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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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New post 23 May 2016, 03:28
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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
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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 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? :-)
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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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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
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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


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

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

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

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New post 20 Apr 2018, 21:00
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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


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

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

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New post 21 Jun 2018, 00:08
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Solution



Given:
    3r≤4s+5
    |s|≤5

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

    |s|≤5
    -5≤ S ≤ 5
    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
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New post 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
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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

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New post 12 Jan 2019, 20:51
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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

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

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New post 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 device


Thank you Chetan2u.
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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

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New post 04 Apr 2019, 02:59
EgmatQuantExpert wrote:

Solution



Given:
    3r≤4s+5
    |s|≤5

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

    |s|≤5
    -5≤ S ≤ 5
    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


Why cant A be the option, r is greater than -5 so -20 cant be the anser?
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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

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