Oct 22 08:00 AM PDT  09:00 AM PDT Join to learn strategies for tackling the longest, wordiest examples of Counting, Sets, & Series GMAT questions Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Current Student
Status: Persevere
Joined: 09 Jan 2016
Posts: 117
Location: Hong Kong
GPA: 3.52

Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
23 May 2016, 03:28
Question Stats:
70% (02:04) correct 30% (02:07) wrong based on 1968 sessions
HideShow timer Statistics
\(3r\leq{4s + 5}\) \(s\leq{5}\) Given the inequalities above, which of the following CANNOT be the value of r? A. –20 B. –5 C. 0 D. 5 E. 20
Official Answer and Stats are available only to registered users. Register/ Login.




Senior SC Moderator
Joined: 22 May 2016
Posts: 3565

Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
28 Aug 2017, 19:26
Quote: nalinnair wrote \(3r\leq{4s + 5}\) \(s\leq{5}\) Given the inequalities above, which of the following CANNOT be the value of r? A. –20 B. –5 C. 0 D. 5 E. 20 Saumya2403 wrote: Hi,
Please elaborate solution to this question, as above given solutions are a bit confusing Saumya2403 , I will try. We have to find which answer choice does not satisfy whatever solution, or range of solutions, that we find for \(r\). The first inequality defines the solutions or ranges of solutions for \(r\) in terms of \(s\). The second inequality defines the solutions for \(s\). So we should find out what to "plug in" for \(s\) first, i.e. find out what \(s\) might be in order to plug it into the first inequality. 1) \(s\leq{5}\) Remove the absolute value bars, and the expression translates to the compound expression \(5\leq{s}\leq{5}\) \(s\) lies between 5 and 5, inclusive. Breaking it down further Case One: \(s\geq {5}\), so we will plug in 5 for \(s\) in the first inequality to test the limits of the possible solutions for \(r\) Case Two: \(s\leq {5}\), so we will plug in 5 for \(s\) 2) Back to the first inequality: \(3r\leq{4s + 5}\) The solutions for \(r\) depend on the solutions for \(s\) that we just found. Case One: if \(s\geq {5}\), then \(3r\leq{4*(5) + 5}\) \(3r\leq {15}\) \(r\leq{5}\) That's one possible range of solutions for \(r\). < (5)Case Two: if \(s\leq {5}\), then \(3r\leq{4(5) + 5}\) \(3r\leq{25}\) \(r\leq{\frac{25}{3}}\), or \(r\leq{8.33}\) That's another range of solutions for \(r\) <0 8.33So the second range of solutions covers the first: < (5)0(8.33) 3) Which answer choice does not lie in that range of solutions? A) –20: that works. 20 is less than 8.33. KEEP B) –5: that works. 5 is less than 8.33. KEEP C) 0: that works. 0 is less than 8.33. KEEP D) 5: that works. 5 is less than 8. KEEP E) 20: that DOES NOT WORK. 20 is greater than 8.33. \(r\) must be LESS than or equal to 8.33. On the number line, 20 lies to the right of where we have defined the solutions for \(r\). 20 CANNOT be the value of \(r\). It's too large. REJECT Answer E Does that make sense?
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Instructions for living a life. Pay attention. Be astonished. Tell about it.  Mary Oliver




Math Expert
Joined: 02 Aug 2009
Posts: 8007

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
23 May 2016, 05:35
nalinnair wrote: \(3r\leq{4s + 5}\) \(s\leq{5}\)
Given the inequalities above, which of the following CANNOT be the value of r?
A. –20 B. –5 C. 0 D. 5 E. 20 Here lets check the range..
we have to check for extremes of s...\(s\leq{5}\)... \(5\leq{s}\leq{5}\)... 1) so lets see for s=5.. \(3r\leq{4s + 5}...........3r\leq{4*(5) + 5}...........................3r\leq{15}....................................r\leq{5}\)............ 1) now lets see for s=5.. \(3r\leq{4s + 5}...........3r\leq{4*(5) + 5}...........................3r\leq{25}....................................r\leq{\frac{25}{3}}\)............ we can see \(r\leq{\frac{25}{3}}\) covers up for BOTH the ranges.. ONLY E does not fit in.... ans E
_________________



