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

 It is currently 16 Nov 2018, 10:13

### 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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### Free GMAT Strategy Webinar

November 17, 2018

November 17, 2018

07:00 AM PST

09:00 AM PST

Nov. 17, 7 AM PST. Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
• ### GMATbuster's Weekly GMAT Quant Quiz # 9

November 17, 2018

November 17, 2018

09:00 AM PST

11:00 AM PST

Join the Quiz Saturday November 17th, 9 AM PST. The Quiz will last approximately 2 hours. Make sure you are on time or you will be at a disadvantage.

# Each of the 30 boxes in a certain shipment weighs either 10

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 12 Oct 2008
Posts: 456
Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

### Show Tags

09 Oct 2009, 22:25
7
22
00:00

Difficulty:

65% (hard)

Question Stats:

65% (02:36) correct 35% (02:40) wrong based on 863 sessions

### HideShow timer Statistics

Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

OA D IMO B
Math Expert
Joined: 02 Sep 2009
Posts: 50619

### Show Tags

09 Oct 2009, 23:07
11
6
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

OA D IMO B

This is a weighted average question: $$Weighted \ average=\frac{weight_1*value_1+weight_2*value_2}{value_1+value_2}$$.

Let $$x$$ be number of 10-pound boxes --> $$\frac{10x + 20*(30-x)}{30}=18$$ --> $$x=6$$;

We have $$6$$ 10-pound boxes and $$24$$ 20-pound boxes.

Let $$y$$ be number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$ --> $$y=20$$.

_________________
Manager
Status: mba here i come!
Joined: 07 Aug 2011
Posts: 210

### Show Tags

21 Aug 2011, 09:59
8
2
total current weight = 30*18 = 540
let's assume that x number of 20lbs boxes will be removed

$$\frac{540-20x}{30-x} = 14$$

x = 20 ............. D
_________________

press +1 Kudos to appreciate posts

##### General Discussion
SVP
Joined: 29 Aug 2007
Posts: 2378

### Show Tags

09 Oct 2009, 22:43
2
1
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

OA D IMO B

x = 20 lb box:

20x + 10(30-x) = 18(30)
x = (540 - 300)/10
x = 24

If k = removed 20 lb box:

20(24-k)+ 10(6) = 14(6+24-k)
480 - 20k + 60 = 420 - 14k
540 - 420 = 6k
540 - 420 = 6k
k = 120/6 = 20
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Manager
Status: Preparing Apps
Joined: 04 Mar 2009
Posts: 88
Concentration: Marketing, Strategy
GMAT 1: 650 Q48 V31
GMAT 2: 710 Q49 V38
WE: Information Technology (Consulting)

### Show Tags

12 Dec 2010, 23:28
dont you have to reduce the average weight of 30 booxes to 14 pounds?
Math Expert
Joined: 02 Sep 2009
Posts: 50619

### Show Tags

13 Dec 2010, 01:47
aalriy wrote:
dont you have to reduce the average weight of 30 booxes to 14 pounds?

Yes, the average weight of the 30 boxes is to be reduced to 14 pounds by removing some of the 20-pound boxes.

We found that there were 6 10-pound boxes and 24 20-pound boxes.

So, if $$y$$ is the number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$, (note that as we are reducing the # of 20-pound boxes (24) by y then the total # of boxes (30) also will be reduced by y) --> $$y=20$$.

Hope it's clear.
_________________
Manager
Status: Preparing Apps
Joined: 04 Mar 2009
Posts: 88
Concentration: Marketing, Strategy
GMAT 1: 650 Q48 V31
GMAT 2: 710 Q49 V38
WE: Information Technology (Consulting)

### Show Tags

13 Dec 2010, 03:43
Bunuel wrote:
So, if $$y$$ is the number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$, (note that as we are reducing the # of 20-pound boxes (24) by y then the total # of boxes (30) also will be reduced by y--> is this implied or an assumption) --> $$y=20$$.

.

See my comment in Red Italics. What if we make an attempt to keep total number of boxes 30 by adding $$y$$ 10-pound boxes. Does this have to be specifically mentioned in the question?
Director
Joined: 03 Sep 2006
Posts: 793

### Show Tags

13 Dec 2010, 04:15
x boxes of 10pounds
y boxes of 20pounds
x+y = 30

$$(10*x+20*y)/30=18$$

$$10x+20y=540$$

Let ? be the number of 20 pound boxes which should be removed in order to get the average weight of 14pounds

$$(10*x+20*y-20*?)/(30-?)=14$$

$$(540-20*?)= (420-14*?)$$

$$? = (120/6) = 20$$
Math Expert
Joined: 02 Sep 2009
Posts: 50619

### Show Tags

13 Dec 2010, 04:53
aalriy wrote:
Bunuel wrote:
So, if $$y$$ is the number of 20-pound boxes that should be removed so that average to become 14 pounds --> $$\frac{10*6+20*(24-y)}{30-y}=14$$, (note that as we are reducing the # of 20-pound boxes (24) by y then the total # of boxes (30) also will be reduced by y--> is this implied or an assumption) --> $$y=20$$.

.

See my comment in Red Italics. What if we make an attempt to keep total number of boxes 30 by adding $$y$$ 10-pound boxes. Does this have to be specifically mentioned in the question?

You should read the stem more carefully: "If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?" So stem mentions that the reduction in average weight should be made by removing some of the 20-pound boxes.

Hope it's clear.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8537
Location: Pune, India

### Show Tags

13 Dec 2010, 21:02
12
1
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

OA D IMO B

Long question stem. Best approach here would be to analyze one sentence at a time.

Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds.

If average of 10 and 20 is 18, the ratio of no. of 10 pound boxes: no of 20 pound boxes is 1:4
(If this concept is not clear, check out http://gmatclub.com/forum/tough-ds-105651.html#p828579)
So out of 30 boxes, 6 are 10 pound boxes and 24 are 20 pound boxes.

If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes

Now average of 10 and 20 pound boxes is to be 14. So ratio of no. of 10 pound boxes:no. of 20 pound boxes = 3:2
The number of 10 pound boxes remain the same so we still have 6 of them. To get a ratio of 3:2, the no of 20 pound boxes must be 4.

how many 20-pound boxes must be removed?
There were 24 of these. Now there are only 4. 24 - 4 = 20 should be removed.
_________________

Karishma
Veritas Prep GMAT Instructor

GMAT self-study has never been more personalized or more fun. Try ORION Free!

Manager
Joined: 02 Nov 2009
Posts: 108

### Show Tags

10 Feb 2012, 05:35
N1+ N2 = 30

10(N1) + 20(N1)
_____________ =18
N1+ N2

----- > solve 10(N1) + 20(N2) = 18(N1) + 18(N2)

So n1 =6 n2 = 24

Now N1 will not change ONLY n2 will change so

reduce the average to 14

10(6) + 20(N2) = 14(6) + 14(N2)

N2 =4

So it has to reduce by 20
Intern
Joined: 23 May 2013
Posts: 19
Location: United Kingdom
WE: Project Management (Real Estate)
Re: Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

### Show Tags

29 May 2013, 04:13
This is how I solved

X+Y =30

from condition given

10X+20Y = 540

therefore X=6 and Y = 24

Condition 2
X= 6 Y changes by "n" and avg weight of 30-n =14

therefore

10*6 + 20(24-n)=14( 30-n)

after solving the above equation we have

n=20

Ans is D
_________________

Correct me If I'm wrong !! looking for valuable inputs

Manager
Joined: 24 Nov 2012
Posts: 157
Concentration: Sustainability, Entrepreneurship
GMAT 1: 770 Q50 V44
WE: Business Development (Internet and New Media)

### Show Tags

31 May 2013, 12:55
1
MBAhereIcome wrote:
total current weight = 30*18 = 540
let's assume that x number of 20lbs boxes will be removed

$$\frac{540-20x}{30-x} = 14$$

x = 20 ............. D

Faster to use this logic
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013
Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013
Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013
GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013
Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013
Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013
GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013
My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

Manager
Joined: 20 Jun 2012
Posts: 82
Location: United States
Concentration: Finance, Operations
GMAT 1: 710 Q51 V25
Re: Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

### Show Tags

10 Jun 2013, 08:24
Each of the 30 boxes in a certain shipment weighs either 10 pounds or 20 pounds, and average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?

A. 4
B. 6
C. 10
D. 20
E. 24

OA D IMO B

Easy question:

suppose 10 pound boxes = x
20 pound boxes = (30-x)

Given in question >> [x*10+(30-x)*20]/30=18

=> [x*10+(30-x)*20] = 540 ----eq. 1 .. Leave it like this.

Now we have to remove some 20 Pound boxes

let 20 pound boxes removed = a
new equation will be >> [x*10+(30-x-a)*20]/(30-a)=14
>> x*10+(30-x)*20 - 20a = 14(30-a)
From eq 1 >> 540 - 20a = 420 -14a
solve for a, and you'll get a=20 ..

I made it look quite big, but I solved it under 2 minutes using this in first try on such question
_________________

Forget Kudos ... be an altruist

Current Student
Joined: 19 Nov 2014
Posts: 3
Location: United States
GMAT 1: 600 Q41 V31
GPA: 3.88
WE: Project Management (Pharmaceuticals and Biotech)
Re: Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

### Show Tags

19 May 2015, 12:25
My solution is as below:

10x+ (30-x)20 = 540
x=6 (no. of 10 pound boxes)
hence 24 is no. of 20 pound boxes

If y is the new no. of 20 pound boxes required to make the average 14 then,

(10*6+20y) /(6+y) = 14

y=4

hence the no. of 20 pound boxes to be removed is 24-4 = 20 boxes
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 12863
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Each of the 30 boxes in a certain shipment weighs either 10  [#permalink]

### Show Tags

11 Apr 2018, 19:08
Hi All,

This question can be solved in a number of different ways - including by TESTing THE ANSWERS. There's a Number Property rule that can actually make that solution a bit faster.

We're told that the average weight is 18 pounds, which makes the total weight 540 pounds (18 pounds x 30 boxes = 540 pounds)

Since the box weights are 10 or 20 pounds each, you can figure out the number of each rather quickly.

If there were 30 20-pound boxes, then the total weight would be 600 pounds.
Since the total weight is 540 pounds, that means that 6 of the boxes weigh 10 pounds instead of 20 pounds.

So, there are:
6 10-pound boxes
24 20-pound boxes

Since we want to remove some number of the 20-pound boxes and get an average weight of 14 pounds, we know that the number of 10-pound boxes MUST BE greater than the number of 20-pound boxes (this is a Weighted Average Number Property). Since we have 6 10-pound boxes, we need to remove at least 19 20-pound boxes.

We can certainly TEST answer D first OR we can note that answer E (24) would eliminate ALL of the 20-pound boxes, which would make the average of the remaining boxes 10 pounds, which is too low (and would prove that answer D was correct).

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****

Re: Each of the 30 boxes in a certain shipment weighs either 10 &nbs [#permalink] 11 Apr 2018, 19:08
Display posts from previous: Sort by