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Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
26 Jun 2017, 01:51
Question Stats:
63% (01:17) correct
36% (01:18) wrong based on 122 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.
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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 \(\frac{2^5*2^52*2^5*5^5}{2*2^5*5^5}*100 = 50%\), while column B is \(\frac{400200}{200}*100 = 100%\). 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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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)/oldPercent 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: \(= \frac{100(200)}{200}\) = 100So, y increased by 100% We get: QUANTITY A: 50QUANTITY B: 100Answer: B Cheers, Brent
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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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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]
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.



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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
03 May 2020, 01:48
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I did some peripheral calculations about percent decreases and percent increases and found out: 1. Percent decrease, especially for real world problems like price cannot be more than 100%. A 100% decrease means the object goes from X to 0. 2. Percent increase can have any value. A 100% increase is doubling the current price.
Now, A and B are similar in the sense that one quantity is twice the other. In A, second quantity is half the first quantity and in B, second quantity is twice the first quantity. But this does not mean that the percent changes are equal.
The percent increase in B is 100%, while the percent decrease in A HAS TO BE LESS THAN 100%, because if it were 100%, the second quantity would be ZERO.




Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5,
[#permalink]
03 May 2020, 01:48





