Last visit was: 05 Sep 2026, 01:04 It is currently 05 Sep 2026, 01:04
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
ayl989
User avatar
Joined: 15 Aug 2005
Last visit: 12 Aug 2007
Posts: 12
Posts: 12
Kudos: 0
Heres a problem, Aribica coffee costs .4 per ounce, robusta 20 Apr 2007, 17:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Heres a problem,
Aribica coffee costs .4 per ounce, robusta .3 per ounce. if the blend of both aribica and robusta costs .33, what is the share of arabica in this blend.

The solution gives the equations,
let A= ounces Arib, R= ounces Robusta
then,
(A*.4+R*.3)/(A+R)= .33

my question is why do you set up this type of equation, whats the principle? I just figured out combinations that would work, which I'm sure is the long way. So, I'm sure this eqn is more efficient, but I'm not sure why its being set up.
TIA.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
goalsnr
User avatar
Joined: 03 Apr 2007
Last visit: 17 Oct 2012
Posts: 625
Own Kudos:
5,181
Given Kudos: 10
Products:
My Reviews
Posts: 625
Kudos: 5,181
Heres a problem, Aribica coffee costs .4 per ounce, robusta 20 Apr 2007, 18:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ayl989
Heres a problem,
Aribica coffee costs .4 per ounce, robusta .3 per ounce. if the blend of both aribica and robusta costs .33, what is the share of arabica in this blend.

The solution gives the equations,
let A= ounces Arib, R= ounces Robusta
then,
(A*.4+R*.3)/(A+R)= .33

my question is why do you set up this type of equation, whats the principle? I just figured out combinations that would work, which I'm sure is the long way. So, I'm sure this eqn is more efficient, but I'm not sure why its being set up.
TIA.
Show more


Let the amt of Arabica coffee in 1 ounce of Blend be x
Let the amt of Arabica coffee in 1 ounce of Blend be y
->x+y =1 .....(1)

Cost of 1 ounce of arabica coffee is .4
Therefore cost of x ounce of arabica cofee is .4x

Cost of 1 ounce of Robustica coffee is .3
Therefore cost of y ounce of arabica cofee is .4y

Given :cost of blend/ounce = .33

->.4x+.3y = .33 ....(2)

Now solve equstions 1 and 2 to get values of x and y
x = .3
y =.7
User avatar
techjanson
User avatar
Joined: 28 Feb 2007
Last visit: 18 May 2010
Posts: 103
Own Kudos:
55
Location: California
Posts: 103
Kudos: 55
Heres a problem, Aribica coffee costs .4 per ounce, robusta 20 Apr 2007, 18:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
goalsnr
ayl989
Heres a problem,
Aribica coffee costs .4 per ounce, robusta .3 per ounce. if the blend of both aribica and robusta costs .33, what is the share of arabica in this blend.

The solution gives the equations,
let A= ounces Arib, R= ounces Robusta
then,
(A*.4+R*.3)/(A+R)= .33

my question is why do you set up this type of equation, whats the principle? I just figured out combinations that would work, which I'm sure is the long way. So, I'm sure this eqn is more efficient, but I'm not sure why its being set up.
TIA.

Let the amt of Arabica coffee in 1 ounce of Blend be x
Let the amt of Arabica coffee in 1 ounce of Blend be y
->x+y =1 .....(1)

Cost of 1 ounce of arabica coffee is .4
Therefore cost of x ounce of arabica cofee is .4x

Cost of 1 ounce of Robustica coffee is .3
Therefore cost of y ounce of arabica cofee is .4y

Given :cost of blend/ounce = .33

->.4x+.3y = .33 ....(2)

Now solve equstions 1 and 2 to get values of x and y
x = .3
y =.7
Show more


But it is not specified that cost of 1 ounce of blend is 0.33
Is it ok to assume that its per ounce price?
User avatar
ayl989
User avatar
Joined: 15 Aug 2005
Last visit: 12 Aug 2007
Posts: 12
Posts: 12
Kudos: 0
Heres a problem, Aribica coffee costs .4 per ounce, robusta 20 Apr 2007, 18:29
Kudos
Add Kudos
Bookmarks
Bookmark this Post
thanks for your explanation, seems like a straight forward method.

I'm still curious though, why does the original explanation
divide, .4A+.3R/(A+R) and make it equal to .33. I think this is a very simple principle, but I seem to be having a mental block on it.
thanks again.
User avatar
querio
User avatar
Joined: 11 Nov 2006
Last visit: 19 Jun 2007
Posts: 67
Own Kudos:
21
Posts: 67
Kudos: 21
Heres a problem, Aribica coffee costs .4 per ounce, robusta 20 Apr 2007, 22:05
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ayl989
Heres a problem,
Aribica coffee costs .4 per ounce, robusta .3 per ounce. if the blend of both aribica and robusta costs .33, what is the share of arabica in this blend.

The solution gives the equations,
let A= ounces Arib, R= ounces Robusta
then,
(A*.4+R*.3)/(A+R)= .33

my question is why do you set up this type of equation, whats the principle? I just figured out combinations that would work, which I'm sure is the long way. So, I'm sure this eqn is more efficient, but I'm not sure why its being set up.
TIA.
Show more


This solution is saying: (total cost of the mix in Pounds/ total weight of the mix in Ounces) = cost/ounce , it gives you the cost per ounce of mix.
There is a direct method recommended by Kaplan to solve this type of questions:
A: amount of Aribica coffee; R: amount of Robusta coffee

(Cost A-Cost Mix) * A = (Cost Mix-Cost R) * R
(0.3-0.33) * A = (0.33-0.4) * R
A/R = 7/3
Once you know the ratio (in this case weight of A to the weight of R) you can calculate the share of A in the total mix.
A/(A+R) = 7/(7+3) = 0.7 --> 70%



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!