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 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 Feb 2021, 07:30
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
62% (03:03) correct
38%
(03:15)
wrong
based on 129
sessions
History
Date
Time
Result
Not Attempted Yet
A boat takes 35 minutes less to travel 28 km downstream than it takes to travel the same distance upstream. If the speed of the boat in still water is 14kmph, what is the speed of the stream ? (in kmph)
(A) 10
(B) 8
(C) 6
(D) 4
(E) 2
(A) 10
(B) 8
(C) 6
(D) 4
(E) 2
03 Feb 2021, 21:18
Kudos
Bookmarks
\(T_1 =\frac{28}{14-W}\)
\(T_2=\frac{28}{14+W}\)
\(T_1-T_2=\frac{35}{60}\)
\(\frac{28}{14-W} - \frac{28}{14+W} = \frac{35}{60}\) In this stage try to plug in the numbers (or POE) and get the answer quickly
\(\frac{1}{14-W} - \frac{1}{14+W} = \frac{35}{60*28}=\frac{1}{48}\)
\(\frac{14+W-(14-W)}{(14^2-W^2)}=\frac{1}{48}\)
\(48*2W=196-W^2\)
\(W^2+96W-196=0\)
\(W^2+98W-2W-196=0\)
\((W+98)(W-2)=0\)
\(W=2,-98\)
Ans E
\(T_2=\frac{28}{14+W}\)
\(T_1-T_2=\frac{35}{60}\)
\(\frac{28}{14-W} - \frac{28}{14+W} = \frac{35}{60}\) In this stage try to plug in the numbers (or POE) and get the answer quickly
\(\frac{1}{14-W} - \frac{1}{14+W} = \frac{35}{60*28}=\frac{1}{48}\)
\(\frac{14+W-(14-W)}{(14^2-W^2)}=\frac{1}{48}\)
\(48*2W=196-W^2\)
\(W^2+96W-196=0\)
\(W^2+98W-2W-196=0\)
\((W+98)(W-2)=0\)
\(W=2,-98\)
Ans E
13 Feb 2021, 07:14
Kudos
Bookmarks
Bunuel
Solution:
Letting r be the speed of the stream in kmph and t be the time, in hours, for the boat to travel upstream, we can create the equations:
(14 - r) * t = 28
and
(14 + r) * (t - 35/60) = 28
From the first equation, we see that t = 28 / (14 - r). Substituting this into the second equation, we have:
(14 + r) * (28 / (14 - r) - 7/12) = 28
Multiplying the equation by 12(14 - r), we have:
(14 + r) *[(28 * 12 - 7(14 - r)] = 28 * 12(14 - r)
Dividing the equation by 7, we have:
(14 + r) * (4 * 12 - (14 - r)) = 4 * 12(14 - r)
(14 + r)(34 + r) = 48(14 - r)
476 + 48r + r^2 = 672 - 48r
r^2 + 96r - 196 = 0
(r - 2)(r + 98) = 0
r = 2 or r = -98
Since r can’t be negative, r = 2.
Answer: E
