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

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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, 21:52
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........  [#permalink]

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New post 05 Aug 2019, 08:44
Simplest way to solve this:

Given | s | <= 5
=> -5 <= s <= 5
=> -20 <= 4s <= 20 ( Sign remains unchanged if the inequalities divided/multiplied/added or subtracted by SAME +VE NUMBER)
=> -15 <= 4s + 5 <= 25
=> -5 <= (4s + 5)/3 <= 8.33

Now we know that r <= (4s+5)/3
so r's range can be r <= -5, -5 <=r <= (4s+5)/3 <=8.33

Now we know that r can NEVER exceed 8.33 -> 20 is our ANSWER.

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

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New post 20 Aug 2019, 08:42
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


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

\(|s|\leq{5}\) --> \({-5}\leq s\leq{5}\)

MAX S = 5, so \(3r-5\leq{20}\)
Testing E, r = 20 gives us a false statement \(55\leq{20}\)
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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New post 30 Nov 2019, 23:45
generis wrote:
Quote:
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,



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? :-)


Why cant we insert value of S in 3r<=4s+5 equation and check if s is between -5 and 5? when i do this, i got 13.75<=s for (a). Which I thought that was the answer
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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New post 23 Dec 2019, 07:08
Top Contributor
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, w hey buddye can readily see that r CANNOT equal 20

Answer: E

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

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New post 21 Mar 2020, 14:40
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........   [#permalink] 21 Mar 2020, 14:40

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