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# Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of

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Manager
Joined: 16 Oct 2016
Posts: 192
Location: India
Concentration: General Management, Healthcare
GMAT 1: 640 Q40 V38
GMAT 2: 680 Q48 V35
GPA: 3.05
WE: Pharmaceuticals (Health Care)
Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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09 Jan 2019, 06:54
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25% (medium)

Question Stats:

80% (01:04) correct 20% (00:54) wrong based on 20 sessions

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Given that $$10^\frac{48}{100}$$ = x, $$10^\frac{7}{10}$$ = y and$$x^z$$ = $$y^2$$, then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36

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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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09 Jan 2019, 07:10
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Top Contributor
fitzpratik wrote:
Given that $$10^\frac{48}{100}$$ = x, $$10^\frac{7}{10}$$ = y and$$x^z$$ = $$y^2$$, then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36

GIVEN: $$10^\frac{48}{100}$$ = x
Rewrite as: $$10^{0.48}$$ = x

GIVEN: $$10^\frac{7}{10}$$ = y
Rewrite as: $$10^{0.7}$$ = y

GIVEN: $$x^z$$ = $$y^2$$

Replace with above values to get: $$(10^{0.48})^z$$ = $$(10^{0.7})^2$$

Apply Power of a Power Law to get: $$10^{0.48z}=10^{1.4}$$

This means: $$0.48z = 1.4$$

Divide both sides by 0.48 to get: $$z = \frac{1.4}{0.48}$$

NOTE: Since $$\frac{1.4}{0.5}=2.8$$, we know that $$\frac{1.4}{0.48}$$ is a little bit greater than 2.8

Check the answer choices . . .

Cheers,
Brent
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Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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09 Jan 2019, 07:29
fitzpratik wrote:
Given that $$10^\frac{48}{100}$$ = x, $$10^\frac{7}{10}$$ = y and$$x^z$$ = $$y^2$$, then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36

I just ball parked in this question.
$$10^\frac{48}{100}$$ = x, $$10^\frac{7}{10}$$ = y

$$x^z$$ = $$y^2$$

$$10^\frac{48}{100}$$ = x
$$10^\frac{1}{2}$$ = x

$$x^z$$ = $$y^2$$

$$10^\frac{z}{2}$$ = $$10^\frac{14}{10}$$
$$\frac{z}{2}$$= $$\frac{7}{5}$$

z = 2.85

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Location: Pune, India
Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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09 Jan 2019, 08:00
fitzpratik wrote:
Given that $$10^\frac{48}{100}$$ = x, $$10^\frac{7}{10}$$ = y and$$x^z$$ = $$y^2$$, then the value of z is closest to?

A. 1.45

B. 1.88

C. 2.9

D. 3.7

E. 4.36

$$y = 10^{\frac{7}{10}}$$

$$y^2 = 10^{\frac{7}{10} * 2} = 10^{\frac{7}{5}}$$

$$x = 10^{\frac{48}{100}} = 10^{\frac{12}{25}}$$

$$x^z = 10^{\frac{12}{25}}^z = 10^{\frac{7}{5}} = y^2$$

So $$(\frac{12}{25})z = \frac{7}{5}$$

$$z = \frac{35}{12} = 2.9$$

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Intern
Joined: 16 Oct 2011
Posts: 49
GMAT 1: 620 Q37 V38
Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of  [#permalink]

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09 Jan 2019, 09:32
We get

10^(48z/100) =10^14/10

Equating exponents (since we are in base 10) gives 48z/100 =14/10
then 480z=1400
and z=1400/480. 480*3=1440, therefore the answer is a little less than 3. C
Re: Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of &nbs [#permalink] 09 Jan 2019, 09:32
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# Given that 10^0.48 = x, 10^0.70 = y and x^z = y^2, then the value of

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