In a mixture of milk and water, the proportion of milk by weight is 60%. If from the 80 gm mixture, 20 gm of mixture is taken out and 6 gm of pure water is added to the mixture then find the ratio of milk and water in the new mixture.

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**Solution:**

1. Initial mixture weight = 80 gm
2. Milk proportion = 60%, so milk = $80 	imes \frac{60}{100} = 48$ gm and water = $80 - 48 = 32$ gm.

3. When 20 gm of mixture is taken out, the ratio of milk to water is 60:40 or 3:2.

4. Milk removed = $20 	imes \frac{3}{5} = 12$ gm
5. Water removed = $20 	imes \frac{2}{5} = 8$ gm

6. Milk left = $48 - 12 = 36$ gm
7. Water left = $32 - 8 = 24$ gm

8. Now, 6 gm of pure water is added, so new water quantity = $24 + 6 = 30$ gm

9. New ratio of milk to water = $\frac{36}{30} = \frac{6}{5}$

**Answer:** 6 : 5

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