There are two bags A and B having balls of colors red, blue, green and yellow. They have 3 and 4 colored balls (not necessarily in the same order). Only one of the bags has yellow color balls. In bag A difference between red and blue color balls is 6. In both bags, the number of green balls is half of the red color balls. The difference between red and blue balls of bag B is 5. The yellow balls are 3 more than the number of green balls in the bag containing yellow balls. Average number of balls in bag A and B are 48 & 37 respectively.

Quantity I: Find the difference between blue balls of A and red balls of bag B.
Quantity II: Difference between yellow balls of bag containing it and number of green balls in bag B.

Which of the following is true?

Question

There are two bags A and B having balls of colors red, blue, green and yellow. They have 3 and 4 colored balls (not necessarily in the same order). Only one of the bags has yellow color balls. In bag A difference between red and blue color balls is 6. In both bags, the number of green balls is half of the red color balls. The difference between red and blue balls of bag B is 5. The yellow balls are 3 more than the number of green balls in the bag containing yellow balls. Average number of balls in bag A and B are 48 & 37 respectively.

Quantity I: Find the difference between blue balls of A and red balls of bag B.
Quantity II: Difference between yellow balls of bag containing it and number of green balls in bag B.

Which of the following is true?

Accepted Answer

**Solution:**

1. Let red balls in bag A be $2a$. Then green balls in bag A = $a$ (half of red).
2. Difference between red and blue balls in bag A is 6, so blue balls in bag A = $2a - 6$.
3. Yellow balls in bag A = $3 + a$ (since only one bag has yellow balls and yellow balls are 3 more than green balls in that bag).

4. Average number of balls in bag A is 48, so total balls in bag A = $48 	imes 3 = 144$ (since bag A has 3 colors).

5. Total balls in bag A:
   $$2a + a + (2a - 6) + (3 + a) = 6a - 3$$
   But bag A has 3 colors, so yellow balls are not in bag A. Hence yellow balls belong to bag B.

So for bag A (3 colors): red = $2a$, green = $a$, blue = $2a - 6$
   Total balls in bag A = $2a + a + (2a - 6) = 5a - 6$

Given average = 48, total balls = $48 	imes 3 = 144$

So,
   $$5a - 6 = 144 \implies 5a = 150 \implies a = 30$$

6. For bag B (4 colors): red = $2b$, green = $b$, blue = $2b - 5$, yellow = $3 + b$

7. Average number of balls in bag B is 37, so total balls in bag B = $37 	imes 4 = 148$

8. Total balls in bag B:
   $$2b + b + (2b - 5) + (3 + b) = 6b - 2$$

Given total balls = 148,
   $$6b - 2 = 148 \implies 6b = 150 \implies b = 25$$

9. Now calculate quantities:

- Quantity I = Blue balls of A - Red balls of B
     $$= (2a - 6) - 2b = (2 	imes 30 - 6) - 2 	imes 25 = (60 - 6) - 50 = 54 - 50 = 4$$

- Quantity II = Yellow balls - Green balls of B
     $$= (3 + b) - b = 3 + 25 - 25 = 3$$

10. Compare:

Quantity I = 4

Quantity II = 3

So, Quantity I > Quantity II

**Answer:** (a) Quantity I > Quantity II

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