Mixtures and Proportions for CAT: Part 5 (More Solved Examples)

[@nikitadas9671]

beautiful!Thank you so much!

[@yashsharma2198]

Thanks sir ! Your lectures were really good, a lot focus on concepts rather than formulas. Now i am able to solve the questions after watching the videos and no other faculty i guess have ever taught with such basics

[@nitishkunwar7891]

Hats off to you sir! You made the most confusing topic easy by clearing the fundamentals.

[@ankitapriyadarsineerout2292]

Thanks sir

[@anshmundhra9164]

Is this Content relevant for Cat 23 ?

[@muhammedhashirp7178]

Chandreshekar Sir Thanks alot

[@ArvindYadav-dv4wi]

Thanks alot sir..😊

[@competativeassam]

sir please provide a lecture series on probability. i have completed all the lessons of you. you are really awesome... thnx a lot

[@_takshzila]

You can watch Probability videos in our course takshzila.learnyst.com/learn/PnC-Probability?. The videos there too are FREE.

[@vanditshrivastava3571]

Thanks uncle 🙂🙂🙂

[@_takshzila]

:)

[@mahalakshmi-sx8tz]

There is a mixture of milk and water in which water is 40%, now 20% of the mixture is replaced by water. Again 30% of the mixture is replaced with water,now 10% of the mixture is taken out and replaced by water which is 10% of the current mixture. Again 10% of the mixture is taken out and replaced by water which is 20% of the current mixture. What is the approx % of milk in the final mixture ? Sir plz help me. How to approach it.

[@_takshzila]

Whenever replacement is done, we are adding pure water. Hence, whenever water is added, milk % will vary inversely with total volume. First time, when water is added, total volume changes from 80k to 100k. Thus, milk % becomes 80k/100k i.e. 4/5 times. Second time, when water is added, total volume changes from 70k to 100k. Thus, milk % becomes 70k/100k i.e. 7/10 times. Third time when water is added, total volume changes from 100k to 110k. Thus, milk % becomes 100k/110k i.e. 10/11 times. Fourth time when water is added, total volume changes from 100k to 120k. Thus, milk % becomes 100k/120k i.e. 5/6 times. Thus, final milk % = 40% × 4/5 × 7/10 × 10/11 × 5/6 = 560/33 = 16.7%

[@priyasrivastav8889]

@@_takshzila 😍

[@riyazkhanpatan4602]

@@_takshzila Hi Sir, I didn't get this logic " now 10% of the mixture is taken out and replaced by water which is 10% of the current mixture." from the question, whenever they are taking out 10% of solution then it has to be 90k right but why you had taken as 100k ??

[@DOTSTUDY]

@@riyazkhanpatan4602 yeah ..sir please explain..

[@varshinimalayil]

Sir, Why isn't the factor 9/8 for the last sum when it is added and removed??

[@nikitadas9671]

the factor is=(Vol. before addition/ Vol. after addition), In last ques, initially Volume was 80 Lts, (before addition) and after addition of 10 Lts of water, it became 90 Lts. So, it is 80/90 = 8/9.

[@harishkumarvarshney63]

Sir in example 4 why we are not taking initial milk proportion as 60 ?

[@_takshzila]

Ratio of milk and water at start is 5 : 3 ⇒ proportion of milk = 5/(5+3) = 5/8. In percent, it will be 5/8 * 100 = 62.5% Ratio of milk and water at end is 2 : 3 ⇒ proportion of milk = 2/(2+3) = 2/5. In percent, it will be 2/5 * 100 = 40% I am not sure, where are you getting 60.

[@harishkumarvarshney63]

@@_takshzila actually sir what I m not getting is why we have taken intial milk proportion as 1 ? Plz help me out Sir

[@_takshzila]

@@harishkumarvarshney63 At start it is 60 litres of milk AND NOT SOLUTION. Thus, at start it is PURE milk. Milk proportion = milk/total. Since total = milk, initial proportion = 1.

[@harishkumarvarshney63]

@@_takshzila ohk.. thanks a lot Sir