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A doctor prescribed 18 cubic centimeters of a certain drug t

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by BTGmoderatorDC » Wed Jan 30, 2019 5:32 am

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Answers

A

B

C

D

E

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Difficulty—

A doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds. If the typical dosage is 2 cubic centimeters per 15 pounds of the body weight, by what percent was the prescribed dosage greater than the typical dosage?

A) 8%
B) 9%
C) 11%
D) 12.5%
E) 14.8%

OA D

Source: Official Guide
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by Brent@GMATPrepNow » Wed Jan 30, 2019 5:40 am
BTGmoderatorDC wrote:A doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds. If the typical dosage is 2 cubic centimeters per 15 pounds of the body weight, by what percent was the prescribed dosage greater than the typical dosage?

A) 8%
B) 9%
C) 11%
D) 12.5%
E) 14.8%

OA D

Source: Official Guide
We can use equivalent ratios to determine what the dosage SHOULD have been.
We'll use the ratio of drug dosage (in cubic centimeters)/body weight (in pounds)

Let x = the dosage (in cubic centimeters) the patient SHOULD have received

We get: 2/15 = x/120
Cross multiply to get: 15x = (2)(120)
Simplify: 15x = 240
Solve: x = 16
So, the patient SHOULD have received a dosage of 16 cubic centimeters
Instead, the patient was prescribed a dosage of 18 cubic centimeters

Percent increase = 100(new - original)/original
= 100(18 - 16)/16
= 200/16
= 12.5

Answer: D

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Brent
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by [email protected] » Wed Jan 30, 2019 7:14 pm
Hi All,

This question is essentially about working through the necessary arithmetic. Thankfully, the math itself isn't that tough, but you do have to stay organized to make sure that you're setting up the proper calculations.

We're told that a typical dose of a drug is 2 cm^3 per 15 pounds. With a 120 pound person, there are 8 15-pound "sets", so the typical dose would be 8(2 cm^3) = 16 cm^3. Since the doctor prescribed 18 cm^3 of the drug, the dose was clearly larger than the typical dose. The question asks by what percent greater was that dose.

Since 18 is "2 more" than 16, we can know that the prescribed dose was 2/16 = 1/8 = 12.5% greater than it typically would be.

Final Answer: D


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by fskilnik@GMATH » Thu Jan 31, 2019 4:44 am
BTGmoderatorDC wrote:A doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds. If the typical dosage is 2 cubic centimeters per 15 pounds of the body weight, by what percent was the prescribed dosage greater than the typical dosage?

A) 8%
B) 9%
C) 11%
D) 12.5%
E) 14.8%
Source: Official Guide
Perfect opportunity for UNITS CONTROL, one of the most powerful tools carefully explained in our course!
$${\rm{atypical}}\,\,\left( {120\,\,{\rm{pounds}}} \right)\,\,\,:\,\,\,18\,{\rm{c}}{{\rm{m}}^3}$$
$${\rm{typical}}\,\,\,{\rm{:}}\,\,\,120\,{\rm{pounds}}\,\, \cdot \,\,\left( {{{2\,{\rm{c}}{{\rm{m}}^3}} \over {15\,{\rm{pounds}}}}} \right)\,\,\, = \,\,\,16\,\,{\rm{c}}{{\rm{m}}^{\rm{3}}}$$
$$16\mathop \to \limits^{?\, = \,\Delta \% } 18\,\,\,\,\left[ {{\rm{c}}{{\rm{m}}^{\rm{3}}}} \right]$$
$${\rm{?}}\,\, = \,\,\Delta \% \,\, = \,\,{{18 - 16} \over {16}}\,\,\, = \,\,\,{1 \over 8} \cdot 100\% \,\,\, = \,\,\,12.5\% $$

The correct answer is therefore (D).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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by Scott@TargetTestPrep » Fri Feb 01, 2019 5:22 pm
BTGmoderatorDC wrote:A doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 120 pounds. If the typical dosage is 2 cubic centimeters per 15 pounds of the body weight, by what percent was the prescribed dosage greater than the typical dosage?

A) 8%
B) 9%
C) 11%
D) 12.5%
E) 14.8%

OA D

Source: Official Guide
We are given that the typical dosage is (2 cm^3)/(15 lbs). Thus a patient with a weight of 120 lbs should be prescribed 120 lbs x (2 cm^3)/(15 lbs) = 240/15 = 16 cm^3 of the drug. Since the actual prescribed dosage was 18 cm^3, the percent increase was

(18 - 16)/16 x 100 = 2/16 x 100 = 1/8 x 100 = 12.5 percent

Answer: D

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