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`.
The difference between the number of Green balls in bag A and bag C is how much more/less per cent than the sum of the number of Black balls in bag A and bag C together?

Options

A. `100%`
B. `95%`
C. `97.5%`
D. `102.5%`

Answer

To solve the problem step by step, we will analyze the information provided for each bag and calculate the required values.

### Step 1: Determine the number of balls in Bag A
Let:
- \( y \) = number of Yellow balls in Bag A
- \( g \) = number of Green balls in Bag A

From the problem, we know:
1. \( g = y + 4 \) (the number of green balls is 4 more than the number of yellow balls)
2. The probability of picking a Black ball from Bag A is \( \frac{5}{13} \).

The total number of balls in Bag A can be expressed as:
\[ \text{Total in A} = y + g + b \]
where \( b \) is the number of Black balls.

Using the probability of picking a Black ball:
\[ \frac{b}{y + g + b} = \frac{5}{13} \]

Rearranging gives:
\[ 13b = 5(y + g + b) \]
\[ 13b = 5(y + (y + 4) + b) \]
\[ 13b = 5(2y + 4 + b) \]
\[ 13b = 10y + 20 + 5b \]
\[ 8b = 10y + 20 \]
\[ b = \frac{10y + 20}{8} = \frac{5y + 10}{4} \]

### Step 2: Relate \( y \) and \( g \)
We are also told that \( y \) is \( 18 \frac{2}{11}\% \) less than \( g \):
\[ y = g \left(1 - \frac{18 \frac{2}{11}}{100}\right) \]
Converting \( 18 \frac{2}{11} \) to a fraction:
\[ 18 \frac{2}{11} = \frac{200}{11} \]
Thus:
\[ y = g \left(1 - \frac{200}{1100}\right) = g \left(\frac{900}{1100}\right) = g \left(\frac{9}{11}\right) \]

### Step 3: Substitute \( g \) in terms of \( y \)
From \( g = y + 4 \):
\[ g = \frac{11y}{9} \]
Setting the two expressions for \( g \) equal:
\[ y + 4 = \frac{11y}{9} \]
Multiplying through by 9 to eliminate the fraction:
\[ 9y + 36 = 11y \]
\[ 2y = 36 \]
\[ y = 18 \]

Now substituting back to find \( g \):
\[ g = y + 4 = 18 + 4 = 22 \]

### Step 4: Determine the number of Black balls in Bag A
Using the value of \( y \):
\[ b = \frac{5y + 10}{4} = \frac{5(18) + 10}{4} = \frac{90 + 10}{4} = \frac{100}{4} = 25 \]

### Step 5: Total number of balls in Bag A
\[ \text{Total in A} = y + g + b = 18 + 22 + 25 = 65 \]

### Step 6: Determine the number of Yellow balls in Bag B
In Bag B, the number of Yellow balls is \( 22 \frac{2}{9}\% \) more than in Bag A:
\[ \text{Yellow in B} = 18 \left(1 + \frac{22 \frac{2}{9}}{100}\right) \]
Converting \( 22 \frac{2}{9} \) to a fraction:
\[ 22 \frac{2}{9} = \frac{200}{9} \]
Thus:
\[ \text{Yellow in B} = 18 \left(1 + \frac{200}{900}\right) = 18 \left(\frac{1100}{900}\right) = 22 \]

### Step 7: Determine the number of Green balls in Bag B
Let \( g_B \) be the number of Green balls in Bag B. The total number of balls in Bag B is 75, and the probability of picking two Green balls is \( \frac{4}{37} \):
\[ \frac{g_B}{75} \cdot \frac{g_B - 1}{74} = \frac{4}{37} \]
Cross-multiplying gives:
\[ 37g_B(g_B - 1) = 4 \cdot 75 \cdot 74 \]
\[ 37g_B^2 - 37g_B = 22200 \]
Rearranging gives:
\[ 37g_B^2 - 37g_B - 22200 = 0 \]

### Step 8: Solve 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:
\[ 1369 + 3284400 = 3285769 \]
Taking the square root:
\[ \sqrt{3285769} = 1813 \]

Thus:
\[ g_B = \frac{37 \pm 1813}{74} \]
Calculating the positive root:
\[ g_B = \frac{1850}{74} = 25 \]

### Step 9: Determine the number of Black balls in Bag B
Total balls in Bag B = 75:
\[ b_B = 75 - (22 + 25) = 28 \]

### Step 10: Determine the number of balls in Bag C
In Bag C, the ratio of Green to Black balls is \( 7:5 \) and the total is 36:
Let \( 7x + 5x = 36 \) gives \( 12x = 36 \) thus \( x = 3 \).
- Green balls = \( 7x = 21 \)
- Black balls = \( 5x = 15 \)

### Step 11: Calculate the difference in Green balls
Difference in Green balls between Bag A and Bag C:
\[ 22 - 21 = 1 \]

### Step 12: Calculate the sum of Black balls
Sum of Black balls in Bag A and Bag C:
\[ 25 + 15 = 40 \]

### Step 13: Calculate the percentage difference
Percentage difference:
\[ \frac{1}{40} \times 100 = 2.5\% \]

### Final Answer
The difference between the number of Green balls in Bag A and Bag C is **2.5%** less than the sum of the number of Black balls in Bag A and Bag C together.