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The Problem description is not precise, could you provide clear problem solving strategy.

Thanks.

If 2s > 8 and 3t < 9, which of the following could be the value of s-t?

I. -1 II. 0 III. 1

A. None B. I only C. II only D. III only E. II and III

we have two equations.. 2s > 8 .. s>4, so' s' can be 4.1,5,5.05 etc and 3t < 9 .. t<3 so 't'can be 2.9999, 2,-1 etc..

if we take the lowest possible difference between s and t, we will take lowest value of s, which is just above 4 and highest value of t, which is just lower to 3.. s-t >4.0000000001 -2.999999999 .... so s-t>1 therefore all values of -1,0,1 are not possible ans none A hope it helped
_________________

I understand how we get s>4 and t<3, but I'm having a hard time wrapping my head around why t cannot be negative. Can someone please explain?

We don't need to check when t<0 but for the sake of your question, even if t <0 ---> -t>0 and you know that s>0, giving you s-t > 1 for all values of s and t as s>1

Any positive quantity added to a quantity >1 will give you the sum as >1

If 2s > 8 and 3t < 9, which of the following could be the value of s-t?

I. -1 II. 0 III. 1

A. None B. I only C. II only D. III only E. II and III

Given: 2s > 8 Divide both sides by 2 to get: s > 4

Given: 3t < 9 Divide both sides by 3 to get: t < 3

NOTE: If we have two inequalities with the inequality symbols facing in the same direction, we can add the inequalities to learn something new.

So, take t < 3 and multiply both sides by -1 to get: -t > -3[aside: when we divide or multiply both sides of an inequality by a NEGATIVE value, we mist REVERSE the symbol]

We now have: s > 4 -t > -3

When we ADD these two inequalities, we get: s - t > 1

If s - t > 1, then: I) s - t CANNOT equal -1 II) s - t CANNOT equal 0 III) s - t CANNOT equal 1

If 2s > 8 and 3t < 9, which of the following could be the value of s-t?

I. -1 II. 0 III. 1

A. None B. I only C. II only D. III only E. II and III

s>4

t<3 (multiply by -1) --> -t>-3

Add both of them

Thus s-t >4-3

or s-t>1. As none of the options given are greater than 1, the answer is none. A.

pushpitkc any idea why do we multiply t<3 by -1 and s>4 leave as it is ?

thank you

Hi dave13 - I hope you have been slaying Quant dragons. At the least, dump some water on their heads. I'm going to expand a little on pushpitkc 's good answer. BTW, I have to add little dots at times to get terms to line up.

We multiply one of the inequalities by -1 because they have signs that point in different directions. The rule: you cannot add inequalities unless their signs point in the SAME direction.

Another rule: multiplying an inequality by any negative number changes the direction of the sign. Another rule: Multiplying by -1 changes the sign but leaves the numbers and variables the same except with opposite signs.

We are asked to find \(s-t\). If we multiply \((t<3)\)by \(-1,\) we can make \(t\) negative (hang on) AND flip its sign so it points the same way as that of \(s\) We've isolated \(s\) and \(t\) to get: \(s>4\) and \(t<3\)

One sign MUST change so we can add. We change the \(t\) inequality because we will get a sign flip AND a MINUS \(t\) \(s\) + \((-t)\)? Is \((s-t)\)

Mulitply \((t<3)\) by (-1). SIGN flips. \((-1*t)>(-1*3)\) \(-t > -3\)

Now add the two inequalities ··\((s > 4)\) +\((-t>-3)\) ------------------ \(s - t> 1\)

No answer choices are greater than 1. The answer is A. Hope that helps.

Technically, we can subtract (NOT add) inequalities with different signs. The sign on top controls. ··\((t<3)\) -\((s>4)\) ============ \(t-s<-1\) ....Oh yay. Now we get to multiply by -1. We need (s-t), not (t-s). \((-1*t)-(-1*s)<(-1*-1)\) \(-t + s>1\) \(s-t>1.\) Trust me. ADD. _________________

In the depths of winter, I finally learned that within me there lay an invincible summer. -- Albert Camus, "Return to Tipasa"

gmatclubot

If 2s > 8 and 3t < 9 &nbs
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02 Jun 2018, 21:19