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Difficulty: 555-605 Level,    Absolute Values,    Inequalities,                            
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
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 the val [#permalink]
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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 the val [#permalink]
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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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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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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
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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 the val [#permalink]
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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 the val [#permalink]
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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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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
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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 the val [#permalink]
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?
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
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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 the val [#permalink]
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 the val [#permalink]
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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"?
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
Expert Reply
AlN wrote:
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


Assuming s= -5, we get
\(r\leq{-5}\)

From where do you get \(r\geq{-5}\)?
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
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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


Let's see what happens when we minimize and maximize the value of s

GIVEN: |s| ≤ 5
So s = 5 is the GREATEST possible value of s
And s = -5 is the LEAST possible value of s

If s = 5, we get: 3r ≤ (4)(5) + 5
Simplify to get: 3r ≤ 25
Divide both sides by 3 to get: r ≤ 8.3333..

At this point, we don't have to explore what happens when we minimize the value of s, we can readily see that r CANNOT equal 20

Answer: E

Cheers,
Brent

Originally posted by BrentGMATPrepNow on 23 Dec 2019, 08:08.
Last edited by BrentGMATPrepNow on 19 Jul 2020, 10:48, edited 1 time in total.
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
Expert Reply
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


We rewrite the absolute value inequality for s as -5 ≤ s ≤ 5. If s = -5 (i.e., the smallest value it can be); then we have:

3r ≤ 4(-5) + 5

3r ≤ -15

r ≤ -5

If s = 5 (i.e., the largest value it can be); then we have:

3r ≤ 4(5) + 5

3r ≤ 25

r ≤ 25/3 = 8⅓

Therefore, we see that r can be any of the values in the given choices except 20.

Answer: E
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
Use hit and trial

Pick the largest value of r from the options, i.e. 20
To satisfy the equation containing largest value of r, we have to select the largest value of s, which will be 5 in this case.

Now, 3r=60, s=5 ->4s+5=25

Is 60<25? - No

So, r=20 can't be an acceptable value
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Re: Given the inequalities above, which of the following CANNOT be the val [#permalink]
Expert Reply
Here is my solution to the problem in a video. It basically comes down to the fact that the r is not min bound, but there is a max bound for r (8.33) .. so the value E 20 is not possible for 20.

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