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555-605 (Medium)|   Absolute Values|   Inequalities|                           
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KarishmaB
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Given the inequalities above, which of the following CANNOT be the val 26 Aug 2021, 21:29
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VeritasKarishma
nalinnair
\(3r\leq{4s + 5}\)
\(|s|\leq{5}\)

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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Quote:

If we plug r= -20, we get s≥−16.25
s ≥− 16.25

But as per the second condition −5≥s≤5
The derived range of s does not satisfies this condition completely.
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When we put r = -20, we get that s must be greater than or equal to -16.25.
If s lies between -5 and 5, we know that s will be greater than -16.25.
Hence there is no conflict among the inequalities.


When we put r = 20, we get s >= 13.75
But s must be between -5 and 5. It cannot be greater than or equal to 13.75.
Hence, r cannot be 20.
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Given the inequalities above, which of the following CANNOT be the val 17 Sep 2022, 01:19
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KarishmaB

If s is between -5 AND 5, why we are taking "Union" of values for r ( r<-5 and r<8.33) instead of intersection.
I am confused with interpreting the inequality r<=4s+5/3 in terms of min value of r.
Even though s is bounded by min value (-5), r is not bounded by a min value. Why?

Can you explain using graphical representation of inequality for r?

Thank you for your help.
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Given the inequalities above, which of the following CANNOT be the val 18 Sep 2022, 22:08
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Sneha2021
KarishmaB

If s is between -5 AND 5, why we are taking "Union" of values for r ( r<-5 and r<8.33) instead of intersection.
I am confused with interpreting the inequality r<=4s+5/3 in terms of min value of r.
Even though s is bounded by min value (-5), r is not bounded by a min value. Why?

Can you explain using graphical representation of inequality for r?

Thank you for your help.
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We are given:

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

For every value of s, we will get a different inequality.

If s = -5, we get r <= -5
If s = -4, we get r <= -11/3
...
If s = 0, we get r <= 5/3
...
If s = 5, we get r <= 25/3

So note that if we take the value of r to be -5 or less than -5, it will satisfy all these inequalities. That is, no matter what the allowed value of s, \(r\leq{\frac{(4s + 5)}{3}}\)

Is there is minimum value r must take? No. It should just not be greater than -5.

Think of it this way: Given a <= -10
What is the minimum value of a here? We don't have a minimum value of a. a could take any value less than -10. It's maximum value is -10.
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Given the inequalities above, which of the following CANNOT be the val 22 Sep 2023, 13:59
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The maximum value of S would be 5 (Since we know that ABS(S) <= 5)
So the maximum value of 4S + 5 would be 25
r cannot be greater than 25...
r = 20 is the only answer choice that wouldn't work.
For anything below 3*r = 25, it is possible
that's what I understood from the question, not sure tho
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Given the inequalities above, which of the following CANNOT be the val 27 Aug 2024, 07:47
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The inequality is \( r \leq \frac{4s + 5}{3} .\)

There is no lower limit for r, but there is an upper limit.

Hence, r cannot be equal to the largest value out of the given options.

To understand, If r cannot be 5, it cannot be 20 also. This is because there is an upper limit for value of r.

Since we cannot have more than one correct answer, r CANNOT be the largest value 20.

Correct answer is E.
nalinnair
\(3r\leq{4s + 5}\)
\(|s|\leq{5}\)

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 the val 10 May 2025, 10:35
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\(|s|≤5 \implies -5 \leq s \leq 5\\
\\\\
\frac{3r - 5}{4} \leq s \implies \frac{3r - 5}{4} \leq s \leq 5 \implies 3r \leq \frac{25}{3}\)

i.e r can't take the value 20
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Given the inequalities above, which of the following CANNOT be the val 03 Nov 2025, 03:05
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You are equating r and (4s+5)/3 and thus substituting r with that expression. But they are not equal, all we know is r is < (4s+5)/3, thus we can only substitute value of s in the equation.
taniad
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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Given the inequalities above, which of the following CANNOT be the val 23 Mar 2026, 05:57
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I recommend using value of r in the options rather than using range of s. It will be quicker
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