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

It is currently 19 Oct 2018, 21:23

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

A pumpkin patch contains x pumpkins that weigh 10 pounds each and y

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 24 Jul 2015, 00:52
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

55% (02:40) correct 45% (02:22) wrong based on 170 sessions

HideShow timer Statistics

A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.

Kudos for a correct solution.

_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
User avatar
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 422
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 24 Jul 2015, 02:14
Hello,

I don't think I have found it. But I want to be notified by the answer, so I will only post my reasoning. I will go for E.

The stem gives us this information:
x(10) + y(r) = 12 (x+y) --> just diretly moved x+y from the denominator so it is easier to read.
On the side, this also tells us that x+y are multiples of 12, even though I wasn't able to use this anywhere..

[1] Based on the first stem, we know that the heavier pumpkins are 5 more than the lighter ones, h = l+5.
However, we don't know if r>10 or r<10. It cannot be 10 because then there wouldn't be any heavier pumkins. So, NS.

[2] Says that the weight of the heavier pumkins is 3 more than their number.
We still don't know if r is more or less than 10 and we also don't know x and y (so the number of each type of pumkins). So, NS.

[1,2] Tell us thatt the weight of the heavier pumkins is l+5+3=l+8.

What I did then is replace x+y with l+h, because it is basically the same. So, this gives:
x+y = l+h = (l+5)(l+8)+l(r).

However, we still don't know if x or y are the heavier ones. So, in the end we still end up with r, y, x and l to solve for. So, NS.

So, I think we need to know something more to solve it.
Manager
Manager
User avatar
Joined: 26 Dec 2011
Posts: 114
Schools: HBS '18, IIMA
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 24 Jul 2015, 03:52
2
1
Bunuel wrote:
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.

Kudos for a correct solution.


Solution -

Average is greater than 10, it leads to r > 10.

Average --> (10x+ry)/(x+y)=12 --> r=(2x/y)+12?

Stmt1 - Given that, y=x+5.

r = (2x/y)+12 = (2*(y-5)/y)+12. We do not know the value of Y. Not Sufficient.

Stmt2 - Given that, r = 3+y. We do not know the value of Y. Not Sufficient.

1+2 - r = 3+y and y = x+5 --> r-3 = x+5 --> y = r-3 and x = r-8

Substitute x and y in r = (2x/y)+12 --> 2*(r-8) = (r-3)*(r-12) --> r^2-17r+52=0 --> (r-13)*(r-4) = 0 --> r =13 or 4.

The value of r = 13. Sufficient.

ANS C.
_________________

Thanks,
Kudos Please

Current Student
avatar
S
Joined: 20 Mar 2014
Posts: 2633
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Premium Member Reviews Badge
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 24 Jul 2015, 04:12
1
pacifist85 wrote:
Hello,

I don't think I have found it. But I want to be notified by the answer, so I will only post my reasoning. I will go for E.

The stem gives us this information:
x(10) + y(r) = 12 (x+y) --> just diretly moved x+y from the denominator so it is easier to read.
On the side, this also tells us that x+y are multiples of 12, even though I wasn't able to use this anywhere..

[1] Based on the first stem, we know that the heavier pumpkins are 5 more than the lighter ones, h = l+5.
However, we don't know if r>10 or r<10. It cannot be 10 because then there wouldn't be any heavier pumkins. So, NS.


[2] Says that the weight of the heavier pumkins is 3 more than their number.
We still don't know if r is more or less than 10 and we also don't know x and y (so the number of each type of pumkins). So, NS.


[1,2] Tell us thatt the weight of the heavier pumkins is l+5+3=l+8.

What I did then is replace x+y with l+h, because it is basically the same. So, this gives:
x+y = l+h = (l+5)(l+8)+l(r).

However, we still don't know if x or y are the heavier ones. So, in the end we still end up with r, y, x and l to solve for. So, NS.

So, I think we need to know something more to solve it.



I am critiquing your logic for statement 1. If you are given that h=l+5, you can translate it to say y=x+5 as the average of x pumpkins is 10 and the overall is 12. So x can not be the heavier ones. The heavier ones have to be 'y'. On its own, this statement is still not sufficient as :

x(10) + y(r) = 12 (x+y) and y=x+5 ---> ry = 14y+10. Still not sufficient.

Again, for your statement 2, heavier = y , lighter = x

Thus, per statement 2, r=y+3. Thus on its own, not sufficient.

Combining, you will be able to find out the value of r.

Hope this helps.
Senior Manager
Senior Manager
avatar
Joined: 27 Dec 2013
Posts: 251
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 24 Jul 2015, 04:15
1
My answer is C.

Stem:

X pumpkins weight 10 pound each-> [Total weight]= 10x

Y pumpking weight r pound each-> [Total weight] = Yr.

total weight of all pumpkins= 10x+Yr.

Average weight of all pumpins = (10x+Yr) / X+Y.

