Read the passage carefully and answer the question. There are two bags A and B…
2023
Read the passage carefully and answer the 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 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.
- A.
Quantity I > Quantity II
- B.
Quantity I < Quantity II
- C.
Quantity I ≥ Quantity II
- D.
Quantity I ≤ Quantity II
- E.
Quantity I = Quantity II or can’t be determined
Show answer & explanation
Correct answer: A
Concept
In a Quantity I vs Quantity II comparison, you first translate every word-clue of the data set into algebraic equations, solve for each unknown count, and only then compare the two requested quantities. The governing relations here are: for each bag green = red ÷ 2; the |red − blue| gap is fixed per bag; the yellow-bag has exactly one extra colour and its yellow = green + 3; and the stated average equals total balls ÷ number of distinct colours present in that bag.
Setting up the bags
“They have 3 and 4 coloured balls” means one bag uses 3 colours and the other uses 4 colours; since only one bag holds yellow, that bag is the 4-colour bag and the other holds only red, blue, green. Average × (number of colours) gives the bag total:
Bag A average 48 over 3 colours → total = 48 × 3 = 144 (red, blue, green only).
Bag B average 37 over 4 colours → total = 37 × 4 = 148 (red, blue, green, yellow — the yellow bag).
Solving Bag A (red, blue, green)
Let red = R, so green = R ÷ 2, and |R − blue| = 6.
Total: R + blue + R÷2 = 144.
Taking blue = R − 6 (so red exceeds blue): R + (R − 6) + R÷2 = 144 → 2.5R = 150 → R = 60.
So red = 60, blue = 54, green = 30 (60 + 54 + 30 = 144 ✓).
Solving Bag B (red, blue, green, yellow)
Let red = R, green = R ÷ 2, |R − blue| = 5, yellow = green + 3 = R÷2 + 3.
Total: R + blue + R÷2 + (R÷2 + 3) = 148.
Taking blue = R − 5: R + (R − 5) + R÷2 + R÷2 + 3 = 148 → 3R − 2 = 148 → 3R = 150 → R = 50.
So red = 50, blue = 45, green = 25, yellow = 28 (50 + 45 + 25 + 28 = 148 ✓).
Computing the two quantities
Quantity I = | blue of A − red of B | = | 54 − 50 | = 4.
Quantity II = | yellow of the yellow-bag − green of B | = | 28 − 25 | = 3.
Uniqueness of the solution
For each bag, the other branch of the |red − blue| equation (taking blue = red + gap, so blue exceeds red) makes the totals fail to give whole-number counts, so it is rejected. This is why each bag has exactly one valid integer solution.
Cross-check and result
Both bags independently satisfy every clue (green = half of red, the fixed red−blue gaps of 6 and 5, yellow = green + 3, and the two averages), so the counts are unique. Comparing the values, 4 is greater than 3, hence Quantity I exceeds Quantity II.