Question

A doctor orders 20 grams of a 52% solution of a certain medicine. The pharmacist has only bottles of 40% solution and bottles of 70% solution. How much of each must he use to obtain the 20 grams of the 52% solution?

Answer 1

Try, please, this:
1) Imagine, that pharmacist used 'x' gr. of 40% solution and 'y' of 70% solution.
Together (20 gr. of 52% sol.) it will be x+y=20.
2) From other side, the pharmacist combined 40x and 70y gramms, after obtaining together it is 52(x+y) gramms. In other words 40x+70y=52(x+y)
3) 
$\left \{ {{x+y=20} \atop {40x+70y=52(x+y)}} \right. \ =\ \textgreater \ \ \left \{ {{y=8} \atop {x=12}} \right.$
Answer: 8 gr. of 70% solution and 12 gr. of 40%.

Answer 2

Answer:

12 grams of 40% solution and 8 grams of 70% solution are mixed to get 20 grams of 52% solution.

Step-by-step explanation:

Given : A doctor orders 20 grams of a 52% solution of a certain medicine. The pharmacist has only bottles of 40% solution and bottles of 70% solution.

To find : How much of each must he use to obtain the 20 grams of the 52% solution?

Solution :

Let x be the required amount.

x amount is used to obtain 20 grams = 20-x

According to question,

x grams of 40% and (20-x) grams of 70% is equal to 20 grams of 52%.

$x(40\%)+(20-x)(70\%)=20(52\%)$

$0.4x+20\times 0.7-0.7x=20\times 0.52$

$14-0.3x=10.4$

$0.3x=3.6$

$x=\frac{3.6}{0.3}$

$x=12$

Therefore, The required amount is 12 grams.

and 20-x=20-12=8 grams

i.e, 12 grams of 40% solution and 8 grams of 70% solution are mixed to get 20 grams of 52% solution.