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Trial of a new drug

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by adamz » Thu Nov 24, 2011 5:03 pm
In a trial of a new drug, patients received a progressively growing dosage of the drug for a few days. On the first day, each patient received 15 milligrams of Drug. On each of the following days, the daily dosage was k milligrams greater than the dosage received the day before, reaching a dosage of 43 milligram on the last day of the trial. For how many days did the trial last, if each patient received a total amount of 145 milligrams of drug during the whole trial ?

(a) 4
(b) 5
(c) 6
(d) 7
(e) 9

OA:B
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by HSPA » Thu Nov 24, 2011 6:43 pm
Arth Series:

A = 15, d= k; An = 43; n/2 * (2A+ (n-1)d) = 145.. All these are sufficent to find n.
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by chufus » Thu Nov 24, 2011 8:01 pm
There is definitely something wrong with the question.
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by vishal.pathak » Thu Nov 24, 2011 8:58 pm
adamz wrote:In a trial of a new drug, patients received a progressively growing dosage of the drug for a few days. On the first day, each patient received 15 milligrams of Drug. On each of the following days, the daily dosage was k milligrams greater than the dosage received the day before, reaching a dosage of 43 milligram on the last day of the trial. For how many days did the trial last, if each patient received a total amount of 145 milligrams of drug during the whole trial ?

(a) 4
(b) 5
(c) 6
(d) 7
(e) 9

OA:B
Assume that the no of days = n

Day 1: 15
Day 2: 15 + k
Day n: 15 + (n-1)k

15 + (n-1)k = 43 .......... (1)
If we sum all these terms, we will have 15n, k*(The sum of an AP). The AP will be (1 + 2 + .. + (n-1))

so
15n + k(n-1)(2 + (n-2))/2 = 145 ........... (2)(I have used the formula for the sum of an AP)

Putting the value of (n-1)k from eqn (1) in eqn(2), we get,

15n + 14n = 145
so n = 5. Therefore, answer is B
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by chufus » Fri Nov 25, 2011 1:30 am
Yup B is correct. Sorry for my earlier post.
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by neelgandham » Fri Nov 25, 2011 2:30 am
From the question, The amount(in milligrams) of daily dosage of the drug is in AP.

(number of terms/2)*(First Term + Last Term) = Sum of the Terms
(number of days/2)*(First Dosage + Last Dosage) = Total amount
(number of days/2)*(15+43) = 145
(number of days)*(15+43)/2 = 145
(number of days)*29 = 145
(number of days = 5)

Answer : B
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by Tani » Fri Nov 25, 2011 7:32 am
Try backsolving this one.
The dosage grew from 15 to 43 units, therefore the incremental dose on the last day was 28 units.
This means the total number of days in the study minus one (allowing the start day) has to be a factor of 28. Looking at our answer choices, only B works.
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by gmatpup » Fri Nov 25, 2011 7:51 am
Tani Wolff - Kaplan wrote:Try backsolving this one.
The dosage grew from 15 to 43 units, therefore the incremental dose on the last day was 28 units.
This means the total number of days in the study minus one (allowing the start day) has to be a factor of 28. Looking at our answer choices, only B works.

I did this exact thing!

I got the answer B by doing the following:

Since 43-15 = 28, again, we have to look for a factor of 28. I used 7.

Thus:

15 (add 7), 22 (add 7), 29 (add 7), 36 (add 7), 43 (add 7)

Add this up and we do end up getting the 145

So now we can see that it took 5 days :)

I hope this helps!
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by Tani » Fri Nov 25, 2011 8:00 am
Good work!

My most difficult teaching task is often convincing my strongest math students to use backsolving and picking numbers. Even if you can solve tough problems using sophisticated math, you can often save time and gain confidence by using simple strategies. Also, when faced with a tricky problem you will have more tools at your disposal.
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