Study the given information and answer the following questions.
There are three bags A, B and C. In each bag there are three types of coloured balls - Yellow, Green and Black In bag A the number of Yellow balls is y and the number of Green balls is g. The number of green balls is 4 more than the number of Yellow balls. When one ball is picked at random then the probability of getting Black ball is `5//13` . The value of y is `18 2/11%` less than g. In bag B, the number of Yellow balls is `22 2/9%` more than that of bag A. If two balls are picked at random from bag B then the probability of getting both Green colour balls is `4//37`. The total number of balls in bag B is 75. In bag C, the ratio of the number of Green balls to the number of Black balls is 7:5. The total number of Green and Black balls is 36. If one of the balls, is picked at random t hen the probability of getting one Yellow ball is `7//13`.
x number of Yellow balls from bag B are taken and placed into bag A and `20%` of black balls from bag A are taken and placed into bag B. If we pick one ball from bag B then the probability that the ball is of Black colour is 11/26.Then find the value of x.

Options

A. 5
B. 3
C. 2
D. 6

Answer

To solve the problem step by step, we will break down the information provided and calculate the required values systematically.

### Step 1: Determine the number of Yellow and Green balls in Bag A

Given:
- Let the number of Yellow balls in Bag A be \( y \).
- The number of Green balls in Bag A is \( g \).
- It is stated that the number of Green balls is 4 more than the number of Yellow balls:
\[ g = y + 4 \]

### Step 2: Set up the equation based on the probability of picking a Black ball

The probability of picking a Black ball from Bag A is given as \( \frac{5}{13} \). The total number of balls in Bag A is:
\[ y + g + b = y + (y + 4) + b = 2y + 4 + b \]
Let \( b \) be the number of Black balls. The probability of picking a Black ball can be expressed as:
\[ \frac{b}{2y + 4 + b} = \frac{5}{13} \]

Cross-multiplying gives:
\[ 13b = 5(2y + 4 + b) \]
Expanding this, we get:
\[ 13b = 10y + 20 + 5b \]
Rearranging gives:
\[ 8b = 10y + 20 \quad \Rightarrow \quad b = \frac{10y + 20}{8} = \frac{5y + 10}{4} \]

### Step 3: Relate \( y \) and \( g \) using the percentage information

We know that \( y \) is \( 18 \frac{2}{11}\% \) less than \( g \). This can be expressed as:
\[ y = g \left(1 - \frac{18 \frac{2}{11}}{100}\right) \]
Calculating \( 18 \frac{2}{11}\% \):
\[ 18 \frac{2}{11} = \frac{200}{11} \quad \Rightarrow \quad \frac{200}{11 \times 100} = \frac{2}{11} \]
Thus:
\[ y = g \left(1 - \frac{2}{11}\right) = g \left(\frac{9}{11}\right) \]
Substituting \( g = y + 4 \):
\[ y = (y + 4) \left(\frac{9}{11}\right) \]
Expanding gives:
\[ 11y = 9y + 36 \quad \Rightarrow \quad 2y = 36 \quad \Rightarrow \quad y = 18 \]
Then, substituting back to find \( g \):
\[ g = 18 + 4 = 22 \]

### Step 4: Find the number of Black balls in Bag A

Substituting \( y = 18 \) into the equation for \( b \):
\[ b = \frac{5(18) + 10}{4} = \frac{90 + 10}{4} = \frac{100}{4} = 25 \]

### Step 5: Summary of Bag A

- Yellow balls \( y = 18 \)
- Green balls \( g = 22 \)
- Black balls \( b = 25 \)

Total balls in Bag A:
\[ 18 + 22 + 25 = 65 \]

### Step 6: Calculate the number of Yellow balls in Bag B

In Bag B, the number of Yellow balls is \( 22 \frac{2}{9}\% \) more than Bag A:
\[ \text{Percentage} = \frac{200}{9} \quad \Rightarrow \quad \text{More than} = 1 + \frac{200}{900} = \frac{1100}{900} = \frac{11}{9} \]
Thus:
\[ \text{Yellow balls in Bag B} = 18 \times \frac{11}{9} = 22 \]

### Step 7: Probability of picking two Green balls from Bag B

Given that the total number of balls in Bag B is 75, we need to find the number of Green balls. Let \( g_B \) be the number of Green balls in Bag B:
\[ P(\text{both Green}) = \frac{g_B}{75} \cdot \frac{g_B - 1}{74} = \frac{4}{37} \]
Cross-multiplying gives:
\[ 37g_B(g_B - 1) = 4 \cdot 75 \cdot 74 \]
Calculating the right side:
\[ 4 \cdot 75 \cdot 74 = 22200 \]
Thus:
\[ 37g_B^2 - 37g_B - 22200 = 0 \]

### Step 8: Solving the quadratic equation

Using the quadratic formula:
\[ g_B = \frac{-(-37) \pm \sqrt{(-37)^2 - 4 \cdot 37 \cdot (-22200)}}{2 \cdot 37} \]
Calculating the discriminant:
\[ 1369 + 4 \cdot 37 \cdot 22200 = 1369 + 3284400 = 3285769 \]
Taking the square root:
\[ \sqrt{3285769} = 1813 \]
Thus:
\[ g_B = \frac{37 \pm 1813}{74} \]
Calculating the two possible values:
1. \( g_B = \frac{1850}{74} \approx 25 \)
2. \( g_B = \frac{-1776}{74} \) (not valid)

### Step 9: Summary of Bag B

- Yellow balls \( = 22 \)
- Green balls \( = 25 \)
- Black balls \( = 75 - 22 - 25 = 28 \)

### Step 10: Summary of Bag C

Given the ratio of Green to Black balls is \( 7:5 \) and the total is 36:
Let \( 7x + 5x = 36 \Rightarrow 12x = 36 \Rightarrow x = 3 \)
Thus:
- Green balls \( = 21 \)
- Black balls \( = 15 \)

### Step 11: Transfer of balls and final calculation

Let \( x \) Yellow balls from Bag B be transferred to Bag A, and \( 20\% \) of Black balls from Bag A to Bag B.

After transferring:
- Bag A: Yellow \( = 18 - x \), Green \( = 22 \), Black \( = 25 - 0.2 \cdot 25 = 20 \)
- Bag B: Yellow \( = 22 - x \), Green \( = 25 \), Black \( = 28 + 5 = 33 \)

Total balls in Bag B:
\[ (22 - x) + 25 + 33 = 80 - x \]
Probability of picking a Black ball from Bag B:
\[ P(\text{Black}) = \frac{33}{80 - x} = \frac{11}{26} \]
Cross-multiplying gives:
\[ 33 \cdot 26 = 11(80 - x) \Rightarrow 858 = 880 - 11x \]
Rearranging gives:
\[ 11x = 880 - 858 = 22 \Rightarrow x = 2 \]

### Final Answer

The value of \( x \) is \( \boxed{2} \).