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Re: r + s + t > 1 and 0 > s + t [#permalink]
1
Given that r + s + t > 1 and 0 > s + t

0 > s + t
=> s + t < 0
Multiplying both the sides by -1 we get (Sign of inequality will reverse)

-s -t > 0
r + s + t > 1

Adding both of them we get
-s -t + r + s + t > 0 + 1
=> r > 1

Clearly, Quantity B(r) > Quantity A(1)

So, Answer will be B
Hope it helps!

Watch the following video to learn the Basics of Inequalities

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r + s + t > 1 and 0 > s + t [#permalink]
1
\(r + s + t > 1\) and \(0 > s + t\)

Let us take the extreme case where \(s + t\) is closest to zero, that is \(s + t = -0.000000001\)(and lesser). In fact, we can extrapolate and assume the extreme value of \(s+ t = 0\)

In which case

\(r + s + t > 1\)

becomes

\(r + 0 > 1\)

Therefore,

\(r > 1\).

For all other values of \(s + t\),\( r\) will be much greater than 1.

The answer is Choice B
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r + s + t > 1 and 0 > s + t [#permalink]
1
s+t is negative
In the 1st ineqaulity,
If we're adding a negative to r which is the same as subtracting something from r and it is still greater than 1, then r must be greater than 1.

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