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Re: Given the inequalities above, which of the following CANNOT be the val
[#permalink]
26 Aug 2021, 21:29
26 Aug 2021, 21:29
Expert Reply
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
\(|s|\leq{5}\)
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)
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.
s ≥− 16.25
But as per the second condition −5≥s≤5
The derived range of s does not satisfies this condition completely.
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.
Re: Given the inequalities above, which of the following CANNOT be the val
[#permalink]
17 Sep 2022, 01:19
17 Sep 2022, 01:19
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.
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.
Given the inequalities above, which of the following CANNOT be the val
[#permalink]
18 Sep 2022, 22:08
18 Sep 2022, 22:08
Expert Reply
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.
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.
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.
