Is \(\frac{7+x}{8+x} ≥ \frac{7}{8}\)?

The question transforms to: is \(\frac{x}{x+8}\geq{0}\). This equation will hold true for \(x<-8\) and \(x\geq{0}\)

(1) \(x \leq 0\). Not sufficient.
(2) \(x \geq 0\). Sufficient.

Answer: B.
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7/8= 7*0.125=0.875

(1) x≤0x≤0
say x=-4 ; 3/4=0.75
Meaning for all values of x <0 the expression value will be less than 7/8 and for x=0 it will be equal
Hence, not sufficient

(2) x≥0

For X=0 value RHS=LHS
For Values of x>0

expression will move closer to 1 i.e 9/10 < 11/12 < 12/13 < 13/14 ....etc


Hence, Sufficient
Answer should be B in my opinion

Given: \(\frac{7+x}{8+x} ≥ \frac{7}{8}\)
Cross-multiply to get:
\(8*(7+x) ≥ 7*(8+x)\)
\(56 + 8x ≥ 56 + 7x\)
\(8x ≥ 7x\)
\(x ≥ 0\) - Option(B)
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Given , 7 + x / 8 + x = > 7 /8 ?
At first we need to completely simplify the above equation by bringign the entire equation to one side.
Fianlly we get the equation as x / 8 x + 64 => 0 ?

Statement 1: x <= 0
Putting x as 0 we can get 0 /64 = 0
now putting x as -1 we get (-1/ 56) which is less than 0
Hence the statement has two contrasting values.
Not sufficient.

Statement 2 : x => 0
putting x = 0
we are getting 0/ 64 = 0
putting x as 1, we are still getting a positive number which is greater than 0
Hence sufficient

IMO B