Intern
Joined: 01 May 2015
Posts: 36

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
02 Jun 2016, 01:04
chetan2u wrote: we can see \(r\leq{\frac{25}{3}}\) covers up for BOTH the ranges.. This is where I get confused. Isn't r <= 5 actually covering for both the ranges?



Math Expert
Joined: 02 Aug 2009
Posts: 8007

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
02 Jun 2016, 06:41
sauravpaul wrote: chetan2u wrote: we can see \(r\leq{\frac{25}{3}}\) covers up for BOTH the ranges.. This is where I get confused. Isn't r <= 5 actually covering for both the ranges? Hi.. No, it doesn't .. check the two ranges.. 1) x<=5 and 2) x<=25/3...... from 1st x can be 7, 10 etc, they will be there in x<=25/3.. But x<=25/3 can have x as 0,2,5... it will NOT fall in x<=5.... so we require x<=25/3 to cover BOTH ranges
_________________



Current Student
Joined: 18 Oct 2014
Posts: 801
Location: United States
GPA: 3.98

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
02 Jun 2016, 06:53
nalinnair wrote: \(3r\leq{4s + 5}\) \(s\leq{5}\)
Given the inequalities above, which of the following CANNOT be the value of r?
A. –20 B. –5 C. 0 D. 5 E. 20 3r4s<=5 1 5>=s>= 52 Now lets put answer options in equation 1 keeping equation 2 in mind r=20. Even if s takes the smaller value 5 or +5, it will always be less than 5 r= 5. If s is +5 than the value of equation 1 will be less than 5 r=0. if s=+5 than value of equation 1 will be less than 5 r=5. if s= +5 than value of equation 1 will be less than 5 r=20. it doesn't matter weather the value of s is maximum or minimum, the value of equation 1 will always be greater than 5 E is the answer
_________________
I welcome critical analysis of my post!! That will help me reach 700+



VP
Joined: 05 Mar 2015
Posts: 1000

Re: Given the inequalities above, which of the following cannot be the val
[#permalink]
Show Tags
19 Jul 2016, 08:07
ameyaprabhu wrote: \(3r\leq{4s+5}\) \(s\leq{5}\) Given the inequalities above, which of the following cannot be the value of r? A) 20 B) 5 C) 0 D) 5 E) 20 The way I solved it is
step 1: 5 <= s <= 5
step 2: substitute s in eq 1. We get, r <= 25 / 3 or r <=  5
I am confused as to where I am going wrong step 1: 5 <= s <= 5 step 2: substitute s in eq 1. We get, r <= 25 / 3 or r <=  5 I am confused as to where I am going wrong[/spoiler][/quote] u r right on your solution so r<=25/3 >8.333 all options come under this range exept 20 whuch is >25/3 So Ans E



Senior Manager
Joined: 15 Jan 2017
Posts: 340

Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
07 Jul 2017, 12:30
I did this question algebraically  personally I find the above explanations harder to absorb other types of methods. I am going through them though But do let me know if my approach is correct, would be really thankful! so we have 3x<= 4s + 5, 4s + 5 must add to a multiple of 3*r So i plugged in values from given to check for incongruities: A. –20 => putting this in 4s +5, it divides, so 3r is a multiple of 4s + 5 B. –5 => again in 4s + 5; 3r<= 21, so its a multiple C. 0 => r = 5/ 3 not a multiple D. 5 = > r <= 25/ 3 E. 20 => r <= 85/ 3 between D and E can we check for the middle value => 85/3  25/3 = 60/3 = 20 so then R cannot be 20;based on the options below If we equate it to C, we again get > r = 25/3  5/ 3 = 20/3 non integer value and 85/ 3  5/3 = 80/3.So none of these non multiple forms except for D and E give R, hence E is the answer.



