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Fittoniya [83]
3 years ago
9

To decrease an amount by 18%, what single multiplier would you use?

Mathematics
2 answers:
suter [353]3 years ago
8 0
I believe you would calculate this like so;

We want 18% less than 100% of that amount therefore the single multiplier we would use is as follows:

(100% - 18%) = 82% 

Which alternatively = 0.82

Does that make sense? If not I can explain it a bit more. Please let me know. 
Juli2301 [7.4K]3 years ago
4 0

Answer:

0.82

Step-by-step explanation:

We are asked to find the single multiplier used to show an amount decreased by 18%.

Since amount is decreasing by 18%, this means that 82% (100-18) of the amount is remaining.

The multiplier will be in decimal form, so we will convert 82% to decimal as:

\frac{82}{100}=0.82

Therefore, our required single multiplier is 0.82.

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How much pure acid should be mixed with 6 gallons of a 20% acid solution in order to get a 90% acid solution?
beks73 [17]
So, how much acid is there in 6 gallons?  well is 20% acid or (20/100), so the amount of acid in it just (20/100) * 6 or 1.2, the rest is say water.

now, if we want a 90% solution, and say we add "y" gallons, how much acid is in it?  well (90/100) * y, or 0.9y.

now let's add "x" gallons of pure acid, now, pure acid is just pure acid, so is 100% acid, how much acid is there in it?  (100/100) * x, or 1x or just x.

we know whatever "x" and "y" amounts are, they -> x + 6 = y

and we also know that x + 1.2 = 0.9y

\bf \begin{array}{lccclll}
&\stackrel{gallons}{acid}&\stackrel{acid~\%}{quantity}&\stackrel{acid~gallons}{quantity}\\
&------&------&------\\
\textit{pure acid}&x&1.00&x\\
\textit{20\% sol'n}&6&0.20&1.2\\
------&------&------&------\\\
mixture&y&0.90&0.9y
\end{array}
\\\\\\
\begin{cases}
x+6=\boxed{y}\\
x+1.2=0.9y\\
----------\\
x+1.2=0.9\left( \boxed{x+6} \right)
\end{cases}
\\\\\\
x+1.2=0.9x+5.4\implies x-0.9x=5.4-1.2\implies 0.1x=4.2
\\\\\\
x=\cfrac{4.2}{0.1}\implies x=\stackrel{gallons}{42}
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