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

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

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23 May 2016, 03:28
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$$3r\leq{4s + 5}$$
$$|s|\leq{5}$$

Given the inequalities above, which of the following CANNOT be the value of r?

A. –20
B. –5
C. 0
D. 5
E. 20
[Reveal] Spoiler: OA

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

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23 May 2016, 05:35
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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

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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02 Jun 2016, 01:04
chetan2u wrote:
we can see $$r\leq{\frac{25}{3}}$$ covers up for BOTH the ranges..

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

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

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02 Jun 2016, 06:41
sauravpaul wrote:
chetan2u wrote:
we can see $$r\leq{\frac{25}{3}}$$ covers up for BOTH the ranges..

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

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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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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

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02 Jun 2016, 06:53
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-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

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Re: Given the inequalities above, which of the following cannot be the val [#permalink]

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19 Jul 2016, 08:07
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ameyaprabhu 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

[Reveal] Spoiler:
The way I solved it is

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

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

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

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07 Jul 2017, 12:30
I did this question algebraically - personally I find the above explanations harder to absorb other types of methods. I am going through them though But do let me know if my approach is correct, would be really thankful!
so we have 3x<= 4s + 5, 4s + 5 must add to a multiple of 3*r
So i plugged in values from given to check for incongruities:
A. –20 => putting this in 4s +5, it divides, so 3r is a multiple of 4s + 5
B. –5 => again in 4s + 5; 3r<= 21, so its a multiple
C. 0 => r = 5/ 3 not a multiple
D. 5 = > r <= 25/ 3
E. 20 => r <= 85/ 3
between D and E can we check for the middle value => 85/3 - 25/3 = 60/3 = 20
so then R cannot be 20;based on the options below
If we equate it to C, we again get --> r = 25/3 - 5/ 3 = 20/3 non integer value and 85/ 3 - 5-/3 = 80/3.So none of these non multiple forms except for D and E give R, hence E is the answer.

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

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28 Aug 2017, 11:45
Hi,

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

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

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28 Aug 2017, 19:26
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Quote:
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

Saumya2403 wrote:
Hi,

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

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 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

Does that make sense?

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Given the inequalities above, which of the following CANNOT be........   [#permalink] 28 Aug 2017, 19:26
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