A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.
A) 20%
B) 30.72%
C) 40.23%
D) 52.15%
E) 63.2%
[spoiler](B)[/spoiler]
@GMATGuruNY
A vessel ( capacity 10 ltrs) is filled entirely with petrol
This topic has expert replies
 shivanshchauhan94
 Newbie  Next Rank: 10 Posts
 Posts: 3
 Joined: Sat Jul 29, 2017 11:28 pm
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats

 Legendary Member
 Posts: 2214
 Joined: Fri Mar 02, 2018 2:22 pm
 Followed by:5 members
$$Initial\ volume\ of\ petrol=\frac{3}{5}\cdot10=\frac{30}{5}=6$$
Let the volume of petrol after 3 replacement = x
$$x=6\left(1\frac{2}{10}\right)^{^3}$$
$$x=6\left(\frac{102}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{8}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{4}{5}\right)^{^{^3}}$$
$$x=6\cdot\frac{64}{125}=6\cdot0.512=3.072$$
$$\%\ of\ petrol\ in\ the\ final\ solution=\frac{3.072}{10}\cdot100\ =30.72\%$$
Hence, option B is the correct answer.
Thanks
Let the volume of petrol after 3 replacement = x
$$x=6\left(1\frac{2}{10}\right)^{^3}$$
$$x=6\left(\frac{102}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{8}{10}\right)^{^{^3}}$$
$$x=6\left(\frac{4}{5}\right)^{^{^3}}$$
$$x=6\cdot\frac{64}{125}=6\cdot0.512=3.072$$
$$\%\ of\ petrol\ in\ the\ final\ solution=\frac{3.072}{10}\cdot100\ =30.72\%$$
Hence, option B is the correct answer.
Thanks
GMAT/MBA Expert
 Scott@TargetTestPrep
 GMAT Instructor
 Posts: 7115
 Joined: Sat Apr 25, 2015 10:56 am
 Location: Los Angeles, CA
 Thanked: 43 times
 Followed by:29 members
We can represent the initial amount of petrol and kerosene in the mixture by 3x and 2x, for some positive number x. Since 3x + 2x = 10, we find x = 2 and thus, there are 3 * 2 = 6 liters of petrol and 2 * 2 = 4 liters of kerosene in the mixture initially.shivanshchauhan94 wrote:A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.
A) 20%
B) 30.72%
C) 40.23%
D) 52.15%
E) 63.2%
[spoiler](B)[/spoiler]
Since 2 liters of mixture is replaced by 2 liters of kerosene, the total mixture is always 10 liters.
Notice that the number of liters of kerosene decreases by the amount of kerosene in the 2 liters of mixture and increases by 2 liters after each iteration, whereas the amount of petrol decreases by the amount of petrol in 2 liters of mixture after each iteration and never increases; therefore, it will be easier to calculate the amount of petrol in the mixture after each iteration.
After the first iteration, 2 * (6/10) = 1.2 liters of petrol is removed and thus, there are 6  1.2 = 4.8 liters of petrol remaining in the mixture.
After the second iteration, 2 * (4.8/10) = 0.96 liters of petrol is removed and thus, there are 4.8  0.96 = 3.84 liters of petrol remaining in the mixture.
After the third iteration, 2 * (3.84/10) = 0.768 liters of petrol is removed and thus, there are 3.84  0.769 = 3.072 liters of petrol remaining in the mixture.
Thus, after the third iteration, the mixture contains (3.072/10) * 100 = 30.72% petrol.
Answer: B
Scott WoodburyStewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
 GMATGuruNY
 GMAT Instructor
 Posts: 15539
 Joined: Tue May 25, 2010 12:04 pm
 Location: New York, NY
 Thanked: 13060 times
 Followed by:1906 members
 GMAT Score:790
Since the question stem asks for a PERCENTAGE, the original amount of petrol is irrelevant.shivanshchauhan94 wrote:A vessel ( capacity 10 ltrs) is filled entirely with petrol and kerosene in the ratio 3:2. Two ltrs of this solution is removed and replaced with 2 ltrs of kerosene. This procedure is repeated thrice. Find the % of petrol in the resulting solution.
A) 20%
B) 30.72%
C) 40.23%
D) 52.15%
E) 63.2%
In the original solution, petrol:kerosene = 3:2, implying that petrol constitutes 3 of every 5 liters.
Thus, petrol = 3/5.
Each time 2 liters are removed from the 10liter vessel, 8 liters remain.
Since 8/10 = 4/5, left in the vessel after each removal process will be 4/5 of the amount of petrol.
Since the 4/5 process is performed THREE TIMES upon the original 3/5 portion of petrol, we get:
3/5 * 4/5 * 4/5 * 4/5 = 192/625 â‰ˆ 200/600 = 1/3 â‰ˆ 30%
The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 testtakers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia  a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua  a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 testtakers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia  a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua  a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3