Mixture A contains a mixture of chemical S and it contains 60% of chemical S. Mixture B contains the mixture of chemical U and T in the ratio of 7:2. If two mixtures A and B are mixed together in the ratio of 5:3, then find the ratio of chemical S to T in the resultant mixture?
Answer (1)
To solve this problem, we need to find out the ratio of chemical S to T in the resultant mixture when mixtures A and B are mixed together in a given ratio.
Understand the components of the mixtures:
Mixture A consists of 60% chemical S.
Mixture B consists of chemicals U and T in the ratio of 7:2.
Mixing Mixtures A and B: Mixtures A and B are mixed in the ratio of 5:3. This means that for every 5 parts of mixture A, 3 parts of mixture B are used.
Calculate the overall composition of chemical S in the resultant mixture:
Since Mixture A is 60% chemical S, in 5 parts of Mixture A, the amount of chemical S = 0.6 × 5 = 3 parts.
Mixture B does not contain chemical S, so the contribution of chemical S from Mixture B is 0.
Calculate the overall composition of chemical T in the resultant mixture:
In Mixture B, the ratio of U to T is 7:2. This implies that for every 9 parts of Mixture B, 2 parts are chemical T.
Therefore, in 3 parts of Mixture B, the amount of chemical T = 9 2 × 3 = 3 2 parts.
Determine the ratio of chemical S to T in the resultant mixture:
Chemical S = 3 parts (from Mixture A)
Chemical T = 3 2 parts (from Mixture B)
Therefore, the ratio of chemical S to T = 3 : 3 2 = 3 ÷ 3 2 = 3 × 2 3 = 2 9
This simplifies to 9:2.
Thus, the ratio of chemical S to T in the resultant mixture is 9 : 2 .
