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805+ Level|   Absolute Values|   Inequalities|   Number Properties|   Roots|         
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 01 Sep 2020, 08:00
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D
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95% (hard)

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28% (01:58) correct 72% (02:03) wrong based on 523 sessions

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If x ≠ 5 and x is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\)?


(1) |y + 5| < 1 - x

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\)



 

This question was provided by GMAT Club
for the Heroes of Timers Competition

 




M36-105

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Bunuel
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 15 Nov 2021, 03:25
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Bunuel
If x ≠ 5 and x is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\)?


(1) |y + 5| < 1 - x

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\)



 

This question was provided by GMAT Club
for the Heroes of Timers Competition

 




M36-105
Show more

Official Solution:


If \(x ≠ 5\) and \(x\) is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\) ?

If \(x > 0\), then \(|x|=x\) and \(\frac{|x| + 5}{5 - x}=\frac{x + 5}{5 - x}\)

If \(x < 0\), then \(|x|=-x\) and \(\frac{-x + 5}{5 - x}=1\)

So, if \(x\) is positive, then the value of \(\frac{|x| + 5}{5 - x}\) depends on exact value of \(x\) but if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1.

(1) \(|y + 5| < 1 - x\):

The left hand side (\(|y+5|\)), is the absolute value of a number, so it must be positive or 0. Thus, \(0 < 1 - x\). This gives \(x < 1\). Since given that \(x\) is a non-zero integer, then \(x\) must be negative integer (-1, -2, ...). As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\):

\(\frac{|x|}{x}=-1\);

\(|x|=-x\). This means that \(x\) is negative. As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.


Answer: D
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what Updated on: 02 Sep 2020, 11:14
Originally posted by mSKR on 01 Sep 2020, 08:13.
Last edited by mSKR on 02 Sep 2020, 11:14, edited 1 time in total.
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x is non-zero
x can be +ve or negative

we can only determine value of expression if x is negative


1. 1-x >0 it means x<1. but x is non zero integer, so the only value left for x are less than 0 integgers , hence x is negative
sufficient


2. root of x and cube of x is negative , it can only be possible if x is negative
with x negative we can calculate the value of expression = -1\\
sufficient

hence D
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 01 Sep 2020, 08:15
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If x ≠ 5 and x is a non-zero integer, what is the value of (|x|+5)/(5−x)


(1) |y + 5| < 1 - x
St. RHS of above equation can never be negative so the LHS should be >0 and that is possible if x is negative integer
so the value of (|x|+5)/(5−x) will always be 1
let see ex.
y = 4
|y+5| = |4+5| = 9
for 1-x to be greater than 9 x has to be less than -8 and thus in the eq. it get add up in denominator and because of mod it adds up in numerator as well so the result would be 1
Sufficient


(2) x2/(√x3)^(1/3)=−1
after analysing this statement we can conclude that the only value of x that can satisfy the St. 2 if X< 0
Sufficient

IMO D
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 01 Sep 2020, 08:22
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If x ≠ 5 and x is a non-zero integer, what is the value of \(\frac{|x|+5}{5−x} \)?


(1) |y + 5| < 1 - x

1) it is irrelevant what y is. We know for sure that 1-x is greater than zero, and therefore, x < -1.
for all such values of x,
\(\frac{|x|+5}{5−x} \) = 1
- Sufficient


(2) \(\sqrt{x^2}/\sqrt[3]{x^3}=−1\)
This also tells us the same thing as 1) tells us:
\(\frac{|x|}{ x} = - 1\)
So x is negative.
for all such values of x,
\(\frac{|x|+5}{5−x} \) = 1
- Sufficient

Answer D.
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 01 Sep 2020, 08:31
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If x≠ 5 and x is a non-zero integer, what is the value of (|x|+5)/(5-x)?


(1) |y + 5| < 1 - x
0<|y+5|<1-x
x<1
If x=-1; (|x|+5)/(5-x) = 1
If x=-2; (|x|+5)/(5-x)= 1
In general let x=-a; where a is a positive integer, (|x|+5)/(5-x)= (a+5)/(5+a)=1
SUFFICIENT

(2) square root(x^2)/ cube root(x^3) =-1
x is a negative integer
Let x=-a where a is a positive integer
(|x|+5)/(5-x)= (a+5)/(5+a)=1
SUFFICIENT

IMO D

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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 01 Sep 2020, 09:15
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From 1) 1-x>0
=> X<1 but X can't be 0,
So X is a negative integer...
On putting any neg. integer in the question stem, the result will always be 1.

