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At a particular store, candy bars are normally priced at $1.

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At a particular store, candy bars are normally priced at $1. [#permalink] New post 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] New post 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.

Hence B is sufficient

Hope it's clear
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] New post 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] New post 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] New post 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 :( :?
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] New post 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'
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] New post 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] New post 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
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Re: At a particular store, candy bars are normally priced at $1. [#permalink] New post 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

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


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] New post 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.
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Re: At a particular store, candy bars are normally priced at $1.   [#permalink] 13 Aug 2014, 03:38
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