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Given the inequalities above, which of the following CANNOT be........

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Given the inequalities above, which of the following CANNOT be........ [#permalink]

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\(3r\leq{4s + 5}\)
\(|s|\leq{5}\)

Given the inequalities above, which of the following CANNOT be the value of r?

A. –20
B. –5
C. 0
D. 5
E. 20
[Reveal] Spoiler: OA

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Re: Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 23 May 2016, 05:35
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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
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 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?

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Re: Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 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
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Re: Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 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


3r-4s<=5------ 1
5>=s>= -5-------2

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
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Re: Given the inequalities above, which of the following cannot be the val [#permalink]

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New post 19 Jul 2016, 08:07
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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

[Reveal] Spoiler:
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

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Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 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.

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Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 28 Aug 2017, 11:45
Hi,

Please elaborate solution to this question, as above given solutions are a bit confusing

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Given the inequalities above, which of the following CANNOT be........ [#permalink]

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New post 28 Aug 2017, 19:26
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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--------8.33

So 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 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? :-)

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Given the inequalities above, which of the following CANNOT be........   [#permalink] 28 Aug 2017, 19:26
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Given the inequalities above, which of the following CANNOT be........

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