If 2s > 8 and 3t < 9

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.
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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

Answer: A

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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

Solution :
----------
2s > 8 => s > 4
3t < 9 => t < 3

The given values can be represented on the number line as follows :

--t-----------3---------------4--------s-----------
|----+ve---|---------1-----|-+ve-+|

So,s-t will be greater than 1.
Correct option is A.
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Hi,


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
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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

Consider s= 7, t= -5 or -0.3

Both these cases make it s-t >1.

Hope this helps.
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Thank you. I neglected to utilize the fact that s must be positive when testing cases. Appreciate the help

2s > 8 and 3t < 9

So, s > 4 and t < 3

Assume
s = 4 + ds
t = 3 - dt

So, s - t = (4 + ds) - (3 - dt)
s - t = 1 + (ds + dt)

So, s - t has to be > 1

None of the given option qualifies.

We see that s > 4 and that t < 3. Since s is always greater than t, the difference cannot be -1 or zero.

Furthermore, since s > 4 and t < 3, we see that s and t are more than 1 unit apart, so the difference cannot be 1.

Alternate Solution:

Let’s divide each side of 2s > 8 by 2: s > 4

Let’s divide each side of 3t < 9 by -3, paying attention to change the direction of the inequality since we are dividing by a negative number: -t > -3

Let’s add the two inequalities together: s - t > 1

We see that none of the provided numbers is greater than 1.

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

thank you

dave13 - We have been asked to find the value of s-t. That's the reason for multiplying the inequality involving t with -1.

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
··\((s > 4)\)
+\((-t>-3)\)
------------------
\(s - t> 1\)

Horizontally:
\(s + (-t) = (s-t)\)
\(4 + (-3) = (4-3) = 1\)
The > sign is between, thus \(s - t> 1\)

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.
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visualizing a number line may be helpful.

--------0----1----2----3----4-----

s is somewhere to the right of 4. t is somewhere to the left of 3. We can see the space between them must be greater than 1. All of the choices are not possible.

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