A box contains 12 balls out of which x are black. If 6 more black ball

1.	A box contains 12 balls out of which x are black. If 6 more black balls are put in the box, the probability of drawing a black ball now is double of what it was before. The value of x is ………A) 5 B) 4 C) 3 D) 2
Answer» Correct option is: C) 3 Total balls is 12 in which x are black. \(\therefore\) Probability of drawing a black ball = \(\frac x{12}\) After putting 6 more black balls. Total balls = 12 + 6 = 18 and total black balls = x + 6 \(\therefore\) Probability of drawing a black balls after putting 6 more black balls = \(\frac {x+6}{18}\) According to given conditions, we have \(\frac {x+6}{18}\) = \(\frac {2x}{12}\) \(\Rightarrow\) 12 (x + 6) = 36x (By cross multiplication) \(\Rightarrow\) 12x + 72 = 36x \(\Rightarrow\) 36x -12x = 72 \(\Rightarrow\) 24x = 72 \(\Rightarrow\) x = \(\frac {72}{24}= 3\) Hence, the value of x is 3. Correct option is: C) 3

A box contains 12 balls out of which x are black. If 6 more black balls are put in the box, the probability of drawing a black ball now is double of what it was before. The value of x is ………A) 5 B) 4 C) 3 D) 2

Answer»

Correct option is: C) 3

Total balls is 12 in which x are black.

\(\therefore\) Probability of drawing a black ball = \(\frac x{12}\)

After putting 6 more black balls.

Total balls = 12 + 6 = 18

and total black balls = x + 6

\(\therefore\) Probability of drawing a black balls after putting 6 more black balls = \(\frac {x+6}{18}\)

According to given conditions, we have

\(\frac {x+6}{18}\) = \(\frac {2x}{12}\)

\(\Rightarrow\) 12 (x + 6) = 36x (By cross multiplication)

\(\Rightarrow\) 12x + 72 = 36x

\(\Rightarrow\) 36x -12x = 72

\(\Rightarrow\) 24x = 72

\(\Rightarrow\) x = \(\frac {72}{24}= 3\)

Hence, the value of x is 3.

Correct option is: C) 3

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