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.
May 2008:00 AM PDT
-08:30 AM PDT
What’s in it for you- Live Profile Evaluation Chat Session with Jenifer Turtschanow, CEO, ARINGO. Come with your details prepared and ARINGO will share insights! Pre-MBA Role/Industry, YOE, Exam Score, C/GPA, ECs Post-MBA Role/ Industry & School List.
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
17 Apr 2020, 19:40
Kudos
Bookmarks
[GMAT math practice question]
(Algebra) What is the value of \(x^5 + y^5\)?
1) \(x + y = 3.\)
2) \(xy = 2.\)
(Algebra) What is the value of \(x^5 + y^5\)?
1) \(x + y = 3.\)
2) \(xy = 2.\)
Updated on: 30 May 2021, 02:41
Originally posted by MathRevolution on 19 Apr 2020, 20:37.
Last edited by MathRevolution on 30 May 2021, 02:41, edited 1 time in total.
Last edited by MathRevolution on 30 May 2021, 02:41, edited 1 time in total.
Kudos
Bookmarks
MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary. A standard deviation is the square root of an average of squares minus the square of the average. Thus, C is the answer.
The following is more detailed reasoning.
Assume \(m\) is the average of x1, x2, …, xn and p is the average of x1^2, x2^2, …, xn^2.
Then we have (x1 + x2 + … + xn) / n = m or (x1 + x2 + … + xn) = mn, and we have
(x1^2+x2^2+⋯+xn^2)/n= p or (x1^2+x2^2+⋯+xn^2)= pn.
Then we have
(x1-m)^2+(x2-m)^2+⋯+(xn-m)^2/n
= x1^2 + x2^2 +⋯+ xn^2- 2m(x1 + x2 + ⋯ + xn) + nm^2/n
= np - 2m∙nm + nm^2/n = p - m^2
The standard deviation is \(\sqrt{p-m^2}.\)
Since we have m = 1 and p = 5 from both conditions 1) and 2), we have the standard deviation \(\sqrt{5-1^2}=\sqrt{4}=2.\)
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
Updated on: 30 May 2021, 02:42
Originally posted by MathRevolution on 19 Apr 2020, 20:37.
Last edited by MathRevolution on 30 May 2021, 02:42, edited 1 time in total.
Last edited by MathRevolution on 30 May 2021, 02:42, edited 1 time in total.
Kudos
Bookmarks
MathRevolution
=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.
The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
Since we have \(2\) variables (\(x\) and \(y\)) and \(0\) equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.
Conditions 1) & 2)
We have \(x = 1, y = 2\) or \(x = 2, y = 1.\)
Then we have \(x^5 + y^5 = 33\) for both cases.
Since both conditions together yield a unique solution, they are sufficient.
Therefore, C is the answer.
Answer: C
Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