SUFFICIENT

2)
|X|/X=-1
it means ,i.e.|x|=-x
Hence x<0..
Again we will get 1 as answer for negative integers..
SUFFICIENT

ANS: D

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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 02 Sep 2020, 05:32
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(1) |y + 5| < 1 - x. Since |y + 5| is always greater than or equal to zero, 1-x is always greater than or equal to zero: \(1-x\geq{0}\). Since x is non-zero integer, x can be 0 or negative integer. If x=0, (|x|+5)/(5-x)=(0+5)/(5-0)=5/5=1. If x<0, (|x|+5)/(5-x)=(-x+5)/(5-x)=(5-x)/(5-x)=1 Sufficient.
(2)(\(\sqrt{x^2}\)/\(\sqrt[3]{x^3}\))=-1 ---> Since \(\sqrt{x^2}\)>0, x is smaller than zero. Then (|x|+5)/(5-x)=(-x+5)/(5-x)=(5-x)/(5-x)=1 Sufficient.
Answer is D)
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 17 Jul 2022, 20:27
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Hi Bunuel Is the highlighted part correct, Shouldn't it be \(x<1\) only.

Thanks
Bunuel
Bunuel
If x ≠ 5 and x is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\)?


(1) |y + 5| < 1 - x

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\)



 

This question was provided by GMAT Club
for the Heroes of Timers Competition

 




M36-105

Official Solution:


If \(x ≠ 5\) and \(x\) is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\) ?

If \(x > 0\), then \(|x|=x\) and \(\frac{|x| + 5}{5 - x}=\frac{x + 5}{5 - x}\)

If \(x < 0\), then \(|x|=-x\) and \(\frac{-x + 5}{5 - x}=1\)

So, if \(x\) is positive, then the value of \(\frac{|x| + 5}{5 - x}\) depends on exact value of \(x\) but if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1.

(1) \(|y + 5| < 1 - x\):

The left hand side (\(|y+5|\)), is the absolute value of a number, so it must be positive or 0. Thus, \(0 \leq 1 - x\). This gives \(x \leq 1\). Since given that \(x\) is a non-zero integer, then \(x\) must be negative integer (-1, -2, ...). As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\):

\(\frac{|x|}{x}=-1\);

\(|x|=-x\). This means that \(x\) is negative. As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.


Answer: D
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 18 Jul 2022, 01:45
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stoned
Hi Bunuel Is the highlighted part correct, Shouldn't it be \(x<1\) only.

Thanks
Bunuel
Bunuel
If x ≠ 5 and x is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\)?


(1) |y + 5| < 1 - x

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\)



 

This question was provided by GMAT Club
for the Heroes of Timers Competition

 




M36-105

Official Solution:


If \(x ≠ 5\) and \(x\) is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\) ?

If \(x > 0\), then \(|x|=x\) and \(\frac{|x| + 5}{5 - x}=\frac{x + 5}{5 - x}\)

If \(x < 0\), then \(|x|=-x\) and \(\frac{-x + 5}{5 - x}=1\)

So, if \(x\) is positive, then the value of \(\frac{|x| + 5}{5 - x}\) depends on exact value of \(x\) but if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1.

(1) \(|y + 5| < 1 - x\):

The left hand side (\(|y+5|\)), is the absolute value of a number, so it must be positive or 0. Thus, \(0 \leq 1 - x\). This gives \(x \leq 1\). Since given that \(x\) is a non-zero integer, then \(x\) must be negative integer (-1, -2, ...). As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3}}=-1\):

\(\frac{|x|}{x}=-1\);

\(|x|=-x\). This means that \(x\) is negative. As we deduced above, if \(x\) is negative, then he value of \(\frac{|x| + 5}{5 - x}\) is 1. Sufficient.


Answer: D
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Yes. Edited. The typo. Thank you!
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HOT Competition 1 Sep/8AM: If x 5 and x is a non-zero integer, what 17 May 2023, 01:36
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Bunuel
If x ≠ 5 and x is a non-zero integer, what is the value of \(\frac{|x| + 5}{5 - x}\)?


(1) |y + 5| < 1 - x

(2) \(\frac{\sqrt{x^2}}{\sqrt[3]{x^3=-1\)



 

This question was provided by GMAT Club
for the Heroes of Timers Competition

 




M36-105
Show more


can we do it this way as well
for statement 1
|y+5|<1-x
y can be positive or negative
y positive
y+5<1-x
y+x<-4 (1)
y negative
-y+5<1-x
x-y<-4 (2)
add 1 and 2
X<-2
substitute any integer values <-2 in the reqd equation
answer is 1
so statement 1 suffi
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