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Sub 505 (Easy)|   Algebra|            
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gmatt1476
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = Updated on: 01 May 2020, 11:18
Originally posted by gmatt1476 on 20 Apr 2020, 11:21.
Last edited by Carcass on 01 May 2020, 11:18, edited 1 time in total.
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C
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94% (00:52) correct 6% (01:25) wrong based on 1542 sessions

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If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} =\frac{ w}{z}\), then\( w = \)

A. \(z + 1 \)

B. \(z^2 + 1\)

C. \(-z^2 + 1\)

D. \(-z^2 + z + 1\)

E. \(-2z^2 + 1\)

PS14031.02
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C
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sujoykrdatta
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 20 Apr 2020, 11:28
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gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} = w/z\), then w =

A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1

PS14031.02
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\(z + \frac{1 - 2z^2}{z} = w/z\)

=> \(z(z + \frac{1 - 2z^2}{z}) = w\)

=> \(w = z^2 + (1 - 2z^2)\)

=> \(w = 1 - z^2\)

Answer C
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 28 Apr 2020, 08:38
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gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} = \frac{w}{z}\), then \(w =\)

A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1

PS14031.02
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Given: \(z + \frac{1 - 2z^2}{z} = \frac{w}{z}\)

Eliminate the fractions, multiply both sides of the equation by \(z\) to get: \(z^2 + (1 - 2z^2) = w\)

Simplify to get: \(1 - z^2 = w\)

Rewrite as: \(-z^2 + 1= w\)

Answer: C

Cheers,
Brent
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 29 May 2020, 10:46
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Assume any value of z
Let z=1
=> w = 0
Now check where w= 0 for z=1 in the option
(a) w=2
(b) w=2
(C) w= 0.
(D) w=1
(e) w= -1

Therefore, (c) is the correct option.

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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 30 May 2020, 06:12
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(z^2 + 1-2z^2)/z = w/z

z^2 + 1 -2z^2 = w

1 - z^2 = w

Option C

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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 06 Jul 2020, 13:23
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\(z + \frac{(1 - 2z^2)}{z} = \frac{w}{z}\)
\(w=z^2+1-2z^\)
\(w=1-z^2\)

Answer C
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 11 Feb 2021, 07:59
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SOLUTION:

z+ (1−2z^2/z) = w/z

=> (z^2 + 1-2z^2 )/z = w/z

Cancel out the "w" in the denominators, add the terms in the numerator of the LHS and arrive at

1- z^2 = w (OPTION C)

Hope this helps :thumbsup:
Devmitra Sen(Math)
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 19 Mar 2021, 12:25
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sujoykrdatta
gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} = w/z\), then w =

A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1

PS14031.02


\(z + \frac{1 - 2z^2}{z} = w/z\)

=> \(z(z + \frac{1 - 2z^2}{z}) = w\)

=> \(w = z^2 + (1 - 2z^2)\)

=> \(w = 1 - z^2\)

Answer C
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sujoykrdatta

Could you please break down the following for me: => \(w = z^2 + (1 - 2z^2)\)

=> \(w = 1 - z^2\)

This step is unclear.. Why did the Z^2 disappear?
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 19 Mar 2021, 12:31
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Neghineh
sujoykrdatta
gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} = w/z\), then w =

A. z + 1
B. z^2 + 1
C. -z^2 + 1
D. -z^2 + z + 1
E. -2z^2 + 1

PS14031.02


\(z + \frac{1 - 2z^2}{z} = w/z\)

=> \(z(z + \frac{1 - 2z^2}{z}) = w\)

=> \(w = z^2 + (1 - 2z^2)\)

=> \(w = 1 - z^2\)

Answer C

sujoykrdatta

Could you please break down the following for me: => \(w = z^2 + (1 - 2z^2)\)

=> \(w = 1 - z^2\)

This step is unclear.. Why did the Z^2 disappear?
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You are adding the terms z² and -2z² to get the result -z². The '1' remains as it is.

Posted from my mobile device
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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 15 May 2021, 04:18
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gmatt1476
If z ≠ 0 and \(z + \frac{1 - 2z^2}{z} =\frac{ w}{z}\), then\( w = \)

A. \(z + 1 \)

B. \(z^2 + 1\)

C. \(-z^2 + 1\)

D. \(-z^2 + z + 1\)

E. \(-2z^2 + 1\)

PS14031.02
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Answer: Option C

Video solution by GMATinsight

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If z ≠ 0 and z + (1 - 2z^2)/z = w/z, then w = 05 Nov 2025, 14:30
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