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If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the

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If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 06:38
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If x = \((0.08)^2\), y = \(\frac{1}{(0.08)^2}\) and z = \((1 - 0.08)^2\)-1, which of the following is true?

(A) x = y = z
(B) y < z < x
(C) z < x < y
(D) y < x and x = z
(E) x < y and x = z

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If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 06:49
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Bunuel wrote:
If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the following is true?

(A) x = y = z
(B) y < z < x
(C) z < x < y
(D) y < x and x = z
(E) x < y and x = z


The values of x,y, and z are

x = \(0.0064\)
y = \(\frac{1}{0.0064} = \frac{10000}{64}\)
z = \((0.92)^2 - 1 = 0.8464 - 1 = -0.1536\)

Therefore, Option C(z < x < y) has to be true for the following values.
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If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 06:56
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pushpitkc niks18 Hatakekakashi
amanvermagmat

How about this approach?

On seeing z, I know I can write it as \(a^2\) - \(b^2\) = (a+b) * (a-b)

or (1 - 0.08 + 1 ) * (1 - 0.08 -1 )

The second bracket gives me a negative value as final answer.

Just by looking at x and y , I know these are positive. So only C holds good.
I do not have to care about inequality between x and y.

GMATNinja Sorry to bother you on qaunt forum ;) , but is this approch more efficient than one here at 32:05 min
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If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 07:05
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adkikani wrote:
How about this approach?

On seeing z, I know I can write it as \(a^2\) - \(b^2\) = (a+b) * (a-b)

or (1 - 0.08 -1 ) * (1 - 0.08 +1 )

The first bracket gives me a negative value as final answer.

Just by looking at x and y , I know these are positive. So only C hold good.
I do not have to care about inequality between x and y.


Hey adkikani ,

I completly agree with this approach and I really like it. :thumbup:

This method is also know as smart way of solving any inequality question.
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 07:08
Hi adkikani

I don't see the problem with using the method you are suggesting to find
the value of z. Also, specific to this problem, we might not need to bother
about the inequality between x and y.

But as far as the GMAT is concerned, it always makes sense to find the
value of the x and y in case we have both z < x < y and z < y < x as
answers!
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 07:19
1
pushpitkc wrote:
Bunuel wrote:
If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the following is true?

(A) x = y = z
(B) y < z < x
(C) z < x < y
(D) y < x and x = z
(E) x < y and x = z


The values of x,y, and z are

x = \(0.0064\)
y = \(\frac{1}{0.0064} = \frac{10000}{64}\)
z = \((-0.02)^2 - 1 = 0.0004 - 1 = -0.9996\)

Therefore, Option C(z < x < y) has to be true for the following values.


how do u get z=0.02..........it's given that z=(1-0.08)^2-1...........1-0.08=0.92
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 07:24
selim wrote:
pushpitkc wrote:
Bunuel wrote:
If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the following is true?

(A) x = y = z
(B) y < z < x
(C) z < x < y
(D) y < x and x = z
(E) x < y and x = z


The values of x,y, and z are

x = \(0.0064\)
y = \(\frac{1}{0.0064} = \frac{10000}{64}\)
z = \((-0.02)^2 - 1 = 0.0004 - 1 = -0.9996\)

Therefore, Option C(z < x < y) has to be true for the following values.


how do u get z=0.02..........it's given that z=(1-0.08)^2-1...........1-0.08=0.92


Have corrected my solution. Thanks for notifying, selim
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 07:29
Bunuel wrote:
If x = \((0.08)^2\), y = \(\frac{1}{(0.08)^2}\) and z = \((1 - 0.08)^2\)-1, which of the following is true?

(A) x = y = z
(B) y < z < x
(C) z < x < y
(D) y < x and x = z
(E) x < y and x = z


we have to keep in mind that if we square a fraction it would be even smaller.
start with Z= (1-0.08)^2-1
we don't have calculate anything. just think logically. we will get fractional value inside the bracket. furthermore we are going to square it. the ultimate value will be even more smaller. finally we deduct 1. so, Negative value is a must here. thus it will be the smallest one.

only option C meets the condition.

but for more analyze, y = 1/0.08^2

here ultimate value will be more that 1 . how ? coz u are dividing a integer by a fraction .
x= 0.08^2
we will get a fraction but smaller one , positive one.
so, y>x>z
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 07:34
I agree pushpitkc

but options other than (C) can be eliminated using logic:

two numbers, out of which one is smaller (or negative to be more precise)
CAN NOT BE equal to other one.

Quote:
(A) x = y = z

Nope, I know z is negative and x is positive, rejected.

Quote:
(B) y < z < x

Nope, I know z is negative and y is positive, rejected.

Quote:
(D) y < x and x = z

x and z can not be equal.

Quote:
(E) x < y and x = z

x and z can not be equal.

Hope I am thinking on correct lines ;)
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 09 Mar 2018, 11:54
adkikani wrote:
pushpitkc niks18 Hatakekakashi
amanvermagmat

How about this approach?

On seeing z, I know I can write it as \(a^2\) - \(b^2\) = (a+b) * (a-b)

or (1 - 0.08 + 1 ) * (1 - 0.08 -1 )

The second bracket gives me a negative value as final answer.

Just by looking at x and y , I know these are positive. So only C holds good.
I do not have to care about inequality between x and y.

GMATNinja Sorry to bother you on qaunt forum ;) , but is this approch more efficient than one here at 32:05 min


hey that's a great way to solve the problem but it is problem specific. Great observation .. loved it :)
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the  [#permalink]

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New post 12 Mar 2018, 16:25
Bunuel wrote:
If x = \((0.08)^2\), y = \(\frac{1}{(0.08)^2}\) and z = \((1 - 0.08)^2\)-1, which of the following is true?

(A) x = y = z
(B) y < z < x
(C) z < x < y
(D) y < x and x = z
(E) x < y and x = z


We can observe the following:

Since 0.08 < 1, x = (0.08)^2 will be less than 1 also. However, its reciprocal, y = 1/(0.08)^2 will be greater than 1.

Finally we see that z is negative, since (1 - 0.08)^2 is less than 1. We know that it is less than 1 because (1 - 0.08) is less than 1, and thus, when (1 - 0.08) is squared, the result is also less than 1.

Thus, z < x < y

Answer: C
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Re: If x = (0.08)^2, y = 1/(0.08)^2 and z = (1 - 0.08)^2 -1, which of the &nbs [#permalink] 12 Mar 2018, 16:25
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