Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

1

This post received KUDOS

3

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

32% (03:43) correct
68% (01:58) wrong based on 332 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

1

This post received KUDOS

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, 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]
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]
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.