BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

210 bags contain almonds

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760
by sanju09 » Wed Feb 25, 2009 5:46 am
Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?

A. 256
B. 260
C. 316
D. 320
E. It cannot be determined from the given information.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Sat Jan 24, 2009 3:00 pm
Thanked: 11 times
GMAT Score:730

by BuckeyeT » Wed Feb 25, 2009 6:04 am
D. 320
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Wed Feb 25, 2009 6:05 am
Raisins only = Ro
Peanuts only = Po
Almonds only = Ao
Raisins and peanuts only = RP

Lets say Ao = x, then from the info within the question:
Po = (1/5)*x
RP = x/20
Ro = 10*Po = 2x

The three-set equation for this problem is as follows:

Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
9x/4 = 225
x = 100

Therefore, bags that contain only one item = Ao + Po + Ro = x + x/5 + 2x = 3.2*(100) = 320.

Choose D.

-BM-
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 76
Joined: Sat Jan 24, 2009 3:00 pm
Thanked: 11 times
GMAT Score:730

by BuckeyeT » Wed Feb 25, 2009 6:13 am
Explained answer...

Let r = raison only, a = almond only, p = peanut only
ra = raison/almond only, rp = raison/peanut only, ap = almond/peanut only, rap = raison/almond/peanut

r + a + p + ra + rp + ap + rap = 435
a + ra + ap + rap = 210
r = 10p
a = 20(rp)
p = 1/5 a

Combining the first two statements above, we get
r + a + p + ra + rp + ap + rap = 435
a + ra + ap + rap = 210
=r + p + rp = 225
=r + p + rp = 225

These can all be solved in terms of a from the third, fourth, and fifth statements above.

10(1/5 a) + 1/5 a + 1/20 a = 225
2a + 1/5 a + 1/20 a = 225
40/20 a + 4/20 a + 1/20 a = 225
45/20 a = 225
a = 100.

Solve for r and p.
p = 1/5 a
p = 1/5 (100)
p = 20.

r = 10 p
r = 10 (20)
r = 200.

a + r + p = 100 + 200 + 20 = 320.
D.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Wed Feb 25, 2009 6:13 am
bluementor wrote:Raisins only = Ro
Peanuts only = Po
Almonds only = Ao
Raisins and peanuts only = RP

Lets say Ao = x, then from the info within the question:
Po = (1/5)*x
RP = x/20
Ro = 10*Po = 2x

The three-set equation for this problem is as follows:

Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
9x/4 = 225
x = 100

Therefore, bags that contain only one item = Ao + Po + Ro = x + x/5 + 2x = 3.2*(100) = 320.

Choose D.

-BM-
I don't understand what you did here :oops: could you explain me?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Wed Feb 25, 2009 6:21 am
Ah nevermind the explanation of BuckeyeT is clear to me:D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Wed Feb 25, 2009 6:52 am
deepoe wrote:I don't understand what you did here :oops: could you explain me?
First I define the variables that I will use in this problem:
Raisins only = Ro
Peanuts only = Po
Almonds only = Ao
Raisins and peanuts only = RP

From the question, the relationship between the variables are given. First I assume Ao to be x. Then, I use this as the basis for the relationship to the other variables.

Info 1: The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds
Po = (1/5)*Ao
Po = (1/5)*x

Info 2: The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts

Ao = 20*RP
x = 20*RP
RP = x/20

Info 3: The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts

Ro = 10*Po
Ro = 10*(1/5)*x … (we use Info 1 to substitute Po)
Ro = 2x

Info 4: 210 bags contain almonds

210 = All the bags that contain almonds
210 = Almonds_only + (Almonds_and_Peanuts) + (Almonds_and_Raisins) + (Almonds_Peanuts_Raisins)

The diagram attached is a three-set representation of this problem. The 210 bags containing almonds is marked by the green region.

Therefore, the three set equation that can be derived from this diagram is:

Total = Total_almonds_(green region) + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
9x/4 = 225
x = 100

Therefore, bags that contain only one item = Ao + Po + Ro = x + x/5 + 2x = 3.2*(100) = 320.

Did this help? If not, let me know which part bothers you.

-BM-
Attachments
Peanuts Almonds Raisins.doc
(21 KiB) Downloaded 119 times
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 319
Joined: Wed Feb 04, 2009 10:32 amFollowed by: 9 members
Location: Delhi
Thanked: 84 times

by sureshbala » Wed Feb 25, 2009 7:27 am
Folks, believe me I am really shocked to see many equations here....
Anyway, here is my solution which is quite quick enough I guess...


Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 258
Joined: Thu Aug 07, 2008 5:32 am
Thanked: 16 times

Re: 210 bags contain almonds

by x2suresh » Wed Feb 25, 2009 7:35 am
sanju09 wrote:Each of 435 bags contains at least one of the following three items: raisins, almonds, and peanuts. The number of bags that contain only raisins is 10 times the number of bags that contain only peanuts. The number of bags that contain only almonds is 20 times the number of bags that contain only raisins and peanuts. The number of bags that contain only peanuts is one-fifth the number of bags that contain only almonds. 210 bags contain almonds. How many bags contain only one kind of item?

A. 256
B. 260
C. 316
D. 320
E. It cannot be determined from the given information.



R+A+P+RA+PA+RP+RAP=435 -->(1)

R=10P
A=20RP
P=1/5*A
A+RA+PA+RAP=210 -->(2)

FROM 1 AND 2

R+P+RP =435-210 = 225

10P+P+1/4*P = 225
-->P=20

R+A+P = 200+100 +20 =320
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Wed Feb 25, 2009 1:42 pm
bluementor how did you calculate the 9x?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Wed Feb 25, 2009 4:09 pm
deepoe wrote:bluementor how did you calculate the 9x?
Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
435 - 210 = x(2 + 1/5 + 1/20)
225 = x(2 + 4/20 + 1/20)
225 = x(2 + 5/20)
225 =x(2 + 1/4)
225 = x(9/4)
x = 100

hth

-BM-
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 151
Joined: Wed Feb 11, 2009 2:45 am
Thanked: 3 times

by deepoe » Thu Feb 26, 2009 12:39 am
bluementor wrote:
deepoe wrote:bluementor how did you calculate the 9x?
Total = Total_almonds + Ro + Po + RP
435 = 210 + 2x + x/5 + x/20
435 - 210 = x(2 + 1/5 + 1/20)
225 = x(2 + 4/20 + 1/20)
225 = x(2 + 5/20)
225 =x(2 + 1/4)
225 = x(9/4)
x = 100

hth

-BM-
Thanks mate!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Fri Feb 27, 2009 12:41 amFollowed by: 3 members
Thanked: 44 times

by kamu » Tue Mar 03, 2009 2:54 am
sureshbala wrote:Folks, believe me I am really shocked to see many equations here....
Anyway, here is my solution which is quite quick enough I guess...


Image
Excellent.
Thanks.

Please could you explain how to use Venn Diagrams in different circumstances.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 40
Joined: Sat Jul 18, 2009 1:11 pm
Thanked: 4 times
GMAT Score:640

by ironsferri » Fri Jun 04, 2010 12:27 pm
what is the difficulty level of this question????
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1