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# Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5,

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Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  26 Jun 2017, 01:51
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66% (01:17) correct 33% (01:24) wrong based on 71 sessions

Suppose $$x_1 = (2)$$$$(10^5)$$, $$x_2 =$$ $$(2^5)(5^5)$$, $$y_1 = 200$$, and $$y_2 = 400$$ and $$x_1$$ and $$x_2$$ represent the value of $$x$$ in years 1 and 2, respectively, and $$y_1$$ and $$y_2$$ represent the value of $$y$$ in years 1 and 2, respectively.

 Quantity A Quantity B The percent decrease in x from year 1 to year 2 The percent increase in y from year 1 to year 2

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Last edited by GreenlightTestPrep on 01 May 2019, 05:29, edited 6 times in total.
Edited by Carcass
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  22 Sep 2017, 04:50
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Column A is $$\frac{2^5*2^5-2*2^5*5^5}{2*2^5*5^5}*100 = -50%$$, while column B is $$\frac{400-200}{200}*100 = 100%$$. Even taking absolute values, column B is greater!
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  27 Sep 2017, 08:55
can anyone suggest fast approach?
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  30 Apr 2019, 14:03
Can we not assume since A is going down percentage? it will be a negative value thus it can never be greater percentage increase (100) of B?
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  01 May 2019, 05:34
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Expert's post
Carcass wrote:

Suppose $$x_1 = (2)$$$$(10^5)$$, $$x_2 =$$ $$(2^5)(5^5)$$, $$y_1 = 200$$, and $$y_2 = 400$$ and $$x_1$$ and $$x_2$$ represent the value of $$x$$ in years 1 and 2, respectively, and $$y_1$$ and $$y_2$$ represent the value of $$y$$ in years 1 and 2, respectively.

 Quantity A Quantity B The percent decrease in x from year 1 to year 2 The percent increase in y from year 1 to year 2

Percent change = 100(new - old)/old

Percent decrease in x $$= \frac{100(2^5)(5^5) - (2)(10^5)}{(2)(10^5)}$$

Notice that $$10^5 = [(2)(5)]^5 = (2^5)(5^5)$$

So, we get: $$= \frac{100(2^5)(5^5) - (2^1)(2^5)(5^5)}{(2^1)(2^5)(5^5)}$$

Simplify to get: $$= \frac{100(2^5)(5^5) - (2^6)(5^5)}{(2^6)(5^5)}$$

Factor the numerator to get: $$= \frac{(100)(2^5)(5^5)[1 - 2]}{(2^6)(5^5)}$$

Simplify to get: $$= \frac{(100)[1 - 2]}{2}$$

Simplify to get: $$= \frac{-100}{2}$$

$$= -50$$

So, x decreased by 50%

-----------------------------------

Percent decrease in y $$= \frac{100(400 - 200)}{200}$$

Simplify to get: $$= \frac{100(200)}{200}$$

= 100

So, y increased by 100%

-----------------------------------

We get:
QUANTITY A: 50
QUANTITY B: 100

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  01 May 2019, 05:57
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Expert's post
jackwang1208 wrote:
Can we not assume since A is going down percentage? it will be a negative value thus it can never be greater percentage increase (100) of B?

That's a great idea. However, the question asks for the percent DECREASE of x.
So, we wouldn't say the decrease = -50%
Instead, we say the decrease = 50%

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com

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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  04 May 2019, 03:36
A real good for application of exponents rule.
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]  11 Jul 2019, 22:55
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bim1946 wrote:
can anyone suggest fast approach?

x1 = 2*10^5
x2 = 2^5 * 5^5 = 10^5

So 50% decrease.
Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5,   [#permalink] 11 Jul 2019, 22:55
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