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allsm [11]
2 years ago
6

How many ounces of a 15​% alcohol solution must be mixed with 4 ounces of a 20​% alcohol solution to make a 17​% alcohol​ soluti

on?
Mathematics
2 answers:
dedylja [7]2 years ago
8 0

Answer:

Therefore 2 ounces of 11% alcohol solution must be mixed with 4 ounces of 20%alcohol to make a 17%alcohol solution

Step-by-step explanation:

Let x ounces of 11% alcohol must be mixed with the 2 ounces of 20% alcohol.

The amount of alcohol in 4 ounces solution is=   ounces

                                                             = 0.8 ounces

The amount of alcohol in x ounces  solution is=   ounces

                                                                             = 0.11 x ounces

Total amount of alcohol is = (4+x) ounces

According to the problem,

Therefore 2 ounces of 11% alcohol solution must be mixed with 4 ounces of 20%alcohol to make a 17%alcohol solution

sukhopar [10]2 years ago
7 0

Answer:

6 ounces

Step-by-step explanation:

Let x = the number of ounces of the 15% alcohol solution needed to be mixed with the 4 ounces of 20% alcohol solution to obtain the 17% alcohol solution.

After converting the percentages to their decimal equivalents, set up the necessary equation as follows:

0.15x+0.20(4) = (x+4)(0.17) Simplify and solve for x.

0.15x+0.8 = 0.17x+0.68 Subtract 0.15x from both sides.

0.8 = 0.02x+0.68 Now subtract 0.68 from both sides.

0.12 = 0.02x Finally, divide both sides by 0.02

x = 6

You need to mix 6 ounces of 15% alcohol solution with 4 ounces of 20% alcohol solution to obtain 10 (x+4)ounces of 17% alcohol solution.

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\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~
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