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.
Sep 1008:00 PM PDT
-09:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
Sep 0812:30 AM EDT
-01:30 AM EDT
Looking for your GMAT motivation to break through the score plateau? Pragati improved her score by massive 160 points with strategic guidance and hard-work! Find out how personalized mentorship and a strong mindset can turn GMAT struggles into success.
Sep 1001:00 AM EDT
-02:00 AM EDT
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.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
31 May 2017, 00:54
Kudos
Bookmarks
Hi, some where I found wonderful explanation on multiple rate problem, but they lack any further explanation. It would be grateful if the quant wizards can explain them with examples.
If A can finish a work in X time and B can finish the
same work in Y time then both of them together can
finish that work in (X*Y)/ (X+Y) time.
If A can finish a work in X time and A & B together can
finish the same work in S time then B can finish that
work in (XS)/(X-S) time.
If A can finish a work in X time and B in Y time and C in
Z time then all of them working together will finish the
work in (XYZ)/ (XY +YZ +XZ) time
If you found this post great, kindly appreciate me with an kudos
If A can finish a work in X time and B can finish the
same work in Y time then both of them together can
finish that work in (X*Y)/ (X+Y) time.
If A can finish a work in X time and A & B together can
finish the same work in S time then B can finish that
work in (XS)/(X-S) time.
If A can finish a work in X time and B in Y time and C in
Z time then all of them working together will finish the
work in (XYZ)/ (XY +YZ +XZ) time
If you found this post great, kindly appreciate me with an kudos
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
31 May 2017, 07:54
Kudos
Bookmarks
Since the work done is same throughout the question so for simplicity work done can be taken as 1
Rate*Time = Work
Let Rate of A be A ,Rate of B be B and Rate of C be C
If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
B can finish the same work in Y
=> B * Y = 1 (Work =1)
=> B = 1/Y
When they both work together then their combined rate = Sum of their rates = A + B
(A+B)*T = 1 (T is the time they take to complete the work)
(1/X + 1/Y)*T =1
=> (X+Y)*T/XY = 1
=> T= XY/(X+Y)
If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
A & B together can finish the same work in S time
=> (A+B)*S = 1
=> A+B = 1/S
=> B = 1/S - A = 1/S - 1/X = (X-S)/XS
Let time taken by B to finish the work = T
B * T = 1
=> T = 1/B = 1/ (X-s)/XS = XS/(X-S)
If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
A can finish a work in X time
=> A = 1/X (Same as above)
B in Y time and C in Z time
=> B = 1/Y and C = 1/Z
If they are working together then
(A+B+C)*T = 1
=> (1/X + 1/Y + 1/Z)*T=1
=> (YZ + XZ + XY)*T/XYZ = 1
=> T = XYZ/(YZ + XZ + XY)
Hope it helps!
Rate*Time = Work
Let Rate of A be A ,Rate of B be B and Rate of C be C
If A can finish a work in X time and B can finish the same work in Y time then both of them together can finish that work in (X*Y)/ (X+Y) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
B can finish the same work in Y
=> B * Y = 1 (Work =1)
=> B = 1/Y
When they both work together then their combined rate = Sum of their rates = A + B
(A+B)*T = 1 (T is the time they take to complete the work)
(1/X + 1/Y)*T =1
=> (X+Y)*T/XY = 1
=> T= XY/(X+Y)
If A can finish a work in X time and A & B together can finish the same work in S time then B can finish that work in (XS)/(X-S) time.
A can finish a work in X time
=> A * X = 1 (Work =1)
=> A = 1/X
A & B together can finish the same work in S time
=> (A+B)*S = 1
=> A+B = 1/S
=> B = 1/S - A = 1/S - 1/X = (X-S)/XS
Let time taken by B to finish the work = T
B * T = 1
=> T = 1/B = 1/ (X-s)/XS = XS/(X-S)
If A can finish a work in X time and B in Y time and C in Z time then all of them working together will finish the work in (XYZ)/ (XY +YZ +XZ) time
A can finish a work in X time
=> A = 1/X (Same as above)
B in Y time and C in Z time
=> B = 1/Y and C = 1/Z
If they are working together then
(A+B+C)*T = 1
=> (1/X + 1/Y + 1/Z)*T=1
=> (YZ + XZ + XY)*T/XYZ = 1
=> T = XYZ/(YZ + XZ + XY)
Hope it helps!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
