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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
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]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



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]
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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]
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]
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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]
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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.
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
This is my answer;
x1 = (2)*(10)^5 ==> 2,000,000
x2 = (2)^5*(5)^5= (2*5)^5 ==> 1,000,000
y1= 200
y2=400

Quantity A the decrease % in x1 and x2
% change= (new-old)/old = (1,000,000-2,000,000)/2,000,000 = -50% or decrease by 50%

Quantity B the increase % in y1 and y2
% change= (400-200)/200= 1 or 100% increase.

Therefore B is bigger
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
bim1946 wrote:
can anyone suggest fast approach?


For QA: X2 = 2^5 * 5^5, using exponent rules you can simplify to 10^5
Since X1 = 2 * 10^5 we can deduce that from x1 to x2 there is a 50% reduction since we are removing 2 as the multiplier from year 1 to year 2

To approbate this, 2 * 10^5 = 200,000 and 10^5 = 100,000; a 50% reduction as stated in my hypothesis above


Hope that helps!
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
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Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
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