A mixture of two chemicals A and B is prepared in the ratio 7:2. If there are 81

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Question: A mixture of two chemicals A and B is prepared in the ratio 7:2. If there are 81 grams of the mixture, how much chemical B is present?

Options:

Correct Answer: 18 grams

Solution:

In a 7:2 ratio, the total parts = 7 + 2 = 9. Chemical B = (2/9) * 81 = 18 grams.

A mixture of two chemicals A and B is prepared in the ratio 7:2. If there are 81 grams of the mixture, how much chemical B is present?

A mixture of two chemicals A and B is prepared in the ratio 7:2. If there are 81 grams of the mixture, how much chemical B is present?

Step 1: Understand the ratio of chemicals A and B, which is 7:2.
Step 2: Add the parts of the ratio together: 7 (for A) + 2 (for B) = 9 total parts.
Step 3: Determine how much one part is worth by dividing the total weight of the mixture (81 grams) by the total parts (9): 81 grams / 9 parts = 9 grams per part.
Step 4: Find the amount of chemical B by multiplying the number of parts for B (which is 2) by the weight of one part (9 grams): 2 parts * 9 grams/part = 18 grams.
Step 5: Conclude that there are 18 grams of chemical B in the mixture.

Ratio and Proportion – Understanding how to divide a total quantity based on a given ratio.
Basic Arithmetic – Performing multiplication and division to find the quantity of a component in a mixture.