mixtures-and-alligations

Mixtures and Alligations

Mixtures and Alligations: What is a mixture? Solved Exercises with Answers for Questions on Mixtures and Alligations

Mixtures and Alligations: What is a mixture? Solved Exercises with Answers for Questions on Mixtures and Alligations

    1 Tips

learning-pundits-content-team

Learning Pundits Content Team

Written on Sep 29, 2017 11:54:46 PM

2 TIPS on cracking Aptitude Questions on Mixtures and Alligations

Looking for Questions instead of tips? - You can directly jump to  Aptitude Test Questions on Mixtures and Alligations

Tip #1:

Volume of a substance in the final solution = Sum of volumes of the substance in the ingredients


Question: How many liters of pure alcohol must be added to a 100 liter solution that is 20% alcohol in order to produce a solution that is 25% alcohol?

Solution:

Here, let us add X liters of 100% alcohol to 100 liters of 20% alcohol to produce 25% alcohol. Then,

Volume of alcohol in the original 100 liter solution = 20% x 100 liters = 20 liters

Volume of alcohol in the pure alcohol solution = X liters

Total volume of alcohol in the final solution = 25% x (100 + X) liters

Volume of alcohol in the final solution = Sum of the volumes of alcohol in the two liquids being mixed

0.25 x (100 + X) liters = 20 + X liters

=   X = 20 / 3 liters 

Are you engaged in a Job Search? - You can get your Resume/ CV reviewed for free and then apply for jobs/ internships.

Tip #2: From a solution of volume V containing x percent of a substance, if y liters are replaced by water, the new percentage of the substance in the solution = x (V-y)/ V



Question: 8 liters are drawn from a cask full of pure wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of water is 16: 65. How much wine did the cask hold originally?

Solution:

Let Vw be the volume of wine in the cask originally.

Let Vt be the total volume of liquid in the cask.

Let X be the percentage of wine in the cask originally.

X = original volume of wine/ total volume of liquid in the cask = Vw / Vt

When 8 liters are drawn out, the volume of wine is reduced by 8 X liters while the total volume of liquid remains unchanged as it is re-filled with water.

Let X’ be the new percentage of wine in the cask after this operation

X’  = (original volume of wine – 8 X) / total volume of liquid in the cask

     = [Vw – 8 (Vw/ Vt)] / Vt

     = X (Vt – 8) / Vt

After 4 such replacement operations, X’’’’ = X [(Vt – 8)/ Vt] ^ 4

From the problem, X’’’’ = 16/ (16 + 65) = 16/ 81

Also, since originally the cask was full of pure wine, X = 1

[(Vt – 8)/ Vt] ^ 4 = 16/ 81

=> Vt = 24 liters

Are you interested in getting Certificates to boost your Resume? Participate in our Online Grammar and Aptitude Contests. It only takes 20 mins. All participants get Participation Certificates while the top 100 winners get Amazon Cash Vouchers every week. Participate NOW!


As a gesture of support, please follow us on Facebook and YouTube

Become a Campus Ambassador for LearningPundits - Promote our Weekly Online Contests to other students in your Campus via Posters, Facebook, WhatsApp, Email and face to face communication. You will receive a stipend based on your performance and an Internship Certificate to boost your Resume. Email your Resume to support@learningpundits.com
mixtures-and-alligations

Mixtures and Alligations

Mixtures and Alligations: What is a mixture? Solved Exercises with Answers for Questions on Mixtures and Alligations

{{sectionSubheading}}

    {{sectionContentCount}} {{contentCountNoun}}



Weekly Contests Leaderboard


Rank - {{getRank($index,weeklyWinner)}}: {{weeklyWinner.userName}}

Loading...

« Previous Next »

Subscribe to our RSS Feed