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Is \(x = 4?\)

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Is \(x = 4?\)

by Vincen » Tue Nov 26, 2019 1:07 am
Is \(x = 4?\)

\((1)\quad |x + 2| < 10\)

\((2) \quad |x + 5| > 10\)

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
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Source: — Data Sufficiency |

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by Jay@ManhattanReview » Tue Nov 26, 2019 9:27 pm
Vincen wrote:Is \(x = 4?\)

\((1)\quad |x + 2| < 10\)

\((2) \quad |x + 5| > 10\)

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
Let's take each statement one by one.

\((1)\quad |x + 2| < 10\)

=> x + 2 < 10 => x < 8; taking plus sign
=> x + 2 > - 10 => x > -12; taking minus sign

Thus, -12 < x < 8. x can have any value between -12 and 8, including 4. So, x is not necessarily equal to 4. Insufficient.

\((2) \quad |x + 5| > 10\)

=> x + 5 > 10 => x > 5; taking plus sign
=> x + 5 < -10 => x < -15; taking minus sign

So, x is either less than - 15 or is greater than 5. In either case x ≠ 4. The answer is no. Sufficient.

Another way of checking it by plugging in the value of x in the inequality; we see that x = 4 does not satisfy the inequality \( \quad |x + 5| > 10\).

The correct answer: B

Hope this helps!

-Jay
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by chetan.sharma » Wed Nov 27, 2019 7:52 am
Vincen wrote:Is \(x = 4?\)

\((1)\quad |x + 2| < 10\)

\((2) \quad |x + 5| > 10\)

[spoiler]OA=B[/spoiler]

Source: Veritas Prep
TWO ways..

(I) Just substitute x=4 and see what happens..
\((1)\quad |x + 2| < 10\)
|4+2|<10...6<10..Yes
But x as -4, 0, 1 and so on also fit in
Insuff
\((2) \quad |x + 5| > 10\)
|4+5|>10...9>10..NO
SO we get an answer for sure, and the answer is NO. We are not bothered what is the range or value of x, till we know that it does not include 4.
Suff

(II) Solve for x
\((1)\quad |x + 2| < 10\)
Think of a number line where distance of x from 2 is less than 10, so x lies between 2-10 and 2+10...-8<x<12
It can be 4 or may not be 4.
Insuff

\((2) \quad |x + 5| > 10\)
This says that x is not within 10 units from 5, so x<5-10, that is x<-5 or x>5+10, that is x>15.
So we can see that 4 cannot be a value..

B

NOTE - You can solve the Modulus by CRITICAL point or even by SQUARING two sides as both sides are positive
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