Question

A drug consists of two components. component A and component B. In a medical, two such drug are there. Drug 1 consisting of 1% of component A and 10% of component B and drug 2 consisting of 5% of component A and 20% of B. But there is a problem. The problem is, the drug with exactly 4% of A is needed (component B does not matter). you have to mix-up the two available drug to prepare a new one. Find the percentage of both drug in mixture.

Answer 1

Answer:

A drug consists of two components. component A and component B. In a medical, two such drug are there. Drug 1 consisting of 1% of component A and 10% of component B and drug 2 consisting of 5% of component A and 20% of B. But there is a problem. The problem is, the drug with exactly 4% of A is needed (component B does not matter). you have to mix-up the two available drug to

Answer 2

Given:

Drug1 contains :

component A = 1%

component B = 10%

Drug2 contains :

component A = 5%

component B = 20%

To Find:

%drug1 and % drug2 needed to make a drug with component A= 4%

Solution:

Now, let us assume the % of drug1 needed = x

then the % of drug2 needed = 100-x

Now according to the question,

1% of x + 5% of (100-x) = 4% of total      ( total = 100)

$\dfrac{x}{100} + \dfrac{5(100-x)}{100} = \dfrac{4\times100}{100}$

x+ 500 - 5x = 400

100 = 4x

x = 25

Thus, the % of drug1 in mixture = 25%

         the % of drug2 in mixture = 75%