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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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04 Jul 2013, 09:34
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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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03 Nov 2013, 09:02
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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.

This is written incorrectly, in the actual question (2) states 'D IS divisible by 3'
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25 Jan 2014, 10:18
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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
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13 Aug 2014, 03:38
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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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07 Jul 2016, 00:18
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Expert's post
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.

Responding to a pm:

In the question stem, what does "D and N are integers" imply?

This is how the total cost progresses with each new candy bought:

$1 -$1.50 - $2.50 -$3
$4 -$4.50 - $5.50 -$6
$7 -$7.50 - $8.50 -$9
...

Note that we have integer cost whenever we buy candies in multiples of 4 or 1 more than a multiple of 4.
The total cost is a multiple of 3 for every multiple of 4 total candies (N is even) bought.
It is 1, 4, 7, 10, 13 ... etc for every 4a+1 (N is odd) candies bought.

Question: Is N odd?
If N is odd, D = 1 or 4 or 7 or 10 etc
If N is even, D = 3, or 6 or 9 ...

(1) D is prime.
D can be 3 or 7. In one case, N is even, in the other it is odd.
Not sufficient.

(2) D is not divisible by 3.
D cannot be 3, 6, 9 etc. So N is not even. N must be odd.
Sufficient.

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