Respuesta :
Therefore he took 40 gram of [tex]1^{st}[/tex] type solution and 10 gram of [tex]2^{nd}[/tex] type solution.
Step-by-step explanation:
Given that , A pharmacist 13% alcohol solution another 18% alcohol solution .
Let he took x gram solution of [tex]1^{st}[/tex] type solution
and he took (50-x) gram of [tex]2^{nd}[/tex] type solution.
Total amount of alcohol =[tex][x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ][/tex] gram
Total amount of solution = 50 gram
According to problem
⇔[tex]\frac{ [x\times\frac{13}{100}] +[(50 -x) \times\frac{18}{100} ]}{50}= \frac{14}{100}[/tex]
⇔[tex]\frac{13x +900-18x}{100\times50} =\frac{14}{100}[/tex]
⇔- 5x= 700 - 900
⇔5x = 200
⇔x = 40 gram
Therefore he took 40 gram of [tex]1^{st}[/tex] type solution and (50 -40)gram = 10 gram of [tex]2^{nd}[/tex] type solution.
