A patient was prescribed a medication dose. It was increased by 15%15\% after 33 days, and the new dosage is 23cc23\operatorname{\mathrm{c}\mathrm{c}}. What was the original dosage?

20 cc ⁡ 20\operatorname{\mathrm{c}\mathrm{c}} 20 cc

A patient was prescribed a medication dose. It was increased by 15 % 15\% after 3 3 days, and the new dosage is 23 cc 23\operatorname{\mathrm{c}\mathrm{c}} . What was the original dosage?

For this problem, a patient was prescribed a medication dose, and that dose was increased by 15% after three days. The new dosage is now at twenty three cc's, and we want to know what was the original dosage. So to start solving this problem, let's first identify the quantities that we know and the quantities that we don't know and assign variables to those things that we don't know. So we know that the amount by which it increased was 15%, But one quantity that we don't know is the actual amount of the increase in cc's. So let's go ahead and say that the amount increased is the variable m. We know that the new dosage amount is twenty three cc's and another thing that we don't know is the original dosage. So let's go ahead and call that x. So using these quantities, we can use our knowledge of percents to start building an equation. We know that to find a partial amount like m, which is the part by which our dosage increased, we can take the percent written as a decimal, so 0.15, and multiply by the original amount or our original dosage, which we've called X. And now that we have this equation, we can start solving. But first, we need to get these in terms of one variable because here we have two variables, M and X. So it's another equation that we could build that shows a relationship between these two quantities. Well, we know that the new dosage is twenty three cc's, and that to calculate twenty three cc's, we could take our original amount, x, and add on the amount by which it increased, m. And now we can solve for one of our variables. Let's solve for m, by subtracting x to both sides. So we have twenty three minuteus x is equal to m, and I can now take this part of the equation and plug it in for m into my main equation. So doing that, we get twenty three minuteus x is equal to 0.15 times x. And now that we have only one variable, x, we're ready to solve. So first, let's combine all of our variable terms to one side by adding x, so we get 23 is equal to 0.15 x plus x. And now let's combine like terms, 0.15x plus 1x gives us 1.15x. And now we will isolate by dividing both sides by 1.15, and we get that x is equal to 20. Now remember, x is the variable that represents our original dosage, which is what we're looking for. So we can go ahead and state our answer as the original dosage being twenty cc's.

A patient was prescribed a medication dose. It was increased by 15 % 15\% after 3 3 days, and the new dosage is 23 cc ⁡ 23\operatorname{\mathrm{c}\mathrm{c}} . What was the original dosage?

20 cc ⁡ 20\operatorname{\mathrm{c}\mathrm{c}} 20 cc

26.5 cc ⁡ 26.5\operatorname{\mathrm{c}c} 26.5 cc

21 cc ⁡ 21\operatorname{\mathrm{c}c} 21 cc

18 cc ⁡ 18\operatorname{\mathrm{c}c} 18 cc

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Percent Problem Solving

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Multiple Choice

A patient was prescribed a medication dose. It was increased by $15 percent sign$ after $3$ days, and the new dosage is $23 cc$ . What was the original dosage?

$20\operatorname{\mathrm{c}\mathrm{c}}$

$26.5\operatorname{\mathrm{c}c}$

$21\operatorname{\mathrm{c}c}$

$18\operatorname{\mathrm{c}c}$

Verified step by step guidance

Let the original dosage be represented by the variable $x$ (in cc).

Since the dosage was increased by 15%, the new dosage can be expressed as $x$ plus 15% of $x$, which is $x + 0.15x$.

Combine like terms to write the new dosage as \$1.15x$.

We know the new dosage is 23 cc, so set up the equation: \$1.15x = 23$.

To find the original dosage $x$, divide both sides of the equation by 1.15: $x = \frac{23}{1.15}$.

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