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trex16864
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At a particular store, candy bars are normally priced at $1. 13 Jan 2013, 03:16
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B
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95% (hard)

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35% (03:00) correct 65% (03:04) wrong based on 783 sessions

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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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B
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At a particular store, candy bars are normally priced at $1. 15 Jan 2013, 06:37
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trex16864
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.
Show more


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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At a particular store, candy bars are normally priced at $1. 27 Jun 2013, 21:26
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Can this problem be turned into an algebraic expression?
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avinashrao9
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At a particular store, candy bars are normally priced at $1. 04 Jul 2013, 10:34
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trex16864
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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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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At a particular store, candy bars are normally priced at $1. 04 Jul 2013, 12:45
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avinashrao9

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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trex16864
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.
Show more

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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At a particular store, candy bars are normally priced at $1. 03 Nov 2013, 10:02
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trex16864
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.
Show more


This is written incorrectly, in the actual question (2) states 'D IS divisible by 3'
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At a particular store, candy bars are normally priced at $1. 10 Nov 2013, 20:19
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hfbamafan
Can this problem be turned into an algebraic expression?
Show more

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.

Matthew
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At a particular store, candy bars are normally priced at $1. 25 Jan 2014, 11:18
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Quote:
D=\frac{3}{4}N + \frac{1}{4} (when N is odd)
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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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At a particular store, candy bars are normally priced at $1. 27 Jan 2014, 02:10
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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 \(D=\frac{3(N-1)}{4}+ 1\)
Show more

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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At a particular store, candy bars are normally priced at $1. 13 Aug 2014, 04:38
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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.
:-D
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At a particular store, candy bars are normally priced at $1. 07 Jul 2016, 01:18
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trex16864
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.
Show more


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.

Answer (B)
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At a particular store, candy bars are normally priced at $1. 07 May 2020, 20:09
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trex16864
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.
Show more

From the stem, we can say that the pattern of spending on candy bar will be like: 1+0.50+1+.05+1+....... Statement 1 says the amount spent by Rajiv is a prime. So he can spend $3(here N =4), or 7 (N=9). Not sufficient
Stmnt 2 says D is not divisible by 3. Rajiv can buy 2 bars or 3 bars. Not sufficient

Together, D is a prime number greater than 3. We can try to make a pattern of the spending. For every 2k bars, D will be 1.5k. So for every 4 bar D will be 3k. To make D prime and odd, N always has to be odd.
C is the answer
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At a particular store, candy bars are normally priced at $1. 05 Nov 2024, 15:39
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