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Re: 4<7-x/3 [#permalink]
Carcass can you please explain how?
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Re: 4<7-x/3 [#permalink]
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manipulating the stem we do have that

\(-5>x\) so x must be negative

Quantity A: Maximum value of −(5−x)
⇒ Maximum value of (x−5)
Since x is negative, and we need to get the maximum value of (x−5), the least possible value of x in the magnitude would be eligible.

Say x=−5.001
Maximum value of (x−5)=−5.001−5=−10.001
Quantity B: Maximum value of 2x at x=−5.001
Maximum value of 2x=2×−5.001=−10.002
Thus, Quantity A (=-10.001) is greater than Quantity B (=-10.002).

No matter how close to -5, you choose the value of x

A>B
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Re: 4<7-x/3 [#permalink]
1
Ami2000 wrote:
\(4 < \frac{7-x}{3}\)

Solving the inequality we will get

12 < 7-x
x<-5

Quantity A is Maximum value of -(5-x)
Putting the value of x as -4.9

A = -9.9

Quantity B is the Maximum Value of 2x
2(-4.9)
B = -9.8

Here Quantity B is greater

Quantity A is the Maximum value of -(5-x)
Putting the value of x as -100

A = -105

Quantity B is the Maximum Value of 2x
2(-100)
B = -200

Here Quantity A is greater.

The Answer should be D in my opinion(Which will probably be wrong)



x < -5
x cant be -4.9

That's where you've err'ed
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Re: 4<7-x/3 [#permalink]
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