Events & Promotions
|
|
GMAT 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.
Nov 1912:30 PM EST
-01:30 PM EST
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2001:30 PM EST
-02:30 PM IST
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
02 Jul 2025, 08:59
Kudos
Bookmarks
initial solution= 2% salt which is 98% water
let, initial weight= 100g (2g salt + 98g water)
after adding water
let, water= X g
total salt remains= 2g
2/100+x=0.01
x=100g
final water: 98g + 100g= 198g
final ration( final water: inital water)
198/98= 99/49
ans: D (99:49)
let, initial weight= 100g (2g salt + 98g water)
after adding water
let, water= X g
total salt remains= 2g
2/100+x=0.01
x=100g
final water: 98g + 100g= 198g
final ration( final water: inital water)
198/98= 99/49
ans: D (99:49)
02 Jul 2025, 09:00
Kudos
Bookmarks
Given conditions,
Initial total weight of the solution = W1
Initial weight of salt = S
Initial weight of water = H1
Given that the saline solution is 2% salt by weight initially, S=0.02×W1
The rest is water, so:
H1 =W1−S=0.98W1
Now, more water is added.
Final total weight of the solution = W2
Final weight of salt = S (amount of salt remains the same, only water is added)
Final weight of water = H2
After water is added, the salt concentration decreases to 1% by weight i.e.,S=0.01×W2
We have two expressions for the weight of salt S: 0.02W1 =0.01W2
W1 =2W1
the final weight of water i.e., H2 = W2 −S
Substitute S=0.02W1 and W2 =2W1
H2=2W1 −0.02W1 =(2−0.02)W1 =1.98W1
ratio of the final weight of water / the initial weight of water:
H2/H1 =1.98/0.98 =198/98
So, the ratio is 99/49
.
Initial total weight of the solution = W1
Initial weight of salt = S
Initial weight of water = H1
Given that the saline solution is 2% salt by weight initially, S=0.02×W1
The rest is water, so:
H1 =W1−S=0.98W1
Now, more water is added.
Final total weight of the solution = W2
Final weight of salt = S (amount of salt remains the same, only water is added)
Final weight of water = H2
After water is added, the salt concentration decreases to 1% by weight i.e.,S=0.01×W2
We have two expressions for the weight of salt S: 0.02W1 =0.01W2
W1 =2W1
the final weight of water i.e., H2 = W2 −S
Substitute S=0.02W1 and W2 =2W1
H2=2W1 −0.02W1 =(2−0.02)W1 =1.98W1
ratio of the final weight of water / the initial weight of water:
H2/H1 =1.98/0.98 =198/98
So, the ratio is 99/49
.
02 Jul 2025, 09:02
Kudos
Bookmarks
From the details given,
Salt content does not change; Water is added
1.)Before adding water:
Weight of salt = sw = 2g (assume)
Let Initial weight of water = wi = 98g (assume)
Let Initial weight of water = wf
We are to find out wf/wi
2.)After adding water:
Since we know salt is not added, sw = 2g(as per assumption)
It's mentioned that the salt content becomes 1% by weight of the saline solution
To find final weight of solution,
Let total weights before and after adding water be ti & tf
.01 * tf = 2
∴tf = 200; ∴ wf = tf -2 = 198
∴wf/wi = 198/98
∴Ration = 99:49
Correct option is Option D
Salt content does not change; Water is added
1.)Before adding water:
Weight of salt = sw = 2g (assume)
Let Initial weight of water = wi = 98g (assume)
Let Initial weight of water = wf
We are to find out wf/wi
2.)After adding water:
Since we know salt is not added, sw = 2g(as per assumption)
It's mentioned that the salt content becomes 1% by weight of the saline solution
To find final weight of solution,
Let total weights before and after adding water be ti & tf
.01 * tf = 2
∴tf = 200; ∴ wf = tf -2 = 198
∴wf/wi = 198/98
∴Ration = 99:49
Correct option is Option D
