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.
Apr 2711:00 AM EDT
-12:00 PM EDT
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
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
12 Oct 2017, 10:49
Kudos
Bookmarks
if:
a/b = c/d (a,b,c,d non zero real numbers), it is given that:
1. (a+b)/(a-b) = (c+d)/(c-d)
Clarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)
How did we (algebraically) arrive at this equivalency?
Clarification 2: can we also claim that: (a-b)/(a+b) = (c-d)/(c+d) ?
Thanks
a/b = c/d (a,b,c,d non zero real numbers), it is given that:
1. (a+b)/(a-b) = (c+d)/(c-d)
Clarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)
How did we (algebraically) arrive at this equivalency?
Clarification 2: can we also claim that: (a-b)/(a+b) = (c-d)/(c+d) ?
Thanks
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!
18 Oct 2017, 10:48
Kudos
Bookmarks
Hi sevenplusplus
Clarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)
How did we (algebraically) arrive at this equivalency? - Here the book only gives the rule that can be applied to ratios. Incase a-b=0, a=b, ratio is 1, so there's no need to go for componendo dividendo. It is useful to apply in some problems directly.
\(\frac{a}{b}\) = \(\frac{c}{d}\) =>Adding and subtracting 1 from both sides;
\(\frac{a}{b}\)+1=\(\frac{c}{d}\)+1 = \(\frac{(a+b)}{b}\)=\(\frac{(c+d)}{d}\) -------(1)
\(\frac{a}{b}\)-1=\(\frac{c}{d}\)-1 = \(\frac{(a-b)}{b}\)=\(\frac{(c-d)}{d}\)--------(2)
Dividing (1) by (2) gives the desired result.
Clarification 2: can we also claim that: \(\frac{(a-b)}{(a+b)}\) = \(\frac{(c-d)}{(c+d)}\) ?
Yes. As it is inverse of the ratio.
Please give kudos, if u liked my post!
Clarification 1: Now 1. above can only be true if a is not equal to b or (a-b) is not zero ..right? (this is not mentioned in the book)
How did we (algebraically) arrive at this equivalency? - Here the book only gives the rule that can be applied to ratios. Incase a-b=0, a=b, ratio is 1, so there's no need to go for componendo dividendo. It is useful to apply in some problems directly.
\(\frac{a}{b}\) = \(\frac{c}{d}\) =>Adding and subtracting 1 from both sides;
\(\frac{a}{b}\)+1=\(\frac{c}{d}\)+1 = \(\frac{(a+b)}{b}\)=\(\frac{(c+d)}{d}\) -------(1)
\(\frac{a}{b}\)-1=\(\frac{c}{d}\)-1 = \(\frac{(a-b)}{b}\)=\(\frac{(c-d)}{d}\)--------(2)
Dividing (1) by (2) gives the desired result.
Clarification 2: can we also claim that: \(\frac{(a-b)}{(a+b)}\) = \(\frac{(c-d)}{(c+d)}\) ?
Yes. As it is inverse of the ratio.
Please give kudos, if u liked my post!
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!
