From a bag containing 12 identical blue balls, y identical

The answer is 19, not 18. Notice that we have an inequality sign there.

From a bag containing 12 identical blue balls, y identical yellow balls, and no other balls, one ball will be removed at random. If the probability is less than 2/5 that the removed ball will be blue, what is the least number of yellow balls that must be in the bag?
A. 17
B. 18
C. 19
D. 20
E. 21

\(P(blue) = \frac{12}{12+y}<\frac{2}{5}\);

\(y>18\);

\(y_{min}=19\).

Answer: C.

Hope it's clear.
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It goes like this:
\(\frac{12}{12+y}<\frac{2}{5}\)

Cross multiply: \(60< 24+2y\);

Subtract 24 from both sides: \(36<2y\);

Divide by 2: \(18<y\), which is the same as \(y>18\) (y is greater than 18).

Hope it's clear.
_________________

12/x = 2/5

x = 30 total number of balls.

There must be at least 18 yellow balls in the bag.

It is 18
P(blue) = 12 / (12+Y)
P(blue) < 2/5
solving this: 12 /12+Y = 2/5
y = 18

Can u pls help in explaining the shift in the inequilaties symbol .

I mean 12/12+y < 2/5
so
cross multiplying

12*5 < (12+y)2 , am i missing out on reversing the sign ....
thanks in advance

Other way of doing is substitute the value.

The question says it probability of blue should be less than 2/5 means it should be less than 40%

Also For a constant numerator and if the denominator is increasing the value of the output will be decreasing.

In our scenario, the numerator is constant which is 12

Based on the given the choice the denominator will be 12/29, 12/30, 12/31, 12/32, 12/33

Since numerator is same and denominator doesn't vary we can conclude as 12/29 > 12/30 > 12/31 > 12/32 > 12/33

Lets choose the number which is easy to solve. Hence choosing 12/30 first which is equal to 2/5

Straight away I can go for "C" from the above answer since 12/29 must be greater than 2/5 and 12/31 should hold the least value of yellow balls which will give the probability as less than 2/5

We can create the following inequality:

12/(y+12) < 2/5

60 < 2(y + 12)

30 < y + 12

18 < y

Since y is greater than 18, the least number of yellow balls is 19.

Answer: C
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GMATNinja, Bunuel, can you please explain why probability of removing a blue ball is the same as probability of blue balls in a bag? I cannot understand the relationship between two concepts.
In the solution, it was indicated that P(blue)=12/12+y<2/5. Why we are implying that probability of selecting 12 balls is also less than 2/5?

Bunuel, can you please explain why P(blue)=12/12+y<2/5. Are there any similar problems that may help better understand this concept?

The probability of an event is the number of favorable outcomes divided by the total number of outcomes: P = (the number of favorable outcomes)/(the total number of outcomes).

Here the probability of picking a blue ball is P = (the number of favorable outcomes)/(the total number of outcomes) = 12/(12 + y), because there are 12 blue balls out of total of (12 + y) balls (the favorable outcome is if we pick any of the 12 blue balls, while the total number of outcomes is (12 + y) because we can pick any of the (12 + y) balls).

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Hope it helps.
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You may also see this problem like this.

12=\frac{2}{5}x

it follows x=30.

To find the number of yellow balls, you go on like this: 30-12=18, but the probability of selecting a blue ball must be <2/5, so at least 19 yellow balls must be in the bag.

Bunuel is this correct?

Sorry for the formatting on the fraction i could not solve it

Thanks

Bunuel ScottTargetTestPrep

Seeing as we have an unknown variable (y), how can we decide not to flip the inequality sign while cross multiplying?

Do we assume positive because there is simply no way for a variable that represents the # of balls to be negative?

Thanks in advance!

It is exactly as you said, we know y is nonnegative because it represents the number of yellow balls in a certain bag. In GMAT, if a variable represents a quantity which can never be negative such as the number of students in a classroom, the side length of a triangle, or the age of some tree, the question won't tell you that the variable cannot be negative, but will expect you to assume so.
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Since the number of blue balls are 12 and the probability is less than 2/5
what if exactly the ratio is 2/5 than the ratio would be 12/30 = favourable outcome/total number of outcome
30-12 = 18 yellow balls, but since the ratio is less than 2/5 the least number of yellow balls are 19