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# Of the N candies in a bag, some are peppermint and the rest are

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Manager
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Joined: 06 Dec 2012
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Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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26 Mar 2013, 07:14
2
15
00:00

Difficulty:

35% (medium)

Question Stats:

78% (02:01) correct 22% (02:04) wrong based on 242 sessions

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Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1) If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2) If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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26 Mar 2013, 10:06
3
mun23 wrote:
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1)If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2)If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.

Need help

The number of Peppermint candies = x, Of Spearmint= N-x.

From F.S 1 , we have (N-1)/5 = (x-1)

or 5x - N = 4. Insufficient.

From F.S 2, we have (N-2)/4 = x

or N - 4x = 2. Insufficient.

Taken together, we can get a particular value for N. Sufficient.

C.
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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16 May 2013, 10:52
mun23 wrote:
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1)If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2)If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.

Need help

Stmt1: 2 unknowns one equation
stmt2 : 2 unknowns and one equation

combine both 2 equations and 2 unknowns so C
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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16 Aug 2013, 05:22
mun23 wrote:
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1)If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2)If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.

Need help

can any body help me here, may be I am missing something but even with using both the statements I am not able to get the value of N

P +S= N

1) $$\frac{1(N-1)}{5}= P-1 \rightarrow N= 5P-4$$

2) $$\frac{1(N-2)}{4} = S-2 \rightarrow N= 4S-6$$

Now can anybody show me how to get N ?

Thanks
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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16 Aug 2013, 05:27
1
stne wrote:
mun23 wrote:
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1)If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2)If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.

Need help

can any body help me here, may be I am missing something but even with using both the statements I am not able to get the value of N

P +S= N

1) $$\frac{1(N-1)}{5}= P-1 \rightarrow N= 5P-4$$

2) $$\frac{1(N-2)}{4} = S-2 \rightarrow N= 4S-6$$

Now can anybody show me how to get N ?

Thanks

The correct equation for the second fact statement would be:

$$\frac{1(N-2)}{4} = P$$

Hope this helps
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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16 Aug 2013, 05:51
mau5 wrote:
stne wrote:
mun23 wrote:
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1)If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2)If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.

Need help

can any body help me here, may be I am missing something but even with using both the statements I am not able to get the value of N

P +S= N

1) $$\frac{1(N-1)}{5}= P-1 \rightarrow N= 5P-4$$

2) $$\frac{1(N-2)}{4} = S-2 \rightarrow N= 4S-6$$

Now can anybody show me how to get N ?

Thanks

The above part involves peppermints, not spearmints.

Hope this helps

Thundering Typhoons! Cue for me to take a break.
Thank you +1

1) $$\frac{1(N-1)}{5}= P-1 \rightarrow N= 5P- 4$$ ..1
2) $$\frac{1(N-2)}{4} =P \hs{15} \rightarrow N= 4P+2$$ ..2

4P+2 =5P-4
p=6 ( Put in 1 or 2 )

Then N= 26
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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18 Jun 2015, 06:46
I put E for this question because I mismanaged to translate the word problem into math.

in (2) I couldnt get all the variables to match into one equation. Why is "S" never part of the equation?
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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15 Oct 2016, 16:17
2
1
This problem can easily be solved by following the rule that states that to solve for a variable we need the same number of distinct, linear equations as variables.
I hope you find my approach "elegant"

Question (Given)
N = p + s
N = ?
We have 3 variables and 1 equation

Statement 1
(p - 1)/s = 1/4
When we combine this statement with the given info, we yield 3 variables and 2 equations. Insufficient

Statement 2
(s - 2)/p = 3/1
When we combine this statement with the given info, we yield 3 variables and 2 equations. Insufficient

Therefore, we need to combine both statements to get 3 variables and 3 equations. Of course, no need to do the actual calculation.

Bonus Track: In case you REALLY want to know the value of variables: p = 6 and s = 20, thereby N = 26.
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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13 Nov 2017, 05:56
I am having an issue translating Statement 2 for some reason. I am fine on S1

On Statement 2, could I not do $$\frac{(s-2)}{(n-2)}$$ = $$\frac{3}{4}$$ (in that 3/4 of remaining candies are Spearmint)?
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Re: Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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13 Nov 2017, 06:17
1
okay wrote:
I am having an issue translating Statement 2 for some reason. I am fine on S1

On Statement 2, could I not do $$\frac{(s-2)}{(n-2)}$$ = $$\frac{3}{4}$$ (in that 3/4 of remaining candies are Spearmint)?

yes you can.. it also leads to same..

$$\frac{(s-2)}{(n-2)}$$ = $$\frac{3}{4}$$..
now s-2=(n-2)-p..
substitute
$$\frac{(s-2)}{(n-2)}=\frac{(n-2)-p}{n-2}=\frac{n-2}{n-2}-\frac{p}{n-2}=1-\frac{p}{n-2}=\frac{3}{4}$$...
$$\frac{p}{n-2}=1-\frac{3}{4}=\frac{1}{4}$$
now these is same as the statement II
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Of the N candies in a bag, some are peppermint and the rest are  [#permalink]

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08 Oct 2019, 04:41
mun23 wrote:
Of the N candies in a bag, some are peppermint and the rest are spearmint.What is the value of N?

(1) If 1 peppermint candy were removed from the N candies,1/5 of the remaining candies would be peppermint.

(2) If 2 spearmint candies were removed from the N candies, 1/4 of the remaining candies would be peppermint.

N = P + S = ?

1) (P - 1)/S = 1/4
4P - S = 4
Not Sufficient

2) P/(S-2) = 1/3
3P - S = -2
Not Sufficient

1+2)
2 equations and 2 unknowns. Can be solved to get P + S = 6 + 20 = 26
Sufficient

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Of the N candies in a bag, some are peppermint and the rest are   [#permalink] 08 Oct 2019, 04:41
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