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

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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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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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)

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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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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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

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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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

Hi,

Please elaborate solution to this question, as above given solutions are a bit confusing

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
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Hey AlN,

Can you please specify how are you concluding "r is greater than -5"?
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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

This is where I get confused. Isn't r <= -5 actually covering for both the ranges?

I have the same question ... can anyone help?
Bunel?

Why cant A be the option, r is greater than -5 so -20 cant be the anser?

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

From where do you get \(r\geq{-5}\)?

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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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

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.