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# A fish tank contains orange fish and silver fish. After k more orange

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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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10 Jun 2016, 05:07
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Difficulty:

55% (hard)

Question Stats:

64% (01:31) correct 36% (01:39) wrong based on 137 sessions

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A fish tank contains orange fish and silver fish. After k more orange fish and 2k more silver fish are added, the probability of choosing an orange fish at random is $$\frac{1}{3}$$ . What was the probability of choosing a silver fish before any more fish were added?

(A) $$\frac{1}{4}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) $$\frac{3}{4}$$

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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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10 Jun 2016, 06:30
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AbdurRakib wrote:
A fish tank contains orange fish and silver fish. After k more orange fish and 2k more silver fish are added, the probability of choosing an orange fish at random is $$\frac{1}{3}$$ . What was the probability of choosing a silver fish before any more fish were added?

(A) $$\frac{1}{4}$$

(B) $$\frac{1}{3}$$

(C) $$\frac{1}{2}$$

(D) $$\frac{2}{3}$$

(E) $$\frac{3}{4}$$

Equation (1): $$o + k = 1x$$

Equation (2): $$s + 2k = 2x$$

Multiplying the first Equation by 2: $$2o + 2k = 2x$$. Let´s call this last equation, Equation (3): $$2o + 2k = 2x$$

We can subtract Eq (3) $$-$$ Eq (2): $$2o - s = 0$$, and then $$s = 2o$$.

Then, $$\frac{s}{s+o} = \frac{2o}{3o} = \frac{2}{3}$$.

The correct answer is letter (D).
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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10 Jun 2016, 09:02
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very easy question. Straight D.

P(silver fish after addition)=1-1/3 = 2/3
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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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Updated on: 10 Jun 2016, 11:15
1
new additions of Orange Fish (OF) to Silver Fish (SF) to the fish tank are in the ratio $$\frac{k}{2k}$$ = $$\frac{1}{2}$$.

As the probability of choosing an OF is 1/3, there must 1 OF in 3 fishes or there must 1 OF for every 2 SF or the ratio of OF to SF = 1:2. So, even after adding fishes in the ratio of 1 OF for 2 SF, the ratio of OF:SF is still 1:2. This is only possible, if the original ratio of OF to SF is also 1:2.

So, if the probability of picking OF is 1/3, the probability of picking SF is 2/3 (both before and after addition of k OF and 2k SF).

So, D.

Hope that helps.

Originally posted by unverifiedvoracity on 10 Jun 2016, 09:26.
Last edited by unverifiedvoracity on 10 Jun 2016, 11:15, edited 1 time in total.
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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10 Jun 2016, 09:47
It is not a 700 level question !
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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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10 Jun 2016, 19:56
1
(o+k)/(s+2k)=1/2
s/o=2:1
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A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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15 Jun 2016, 13:28
hsbinfy wrote:
very easy question. Straight D.

P(silver fish after addition)=1-1/3 = 2/3

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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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24 Jun 2016, 08:45
msk0657 wrote:
hsbinfy wrote:
very easy question. Straight D.

P(silver fish after addition)=1-1/3 = 2/3

But very easy mate.... ^^
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Re: A fish tank contains orange fish and silver fish. After k more orange  [#permalink]

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18 Jul 2018, 04:52

Solution

Given:

• A fish tank contains orange fish and silver fish.
• After k more orange fish and 2k more silver fish are added into the tank, the probability of choosing an orange fish at random is $$\frac{1}{3$$.
}

To find:

• The probability of choosing a silver fish before any more fish were added?

Approach and Working:

Let there were ‘x’ orange fish and ‘y’ silver fish in tank before some more fishes were added in the tank.

• Thus, after adding k more orange fish and 2k more silver fish into the tank,
o Total orange fish= x+k
o Total silver fish= y+2k
o Total fish in the tank= x+y+3k

• Hence, the probability of choosing an orange fish= $$\frac{x+k}{(x+y+3k)}$$
o $$\frac{x+k}{(x+y+3k)} = \frac{1}{3}$$
o 3x+3k= x+y+3k
o 2x= y

• Thus, the probability of choosing a silver fish before any more fish were added = $$\frac{y}{x+y}$$
o $$\frac{2x}{x+2x}= \frac{2x}{3x}= \frac{2}{3}$$

Hence, the correct answer is option D.

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Re: A fish tank contains orange fish and silver fish. After k more orange &nbs [#permalink] 18 Jul 2018, 04:52
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