Given avg= 12.

10x+yr = 12x+ 12 Y.

=>Yr= 2x+ 12Y.- equation 1. [This is a 3 variable equation, hence we need 3 equations to solve]


Option 1: There are five more heavier pumpkins than lighter pumpkins

From the stem we know that Average of X pumpkins will be 10. From the statment, as the average gone up, the average of r pumpkins must be higher.

So statement 1 can be written as y= x+5- Equation 2.

and

Statement 2 can be written as r=3+Y - Equation 3.


so either statment is not sufficient, but both are required.

Also by solving Eq2 and Eq3, we can establish a relationship between X and r and that become ours equ 4.


Hope its correct.


Bunuel wrote:
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.

Kudos for a correct solution.

_________________

Kudos to you, for helping me with some KUDOS.

Intern
Intern
User avatar
Joined: 08 Jul 2012
Posts: 49
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 24 Jul 2015, 10:29
1
Bunuel wrote:
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.

Kudos for a correct solution.


given: x have 10 pounds each and y have r pounds each and avg. wt is 12 pounds, hence r>10
12 = (10x+ry) / (x+y), rearranging we get r=12+2x/y or r-12=2x/y

stmt1 : y = x+5...clearly insufficient, r-12=2x/(x+5), x is unknown

stmt2 : r = 3+y...insufficient, as y is unknown

stmt1+stmt2 : from 1 we get x=y-5 & from 2 we get y=r-3

hence, x=r-3-5=r-8
now we have x&y both in terms of r,
hence r=12+2(r-8)/(r-3)
this gives, (r-13)(r-4)=0
as r>10 (given) ...r=13

Ans. C
_________________

Our greatest weakness lies in giving up. The most certain way to succeed is always to try just one more time. - Thomas A. Edison

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 50002
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 26 Jul 2015, 12:51
Bunuel wrote:
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.

Kudos for a correct solution.


800score Official Solution:

The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier.
Writing an equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. This equation has three unknowns.
Solve for r to get:
r = 12 + (2x/y)

Statement (1) defines the relationship between x and y as y = x + 5 or y = x – 5. Plug in the formula for r to get
r = 12 + (2(y – 5)/y)
r = 14 – 10/y
Plug in couple values of y (greater than 5) to see that they yield different results for r.
Statement (1) is not sufficient.

Statement (2) gives the relationship between y and r as r = y + 3. This isn't enough information as you can plug it into the original equation and see that too many variables remain. Plug in couple values for y to see that we get different possible valuee of r and x.

Using the given with the equations from (1) and (2), the system is
(10x + ry) / (x + y) = 12
y = x + 5
r = y + 3


Rewrite r = y + 3 as r – 3 = y.
Substitute r – 3 = y into y = x + 5 and get r – 3 = x + 5 so r = x + 8.
This gives an equation for y in terms of x, and an equation for r in terms of x.
Substitute these equations into the original equation.
(10x + ry) / (x + y) = 12. Substitute y = x + 5 and r = x + 8.
(10x + (x + 8)(x + 5)) / (x + (x + 5)) = 12
10x + (x² + 13x + 40) = 12(2x + 5)
x² + 23x + 40 = 24x + 60
x² – x – 20 = 0
(x – 5)(x + 4) = 0
x = 5 or x = -4 = 0
So there are 5 of the smaller pumpkins.
Using y = x + 5, y = 5 + 5 = 10, so there are 10 of the larger pumpkins.
And using r = y + 3, r = 13, the larger pumpkins weigh 13 pounds.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 04 Jul 2015
Posts: 3
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 04 Aug 2015, 16:12
Bunuel wrote:
Bunuel wrote:
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.

Kudos for a correct solution.


800score Official Solution:

The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier.
Writing an equation for the average weight of the pumpkins is (10x + ry) / (x + y) = 12. This equation has three unknowns.
Solve for r to get:
r = 12 + (2x/y)

Statement (1) defines the relationship between x and y as y = x + 5 or y = x – 5. Plug in the formula for r to get
r = 12 + (2(y – 5)/y)
r = 14 – 10/y
Plug in couple values of y (greater than 5) to see that they yield different results for r.
Statement (1) is not sufficient.

Statement (2) gives the relationship between y and r as r = y + 3. This isn't enough information as you can plug it into the original equation and see that too many variables remain. Plug in couple values for y to see that we get different possible valuee of r and x.

