Question

A chemical factory produces two kinds of unnatural amino acids: acid A and acid B. Of the acids produced by the factory last year, 1/3 were acid A and the rest were acid B. If it takes 2/5 as many hours to produce acid B per unit as it does to produce acid A per unit, then the number of hours it took to produce the acid B last year was what fraction of the total number of hours it took to produce all the acids?

• A. 2/5
• B. 4/9
• C. 17/35
• D. 1/2
• E. 5/9

Hint:

In problems involving multiple ratios, it's often helpful to assign simple, concrete numbers that satisfy the ratios. For instance, assume 100 units total were produced. Then 1/3 (approx 33) is not a good number. Instead, use the denominators. Assume 3 units total. So, 1 unit of A and 2 units of B. Assume it takes 5 hours for 1 unit of A. Then it takes (2/5)*5 = 2 hours for 1 unit of B. Total time for A = \(1 \times 5 = 5\) hours. Total time for B = \(2 \times 2 = 4\) hours. Total time for all acids = \(5+4=9\) hours. Fraction of time for B = \(4/9\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem requires us to work with fractions and ratios to determine a part-to-whole relationship. We need to find the ratio of the time spent producing acid B to the total time spent producing both acids.
Step 2: Detailed Explanation:
Let's define our variables:


Let \(U_A\) and \(U_B\) be the number of units of acid A and acid B produced.
Let \(H_A\) and \(H_B\) be the number of hours required to produce one unit of acid A and acid B, respectively.
Let \(T_A\) and \(T_B\) be the total hours spent producing acid A and acid B.
From the problem statement:
1. Ratio of units produced: 1/3 of the acids were acid A, which means 2/3 were acid B. The ratio of units is \(U_A : U_B = 1/3 : 2/3 = 1:2\). So, for every 1 unit of A, 2 units of B are produced. Let's say \(U_A = k\) and \(U_B = 2k\) for some constant \(k\).
2. Ratio of hours per unit: It takes 2/5 as many hours for B as for A. So, \(H_B = \frac{2}{5} H_A\). Let's say \(H_A = 5h\), then \(H_B = 2h\) for some constant \(h\).
Now, calculate the total hours for each acid:


Total hours for A: \(T_A = U_A \times H_A = k \times 5h = 5kh\)
Total hours for B: \(T_B = U_B \times H_B = 2k \times 2h = 4kh\)
The total hours for all acids is \(T_{Total} = T_A + T_B = 5kh + 4kh = 9kh\).
The question asks for the fraction of total hours that was for acid B.
\[ \text{Fraction for B} = \frac{T_B}{T_{Total}} = \frac{4kh}{9kh} \]
The variables \(k\) and \(h\) cancel out.
\[ \text{Fraction for B} = \frac{4}{9} \]
Step 3: Final Answer:
The number of hours to produce acid B was 4/9 of the total number of hours. This corresponds to option (B).