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.
Jul 0901:00 PM EDT
-02:00 PM EDT
More Target Test Prep students are earning 100th percentile GMAT scores. Don’t settle for less when the Target Test Prep course gives you everything you need to score high on the GMAT. Start your 5-day free trial today.
Jul 1206:30 AM PDT
-08:30 AM PDT
Join our webinar to master GMAT RC Main Point Questions! Learn effective strategies to decipher passage themes and purposes. Senior Verbal Expert- Sunita Singhvi will equip you with tools to navigate complexities and avoid common traps set by the GMAT.
Jul 1311:00 AM EDT
-12:30 PM 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.
Jul 1508:00 PM PDT
-09:00 PM PDT
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Jul 1611:00 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.
22 Jun 2021, 08:45
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (02:02) correct
43%
(02:20)
wrong
based on 175
sessions
History
Date
Time
Result
Not Attempted Yet
If \(p + q + r = 0\), where \(p ≠ q ≠ r\), then \(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq}\) ?
(A) 0
(B) 1
(C) –1
(D) pqr
(E) p + q + r
(A) 0
(B) 1
(C) –1
(D) pqr
(E) p + q + r
22 Jun 2021, 09:09
Kudos
Bookmarks
Asked: If \(p + q + r = 0\), where \(p ≠ q ≠ r\), then \(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq}\) ?
Let p = 3; q = -2; r= -1
p + q + r = 0
\(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq} = \frac{9}{20} +\frac{ 4}{5} + \frac{1}{(-4)} = \frac{9}{20} + \frac{16}{20} - \frac{5}{20} = \frac{20}{20} = 1\)
(A) 0; Incorrect
(B) 1; Correct
(C) –1; Incorrect
(D) pqr; par = 6; Incorrect
(E) p + q + r=0; Incorrect
IMO B
Let p = 3; q = -2; r= -1
p + q + r = 0
\(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq} = \frac{9}{20} +\frac{ 4}{5} + \frac{1}{(-4)} = \frac{9}{20} + \frac{16}{20} - \frac{5}{20} = \frac{20}{20} = 1\)
(A) 0; Incorrect
(B) 1; Correct
(C) –1; Incorrect
(D) pqr; par = 6; Incorrect
(E) p + q + r=0; Incorrect
IMO B
22 Jun 2021, 11:38
Kudos
Bookmarks
Bunuel
If \(p + q + r = 0\), where \(p ≠ q ≠ r\), then \(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq}\) ?
(A) 0
(B) 1
(C) –1
(D) pqr
(E) p + q + r
(A) 0
(B) 1
(C) –1
(D) pqr
(E) p + q + r
Solution :
We can answer this question by plugging in values of p. q and r.
Let p = 1, q = 0 and r = -1 such that p + q + r = 0 and p ≠ q ≠ r
Then, we get :
\(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq}\)
= \(\frac{1^2}{2 + 0} + 0 + \frac{(-1)^2}{2 + 0 }\)
= \(\frac{1}{2 }+ \frac{1}{2}\)
= \(1\)
Thus, we can eliminate options A, D and E as their value is each equal to be 0
Now, if p = 1, q = 2 and r = -3, then we have :
\(\frac{p^2}{2p^2 + qr} + \frac{q^2}{2q^2 + pr} + \frac{r^2}{2r^2 + pq}\)
= \(\frac{1^2}{2*(1)^2 - 6} + \frac{2^2}{2 * (2)^2 - 3} + \frac{(-3)^2}{2 * (-3)^2 + 2}\)
= \(\frac{1}{-4 }+ \frac{4}{5}\) + \(\frac{9}{20}\)
= \(1\)
Hence, the correct answer is Option B.
