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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] New post 26 Jun 2017, 01:51
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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] New post 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] New post 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] New post 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] New post 01 May 2019, 05:34
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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

Answer: B

Cheers,
Brent
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink] New post 01 May 2019, 05:57
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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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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink] New post 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] New post 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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Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5,

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