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

Question

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

generis · Accepted Answer

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?

chetan2u · Answer

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

KarishmaB · Answer

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)

sauravpaul · Answer

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

chetan2u · Answer

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

Divyadisha · Answer

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

rohit8865 · Answer

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 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

Saumya2403 · Answer

Hi,

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

EgmatQuantExpert · Answer

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

taniad · Answer

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

priarav · Answer

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

chetan2u · Answer

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

AlN · Answer

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

EgmatQuantExpert · Answer

Hey  AlN ,

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

KarishmaB · Answer

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

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

BrentGMATPrepNow · Answer

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

ScottTargetTestPrep · Answer

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

sinhasuh · Answer

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

gmatbyexample · Answer

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.

https://www.youtube.com/watch?v=38zjGAwBfaQ

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