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Are there more than five red chips on the table?

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Are there more than five red chips on the table? [#permalink]

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Are there more than five red chips on the table?

(1) The probability of drawing a non-red chip from the pile of 100 chips is 24/25.
(2) The probability of drawing a red chip is 1/25
[Reveal] Spoiler: OA

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Re: Are there more than five red chips on the table? [#permalink]

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New post 09 May 2017, 06:07
stat1: from 100 chips probability of non red is 24/25 = 96/100.. hence 4 red chips
suff

stat2: not suff,,,different ratios possible

ans A

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Re: Are there more than five red chips on the table? [#permalink]

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New post 10 May 2017, 01:06
how about E

as we dont know how many piles of chips are there on the table.

Waiting for OA

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Re: Are there more than five red chips on the table? [#permalink]

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New post 15 May 2017, 18:47
Bunuel wrote:
Are there more than five red chips on the table?

(1) The probability of drawing a non-red chip from the pile of 100 chips is 24/25.
(2) The probability of drawing a red chip is 1/25



Statement 1:-
The probability of drawing a non red chip is given as :- 24/25=96/100.

Since it is given that there are 100 chips. Number of non-red chips =96.
Hence number of red chips= 4.
Statement 1 is sufficient.


Statement 2:-
Probability of drawing of red chip is 1/25. But we do not have any idea about the total number of chips.
Hence Statement 1 is insufficient.

Hence the solution is :- Option A.

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Re: Are there more than five red chips on the table? [#permalink]

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New post 21 Nov 2017, 12:07
Hi Bunuel,

How can answer be A, as we don't know how many piles will be on the table. If there will be only one pile of chips on the table, it is sufficient, otherwise, we won't be to tell.

Please clarify

Thanks

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Re: Are there more than five red chips on the table? [#permalink]

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New post 21 Nov 2017, 12:14
Bunuel wrote:
Are there more than five red chips on the table?

(1) The probability of drawing a non-red chip from the pile of 100 chips is 24/25.
(2) The probability of drawing a red chip is 1/25


S1) P(Non-red) = \(\frac{24}{25}\)
=> P(Red) = \(\frac{1}{25}\)

Total chips = 100
=> No. of red chips = 4
Sufficient.

S2) P(Red) = \(\frac{1}{25}\)
We don't know the total number of chips
=> There could be 4 red chips, if total chips is 100
=> There could be 8 red chips, if total chips is 200

Insufficient.

A is the answer.
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Re: Are there more than five red chips on the table? [#permalink]

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New post 22 Nov 2017, 10:16
hellosanthosh2k2 wrote:
Hi Bunuel,

How can answer be A, as we don't know how many piles will be on the table. If there will be only one pile of chips on the table, it is sufficient, otherwise, we won't be to tell.

Please clarify

Thanks


Hi

I agree that number of piles is not mentioned. But statement 1 says '.. from the pile of 100 chips..'. I think we can safely assume here that there is only one pile, which is being talked about. Had there been more piles, question would have mentioned something like 'Are there more than 5 red chips on the table where various piles of chips are kept.' or something like that.

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Re: Are there more than five red chips on the table? [#permalink]

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New post 23 Nov 2017, 03:45
@
Bunuel wrote:
Are there more than five red chips on the table?

(1) The probability of drawing a non-red chip from the pile of 100 chips is 24/25.
(2) The probability of drawing a red chip is 1/25


Bunuel

Please confirm-I dont think we can assume that there are 100 chips. 1 only gives us the probability-we only go by what data is given-no assumptions
I think E should be the answer.

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Re: Are there more than five red chips on the table? [#permalink]

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New post 24 Nov 2017, 15:17
Bunuel wrote:
Are there more than five red chips on the table?

(1) The probability of drawing a non-red chip from the pile of 100 chips is 24/25.
(2) The probability of drawing a red chip is 1/25


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The last step of the VA method is to make the number of variables and the equations equal (from the whole question including the original condition, conditions 1) and 2) )

Assume r is the number of re chips and n is the number of non-red chips.
Since we have 2 variables and 0 equation, we need 2 equations.

Condition 1)
r + n = 100, r / ( r + n ) = 24/25
Since we have 2 equations in the condition 1), we can get n = 96.
Thus r = 4.
The answer is No.
This is sufficient by CMT (Common Mistake Type) 1.

Condition 2)
r / ( r + n ) = 1/25
r = 1, n = 24 / r = 10, n = 240.
This is not sufficient.

Therefore, the answer is A.

Normally, in problems which require 2 or more additional equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: Are there more than five red chips on the table?   [#permalink] 24 Nov 2017, 15:17
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