GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
A nurse has 80 ml of a solution that has a concentration of 5% Drug X
[#permalink]
23 Mar 2021, 05:49
23 Mar 2021, 05:49
Expert Reply
5
Bookmarks
00:00
Question Stats:
80% (01:47) correct
20% (02:14) wrong based on 35 sessions
Hide Show timer Statistics
A nurse has 80 ml of a solution that has a concentration of 5% Drug X in water. How many milliliters of pure water containing none of the drug must the nurse add to dilute the concentration to 2% Drug X in water?
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Show: :: OA
120 ml
Kudos for the right answer and explanation
Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A nurse has 80 ml of a solution that has a concentration of 5% Drug X
[#permalink]
23 Mar 2021, 08:40
23 Mar 2021, 08:40
Carcass wrote:
A nurse has 80 ml of a solution that has a concentration of 5% Drug X in water. How many milliliters of pure water containing none of the drug must the nurse add to dilute the concentration to 2% Drug X in water?
Show: :: OA
120 ml
Let us make an equation for the Water:
95%(80) + W = 98%(80 + W)
\(76 + W = 78.4 + 0.98W\)
\(0.02W = 2.4\)
\(W = 120\)
General Discussion
Re: A nurse has 80 ml of a solution that has a concentration of 5% Drug X
[#permalink]
23 Mar 2021, 06:51
23 Mar 2021, 06:51
3
80 ml of a solution that has a concentration of 5% Drug X in water.
So, Drug X = 80 * 0.05 = 4 ml
This 4 ml needs to be 2% in the new solution.
So, the new solution will be = 4 /0.02 = 200 ml
Finally, the amount of water needed = 200 - 80 = 120 ml
So, Drug X = 80 * 0.05 = 4 ml
This 4 ml needs to be 2% in the new solution.
So, the new solution will be = 4 /0.02 = 200 ml
Finally, the amount of water needed = 200 - 80 = 120 ml
