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A bag contains only red and white ball. How many red balls does it hav

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chetan2u
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A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  21 Jan 2016, 07:33
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A bag contains only red and white ball. How many red balls does it have?

(1) The probability of picking two white ball one after another without replacement is 14/33.
(2) The probability of picking a white ball from the bag is 2/3.
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RatneshS
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  21 Jan 2016, 08:46
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From 1:
(W/R+W)*((W-1)/(R+W-1))=14/33
Not sufficient


From2
white=W/R+W=2/3
Not sufficient

adding 1 and 2:
(2/3)*((w-1)/(R+W-1))=14/33=7/11
11w-11=7(R+w)-7
on solving 4W=7R+4-------eqn 1

from stmt 2, 3w=2R+2W or W=2R------eqn 2

from eqn 1 and 2

R=4 W=8
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  21 Jan 2016, 11:59
A
1. w=8, r=4 (it is tricky one)
2. it is obviously insufficient
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  21 Jan 2016, 12:02
What is QA, plz?
I am not sure that A is the correct....
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  22 Jan 2016, 23:02
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Bumping for further discussion..
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  23 Jan 2016, 00:20
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St1: (w/r+w) * (w-1/r+w-1) = 14/33
33*w*(w-1) = 14*(r+w)*(r+w-1) --> Not sufficient as we have two unknowns and there can be values other than r = 4 and w = 8 that can satisfy this equation

St2: w/r+w = 2/3 --> 3w = 2r + 2w --> w = 2r --> Not Sufficient as we get many values for r and w

Combining St1 and St2,
33*(4r^2 - 2r) = 14*(3r)*(3r - 1)
132r^2 - 66r = 126r^2 - 42r
6r^2 - 24r = 0
r^2 - 4r = 0 --> r = 0 or r = 4
Since r cannot be 0, value of r = 4 and w = 8
St1 and St2 are together sufficient

Answer: C
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  23 Jan 2016, 04:54
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Took the help of excel to figure out what other values may be possible for Statement 1 apart from r = 4 and w = 8.
Eventually it turned out that w = 36 and r = 19 is also a possible answer.
Thus St1 is surely not sufficient enough.

Answer is C
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  12 Feb 2021, 21:12
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A bag contains only red and white ball. How many red balls does it have?

(1) The probability of picking two white ball one after another without replacement is 14/33.

P(2 white) = 14/33 = (w / w+r)*(w-1 / w+r -1)

Insufficient.

(2) The probability of picking a white ball from the bag is 2/3.

w = 2/3
r = 1/3
Insufficient

C: Clearly sufficient since we have w and r which we can substitute.

Answer is C.

Please give me a thumbs up if you like my solution.
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Re: A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  07 Aug 2021, 22:10
1 should be sufficient. Therefore OA should be A and not C
w*(w-1)/(r+w) (r+w-1) = 14/33
splitting into prime factors
14/33 = 2*7/3*11
Only possibility of splitting is 2/3 and 7/11
greater possibility is single pick so...
w/(r+w) = 2/3 and (w-1)/(r+w-1)=7/11
rest follows as mentioned in earlier posts
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A bag contains only red and white ball. How many red balls does it hav [#permalink] New post  07 Aug 2021, 22:32
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prathinmba wrote:
1 should be sufficient. Therefore OA should be A and not C
w*(w-1)/(r+w) (r+w-1) = 14/33
splitting into prime factors
14/33 = 2*7/3*11
Only possibility of splitting is 2/3 and 7/11
greater possibility is single pick so...
w/(r+w) = 2/3 and (w-1)/(r+w-1)=7/11
rest follows as mentioned in earlier posts



Why should you restrict to prime factors?

\(\frac{w(w-1)}{t(t-1)}=\frac{2*7}{3*11}\)

We have to convert the numerator and denominator as multiples of two consecutive numbers.

So, \(\frac{w(w-1)}{t(t-1)}=\frac{2*7}{3*11}=\frac{2*4*7}{3*4*11}=\frac{8*7}{12*11}\).
Thus white balls are 8 and total are 12.
P=\(\frac{8}{12}*\frac{7}{11}=\frac{14}{33}\)

Also, \(\frac{w(w-1)}{t(t-1)}=\frac{2*7}{3*11}=\frac{2*18*7*5}{3*18*11*5}=\frac{36*35}{55*54}\).
Thus white balls are 36 and total are 55.
P=\(\frac{36}{55}*\frac{35}{54}=\frac{14}{33}\)
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A bag contains only red and white ball. How many red balls does it hav [#permalink]
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