 It is currently 24 May 2020, 18:00 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### 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
Author Message
TAGS:
Founder  Joined: 18 Apr 2015
Posts: 11096
Followers: 237

Kudos [?]: 2769 , given: 10521

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

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.
[Reveal] Spoiler: OA

_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

Last edited by GreenlightTestPrep on 01 May 2019, 05:29, edited 6 times in total.
Edited by Carcass Director Joined: 03 Sep 2017
Posts: 518
Followers: 2

Kudos [?]: 412  , 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 [?]: 4 , 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: 3244
Followers: 124

Kudos [?]: 3619  , given: 61

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: 3244
Followers: 124

Kudos [?]: 3619  , given: 61

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. Manager Joined: 04 Apr 2020
Posts: 71
Followers: 0

Kudos [?]: 23  , given: 16

Re: Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5, [#permalink]
2
KUDOS
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
Display posts from previous: Sort by

# Suppose x1 = 2 x 10^5, x2 = 2^5 x 5^5,  Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group Kindly note that the GRE® test is a registered trademark of the Educational Testing Service®, and this site has neither been reviewed nor endorsed by ETS®.