GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Dec 2018, 19:40

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

Events & Promotions

Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• FREE Quant Workshop by e-GMAT!

December 16, 2018

December 16, 2018

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

M27-11

Author Message
TAGS:

Hide Tags

Intern
Joined: 30 May 2017
Posts: 8

Show Tags

14 May 2018, 21:40
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 29 Apr 2018
Posts: 4
Location: India
GMAT 1: 750 Q50 V42
GPA: 3.3

Show Tags

31 May 2018, 23:01
I think this is a high-quality question and I agree with explanation. Please paraphrase the questions, started working simultaneously and independently seems like they started working at the same time but independently on separate cars.
Manager
Status: EAT SLEEP GMAT REPEAT!
Joined: 28 Sep 2016
Posts: 168
Location: India

Show Tags

01 Jun 2018, 08:28
Quote:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now, if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

Bunuel

_________________

Regards,

Math Expert
Joined: 02 Sep 2009
Posts: 51218

Show Tags

02 Jun 2018, 02:51
Quote:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now, if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

Bunuel

Bonnie can paint a stolen car in $$x$$ hours, and Clyde can paint the same car in $$y$$ hours. They start working simultaneously and independently at their respective constant rates at 9:45am. If both $$x$$ and $$y$$ are odd integers, is $$x=y$$?
_________________
Manager
Joined: 13 May 2017
Posts: 101
Location: Finland
Concentration: Accounting, Entrepreneurship
GPA: 3.14
WE: Account Management (Entertainment and Sports)

Show Tags

24 Sep 2018, 23:46
I approached statement 1 the same way as arhumsid did before:

"S1: $$x^2+y^2 \lt 12$$
Since we already know that x and y are Odd integers, there's not much left to check for in the above condition and since adding two squares will very soon pass such a small number as 12, we can do the manual work here-
x=1; y=1 => Yes (both equal and sum of squares less than 12)
x=1; y=3 => No (both unequal and still the sum of squares less than 12)"

=>statement 1 is insufficient alone to answer

Statement 2: Bonnie and Clyde complete the painting of the car at 10:30am

From the given details we know that Bonnie and Clyde worked for 45minutes to finish painting one car. We need to prove whether x=y.

If x=y=1 then Bonnie and Clyde working together would finish painting a car in 1/2hr (30min < 45min)
If x=y=3 then Bonnie and Clyde working together with the same 3hrs/car rate would finish painting the car in 1.5hr (90min > 45 min)

We took the smallest pozitive odd integers to make x and y hr/car rate equal and we saw that with 1hr/car=x=y Bonnie and Clyde would finish painting the car less than 45 min and with a 3hr/car rate they would finish painting car more than 45 min.

If the precondition wouldn't hold on odd integer condition, x=y=45min*2=1.5hr. Because of the given precondition, x and y cannot be equal.

=> Therefore, Bonnie and Clyde must work with different hr/car rate => statement 2 is sufficient alone to answer
Intern
Joined: 24 Apr 2016
Posts: 27

Show Tags

16 Oct 2018, 18:48
Bunuel wrote:
Official Solution:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now, if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) $$x^2+y^2 \lt 12$$. It's possible that $$x$$ and $$y$$ are odd and equal each other if $$x=y=1$$, but it's also possible that $$x=1$$ and $$y=3$$ (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am. They complete the job in $$\frac{3}{4}$$ of an hour (45 minutes), since it's not $$\frac{odd}{2}$$ then $$x$$ and $$y$$ are not equal. Sufficient.

Why does it matter whether they complete the job in $$\frac{odd}{2}$$ in order for $$x=y$$?
Math Expert
Joined: 02 Sep 2009
Posts: 51218

Show Tags

16 Oct 2018, 19:54
dracobook wrote:
Bunuel wrote:
Official Solution:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now,if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) $$x^2+y^2 \lt 12$$. It's possible that $$x$$ and $$y$$ are odd and equal each other if $$x=y=1$$, but it's also possible that $$x=1$$ and $$y=3$$ (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am. They complete the job in $$\frac{3}{4}$$ of an hour (45 minutes), since it's not $$\frac{odd}{2}$$ then $$x$$ and $$y$$ are not equal. Sufficient.

Why does it matter whether they complete the job in $$\frac{odd}{2}$$ in order for $$x=y$$?

This is explained in highlighted part above.
_________________
Intern
Joined: 24 Apr 2016
Posts: 27

Show Tags

17 Oct 2018, 15:52
Thank you!

Bunuel wrote:
dracobook wrote:
Bunuel wrote:
Official Solution:

Bonnie and Clyde, when working together, complete the painting of the car in $$\frac{xy}{x+y}$$ hours (sum of the rates equal to the combined rate or reciprocal of total time: $$\frac{1}{x}+\frac{1}{y}=\frac{1}{T}$$. From that we can get that $$T=\frac{xy}{x+y}$$). Now,if $$x=y$$, then the total time would be: $$\frac{x^2}{2x}=\frac{x}{2}$$, since $$x$$ is odd, then this time would be $$\frac{odd}{2}$$: 0.5 hours, 1.5 hours, 2.5 hours, ...

(1) $$x^2+y^2 \lt 12$$. It's possible that $$x$$ and $$y$$ are odd and equal each other if $$x=y=1$$, but it's also possible that $$x=1$$ and $$y=3$$ (or vise-versa). Not sufficient.

(2) Bonnie and Clyde complete the painting of the car at 10:30am. They complete the job in $$\frac{3}{4}$$ of an hour (45 minutes), since it's not $$\frac{odd}{2}$$ then $$x$$ and $$y$$ are not equal. Sufficient.

Why does it matter whether they complete the job in $$\frac{odd}{2}$$ in order for $$x=y$$?

This is explained in highlighted part above.
Re: M27-11 &nbs [#permalink] 17 Oct 2018, 15:52

Go to page   Previous    1   2   [ 28 posts ]

Display posts from previous: Sort by

M27-11

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.