.

Any integer can only have 3 values for remainder when divided by 3, namely (0,1,2).
Hence, any integer which is not a multiple of 3 can be represented as $$3*k+1$$ or $$3*k+2$$, for some positive integer k(k=0 for 1 and 2).

Also,for D=1,N=1(odd),D=3,N=4(even).

Hence,any spending which is a multiple of 3-->$$3*k$$ will always yield --> even # of candy bars(as it is a multiple of 4)

Any spending in the form $$3*k+1$$--> # of bars is $$even+1 -->odd$$.

From F.S 1, for D = 7 , we can represent 7 as $$3*2+1$$ --> # of bars is $$4*2+1$$= 9 bars(odd)

Again, for D = 3 dollars, we anyways know that N=4(even). Thus, as we get both possibilities,this statement is Insufficient.

From F.S 2: As we know that D is not divisible by 3, he would always get an odd no of bars as discussed above.Sufficient.

Hope this helps.

B.
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] 03 Nov 2013, 09:02 trex16864 wrote: At a particular store, candy bars are normally priced at$1.00 each. Last week, the store offered a promotion under which customers purchasing one candy bar at full price could purchase a second candy bar for $0.50. A third candy bar would cost$1.00, a fourth would cost $0.50, and so on. If, in a single transaction during the promotion, Rajiv spent D dollars on N candy bars, where D and N are integers, is N odd? (1) D is prime. (2) D is not divisible by 3. This is written incorrectly, in the actual question (2) states 'D IS divisible by 3' Intern Joined: 21 Dec 2012 Posts: 6 Followers: 0 Kudos [?]: 9 [0], given: 2 Re: At a particular store, candy bars are normally priced at$1. [#permalink]  10 Nov 2013, 19:19
hfbamafan wrote:
Can this problem be turned into an algebraic expression?

Hey bamafan,

You can turn this into a system of equations as follows:

$$D=\frac{3}{4}N$$ (when N is even)
$$D=\frac{3}{4}N + \frac{1}{4}$$ (when N is odd)

The nice thing about this is you can easily see for N to be an even integer, D must be divisible by three:

$$\frac{4D}{3} = N$$ (when N is even)

So that shows that the second case is sufficient. For the first case the odd formula can be rearranged as follows:

$$\frac{4D-1}{3} = N$$ (when N is odd)

From the first equation, D must be divisible by three to be even. D = 3 is prime and fits this rule, so an even N can be created.
From the second equation, N is whole number if D = 7, 13, etc., so N can also be odd when D is prime. Therefore, the first case is insufficient.

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Re: At a particular store, candy bars are normally priced at $1. [#permalink] 25 Jan 2014, 10:18 Quote: D=\frac{3}{4}N + \frac{1}{4} (when N is odd) Hi Can someone please help me understand how we arrived at this expression for N = odd According to my understanding it should be D=\frac{3(N-1)}{4}+ 1 Math Expert Joined: 02 Sep 2009 Posts: 29210 Followers: 4754 Kudos [?]: 50367 [0], given: 7544 Re: At a particular store, candy bars are normally priced at$1. [#permalink]  27 Jan 2014, 01:10
Expert's post
Rohan_Kanungo wrote:
Quote:
$$D=\frac{3}{4}N + \frac{1}{4}$$ (when N is odd)

Hi

According to my understanding it should be $$D=\frac{3(N-1)}{4}+ 1$$

Both equations are the same: $$D=\frac{3(N-1)}{4}+ 1=\frac{3N}{4}-\frac{3}{4}+1=\frac{3N}{4}+\frac{1}{4}$$.

Hope it's clear.
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] 13 Aug 2014, 03:38 We are given that each odd candy costs$1.00 and each even candy costs $0.50. We can have 2 conditions: Case1: N is even So the total cost of all candies would be (1)*(N/2) + (0.5)*(N/2) = 3N/4 = D Case 2: N is odd Total cost is [(1)*{(N+1)/2} + (0.5)*{(N-1)/2}] = (3N+1)/4 = D St 1: D is prime N=4 (in case 1 where N is even) gives D =3 N=9 (in case 2 where N is odd) gives D = 7 So we get prime values for D from both conditions, hence INSUFFICIENT. St 2: D is not divisible by 3 Case 1 clearly shows D should be divisible by 3. Thus we can reject this. Case 2 clearly shows, D is not divisible by 3. Hence, SUFFICIENT. Therefore, B. Thanks. GMAT Club Legend Joined: 09 Sep 2013 Posts: 6220 Followers: 346 Kudos [?]: 71 [0], given: 0 Re: At a particular store, candy bars are normally priced at$1. [#permalink]  03 Sep 2015, 16:40
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