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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. # Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5,  Question banks Downloads My Bookmarks Reviews Important topics
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Founder  Joined: 18 Apr 2015
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Kudos [?]: 1448 , given: 6583

Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
Expert's post 00:00

Question Stats: 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

_________________

Last edited by GreenlightTestPrep on 01 May 2019, 05:29, edited 6 times in total.
Edited by Carcass Director Joined: 03 Sep 2017
Posts: 520
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Kudos [?]: 362  , given: 66

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
1
KUDOS
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!
Intern Joined: 26 Sep 2017
Posts: 31
Followers: 0

Kudos [?]: 3 , given: 19

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
can anyone suggest fast approach?
Intern Joined: 28 Apr 2019
Posts: 7
Followers: 0

Kudos [?]: 3 , given: 1

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? GRE Instructor Joined: 10 Apr 2015
Posts: 2162
Followers: 64

Kudos [?]: 1975  , given: 19

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
1
KUDOS
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
_________________

Brent Hanneson – Creator of greenlighttestprep.com  GRE Instructor Joined: 10 Apr 2015
Posts: 2162
Followers: 64

Kudos [?]: 1975  , given: 19

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
1
KUDOS
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
_________________

Brent Hanneson – Creator of greenlighttestprep.com Intern Joined: 28 Apr 2019
Posts: 24
Followers: 0

Kudos [?]: 9 , given: 4

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
A real good for application of exponents rule. Intern Joined: 11 Jul 2019
Posts: 1
Followers: 0

Kudos [?]: 1  , given: 0

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
1
KUDOS
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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