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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Ratios

Expert replies
by Thrills4ever » Sat May 14, 2011 2:21 pm
Hey Guys,

So this problem is pretty straightforward but i wanted to know if there's an easier way to solve it?

A mixture of red beans and black beans is to be prepared. The price of red beans is $2 per pound, and the price of black beans is 3$ per pound. What is the ratio of red beans to black beans if the mixture is to be sold for $2.75 per pound?

(A) 1:3
(B) 1:2
(C) 2:3
(D) 1:1
(E) 3:1

All help is really appreciated
Join the discussion
Source: — Problem Solving |

by pankajks2010 » Sat May 14, 2011 5:30 pm
Let the required ratio be 1 : x

Now, as per the given fact, we can make the following equation:

(2+3x)/(1+x)=2.75; simplifying, this we get; 0.25x=0.75
Thus, x=3

A
Last edited by pankajks2010 on Sat May 14, 2011 7:13 pm, edited 1 time in total.
Join the discussion

by GMATGuruNY » Sat May 14, 2011 6:53 pm
Thrills4ever wrote:Hey Guys,

So this problem is pretty straightforward but i wanted to know if there's an easier way to solve it?

A mixture of red beans and black beans is to be prepared. The price of red beans is $2 per pound, and the price of black beans is 3$ per pound. What is the ratio of red beans to black beans if the mixture is to be sold for $2.75 per pound?

(A) 1:3
(B) 1:2
(C) 2:3
(D) 1:1
(E) 3:1

All help is really appreciated
For most mixture and weighted average problems, I use alligation.
Alligation is used to determine how much weight must be given to each element in a mixture of 2 elements:

The proportion of each element in the mixture = the distance between the value associated with the mixture and the value associated with the other element.

Proportion of red beans in the mixture = the distance between the black bean price and the mixture price = 3.00-2.75 = .25.
Proportion of black beans in the mixture = the distance between the red bean price and the mixture price = 2.75-2 = .75.

The results above give the ratio of red beans to black beans in the mixture:
Red beans:black beans = (.25): (.75) = 1:3.

The correct answer is A.

The problem above also could be solved with some clever reasoning.
The answer choices represent the ratio of red beans to black beans in the mixture.

The $2.75 selling price of the mixture is closer to the price of the black beans ($3).
Thus, the black beans must be the higher value in the ratio.
Eliminate D and E.

The values in the correct ratio represent how much of each kind of bean is in the mixture.
The sum of the values in the correct ratio represent the total weight of the mixture.
Since the prices of the beans are integer values, when the total weight and the price of the mixture are multiplied, the product should be an integer.

Of the remaining answers, only answer choice A (ratio = 1:3) works:
1+3 = 4.
4*(2.75) = 11.

To confirm that red:black = 1:3 is the correct ratio:
Price of 1 pound of red beans = 1*2 = 2.
Price of 3 pounds of black beans = 3*3 = 9.
Total price = 2+9 = 11.
Total weight = 1+3 = 4 pounds.
Price per pound = 11/4 = 2.75.

The correct answer is A.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion