Of the N candies in a Bag some are peppermint some

[ShyT]

I have covered probabilities pretty extensively, but I struggle with Data Sufficiency questions similar to this one, I can't seem to find anything in the study guides either. I recently took the GMAT and I saw 1 or 2 questions very similar to the one below. This is one that I found in the New Gmat Prep Software.

Of the N candies in a bag some are peppermint and rest are Spearmint. What is the Value of N?

1) If 1 peppermint candies 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 remaining candies would be peppermint.

[Sinta]

Hi Shy,

I'll try to answer as best as I can.

*The Q is about ratio, in my opinion, and not about probability.

*Q Stem Data:
Total is n
Peppermint - P
Spearmint - n-p ( ant not S, since this will only add another factor to find)

From 1 we get: (p-1)/(p-n-1) = 1/4
not sufficient - eliminate A,D

From 2 we get: (p-2)/(p-n-2) = 1/3
{BTW once you see that statment 2 gives pretty much the same kind of info as 1 you can make the same conclusion}
not sufficient - eliminate B

From 1+2: we get two equations with two variables. so how much is n?
WHO CARES?? click D, confirm NEXT!

[ShyT]

Hi Sinta,

I'm struggling to understand the explanation? are the variables in the right order in your fractions?

[ShyT]

The way I came to an answer was very similar to yours, but I was hoping someone can show me a more effective route...

statement 1 says

1 Less peppermint and ratio becomes of 1:4 (P:S)

This means that N is a multiple of 5..+1
6
11
16
21
26

INS

Statement 2 says

2 Less spearmints and the ratio becomes P:S is 1:3

So again N= Multiple of 4..+2

6
10
14
18
22
26

INS

But when 1&2 are together you can see that 6 is a common value for N, and eventually they will meet again at 26.

I can now see that N=26 6 Peppermints and 20 Spearmints,

1) p-1= 5:20 =1:4
2)S-2= 6:18 =1:3

C

Is there an indicator from the ratios or formulas that is derived from the two stems that should identify sufficiency. Even though I know how to come to the right answer, I feel like I might have overlooked a much simpler route.

Thanks

[RonPurewal]

WHO CARES?? click D, confirm NEXT!

actually, if you did this, you would get the problem incorrect. the correct answer option when both statements are used together is (c), not (d).

[RonPurewal]

here are a couple of ways that you can solve this one.

1/
MAKE LISTS
just make lists of possibilities that satisfy the statements. (obviously, you should use some table or chart when you do this on your own paper, but i just wrote out lists because it's rather difficult to format tables or charts on the forum.)

STATEMENT (1)
here's a systematic list of possibilities:
* 5 candies after removal, with 1 peppermint --> 6 and 2 respectively, before removal
* 10 candies after removal, with 2 peppermint --> 11 and 3 respectively, before removal
* 15 candies after removal, with 3 peppermint --> 16 and 4 respectively, before removal
* 20 candies after removal, with 1 peppermint --> 21 and 5 respectively, before removal
* 25 candies after removal, with 1 peppermint --> 26 and 6 respectively, before removal
etc.
tons of possibilities, so, insufficient.
note that you should stop after you find the first two cases. i've just listed more cases here so that the whole post is easier to follow (i.e., so that i don't have to list the rest of the possibilities later, creating a fragmented post).

STATEMENT (2)
here's a systematic list of possibilities:
* 4 candies after removal, with 1 peppermint --> 6 and 1 respectively, before removal
* 8 candies after removal, with 2 peppermint --> 10 and 2 respectively, before removal
* 12 candies after removal, with 3 peppermint --> 14 and 3 respectively, before removal
* 16 candies after removal, with 1 peppermint --> 18 and 4 respectively, before removal
* 20 candies after removal, with 1 peppermint --> 22 and 5 respectively, before removal
* 24 candies after removal, with 6 peppermint --> 26 and 6 respectively, before removal
etc.
tons of possibilities, so, insufficient.
again, you should stop after you find the first two cases, but i'm just putting the whole list in one place so the post is easier to follow.

TOGETHER:
there's only one case that appears in both lists:
* 24 candies after removal, with 6 peppermint --> 26 and 6 respectively, before removal
if you continue the lists far enough, you'll see that there aren't going to be any other common elements. so, this is the only one.
sufficient.

answer (c)

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2/
ALGEBRA

first -- as should be customary for all word problems -- temporarily ignore the actual math, and just take an inventory of what's happening in the problem.
in the problem, we have two unknowns -- the total number of candies and the number of peppermint candies -- that are at first independent of each other. (you don't need another unknown for the number of spearmint candies, because that's just the difference between the aforementioned two quantities.)

because there is no necessary relationship between these quantities at the beginning of the problem, go ahead and designate two variables, say "t" for the total number of candies, and "p" for the number of peppermint candies.

STATEMENT (1)
if one peppermint candy is removed, there are now a total of (t - 1) candies, of which (p - 1) are peppermint.
thus
(1/5)(t - 1) = p - 1
or
(p - 1)/(t - 1) = 1/5
either of which can be rearranged to
t - 1 = 5p - 1
there are infinitely many values of p and t that satisfy this equation, so this statement is not sufficient.

STATEMENT (2)
if 2 spearmint candies are removed, there are now a total of (t - 2) candies, of which p are peppermint.
thus
(1/4)(t - 2) = p
or
(p)/(t - 2) = 1/4
which can be rearranged to
t - 2 = 4p
there are infinitely many values of p and t that satisfy this equation, so this statement is not sufficient.

TOGETHER
from the individual statements, we have both of the following:
t - 1 = 5p - 1
t - 2 = 4p
this is a system that can be solved for unique values of t and p, so, sufficient.

answer (c)

[ShyT]

Thanks Ron,

I really like the algebraic approach, its exactly what I was looking for.

*slight typo for future readers
Statement 1: T-1=5P-5

[jlucero]

Yup. Good catch. And while it's important to know when you should stop on a GMAT DS problem, I love when students work out the math to double check their answer- a great way to continue improving your algebra.