UGC NET Questions (Paper – 1)

Let the total volume be T litres. Initially milk = (7/10)T and water = (3/10)T. When 10 litres of well-mixed solution are removed, milk removed is (7/10) × 10 = 7 litres and water removed is 3 litres. So milk becomes (7/10)T − 7 and water becomes (3/10)T − 3. After adding 10 litres of water, water becomes (3/10)T + 7. The new ratio is [(7/10)T − 7] : [(3/10)T + 7] = 7:5, which simplifies to T = 60.

Option A:
Option A is correct because substituting T = 60 gives initial milk 42 litres and water 18 litres, matching the 7:3 ratio. Removing 10 litres of mixture takes away 7 litres of milk and 3 litres of water, leaving 35 litres of milk and 15 litres of water. After adding 10 litres of water, we have 35 litres of milk and 25 litres of water, giving 35:25 = 7:5, exactly as required.

Option B:
Option B, 50 litres, would give initial milk 35 litres and water 15 litres. Removing 10 litres of mixture would remove 7 litres of milk and 3 litres of water, leaving 28 litres of milk and 12 litres of water, which after adding 10 litres becomes 28:22 = 14:11, not 7:5. Therefore 50 litres does not satisfy the given final ratio.

Option C:
Option C, 70 litres, leads to initial milk 49 litres and water 21 litres. After removing and replacing 10 litres, the final quantities do not simplify to the ratio 7:5; the numbers no longer align with the required proportion. Thus 70 litres is inconsistent with the conditions.

Option D:
Option D, 80 litres, similarly yields initial and final quantities whose ratio becomes something different from 7:5 after the described operation. Checking the arithmetic reveals that the milk and water amounts are not in the required proportion, so 80 litres cannot be the original quantity.

Let the total quantity be T litres. Initially, milk = (5/8)T and water = (3/8)T. When 16 litres are removed from a well-mixed solution, milk removed = (5/8) × 16 = 10 litres and water removed = 6 litres. So the remaining milk is (5/8)T − 10 and the remaining water is (3/8)T − 6. After adding 16 litres of water, water becomes (3/8)T + 10. The new ratio is [(5/8)T − 10] : [(3/8)T + 10] = 5:4. Solving this proportion yields T = 144 litres.

Option A:
Option A, 96 litres, would give initial milk 60 litres and water 36 litres. After removing 16 litres and adding 16 litres of water, the final milk and water quantities would not reduce to the ratio 5:4; a detailed check shows the ratio differs from 5:4. Hence 96 litres does not satisfy the given condition.

Option B:
Option B, 128 litres, leads to initial milk 80 litres and water 48 litres. Following the same removal and replacement process, the resulting milk and water do not maintain a 5:4 ratio. The mathematical equation for the final ratio fails when T = 128, so this value is incorrect.

Option C:
Option C is correct because with T = 144 we get initial milk 90 litres and water 54 litres. Removing 16 litres removes 10 litres of milk and 6 litres of water, leaving 80 litres of milk and 48 litres of water. Adding 16 litres of water gives 80 litres of milk and 64 litres of water, giving 80:64 = 5:4. This exactly matches the new ratio stated.

Option D:
Option D, 160 litres, would create larger starting volumes but the same removal and replacement would not lead to a 5:4 ratio. Solving the final ratio equation with T = 160 gives a contradiction, meaning that this total volume cannot produce the described outcome.