Using the given with the equations from (1) and (2), the system is
(10x + ry) / (x + y) = 12
y = x + 5
r = y + 3


Rewrite r = y + 3 as r – 3 = y.
Substitute r – 3 = y into y = x + 5 and get r – 3 = x + 5 so r = x + 8.
This gives an equation for y in terms of x, and an equation for r in terms of x.
Substitute these equations into the original equation.
(10x + ry) / (x + y) = 12. Substitute y = x + 5 and r = x + 8.
(10x + (x + 8)(x + 5)) / (x + (x + 5)) = 12
10x + (x² + 13x + 40) = 12(2x + 5)
x² + 23x + 40 = 24x + 60
x² – x – 20 = 0
(x – 5)(x + 4) = 0
x = 5 or x = -4 = 0
So there are 5 of the smaller pumpkins.
Using y = x + 5, y = 5 + 5 = 10, so there are 10 of the larger pumpkins.
And using r = y + 3, r = 13, the larger pumpkins weigh 13 pounds.


For 2, why did not you consider 10=x+3?
Manager
Manager
avatar
S
Joined: 13 Mar 2013
Posts: 166
Location: United States
Concentration: Leadership, Technology
GPA: 3.5
WE: Engineering (Telecommunications)
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 02 Sep 2015, 09:06
The original statement says the average weight of a pumpkin is 12 pounds. Since x pumpkins weigh 10 pounds each, the y pumpkins must be heavier.

I didn't understand this part . Please help to understand .

that if one is 10 pound and other is y pound and avg is 12 .
How can i conclude that -- y >10 taking to into account the avg 12.
_________________

Regards ,

Math Revolution GMAT Instructor
User avatar
V
Joined: 16 Aug 2015
Posts: 6390
GMAT 1: 760 Q51 V42
GPA: 3.82
Premium Member
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 03 Sep 2015, 04:50
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.



In the original condition, this question is typical 2by2 question in GMAT test.

(10x+ry/x+y)=12, and thus there are 3 variables (x,y,r) and 1 equation (10x+ry/x+y)=12. We need to match the number of equations and varibles and thus we need 2 more. Since there is 1 each in 1) and 2), we need both of them and thus C is likely the answer, and it turns out that C actually is the answer.

Using both 1) & 2) together, let's start from the assumption that 10>r. x=y+5, 10=x+3, then x=7, y=2, (10*7+2r/9)=12, 70+2r=12*9=108, 2r=38, r=19. Here we assumed that pumpkin1 is more heavier, and we have contradiction.

Assuming 10<r, y=x+5, r=y+3, 10x+ry=12(x+y)=12x+12y, ry=2x+12y, r=y+3. substituting x=y-5 gives us (y+3)y=2(y-5)+12y, y^2+3y=2y-10+12y, y^2-11y+10=0, (y-10)(y-1)=0, y=10, y=1. since y=1 gives us x=-4, this is not our answer, and having y=10 gives us x=5, r=13 which is the answer.

However, solving the question directly such as the above is inappropriate for 2 minute deadline, and it's more efficient to solve through variable approach method. The answer is C.
Attachments

pumpkin.jpg
pumpkin.jpg [ 14.32 KiB | Viewed 2840 times ]


_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
"Unlimited Access to over 120 free video lessons - try it yourself"

Director
Director
User avatar
G
Joined: 24 Nov 2015
Posts: 522
Location: United States (LA)
Reviews Badge
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 17 Mar 2016, 14:38
didn't understand the combine statement 1 and statement 2 part
Math Expert
User avatar
V
Joined: 02 Aug 2009
Posts: 6961
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 17 Mar 2016, 15:06
2
rhine29388 wrote:
didn't understand the combine statement 1 and statement 2 part


Quote:
A pumpkin patch contains x pumpkins that weigh 10 pounds each and y pumpkins that weigh r pounds each. If the average (arithmetic mean) weight of the pumpkins is 12 pounds, what is the value of r?

(1) There are five more heavier pumpkins than lighter pumpkins.
(2) The weight in pounds of each of the heavier pumpkins is 3 more than their number.



Hi,
since you have asked about combined solution, i'll just speak on that..

INFO from Q-


there are x pumpkins of 10 pounds each and y pumpkins that weigh r pounds each..
avg weight =12
so 10x+ry=12(x+y)

Info from statements


1) y=x+5
2) r=y+3= x+5+3=x+8

Combined


we have all values in terms of x..
lets substitute them in 10x+ry=12(x+y)..

so \(10x + (x+8)(x+5) = 12 (x+x+5)\)
=> \(10x + x^2 +13x + 40 = 24x + 60\)..
\(x^2 - x -20 =0\)..
\(x^2 - 5x + 4x -20=0\)..
\(x(x-5)+4(x-5)=0\)..
\((x-5)(x+4)=0\).
so\(x=5,-4\)..
x cannot be -ive so x= 5 and \(r = x + 8 = 13\)


Hope it helps
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


GMAT online Tutor

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 8459
Premium Member
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y  [#permalink]

Show Tags

New post 21 Sep 2018, 15:27
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

GMAT Club Bot
Re: A pumpkin patch contains x pumpkins that weigh 10 pounds each and y &nbs [#permalink] 21 Sep 2018, 15:27
Display posts from previous: Sort by

A pumpkin patch contains x pumpkins that weigh 10 pounds each and y

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.