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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Given the inequalities above, which of the following CANNOT be........

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Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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AlN wrote:
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

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

I am not able to understand the highlighted part . isnt r>=-5

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

From where do you get $$r\geq{-5}$$?
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Karishma
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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Simplest way to solve this:

Given | s | <= 5
=> -5 <= s <= 5
=> -20 <= 4s <= 20 ( Sign remains unchanged if the inequalities divided/multiplied/added or subtracted by SAME +VE NUMBER)
=> -15 <= 4s + 5 <= 25
=> -5 <= (4s + 5)/3 <= 8.33

Now we know that r <= (4s+5)/3
so r's range can be r <= -5, -5 <=r <= (4s+5)/3 <=8.33

Now we know that r can NEVER exceed 8.33 -> 20 is our ANSWER.

Regards,
Rishav
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Joined: 05 Feb 2018
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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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

$$3r\leq{4s + 5}$$ --> $$3r-5\leq{4s}$$

$$|s|\leq{5}$$ --> $${-5}\leq s\leq{5}$$

MAX S = 5, so $$3r-5\leq{20}$$
Testing E, r = 20 gives us a false statement $$55\leq{20}$$
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Joined: 18 Aug 2019
Posts: 5
Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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generis wrote:
Quote:
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,

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

Does that make sense? Why cant we insert value of S in 3r<=4s+5 equation and check if s is between -5 and 5? when i do this, i got 13.75<=s for (a). Which I thought that was the answer
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Joined: 11 Sep 2015
Posts: 4959
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Re: Given the inequalities above, which of the following CANNOT be........  [#permalink]

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Top Contributor
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

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, w hey buddye can readily see that r CANNOT equal 20

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

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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

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.

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