Intern
Joined: 21 Sep 2016
Posts: 36
Location: India
GPA: 3.55

Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
28 Aug 2017, 11:45
Hi,
Please elaborate solution to this question, as above given solutions are a bit confusing



Manager
Joined: 10 Sep 2014
Posts: 77
Location: Bangladesh
GPA: 3.5
WE: Project Management (Manufacturing)

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
20 Apr 2018, 07:36
chetan2u could you please elaborate these 2 lines from your solution "we can see r≤253r≤253 covers up for BOTH the ranges.. ONLY E does not fit in...." VeritasPrepKarishma, Bunuel please share your solution for this problem.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
20 Apr 2018, 21:00
nalinnair wrote: \(3r\leq{4s + 5}\) \(s\leq{5}\)
Given the inequalities above, which of the following CANNOT be the value of r?
A. –20 B. –5 C. 0 D. 5 E. 20 To get the range of r, we need the value of s. But what we have is the range for s. Let's evaluate it: \(s\leq{5}\) This means \(5 \leq s \leq 5\) Check at the extremes. s = 5 gives \(3r\leq{4*5 + 5}\) so we get \(r\leq{5}\) s = 5 gives \(3r\leq{4*5 + 5}\) so we get \(r\leq{8.33}\) Note that any intermediate value of s will ensure that r is less than 8.33. The maximum value that s can take is 5 and corresponding to that, the max value r can take is 8.33. Hence r can never be 20 Answer (E)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Status: Single
Joined: 02 Jan 2016
Posts: 175
Location: India
Concentration: General Management, Accounting
GPA: 4
WE: Law (Manufacturing)

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
20 Jun 2018, 19:51
VeritasPrepKarishma thats a great solution. I plugged in 5 being the greatest number for s, then it seemed only 20 was far beyond. hope thats correct too.



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3078

Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
21 Jun 2018, 00:08
Solution Given: To find: Among the 5 given option, which one cannot be the value of r. Approach and Working: We are given that 3r≤4s+5 Or r≤4s+5/3 Multiplying by 4 and adding by 5 in the above inequality, we get: 15 ≤ 4S+5 ≤ 25 Dividing the above inequality by 3, we get: 5 ≤ 4S+5/3 ≤ 25/3 So, we get the greatest value of 4s+5/3 as 25/3 and smallest value as 5. Now: When 4s+5 is 25/3 r ≤ 25/3 or r ≤8.33 Clearly, from this, r cannot be 20. Hence, option E is the answer. Answer: E
_________________



Manager
Joined: 17 Mar 2018
Posts: 70

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
29 Oct 2018, 02:10
I am still confused on this. I understand 20 is greater than 8.3, which is the limit of r on the right side, but why not 20? Please do help me understand why we are not considering 20 to NOT be a value of r, as the question asks which of the answer choices can NOT be r
Attachments
IMG_8096.JPG [ 1.87 MiB  Viewed 10664 times ]



Intern
Joined: 29 Nov 2015
Posts: 5

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
12 Jan 2019, 20:16
taniad wrote: I am still confused on this. I understand 20 is greater than 8.3, which is the limit of r on the right side, but why not 20?
Please do help me understand why we are not considering 20 to NOT be a value of r, as the question asks which of the answer choices can NOT be r I have the same question ... can anyone help? Bunel?



