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At a particular store, candy bars are normally priced at $1. [#permalink]
13 Jan 2013, 02:16

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Difficulty:

95% (hard)

Question Stats:

32% (03:40) correct
68% (01:58) wrong based on 339 sessions

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?

Re: At a particular store, candy bars are normally priced at $1. [#permalink]
15 Jan 2013, 05:37

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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.

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.

Re: At a particular store, candy bars are normally priced at $1. [#permalink]
04 Jul 2013, 11:45

1

This post received KUDOS

Expert's post

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.

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'

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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