Question

A service station checks Mr. Gittleboro's radiator and finds it contains only 30% antifreeze. If the radiator holds 10 quarts and is full, how much must be drained off and replaced with pure antifreeze in order to bring it up to a required 50% antifreeze?

Answer 1

Answer:

2.86 quarts ( approx )

Step-by-step explanation:

Given,

The initial quantity of the Mr. Gittleboro's radiator that contains 30% antifreeze = 10 quarts,

Let x quarts of pure antifreeze replaced x quarts of Mr. Gittleboro's radiator to bring it up to a required 50% antifreeze,

So, the quantity of 30% antifreeze radiator after drained off x quarts = (10-x) quarts

Also, the quantity of final antifreeze radiator 50% antifreeze = 10 quarts

Thus, we can write,

30% of (10-x) + 100% of x = 50% of 10

30(10-x) + 100x = 500

300 - 30x + 100x = 500

300 + 70x = 500

70x = 200

x = 2.85714285714 quarts ≈ 2.86 quarts

Answer 2

$\boxed{{\mathbf{2}}{\mathbf{.86 quartz}}}$  of $30\%$  antifreeze replaced by pure antifreeze to maintain $50\%$  antifreeze in the radiator.

Further explanation:

Given:

It is given that radiator contains $30\%$  antifreeze and the radiator holds 10 quartz.

Step by step explanation:

Step 1:

Consider $x$  as the number of quartz of pure antifreeze that is replaced by $x$  number of quartz of Mr. Gittleboro’s radiator to maintain the level of $50\%$ antifreeze in the radiator.

It is given that radiator contains $30\%$  antifreeze and the radiator holds 10 quartz.

The amount of $30\%$  antifreeze radiator after drained off $x{\text{ quartz}}$  is $\left({10-x}\right){\text{quartz}}$ .

The radiator only holds ${\text{10 quartz}}$ .

The amount of final antifreeze radiator $50\%$  antifreeze is ${\text{10 quartz}}$ .

Step 2:

The equation for maintain the level of $50\%$  antifreeze in the radiator by replacing the $x$  number of quartz of $30\%$  antifreeze by pure antifreeze as,

$\begin{aligned}30\%{\text{ of}}\left({10-x}\right)+100\% {\text{ of }}x&=50\% {\text{ of 10}}\\\frac{{30}}{{100}}\times\left({10-x}\right)+\frac{{100}}{{100}}{\text{}}\times{\text{}}x&=\frac{{50{\text{ }}}}{{100}}\times{\text{10}}\\{\text{30}}\left({10-x}\right)+100x&=500\\\end{aligned}$

Now simplify the further equation using the distributive property.

$\begin{aligned}{\text{30}}\left({10-x}\right)+100x&=500\\300-30x+100x&=500\\300+70x&=500\\70x&=500-300\\\end{aligned}$

Now simplify the further equation to obtain the value of $x$ .

$\begin{aligned}70x&=500-300\hfill\\70x&=200\hfill\\x&=\frac{{200}}{{70}}\hfill\\x&=2.85714\approx 2.86{\text{ quartz}}\hfill\\\end{aligned}$

Therefore, $2.86{\text{ quartz}}$  of $30\%$  antifreeze replaced by pure antifreeze to maintain $50\%$  antifreeze in the radiator.

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Linear equation

Keywords: radiator, amount, replaced, Gittlebro, antifreeze, drained, maintain, quartz, number, linear equation, percentage, fraction, final antifreeze.