Math Expert
Joined: 02 Aug 2009
Posts: 8007

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
12 Jan 2019, 20:51
priarav wrote: taniad wrote: I am still confused on this. I understand 20 is greater than 8.3, which is the limit of r on the right side, but why not 20?
Please do help me understand why we are not considering 20 to NOT be a value of r, as the question asks which of the answer choices can NOT be r I have the same question ... can anyone help? Bunel? Hi.. The member above is ok with his final equation.. \(5\leq{\frac{4s+5}{3}}\leq{8.3}\) But you cannot substitute r for (4s+5)/3, because r can be LESS than this value too.. So the higher extreme will always stay because .. \(r\leq{\frac{4s+5}{3}}\leq{8.3}\) Also \(5\leq{\frac{4s+5}{3}}\) and \(r\leq{\frac{4s+5}{3}}\), so you cannot compare r and 5 as both can be equal or LESS than (4s+5)/3 Hope it helps Posted from my mobile device
_________________



Intern
Joined: 29 Nov 2015
Posts: 5

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
12 Jan 2019, 21:46
chetan2u wrote: priarav wrote: taniad wrote: I am still confused on this. I understand 20 is greater than 8.3, which is the limit of r on the right side, but why not 20?
Please do help me understand why we are not considering 20 to NOT be a value of r, as the question asks which of the answer choices can NOT be r I have the same question ... can anyone help? Bunel? Hi.. The member above is ok with his final equation.. \(5\leq{\frac{4s+5}{3}}\leq{8.3}\) But you cannot substitute r for (4s+5)/3, because r can be LESS than this value too.. So the higher extreme will always stay because .. \(r\leq{\frac{4s+5}{3}}\leq{8.3}\) Also \(5\leq{\frac{4s+5}{3}}\) and \(r\leq{\frac{4s+5}{3}}\), so you cannot compare r and 5 as both can be equal or LESS than (4s+5)/3 Hope it helps Posted from my mobile deviceThank you Chetan2u.



Manager
Joined: 02 Jan 2017
Posts: 59
Location: India

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
04 Apr 2019, 02:58
VeritasKarishma wrote: nalinnair wrote: \(3r\leq{4s + 5}\) \(s\leq{5}\)
Given the inequalities above, which of the following CANNOT be the value of r?
A. –20 B. –5 C. 0 D. 5 E. 20 To get the range of r, we need the value of s. But what we have is the range for s. Let's evaluate it: \(s\leq{5}\) This means \(5 \leq s \leq 5\) Check at the extremes. s = 5 gives \(3r\leq{4*5 + 5}\) so we get \(r\leq{5}\)s = 5 gives \(3r\leq{4*5 + 5}\) so we get \(r\leq{8.33}\) Note that any intermediate value of s will ensure that r is less than 8.33. The maximum value that s can take is 5 and corresponding to that, the max value r can take is 8.33. Hence r can never be 20 Answer (E) I am not able to understand the highlighted part . isnt r>=5



Manager
Joined: 02 Jan 2017
Posts: 59
Location: India

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
04 Apr 2019, 02:59
EgmatQuantExpert wrote: Solution Given: To find: Among the 5 given option, which one cannot be the value of r. Approach and Working: We are given that 3r≤4s+5 Or r≤4s+5/3 Multiplying by 4 and adding by 5 in the above inequality, we get: 15 ≤ 4S+5 ≤ 25 Dividing the above inequality by 3, we get: 5 ≤ 4S+5/3 ≤ 25/3 So, we get the greatest value of 4s+5/3 as 25/3 and smallest value as 5. Now: When 4s+5 is 25/3 r ≤ 25/3 or r ≤8.33 Clearly, from this, r cannot be 20. Hence, option E is the answer. Answer: EWhy cant A be the option, r is greater than 5 so 20 cant be the anser?



eGMAT Representative
Joined: 04 Jan 2015
Posts: 3078

Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
Show Tags
04 Apr 2019, 03:12
AlN wrote: Why cant A be the option, r is greater than 5 so 20 cant be the anser? Hey AlN, Can you please specify how are you concluding " r is greater than 5"?
_________________




Re: Given the inequalities above, which of the following CANNOT be........
[#permalink]
04 Apr 2019, 03:12



Go to page
1 2
Next
[ 23 posts ]



