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Tips on cracking Aptitude Questions on Mixtures and Alligations

Tips on cracking Aptitude Questions on Mixtures and Alligations [ https://learningpundits.com/module-view/42-mixtures-and-alligations/2-self-assessment-quiz-on-mixtures-and-alligations/ ].<br><br> LearningPundits helps Job Seekers make great CVs [ https://learningpundits.com/module-view/1-cv-preparation-for-freshers/1-cv-writing-tips-for-freshers/ ] , master English Grammar and Vocabulary [ https://learningpundits.com/course/4-english-grammar/ ] , ace Aptitude Tests [ https://learningpundits.com/course/11-mathematical-aptitude/ ], speak fluently in a Group Discussion [ https://learningpundits.com/module-view/6-group-discussion-questions/1-tips-for-speaking-in-a-group-discussion/ ] and perform well in Interviews [ https://learningpundits.com/course/2-personal-interview/ ] We also conduct weekly online contests on Aptitude and English [ https://learningpundits.com/contest ]. We also allow Job Seekers to apply for Jobs [ https://learningpundits.com/applyForJobs ]

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Tips on cracking Aptitude Questions on Mixtures and Alligations

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  1. 2 TIPS on cracking Aptitude Questions on Mixtures and Alligations

  2. 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

  3. 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

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